M. J. Roberts - 7/12/03
Chapter 2 - Mathematical Description of Signals Solutions 1. If g( t) = 7e −2 t − 3 write out and simplify (a)
g( 3) = 7e −9
(b)
g(2 − t) = 7e −2( 2 − t ) − 3 = 7e −7 + 2 t
(c)
t − −11 t 5 g + 4 = 7e 10
(d)
g( jt) = 7e − j 2 t − 3
(e)
(f)
g( jt) + g(− jt) e − j 2t + e j 2t = 7e −3 = 7e −3 cos(2 t) 2 2 − jt − 3 jt − 3 g + g 2 2 e − jt + e jt =7 = 7 cos( t) 2 2
2. If g( x ) = x 2 − 4 x + 4 write out and simplify (a)
g( z) = z 2 − 4 z + 4
(b)
g( u + v ) = ( u + v ) − 4 ( u + v ) + 4 = u 2 + v 2 + 2 uv − 4 u − 4 v + 4
(c)
g(e jt ) = (e jt ) − 4 e jt + 4 = e j 2 t − 4 e jt + 4 = (e jt − 2)
(d)
g(g( t)) = g( t 2 − 4 t + 4 ) = ( t 2 − 4 t + 4 ) − 4 ( t 2 − 4 t + 4 ) + 4
2
2
2
2
g(g( t)) = t 4 − 8 t 3 + 20 t 2 − 16 t + 4 (e)
g(2) = 4 − 8 + 4 = 0
3. What would be the numerical value of “g” in each of the following MATLAB instructions? (a)
t = 3 ; g = sin(t) ;
(b)
x = 1:5 ; g = cos(pi*x) ;
(c)
f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ;
0.1411 [-1,1,-1,1,-1]
Solutions 2-1
M. J. Roberts - 7/12/03
0.0247 + 0.0920 + 1 0.0920 − 0.0247 −
j 0.155 j 0.289 j 0.289 j 0.155
4. Let two functions be defined by 1 , sin(20πt) ≥ 0 x1 ( t) = −1 , sin(20πt) < 0
and
t , sin(2πt) ≥ 0 x 2 ( t) = . − t , sin(2πt) < 0
Graph the product of these two functions versus time over the time range, −2 < t < 2 . x(t) 2
-2
t
2 -2
5. For each function, g( t) , sketch g(− t) , − g( t) , g( t − 1) , and g(2t) . (a)
(b)
g(t)
g(t)
4
3
t
2
-1
t
1 -3
g(-t)
g(-t)
4
-g(t)
-g(t)
3
t
-2
3
-1
t
1
2 4
-3
g(t-1)
g(t-1)
g(2t)
4
3
4
1
3
t
t
1
2
t
1 -1
t
-3
g(2t) 3 1
-3
6. A function, G( f ) , is defined by
Solutions 2-2
t
-1 2 1 2 -3
t
M. J. Roberts - 7/12/03
f G( f ) = e − j 2πf rect . 2 Graph the magnitude and phase of G( f − 10) + G( f + 10) over the range, −20 < f < 20 . f + 10 f − 10 − j 2π f +10 G( f − 10) + G( f + 10) = e − j 2π ( f −10) rect + e ( ) rect 2 2 |G( f )| 1
-20
f
20
Phase of G( f ) π
-20
f
20 -π
7. Sketch the derivatives of these functions. (All sketches at end.)
(a)
g( t) = sinc( t)
(b)
g( t) = (1 − e
−t
g′ ( t) =
π 2 t cos(πt) − π sin(πt) πt cos(πt) − sin(πt) = πt 2 (πt) 2 e − t , t ≥ 0 − t g′ ( t) = = e u( t) 0 , t < 0
) u(t) (a)
(b)
x(t)
x(t)
1
1
-4
4
t
-1
-1
dx/dt
1
1 4
-1
t
-1
dx/dt -4
4
t
-1
4
t
-1
8. Sketch the integral from negative infinity to time, t, of these functions which are zero for all time before time, t = 0.
Solutions 2-3
M. J. Roberts - 7/12/03
g(t)
g(t)
1
1 1
2
3
t
1 2
∫ g(t) dt
1
2
3
1
2
3
t
∫ g(t) dt
1
1
1 2
1
2
t
3
t
9. Find the even and odd parts of these functions. (a)
g( t) = 2 t 2 − 3t + 6 2 t 2 − 3t + 6 + 2(− t) − 3(− t) + 6 4 t 2 + 12 g e ( t) = = = 2t 2 + 6 2 2 2
2 t 2 − 3t + 6 − 2(− t) + 3(− t) − 6 −6 t g o ( t) = = = −3t 2 2 2
(b)
π g( t) = 20 cos 40πt − 4 π π 20 cos 40πt − + 20 cos −40πt − 4 4 g e ( t) = 2
Using cos( z1 + z2 ) = cos( z1 ) cos( z2 ) − sin( z1 ) sin( z2 )
g e ( t) =
π π 20 cos( 40πt) cos − − sin( 40πt) sin − 4 4 +20 cos(−40πt) cos − π − sin(−40πt) sin − π 4 4 2
π π 20 cos( 40πt) cos + sin( 40πt) sin 4 4 +20 cos( 40πt) cos π − sin( 40πt) sin π 4 4 g e ( t) = 2
Solutions 2-4
M. J. Roberts - 7/12/03
20 π g e ( t) = 20 cos cos( 40πt) = cos( 40πt) 4 2
π π 20 cos 40πt − − 20 cos −40πt − 4 4 g o ( t) = 2 Using cos( z1 + z2 ) = cos( z1 ) cos( z2 ) − sin( z1 ) sin( z2 ) π π 20 cos( 40πt) cos − − sin( 40πt) sin − 4 4 −20 cos(−40πt) cos − π − sin(−40πt) sin − π 4 4 g o ( t) = 2
g o ( t) =
π π 20 cos( 40πt) cos + sin( 40πt) sin 4 4 −20 cos( 40πt) cos π − sin( 40πt) sin π 4 4 2
20 π g o ( t) = 20 sin sin( 40πt) = sin( 40πt) 4 2 (c)
2 t 2 − 3t + 6 g( t) = 1+ t 2 t 2 − 3t + 6 2 t 2 + 3t + 6 + 1 1− t + t g e ( t) = 2
(2t g e ( t) = g e ( t) =
2
− 3t + 6)(1 − t) + (2 t 2 + 3t + 6)(1 + t) (1 + t)(1 − t) 2
4 t 2 + 12 + 6 t 2 6 + 5 t 2 = 1 − t2 2(1 − t 2 )
2 t 2 − 3t + 6 2 t 2 + 3t + 6 − 1 1− t + t g o ( t) = 2
Solutions 2-5
M. J. Roberts - 7/12/03
(2t g o ( t) =
2
− 3t + 6)(1 − t) − (2 t 2 + 3t + 6)(1 + t) (1 + t)(1 − t) 2
−6 t − 4 t 3 − 12 t 2t 2 + 9 g o ( t) = = −t 1 − t2 2(1 − t 2 ) sin(πt) sin(−πt) + t π −πt = sin(πt) g e ( t) = πt 2
(d)
g( t) = sinc( t)
(e)
g( t) = t(2 − t 2 )(1 + 4 t 2 )
g o ( t) = 0
g( t) = {t (2 − t 2 )(1 + 4 t 2 ) 4 4 3 odd 12312 even
even
Therefore g( t) is odd, g e ( t) = 0 and g o ( t) = t(2 − t 2 )(1 + 4 t 2 ) (f)
g( t) = t(2 − t)(1 + 4 t) g e ( t) =
t(2 − t)(1 + 4 t) + (− t)(2 + t)(1 − 4 t) 2
g e ( t) = 7 t 2
g o ( t) =
t(2 − t)(1 + 4 t) − (− t)(2 + t)(1 − 4 t) 2
g o ( t) = t(2 − 4 t 2 )
10. Sketch the even and odd parts of these functions.
Solutions 2-6
M. J. Roberts - 7/12/03
g(t)
g(t)
1
1
t
1
1
2
t
-1
g e(t)
g e(t)
1
1
t
1
1
2
t
-1
g o(t)
g o(t)
1
1
t
1
1
2
t
-1
(a)
(b)
11. Sketch the indicated product or quotient, g( t) , of these functions. (a)
(b)
1
1
-1 1
-1
t
1
-1
t
-1
g(t)
g(t)
1 Multiplication -1
1
t
1
Multiplication
-1 1
t
-1
g(t)
g(t)
1
1
-1 1 -1
t
-1
1 -1
Solutions 2-7
t
M. J. Roberts - 7/12/03
(c)
(d)
1
1
t
-1
t
1
g(t)
g(t)
Multiplication
1 1
Multiplication
1
t
t
1
g(t)
g(t) -1
1
t
-1
1 -1
t
1
(e)
(f) 1
1
...
...
-1
1
t
1
t
-1
g(t)
-1
g(t)
1
Multiplication
1 Multiplication -1
t
1
t
1 -1
g(t)
g(t)
1
...
... -1
1
1
t
t
1 -1
-1
(g)
(h)
1
1
t
-1
-1
-1
1
t
g(t) Division
1 1
g(t) π
Division
t
1
g(t)
t
g(t) 1
t -1
-1
1
t
12. Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way.
Solutions 2-8
M. J. Roberts - 7/12/03
1
1
−1
−1 even
∫ (2 + t)dt = ∫
(a)
1
1
−1 odd
0
2{ dt + ∫ {t dt = 2 ∫ 2 dt = 4
(b) 1 20
1 20
∫ [4 cos(10πt) + 8 sin(5πt)]dt = ∫
−
1 20
−
1 20
∫
(c)
1 − 20
1 20
4 cos(10πt) dt + 14243 even
1 20
1 20
−
1 20
odd
0
4 t{ cos(10πt) dt = 0 1424 3 odd even 142 43 odd
1 1 10 10 cos(10πt) cos(10πt) (d) ∫ t{ sin(10πt) dt = 2 ∫ t sin(10πt) dt = 2 − t + dt 1 424 3 10π 0 ∫0 10π 1 odd 0 odd − 142 43 10 even 1 10
1 10
1 10 sin(10πt) 1 1 = (10π3t)dt == 2 + 2 ∫1 odd{t sin 1 424 100π (10π ) 0 50π odd − 142 43 10 1 10
even
1
(e)
1
1
[
−t −t −t −t ∫ e{ dt = 2∫ e dt = 2∫ e dt = 2 −e
−1 even
0
]
1 0
= 2(1 − e −1 ) ≈ 1.264
0
1
(f)
∫
−t t{ e{ dt = 0 odd2 even −1 1 3 odd
13. Find the fundamental period and fundamental frequency of each of these functions. (a)
g( t) = 10 cos(50πt)
f 0 = 25 Hz , T0 =
1 s 25
(b)
π g( t) = 10 cos 50πt + 4
f 0 = 25 Hz , T0 =
1 s 25
(c)
g( t) = cos(50πt) + sin(15πt) 1 15 = 0.4 s f 0 = GCD 25, = 2.5 Hz , T0 = 2.5 2
(d)
8
(5π3t)dt = 8 ∫ cos(10πt)dt = ∫ 81sin 424 10π
3π g( t) = cos(2πt) + sin( 3πt) + cos 5πt − 4
Solutions 2-9
M. J. Roberts - 7/12/03
1 3 5 1 f 0 = GCD1, , = Hz , T0 = = 2 s 1 2 2 2 2 14. Find the fundamental period and fundamental frequency of g( t) .
...
g(t)
(a) ...
... 1
t
...
t
1
(b) ...
...
+ t
1
...
...
t
1
(c) ...
...
+
g(t)
t
1
1 s 3
(a)
f 0 = 3 Hz and T0 =
(b)
f 0 = GCD(6, 4 ) = 2 Hz and T0 =
(c)
f 0 = GCD(6, 5) = 1 Hz and T0 = 1 s
1 s 2
15. Plot these DT functions. (a)
2π ( n − 2) 2πn x[ n ] = 4 cos − 3 sin 12 8
,
−24 ≤ n < 24
x[n] 7 -24
24
n
-7
(b)
x[ n ] = 3ne
−
n 5
,
−20 ≤ n < 20 x[n] 6
-20
20 -6 2
(c)
n x[ n ] = 21 + 14 n 3 2
, −5 ≤ n < 5
Solutions 2-10
n
g(t)
M. J. Roberts - 7/12/03
x[n] 2000 -5
n
5 -2000
n
2
− 2πn 16. Let x1[ n ] = 5 cos and x 2 [ n ] = −8e 6 . Plot the following combinations of those 8 two signals over the DT range, −20 ≤ n < 20 . If a signal has some defined and some undefined values, just plot the defined values.
x[n] 40
-20
(a)
x[ n ] = x1[ n ] x 2 [ n ]
20
n
-40
x[n] 20 -20
(b)
x[ n ] = 4 x1[ n ] + 2 x 2 [ n ]
20
n
-40
x[n] 20 -20
(c)
x[ n ] = x1[2 n ] x 2 [ 3n ]
20
n
-40 x[n] 10000 -20
(d)
x[ n ] =
x1[2 n ] x 2 [− n ]
20
n
-50000
x[n] 5 -20
(e)
n n x[ n ] = 2 x1 + 4 x 2 2 3
20
-40
Solutions 2-11
n
M. J. Roberts - 7/12/03
17. A function, g[ n ] is defined by −2 , n < −4 g[ n ] = n , − 4 ≤ n < 1 . 4 , 1≤ n n n Sketch g[− n ], g[2 − n ], g[2n ] and g . 2 g[n] 4 -10
n
10 -4
g[- n]
g[2- n]
4
4 10
-10
10
n
-10
-4
-4
g[2n]
g[n/2]
4
n
4
-10 10
-10
n
10
-4
n
-4
18. Sketch the backward differences of these DT functions.
(a)
(b)
g[n]
g[n]
1
1
-4
20
n
-4
-1
20
n
-1
∆g[n-1]
∆g[n-1]
1
1
-4
20
n
-4
-1
20
n
-1
(c) 2
g[n] = (n/10) 4
-4
∆g[n-1]
20
n
0.5 -4 -0.25
20
n
19. Sketch the accumulation, g[ n ], from negative infinity to n of each of these DT functions.
Solutions 2-12
M. J. Roberts - 7/12/03
(a)
h[ n ] = δ [ n ]
(b)
h[ n ] = u[ n ]
(c)
2πn h[ n ] = cos u[ n ] 16
(d)
2πn h[ n ] = cos u[ n ] 8
(e)
2πn h[ n ] = cos u[ n + 8] 16
(a)
-16
h[n]
h[n]
1
1
g[n]
16
n
(b)
-16
g[n]
1 -16
16
n
-16
16
(c)
16
1
n
(d)
g[n]
-16
-3
16
16
-1 3
n -16
16
-3
h[n] 1
(e)
-16
-1
16
n
g[n] 3 -16
n
g[n]
3 -16
n
h[n]
1 -1
n
16
h[n] -16
16
-3
16
n
20. Find and sketch the even and odd parts of these functions. −
n 4
(a)
g[ n ] = u[ n ] − u[ n − 4 ]
(b)
g[ n ] = e
(c)
2πn g[ n ] = cos 4
(d)
2πn g[ n ] = sin u[ n ] 4
Solutions 2-13
u[ n ]
n
M. J. Roberts - 7/12/03
g[n]
g[n]
1 -10
-1
1 10
n
-10
-1
g [n]
-1
n
-10
-1
-10
10
n
-10
-1
10
-10
-1
10
10
n
10
n
1
n
-10
-1
go[n]
go[n]
1 -1
n
ge[n]
-1
-10
10
1
n
ge[n] 1
n
g[n]
-1
-10
10
o 1
g[n] 1
n
g [n]
o 1
-1
10
1 10
g [n] -10
n
g [n] e
e 1
-10
10
10
n
1
-10
-1
21. Sketch g[ n ]. (a)
(b)
g1[n]
g1[n]
1 -10
1 10
-4
n
20
-1
g2[n]
g2[n]
g[n]
1
g[n]
1
-10 10
n
Multiplication -4
-1
20
(d)
1
1 20
n
-10
-1
10
n
-1
g2[n]
g[n]
g[n]
1 20 -1
Multiplication
g1[n]
g1[n]
-4
n
-1
(c)
-4
n
-1
n
g[n]
1
Multiplication -10
10 -1
Solutions 2-14
n
Multiplication
M. J. Roberts - 7/12/03
(a)
(b)
g[n]
g[n]
1
1
-10 10
n
-4
-1
20
n
-1
(c)
(d)
g[n]
g[n]
1
1
-4
20
n
-10
-1
10
n
-1
22. Find the fundamental DT period and fundamental DT frequency of these functions. (a)
2πn g[ n ] = cos 10
N 0 = 10 , F0 =
1 10
(b)
πn g[ n ] = cos 10
N 0 = 20 , F0 =
1 20
(c)
2πn 2πn g[ n ] = cos + cos 5 7
N 0 = 35 , F0 =
1 35
(d)
g[ n ] = e
j
N 0 = 20 , F0 =
1 20
g[ n ] = e
−j
N 0 = 12 , F0 =
1 12
(e)
2πn 20
2πn 3
+e
−j
+e
2πn 20
−j
2πn 4
23. Graph the following functions and determine from the graphs the fundamental period of each one (if it is periodic). (a)
2πn 2πn g[ n ] = 5 sin + 8 cos 4 6
(b)
14πn 7πn g[ n ] = 5 sin + 8 cos 12 8
(c)
πn −j g[ n ] = Re e jπn + e 3
(d)
n −j g[ n ] = Re e jn + e 3
Solutions 2-15
M. J. Roberts - 7/12/03
(a)
(b)
g[n]
g[n]
12
12
-24
n
24
-24
n
24
-12
-12
N0 = 12
N0 = 24
(c)
(d)
g[n]
g[n]
2
2
24
-24
n
-24
n
24 -2
-2
N0 = 6
Not Periodic
24. Find the signal energy of these signals.
(a)
∞
x( t) = 2 rect ( t)
∫ 2 rect (t)
Ex =
−∞
(b)
dt = 4 ∫ dt = 4 −
∞
∫
−∞
10
A(u( t) − u( t − 10)) dt = A 2 ∫ dt = 10 A 2 2
0
∞
x( t) = u( t) − u(10 − t)
Ex =
∫ u(t) − u(10 − t)
−∞
(d)
2
dt =
0
∞
−∞
10
∫ dt + ∫ dt → ∞
x( t) = rect ( t) cos(2πt) ∞
Ex =
∫ rect (t) cos(2πt)
−∞
2
1 2
dt =
1 2
∫ cos (2πt)dt = 2 ∫ (1 + cos(4πt))dt 2
−
1
1 2
1 12 2 1 1 E x = ∫ dt + ∫ cos( 4πt) dt = 2 2 1 1 − − 2 2 14 4244 3 =0 (e)
1 2
x( t) = A(u( t) − u( t − 10)) Ex =
(c)
2
1 2
x( t) = rect ( t) cos( 4πt)
Solutions 2-16
−
1 2
M. J. Roberts - 7/12/03
∞
Ex =
∫ rect (t) cos(4πt)
2
1 2
1 2
∫ cos (4πt)dt = 2 ∫ (1 + cos(8πt))dt
dt =
1
2
−∞
−
1 2
−
1 2
1 12 2 1 1 E x = ∫ dt + ∫ cos(8πt) dt = 2 2 1 1 − − 2 2 14 4244 3 =0 (f)
x( t) = rect ( t) sin(2πt) ∞
Ex =
∫ rect (t) sin(2πt)
2
1 2
∫ sin (2πt)dt = 2 ∫ (1 − cos(4πt))dt
dt =
−∞
1 2
1
2
−
1 2
−
1 2
1 12 2 1 1 E x = ∫ dt − ∫ cos( 4πt) dt = 2 2 1 1 − − 2 2 14 4244 3 =0 (g)
x[ n ] = A rect N 0 [ n ] ∞
E x = ∑ A rect N 0 [ n ] = A −∞
(h)
x[ n ] = Aδ [ n ]
2
N0
2
∑ (1) = (2N
−N0
0
+ 1) A 2
x[ n ] = Aδ [ n ] ∞
0
E x = ∑ Aδ [ n ] = A 2 ∑ (1) = A 2 2
−∞
(i)
(j) (k)
0
∞
∞
x[ n ] = comb N 0 [ n ]
E x = ∑ comb N 0 [ n ] =
x[ n ] = ramp[ n ]
E x = ∑ ramp[ n ] = ∑ n 2 → ∞
2
−∞
∞
−∞
2
∞ 0
x[ n ] = ramp[ n ] − 2 ramp[ n − 4 ] + ramp[ n − 8]
Solutions 2-17
∑ (1) → ∞
−∞ n = mN 0
M. J. Roberts - 7/12/03 ∞
E x = ∑ ramp[ n ] − 2 ramp[ n − 4 ] + ramp[ n − 8] = (0 2 + 12 + 2 2 + 32 + 4 2 + 32 + 2 2 + 12 + 0 2 ) 2
−∞
E x = 1 + 4 + 9 + 16 + 9 + 4 + 1 = 44 25. Find the signal power of these signals. x( t) = A
(a)
1 T →∞ T
x( t) = u( t)
(b)
Px = lim
∫ −
T 2
T 2 2
A A dt = lim T →∞ T 2
T 2
∫ u(t) −
2
A2 T = A2 ∫T dt = Tlim →∞ T
−
2 T 2
1 1T 1 dt = lim = ∫ T →∞ T T →∞ T 2 2 0
dt = lim
T 2
x( t) = A cos(2πf 0 t + θ )
(c)
Px =
1 Px = lim T →∞ T
T 2
1 T0
T0 2
A cos(2πf 0 t + θ ) dt =
∫
−
2
T0 2
2
A T0
T0 2
∫ cos (2πf t + θ )dt 2
0
−
T0 2
T0 2
T0
A sin( 4πf 0 t + 2θ ) 2 A 1 + cos( 4πf 0 t + 2θ )) dt = Px = ( t + ∫ 2T0 T0 2T0 4πf 0 − T0 − 2
2
2
2
T T sin 4πf 0 0 + 2θ sin −4πf 0 0 + 2θ A2 A 2 2 = T Px = + − 2T0 0 4πf 0 4πf 0 2 14444444 24444444 3 =0 2
∞
(d)
x( t) = A ∑ rect ( t − 2 n ) n =−∞
1 Px = T0
(e)
T0 2
∫
−
T0 2
∞
2
2 1
A A ∑ rect ( t − 2 n ) dt = 2 n =−∞
1 2 2
A ∫ rect (t) dt = 2 −1
∞ 1 x( t) = 2 A − + ∑ rect ( t − 2 n ) 2 n =−∞
Solutions 2-18
2
A2 ∫1 dt = 2
−
2
M. J. Roberts - 7/12/03
1 Px = T0
T0 2
∫
T0 2
−
2
∞ 1 4 A2 1 − + rect ( t) dt 2 A − + ∑ rect ( t − 2 n ) dt = ∫ 2 −1 2 2 n =−∞
2
1
2
1
2
1
1 1 Px = 2 A ∫ − + rect ( t) dt = 4 A 2 ∫ − + rect ( t) dt 2 2 −1 0 2
12 2 2 1 1 2 1 Px = 4 A ∫ dt + ∫ − dt = A 2 2 2 1 0 2 (f)
(g)
(h)
x[ n ] = A
1 Px = lim N →∞ 2 N
x[ n ] = u[ n ]
Px = lim
x[ n ] = A
N −1
∑
n =− N N −1
1 N →∞ 2 N
∑
n =− N
A2 A = lim N →∞ 2 N 2
N −1
A2 (1) = Nlim (2N ) = A 2 ∑ →∞ 2 N n =− N
1 N →∞ 2 N
u[ n ] = lim 2
N −1
N →∞
n =0
∞
∑ rect [n − 8m]
m =−∞
2
N −1
2
∞ A2 7 ∞ 1 0 8 Px = A rect n − m = [ ] ∑ ∑ 2 ∑ ∑ rect 2[n − 8m] 2 N 0 n =− N 0 m =−∞ 2 × 8 n =−8 m =−∞
Px =
(i)
(j)
1
N
∑ (1) = lim 2N = 2
2
2 7 10 A 2 5 A 2 A 2 −6 1 1 1 ( ) + ( ) + ( ) ∑ ∑ ∑= 6 = 2 × 8 = 8 2 × 8 n =−8 n =−2 n
2 1 1 comb N 0 [ n ] = ∑ N 0 n = N0 N0
x[ n ] = comb N 0 [ n ]
Px =
x[ n ] = ramp[ n ]
1 Px = lim N →∞ 2 N
N −1
∑
n =− N
ramp[ n ]
2
1 = lim N →∞ 2 N
N −1
∑n
2
→∞
n =0
26. Using MATLAB, plot the CT signal, x( t) = sin(2πt) , over the time range, 0 < t < 10 , with the following choices of the time resolution, ∆t , of the plot. Explain why the plots look the way they do. (a)
∆t =
1 24
(b)
∆t =
1 12
(c)
∆t =
1 4
(d)
∆t =
1 2
(e)
∆t =
2 3
(f)
∆t =
5 6
Solutions 2-19
M. J. Roberts - 7/12/03
(g)
∆t = 1 ∆t = 1/24
∆t = 1/12
x(t)
x(t)
1
1
10
t
-1
10
t
-1
∆t = 1/4
∆t = 1/2
x(t)
x(t)
1
1
10
t
-1
10
t
-1
∆t = 2/3
∆t = 5/6
x(t)
x(t)
1
1
10
t
-1
10
t
-1
∆t = 1 x(t) 1
10
t
-1
27. Given the function definitions on the left, find the function values on the right. (a)
(b)
π g( t) = 100 sin 200πt + 4 π π π g(0.001) = 100 sin 200π × 0.001 + = 100 sin + = 98.77 5 4 4 g( t) = 13 − 4 t + 6 t 2
g(2) = 13 − 4 (2) + 6(2) = 29 2
(c) g( t) = −5e −2 t e − j 2πt
π
−2 − j 2π − −j 1 4 = −5e 2 e 2 = − j 3.03 g = −5e 4 e 4 1
1
28. Sketch these CT exponential and trigonometric functions.
Solutions 2-20
1
M. J. Roberts - 7/12/03
(a)
g( t) = 10 cos(100πt)
(c)
g( t) = 5e
−
t 10
(b)
g( t) = 40 cos(60πt) + 20 sin(60πt)
(d)
g( t) = 5e 2 cos(2πt)
−
(a)
t
(b)
g(t)
g(t)
10
60
0.04
t
t
0.066667
-10
-60
(c)
g(t)
(d)
g(t)
5
5
8
t
50
t
-5
29. Sketch these CT singularity and related functions. (a)
g( t) = 2 u( 4 − t)
(b)
g( t) = u(2 t)
(c)
g( t) = 5 sgn( t − 4 )
(d)
g( t) = 1 + sgn( 4 − t)
(e)
g( t) = 5 ramp( t + 1)
(f)
g( t) = −3 ramp(2 t)
(g)
g( t) = 2δ ( t + 3)
(h)
g( t) = 6δ ( 3t + 9)
(a)
(c)
(b)
g(t)
(d)
g(t)
g(t)
g(t)
5 2
1 t
4
t
(e)
g(t)
10
-1
(i)
1
1
t
(h) g(t)
g(t)
2
2
t
-6
g( t) = −4δ (2( t − 1))
-3
(j)
t
4
-5
(g)
(f) g(t)
2
t
4
t
-3
1 g( t) = 2 comb t − 2
Solutions 2-21
t
M. J. Roberts - 7/12/03
(k)
g( t) = 8 comb( 4 t)
(l)
t + 1 g( t) = −3 comb 2
(m)
t g( t) = 2 rect 3
(n)
t + 1 g( t) = 4 rect 2
(o)
g( t) = tri( 4 t)
(p)
t − 1 g( t) = −6 tri 2
(i)
(k)
(j)
g(t) 1
2
2
t
...
-2
... 1
(m)
...
... t
t
t
-2
(p)
g(t)
g(t) -1
-1 4
t
1 4
-6
(q)
t g( t) = 5 sinc 2
(r)
g( t) = − sinc(2( t + 1))
(s)
g( t) = −10 drcl( t, 4 )
(t)
t g( t) = 5 drcl , 7 4 (r)
(q) g(t)
g(t) -1
5
t
t
-1
2
(s)
(t)
g(t)
g(t)
10
5
4 -10
3
...
-6
(o) 1
4
3 2
t
1 4
g(t)
2
g(t) -1 1
...
(n)
g(t)
-3 2
(l)
g(t)
g(t)
t -1
Solutions 2-22
8
t
1
3
t
t
M. J. Roberts - 7/12/03
( )
0.1rect t-3 4
-3rect(t-2) 3 2
5 2
0.1
t
(u)
(v)
-3
1
( )
-4tri 3+t 2
3 5
t
4sinc[5(t-3)] 4
(w)
-5 3 -1
t
(x)
-4
-1
1 2 3 4 5 6
4sinc(5t-3) 4 (y)
-1
1 2 3 4 5 6
t
30. Sketch these combinations of CT functions. (a)
g( t) = u( t) − u( t − 1)
(b)
1 g( t) = rect t − 2
(c)
g( t) = −4 ramp( t) u( t − 2)
(d)
g( t) = sgn( t) sin(2πt)
(e)
g( t) = 5e
(f)
g( t) = rect ( t) cos(2πt)
(g)
g( t) = −6 rect ( t) cos( 3πt)
(h)
g( t) = rect ( t) tri( t)
−
t 4
u( t)
Solutions 2-23
t
M. J. Roberts - 7/12/03
g(t)
(a)
1
(c)
(b)
g(t)
(d)
g(t)
g(t)
1
2
t
1
(e)
t -1
(g)
g(t) 1
-1 2
5 t
4
1
-16
(f)
g(t)
t
-8
t
1
4
1 2
t
-1
(h)
g(t) 6
-1 2
-6
1
g(t) 1 1 2
1 2
t -1 2
t
1 2
(i)
1 g( t) = rect ( t) tri t + 2
(j)
1 1 g( t) = u t + ramp − t 2 2
(k)
g( t) = tri2 ( t)
(l)
g( t) = sinc 2 ( t)
(m)
g( t) = sinc( t)
(n)
g( t) =
(o)
1 1 g( t) = rect t + − rect t − 2 2
(p)
t g( t) = ∫ δ (λ + 1) − 2δ (λ ) + δ (λ − 1) dλ −∞
(i)
(k)
(j)
g(t) 1
1 2
t
t
1 2
t
-1
(o)
(p)
g(t)
g(t)
g(t)
1
1
1
1
1 -1 -1 1
1
-1
(n)
g(t)
g(t) 1
1
-1 2
(m)
(l)
g(t)
g(t) 1
-1 2
d (tri(t)) dt
t
1 -1
t
-1
t
Solutions 2-24
1 -1 -1
t
1
t
M. J. Roberts - 7/12/03
6
(q)
2t t 3 tri + 3 rect 3 3
3
, t -3 2
3 2
6
(r)
t t 6 tri rect 3 3
3 t -3 2
3 2
4
4
(s)
t
4 sinc(2t ) sgn( −t )
-1 2
1 2
, (t)
t − 1 2 ramp(t ) rect 2
t 2
-4 3
4
(u)
t − 2 4 tri u( 2 − t ) 2
t 2
t t , (v) 3 rect − 6 rect 4 2
t -2 -1 -3
(w)
t t g( t) = 10 drcl , 5 rect 4 8 (g) g(t) 10
-8
8
-2
t
31. Using MATLAB, for each function below plot the original function and the transformed function. %
Plotting functions and transformations of those functions
close all ; % (a) part tmin = -2 ; tmax = 2 dt = (tmax - tmin)/N g0 = g322a(t) ; g1 = subplot(2,1,1) ; p = ylabel('g(t)') ;
; N = 400 ; ; t = tmin + dt*[0:N]' ; 5*g322a(2*t) ; plot(t,g0,'k') ; set(p,'LineWidth',2) ; grid ;
Solutions 2-25
1 2
M. J. Roberts - 7/12/03
title('Exercise 3.2.2 (a) - g(t) = 10*cos(20*pi*t)*(1-|t|) , -13') ; subplot(2,1,2) ; p = plot(t,g1,'k') ; set(p,'LineWidth',2) ; grid ; xlabel('t') ; ylabel('-3g(4-t)') ; % (c) part figure ; tmin = 0 ; tmax = 96 ; N = 400 ; dt = (tmax - tmin)/N ; t = tmin + dt*[0:N]' ; g0 = g322c(t) ; g1 = g322c(t/4) ; subplot(2,1,1) ; p = plot(t,g0,'k') ; set(p,'LineWidth',2) ; grid ; ylabel('g(t)') ; title('Exercise 3.2.2 (c) - g(t) = Re(exp(j*pi*t) + exp(j*1.1*pi*t))') ; subplot(2,1,2) ; p = plot(t,g1,'k') ; set(p,'LineWidth',2) ; grid ; xlabel('t') ; ylabel('g(t/4)') ; % (d) part figure ; fmin = -20 ; fmax = 20 ; N = 200 ; df = (fmax - fmin)/N ; f = fmin + df*[0:N]' ; G0 = G322d(f) ; G1 = abs(G322d(10*(f-10)) + G322d(10*(f+10))) ; subplot(2,1,1) ; p = plot(f,G0,'k') ; set(p,'LineWidth',2) ; grid ; ylabel('G(f)') ; title('Exercise 3.2.2 (d) - G(f) = |5/(f^2 - j2 + 3)|') ; subplot(2,1,2) ; p = plot(f,G1,'k') ; set(p,'LineWidth',2) ; grid ; xlabel('f') ; ylabel('|G(10(f-10)) + G(10*(f+10))|') ; function y = g322a(t) g = (1-abs(t)).*(-1 < t & t < 1) ; y = 10*cos(20*pi*t).*g ; function y = g322b(t) y = -2*(t <= -1) + 2*t.*(-1 < t & t <= 1) + ... (3-t.^2).*(1 < t & t <= 3) - 6*(t > 3) ; function y = g322c(t) y = real(exp(j*pi*t) + exp(j*1.1*pi*t)) ; function y = G322d(f) y = abs(5./(f.^2 - j*2 + 3)) ;
(a)
g( t) = 10 cos(20πt) tri( t)
5 g(2 t)
Solutions 2-26
vs. t
M. J. Roberts - 7/12/03
Original g(t) 10
-2
2
t
-10
Transformed g(t) 50
-2
2
t
-50
(b)
−2 , t < −1 2 t , − 1 < t < 1 g( t) = 2 3 − t , 1 < t < 3 −6 , t > 3
−3 g( 4 − t)
vs. t
Original g(t) 2 -4
8
t
-6
Transformed g(t) 20
-4
8
t
-10
(c) g( t) = Re(e jπt + e j1.1πt )
t g vs. t 4
Solutions 2-27
M. J. Roberts - 7/12/03
Original g(t) 2
t
100 -2
Transformed g(t) 2
t
100 -2
(d) G( f ) =
G(10( f − 10)) + G(10( f + 10))
5 f − j2 + 3 2
Original g(t) 1.5
-20
20
t
Transformed g(t) 1.5
-20
20
t
32. Let two signals be defined by 1 , cos(2πt) ≥ 1 x1 ( t) = 0 , cos(2πt) < 1
and
2πt x 2 ( t) = sin . 10
Plot these products over the time range, −5 < t < 5 . (a)
x1 (2t) x 2 (− t)
(b)
t x1 x 2 (20 t) 5
(c)
t x1 x 2 (20( t + 1)) 5
(d)
t − 2 x1 x (20 t) 5 2
Solutions 2-28
vs. f
M. J. Roberts - 7/12/03
(a)
(b)
x1(t)x2(t)
x1(t)x2(t)
1
1
-5
5
t
-5
5
-1
t
-1
(c)
(d)
x1(t)x2(t)
x1(t)x2(t)
1
1
-5
5
t
-5
5
-1
t
-1
33. Given the graphical definition of a function, graph the indicated transformation(s). (a)
g(t) 2 1
g( t) → g(2 t)
1
-2
g( t) → −3 g(− t)
t
2 3 4 5 6 -2 g( t) = 0 , t > 6 or t < −2
g(2t) 2 1 -2
2
4
6
2
4
6
t
-2
-3g(-t) 6 3 -4
-2 -6
Solutions 2-29
t
M. J. Roberts - 7/12/03
(b)
g(t) g( t) → g( t + 4 )
2 1 -2
t
1 2 3 4 5 6
t − 1 g( t) → −2 g 2
-2
g( t) is periodic with fundamental period, 4
g(t + 4) 2 1 -2
t
1 2 3 4 5 6 7 -2
-2g( t -1 ) 2 4 2 -2
1 2 3 4 5 6 7 8
t
-4 34. For each pair of functions graphed below determine what transformation has been done and write a correct functional expression for the transformed function. (a)
g(t) 2
2 -2
-1
1 2 3 4 5 6
t
-4 -3 -2 -1 -1
g( t) → g(2 − t)
Solutions 2-30
1 2 3 4
t
M. J. Roberts - 7/12/03
(b)
g(t) 2 -2
1 2 3 4 5 6
t
-2
1 2 3 4 5 6
-1
t
In (b), assuming g( t) is periodic with fundamental period, 2, find two different transformations which yield the same result 1 1 g( t) → − g( t + 1) or g( t) → − g( t − 1) 2 2 35. Sketch the magnitude and phase of each function versus f. (a)
G( f ) = sinc( f )e
−j
πf 8
G( f ) =
(b)
(a)
(b)
|G( f )|
|G( f )|
1
-16
f
-100
Phase of G( f )
f 10
100
f
Phase of G( f )
π
π
16 -π
1+ j
10
16
-16
jf
f
-100
100 -π
(c)
πf f + 1000 − j 500 f − 1000 G( f ) = rect e + rect 100 100
(d)
G( f ) =
1 250 − f 2 + j 3 f
Solutions 2-31
f
M. J. Roberts - 7/12/03
(c)
(d)
|G( f )|
|G( f )|
1
0.02
-1100
1100
f
-50
50
Phase of G( f )
Phase of G( f )
π
π
-1100
1100
f
-50
50
-π
(e)
f
f
-π
G( f ) = comb(100 f ) sinc(25 f )e
j
πf 50
(e) |G( f )| 0.01
-0.2
0.2
f
Phase of G( f ) π
-0.2
0.2
f
-π
36. Graph versus f , in the range, −4 < f < 4 , the magnitude and phase of |X( f )| 1
-4 -3 -2-1
(a)
X( f ) = sinc( f )
1234
f
X( f ) π -4 -3 -2-1 1234 -π
f
Solutions 2-32
M. J. Roberts - 7/12/03
|X( f )| 2
-4 -3 -2-1
(b)
X( f ) = 2 sinc( f )e − j 4 πf
1234
f
X( f ) -4 -3 -2-1 4π -4π 1 2 3 4
f
π
|X( f )| 5
(c)
X( f ) = 5 rect (2 f )e + j 2πf
1 4
f
1 4
X( f ) π 2
f
π 2
|X( f )| 10
(d)
f X( f ) = 10 sin c 2 4
-4 -3 -2-1
1234
f
X( f ) -4 -3 -2-1
1234
f
|X( f )| 5
(e)
X( f ) = j 5δ ( f + 2) − j 5δ ( f − 2)
-4 -3 -2-1
1234
X( f ) π 2
-4 -3 -2-1
(f)
X( f ) = 2 comb( 4 f )e − jπf
f
π 2
1234
f
Solutions 2-33
M. J. Roberts - 7/12/03
|X( f )| 1 2
...
e − jπf X( f ) = 2
... -1 43 21 41
∞
n δ f − ∑ 4 n =−∞
1 1 3 1 4 2 4
f
X( f ) 4π 3π 2π π -4 -3 -2 -1-π -2π -3π -4π
1 2 3 4
f
37. Sketch the even and odd parts of these CT signals. All sketches combined at the end below. (a)
x(t ) = rect(t − 1)
x e ( t) =
rect ( t − 1) + rect ( t + 1) 2
(b)
3 x(t ) = tri t − + tri t + 4
rect ( t − 1) − rect ( t + 1) 2
,
x o ( t) =
,
x 0 ( t) = 0
3 4
3 3 x e ( t) = tri t − + tri t + 4 4 (a)
(b)
xe(t)
xe(t)
1
-3
1
3
t
-3
3
-1
-1
xo(t)
xo(t)
1
-3
1
3
t
-3
-1
(c)
t − 1 x(t ) = 4 sinc 2 π cos t 2 x e ( t) = −8 π ( t 2 − 1)
t
3
t
-1
,
π cos t 2 x o ( t) = −8 t π ( t 2 − 1) Solutions 2-34
M. J. Roberts - 7/12/03
(d)
π x(t ) = 2 sin 4πt − rect(t ) 4
π x e ( t) = −2 sin cos( 4πt) rect ( t) 4
π x o ( t) = 2 cos sin( 4πt) rect ( t) 4
,
(c)
(d)
xe(t)
xe(t)
4
-10
2
10
t
-1
1
-4
-2
xo(t)
xo(t)
4
-10
2
10 -4
t
t
-1
1
t
-2
38. Let the CT unit impulse function be represented by the limit, 1 x δ ( x ) = lim tri , a > 0 . a→0 a a The function, (a)
(b)
(c)
1 x tri has an area of one regardless of the value of a. a a 1 4x What is the area of the function, δ ( 4 x ) = lim tri ? a→0 a a 1 x This is a triangle with the same height as tri but 1/4 times the base a a width. Therefore its area is 1/4 times as great or 1/4. 1 −6 x ? What is the area of the function, δ ( −6 x ) = lim tri a→0 a a 1 x This is a triangle with the same height as tri but 1/6 times the base a a width. (The fact that the factor is “-6” instead of “6” just means that the triangle is reversed in time which does not change its shape or area.) Therefore its area is 1/6 times as great or 1/6. 1 bx What is the area of the function, δ (bx ) = lim tri for b positive and a→0 a a for b negative ? for “b” positive and for “b” negative? It is simply “1/|b|”.
39. Using a change of variable and the definition of the unit impulse, prove that Solutions 2-35
M. J. Roberts - 7/12/03
δ ( a(t − t0 )) =
1 δ (t − t0 ) . a
δ ( x) = 0 , x ≠ 0 ,
[
∞
∫ δ ( x )dx = 1
−∞
]
δ a(t − t0 ) = 0 , where a(t − t0 ) ≠ 0 or t ≠ t0 ∞
[
]
Strength = ∫ δ a(t − t0 ) dt −∞
Let
a(t − t0 ) = λ and ∴ adt = dλ
Then, for a > 0,
∞
Strength = ∫ δ (λ ) −∞
and for a < 0,
−∞
∞
1 1 dλ 1 = ∫ δ (λ )dλ = = a a −∞ a a
−∞
∞
1 1 1 dλ 1 Strength = ∫ δ (λ ) = ∫ δ (λ )dλ = − ∫ δ (λ )dλ = − = a a∞ a −∞ a a ∞ Therefore for a > 0 and a < 0, 1 1 Strength = and δ a(t − t0 ) = δ (t − t0 ) . a a
[
]
40. Using the results of Exercise 39, show that (a) comb( ax ) =
1 a
∞
∑ δ x − a n
n = −∞
∞
From the comb definition, comb( ax ) = ∑ δ ( ax − n ) . −∞
∞ n 1 Then, using the property from Exercise 39, comb( ax ) = ∑ δ a x − = a a −∞
(b)
∞
−∞
the average value of comb( ax ) is one, independent of the value of a The period is 1/a. Therefore t0 +
comb( ax ) =
1 1 a
1 a
1 2a
1 2a
∫ comb(ax )dx = a ∫ comb(ax )dx = a ∫ δ (ax )dx
t0
−
comb( ax ) =
1 2a
1 2a
−
∫ δ ( x )dx = 1 1 2a
Solutions 2-36
−
1 2a
n
∑ δ x − a .
M. J. Roberts - 7/12/03
(c)
a comb function of the form,
1 t comb is a sequence of unit impulses a a
spaced a units apart. ∞ ∞ 1 1 t comb = × a ∑ δ (t − an) = ∑ δ (t − an) a a a n = −∞ n = −∞
and
(d)
even though δ ( at) =
1 1 δ ( t) , comb( ax ) ≠ comb( x ) a a
∞
1 δ ( ax − n ) ≠ ∑ a n =−∞ 1 a
∞
n 1 δ x − ≠ ∑ a a n =−∞
∞
∑ δ ( x − n)
n =−∞
∞
∑ δ ( x − n)
QED
n =−∞
πt 41. Sketch the generalized derivative of g(t ) = 3 sin rect(t ) . 2 1 1 Except at the discontinuities at t = ± , the derivative is either zero, for t > , or it 2 2 1 πt 3π πt is the derivative of 3 sin , cos , for t < . At the discontinuities the generalized 2 2 2 2 derivative is an impulse whose strength is the difference between the limit approached from 3 above and the limit approached from below. In both cases that strength is − . 2
d (g(t)) dt 3π 2 t 3 2 Alternate solution: 1 πt 1 g( t) = 3 sin u t + − u t − 2 2 2 d πt 1 1 πt 1 1 3π g( t)) = 3 sin δ t + − δ t − + cos u t + − u t − ( 2 2 2 2 2 2 2 dt
Solutions 2-37
M. J. Roberts - 7/12/03
d πt π 1 π 1 3π g( t)) = 3 sin − δ t + − 3 sin δ t − + cos rect ( t) ( 2 dt 4 2 4 2 2 2 1 d πt 1 3π g( t)) = −3 cos rect ( t) δ t + + δ t − + ( 2 2 2 2 2 dt 42. Sketch the following CT functions. g( t) = 3δ ( 3t) + 6δ ( 4 ( t − 2))
(a)
g(t) 3 2 1
3 Using the impulse scaling property, g(t ) = δ (t ) + δ (t − 2) , 2
t 2 t g(t ) = 2 comb − 5
(b)
g(t) ∞ ∞ 10 t g(t ) = 2 ∑ δ − − n = 10 ∑ δ (t + 5n) , 5 ... n = −∞ n = −∞
... 5 10 15 20
-10 -5
(c)
t
t g(t ) = comb(t ) rect 11
g(t) 5 1 t ∞ g(t ) = rect ∑ δ (t − n) = ∑ δ (t − n) , 11 n = −∞ n = −5
-5 -4 -3 -2 -1 (d) g(t ) =
t 1 2 3 4 5
t 1 t g( t) = 5 sinc comb 4 2 2 ∞ ∞ 5 t ∞ t t n sinc ∑ δ − n = 5 ∑ sinc δ (t − 2 n) = 5 ∑ sinc δ (t − 2 n) 4 n = −∞ 2 2 4 2 n = −∞ n = −∞
Solutions 2-38
M. J. Roberts - 7/12/03
g(t) 5
t 2 6 10 1 λ − 1 λ g( t) = ∫ comb − comb dλ 2 2 2 −∞ t
(e)
g(t) 1
-2 -1
1
2
t
3
43. What is the numerical value of each of the following integrals? ∞
∫ δ (t ) cos(48πt )dt = cos(0) = 1 , (b)
(a)
−∞
∞
∫ δ (t − 5) cos(πt )dt = cos(5π ) = −1
−∞
20
t 8 3 ∫0 δ (t − 8)Λ 32 dt = Λ 32 = 4 , (d)
(c)
20
t
8
0
3π sin 2 2 ∫−2δ (t − 1.5) sinc(t )dt = sinc(1.5) = 3π = − 3π 2 2 sin 6π ∫−2δ (t − 1.5) sinc(4t )dt = sinc(4 × 1.5) = 6π = 0 2
(e)
(f)
44. What is the numerical value of each of the following integrals? >>>Better figures needed ∞
(a)
∞
∞
∞
∞
∫ comb(t ) cos(48πt )dt = ∫ ∑ δ (t − n) cos(48πt )dt = ∑ ∫ δ (t − n) cos(48πt )dt
−∞
∞
−∞n = −∞
n = −∞ −∞
∞
∞
n = −∞
n = −∞
∫ comb(t ) cos(48πt )dt = ∑ cos(48nπ ) = ∑1 → ∞
−∞
Solutions 2-39
1
∫ δ (t − 8) rect 16 dt = rect 16 = 2
M. J. Roberts - 7/12/03
∞
∫ comb(t ) sin(2πt )dt
(b)
−∞ ∞
∞
∞
∞
∑ ∫ δ (t − n) sin(2πt )dt = ∑ sin(2nπ ) = 0
∫ comb(t ) sin(2πt )dt =
n = −∞ −∞
−∞
n = −∞
4 t − 2 rect(t )dt 4
20
∫ comb
(c)
0
t − 2 rect(2 + 4n) = 0 ∫ comb 4 rect(t )dt = 4n∑ ∫ δ (t − 2 − 4n) rect(t )dt = 4n∑ = −∞ = −∞ ∞
20
∞
20
0
0
∞
2
∫ comb(t ) sinc(t )dt =
(d)
−2
∞
2
∑ ∫ δ (t − n) sinc(t )dt = ∑ sinc(n) = 1
n = −∞ −2
n = −∞
45. Sketch the derivatives of these functions. (a)
g( t) = sin(2πt) sgn( t)
(b)
t g( t) = 2 tri − 1 2
− cos(2πt) , t < 0 g′ ( t) = 2π cos(2πt) , t ≥ 0
1 + t , − 2 < t < 0 g( t) = 1 − t , 0 < t < 2 ⇒ 0 , otherwise
sin(2πt) , cos(2πt) < 0 g′ ( t) = 2π − sin(2πt) , cos(2πt) > 0
g( t) = cos(2πt)
(c)
(a)
(b)
x(t)
1
1 -4
4
t
1
-4
4
t
-1
-1
-1
4
t
-4
t
1
t
dx/dt
1
6
1 -1
dx/dt
dx/dt
-6
(c)
x(t)
x(t)
-4
1 , − 2 < t < 0 g′ ( t) = −1 , 0 < t < 2 0 , otherwise
6
4
t
-1 -6
-1
Solutions 2-40
M. J. Roberts - 7/12/03
46. Sketch the derivatives of these functions. Compare the average values of the magnitudes of the derivatives. 1
... -2
... 2 4 6 8 10 12
t
1
... -2
1 4
-2
... 2 4 6 8 10 12
... 2 4 6 8 10 12
t
1
... -2
t
1 2
... -2
... 2 4 6 8 10 12
t
g'3 (t)
g'2 (t)
g'1 (t)
...
g3(t)
g2(t)
g1(t)
...
2 4 6 8 10 12
t
1
... -2
... 2 4 6 8 10 12
t
Average derivative is zero in each case. 47. A function, g( t) , has this description: It is zero for t < −5. It has a slope of –2 in the range, −5 < t < −2 . It has 1 Hz plus a constant in the shape of a sine wave of unit amplitude and with a frequency of 4 the range, −2 < t < 2 . For t > 2 it decays exponentially toward zero with a time constant of 2 seconds. It is continuous everywhere. Write an exact mathematical description of this function. 0 −10 − 2 t g( t) = sin πt 2 t −6e − 2
, t < −5 , − 5 < t < −2 , −2< t<2 , t>2
(a)
Graph g( t) in the range, −10 < t < 10 .
(b)
Graph g(2t) in the range, −10 < t < 10 .
(c)
Graph 2 g( 3 − t) in the range, −10 < t < 10 .
(d)
t + 1 Graph −2 g in the range, −10 < t < 10 . 2
Solutions 2-41
M. J. Roberts - 7/12/03
g(t) -10
g(2t) 10
t
-10
-8
-2g(( t+1)/2) 10
-16
t
-8
2g(3- t) -10
10
t
16
-10
10
t
48. Find the even and odd parts of each of these CT functions. g( t) = 10 sin(20πt) 10 sin(20πt) + 10 sin(−20πt) 10 sin(20πt) − 10 sin(−20πt) = 10 sin(20πt) g e ( t) = = 0 , x o ( t) = 2 2 (b) g( t) = 20 t 3 3 3 20 t 3 + 20(− t) 20 t 3 − 20(− t) g e ( t) = = 0 , g o ( t) = = 20 t 3 2 2 (c) x( t) = 8 + 7 t 2 2 2 8 + 7 t 2 + 8 + 7(− t) 8 + 7 t 2 − 8 − 7(− t) 2 x e ( t) = = 8 + 7 t , x o ( t) = =0 2 2 (d) x( t) = 1 + t 1 + t + 1 + (− t) 1 + t − 1 − (− t) x e ( t) = = 1 , x o ( t) = =t 2 2 (e) x( t) = 6 t 6 t + 6(− t) 6 t − 6(− t) g e ( t) = = 0 , g o ( t) = = 6t 2 2 (f) g( t) = 4 t cos(10πt) (a)
g e ( t) = g o ( t) = (g)
4 t cos(10πt) + 4 (− t) cos(−10πt) 4 t cos(10πt) + 4 (− t) cos(10πt) = =0 2 2
4 t cos(10πt) − 4 (− t) cos(−10πt) 4 t cos(10πt) − 4 (− t) cos(10πt) = = 4 t cos(10πt) 2 2 cos(πt) πt cos(πt) cos(−πt) cos(πt) cos(πt) + + −πt = πt −πt = 0 g e ( t) = πt 2 2
g( t) =
Solutions 2-42
M. J. Roberts - 7/12/03
cos(πt) cos(−πt) cos(πt) cos(πt) − + t t t π π π πt = cos(πt) − = g o ( t) = πt 2 2 g( t) = 12 +
(h)
g e ( t) =
12 +
sin( 4πt) 4πt
sin( 4πt) sin(−4πt) sin( 4πt) sin( 4πt) 12 + + 12 + + 12 + 4πt = 12 + sin( 4πt) 4πt 4πt −4πt = 4πt 2 2
sin( 4πt) sin(−4πt) sin( 4πt) sin( 4πt) − 12 − − 4 4 4πt = 0 4 t t t π π π − = g o ( t) = 2 2 g( t) = (8 + 7 t) cos( 32πt) 12 +
(i)
g e ( t) = g o ( t) = (j)
(8 + 7t) cos(32πt) + (8 − 7t) cos(−32πt) 2
(8 + 7t) cos(32πt) − (8 − 7t) cos(−32πt) 2
= 8 cos( 32πt) = 7 t cos( 32πt)
g( t) = (8 + 7 t 2 ) sin( 32πt) g e ( t) = g o ( t) =
(8 + 7t ) sin(32πt) + (8 + 7(− t) ) sin(−32πt) 2
2
2
(8 + 7t ) sin(32πt) − (8 + 7(− t) ) sin(−32πt) 2
2
2
=0
= (8 + 7 t 2 ) sin( 32πt)
49. Is there a function that is both even and odd simultaneously? Discuss. The only function that can be both odd and even simultaneously is the trivial signal, x(t ) = 0. Applying the definitions of even and odd functions, 0+0 0−0 x e (t ) = = 0 = s(t ) and x o (t ) = = 0 = s( t ) 2 2 proving that the signal is equal to both its even and odd parts and is therefore both even and odd. 50. Find and sketch the even and odd parts of the CT function, x(t).
Solutions 2-43
M. J. Roberts - 7/12/03
x(t) 2 1
1 2 3 4 5
t
-5 -4 -3 -2 -1 -1
xe(t)
xo(t)
2 1
2 1
1 2 3 4 5
t
-5 -4 -3 -2 -1 -1
1 2 3 4 5
-5 -4 -3 -2 -1 -1
t
51. For each of the following signals decide whether it is periodic and, if it is, find the period. (a) (b) (c) (d)
(e)
(f)
g( t) = 28 sin( 400πt) Periodic. Fundamental frequency = 200 Hz, Period = 5 ms. g( t) = 14 + 40 cos(60πt) Periodic. Fundamental frequency = 30 Hz Period = 33.33...ms. g( t) = 5 t − 2 cos(5000πt) Not periodic. g( t) = 28 sin( 400πt) + 12 cos(500πt) Periodic. Two sinusoidal components with periods of 5 ms and 4 ms. Least common multiple is 20 ms. Period of the overall signal is 20 ms. 2π s g( t) = 10 sin(5 t) − 4 cos( 7 t) Periodic. The Periods of the two sinusoids are 5 2π and s. Least common multiple is 2π . Period of the overall signal is 2π s . 7 g( t) = 4 sin( 3t) + 3 sin 3t Not periodic because least common multiple is infinite.
( )
52. The voltage illustrated in Figure E52 occurs in an analog-to-digital converter. Write a mathematical description of it.
Signal in A/D Converter x(t) 5
-0.1
0.3
t (ms)
Figure E52 Signal occurring in an A/D converter
Solutions 2-44
M. J. Roberts - 7/12/03
t − 5 × 10 −5 t − 1.5 × 10 −4 x( t) = tri + tri 5 × 10 −5 5 × 10 −5 53. A signal occurring in a television set is illustrated in Figure E52. Write a mathematical description of it. Signal in Television x(t) 5 -10
t (µs)
60
-10
Figure E52 Signal occurring in a television set t − 2.5 × 10 −6 x( t) = −10 rect 5 × 10 −6 54. The signal illustrated in Figure E54 is part of a binary-phase-shift-keyed (BPSK) binary data transmission. Write a mathematical description of it. BPSK Signal x(t) 1
t (ms)
4 -1
Figure E54 BPSK signal t − 0.5 × 10 −3 t − 1.5 × 10 −3 8000 t ( π ) sin rect − sin(8000πt) rect 10 −3 10 −3 x( t) = t − 2.5 × 10 −3 t − 3.5 × 10 −3 + sin(8000πt) rect − sin(8000πt) rect 10 −3 10 −3 55. This signal illustrated in Figure E55 is the response of an RC lowpass filter to a sudden change in excitation. Write a mathematical description of it. RC Filter Signal x(t)
4 20
t (ns)
-1.3333
-6
Figure E55 Transient response of an RC filter Solutions 2-45
M. J. Roberts - 7/12/03
00πt) rect ( t − 0.5 × 10 −3 ) − sin(8000πt) rect ( t − 1.5 × 10 −3 ) + sin(8000πt) rect ( t − 2.5 × 10 −3 ) − sin(8000πt) rect ( t − 3.5 × t −4 − x( t) = −4 − 21 − e 3 u( t − 4 )
56. Find the signal energy of each of these signals: 2 rect( −t ) , E =
(a)
1 2
∞
∫ [2 rect(−t )] dt = 4 ∫ dt = 4 2
−∞
rect(8t ) , E =
(b)
−
1 16
∞
∫ [rect(8t )] dt = ∫ dt = 8 2
−∞
t 3 rect , E = 4
(c)
tri(2t ) , E =
(d)
−
∞
1
1 16
2
2
t ∫−∞3 rect 4 dt = 9−∫2 dt = 36 1 2
∞
1 2
∫ [tri(2t )] dt = ∫ (1 − 2t ) dt = ∫ (1 − 2 2t + 2t )dt 2
−∞
−
2
2
1 2
−
1 2
∫ (1 + 4t + 4t )dt + ∫ (1 − 4t + 4t )dt = t + 2t 0
E=
1 2
2
2
−
1 2
2
+4
0
t 3 tri , E = 4
∞
1 2
0
1 2
t t 1 + t − 2t 2 + 4 = 3 − 1 3 0 3 3
3
2
4 2 t t t t = − + ∫ 3 tri 4 dt = 9−∫41 − 4 dt 9−∫41 2 4 4 dt −∞ 4 t 2 t 3 0 t 2 t 3 4 0 t t2 t t2 E = 9 ∫ 1 + + dt + ∫ 1 − + dt = 9 t + + + t − + = 24 2 16 4 48 −4 4 48 0 0 −4 2 16 (f) 2sin(200 πt ) ∞ ∞ ∞ 1 1 2 2 E = ∫ [2 sin(200πt )] dt = 4 ∫ sin (200πt )dt = 4 ∫ − sin( 400πt ) dt 2 2 −∞ −∞ −∞
(e)
2
2
4
∞
(g)
cos( 400πt ) E = 2 t + →∞ 400π −∞ (Hint: First find the signal energy of a signal which approaches an impulse some limit, then take the limit.) t 1 δ (t ) = lim rect a a→0 a
δ (t )
E=
∞
2
a 2
1 t 1 t a lim rect dt = lim 2 ∫ rect dt = lim 2 → ∞ a→0 a a a a→0 a a→0 a a −∞
∫
−
2
Solutions 2-46
M. J. Roberts - 7/12/03
(h)
d (rect (t)) dt
x( t) =
d 1 1 rect ( t)) = δ t + − δ t − ( 2 2 dt ∞
2
1 1 E x = ∫ δ t + − δ t − dt → ∞ 2 2 −∞ t
x( t) =
(i)
1
1
∫ rect (λ )dλ = ramp t + 2 − ramp t − 2
−∞ 1 2
Ex =
∞
2
1 ∫1 t + 2 dt + ∫1 dt → ∞ − 2 4 2 1 4244 3 { infinite
finite
x( t) = e( −1− j 8π ) t u( t)
(j)
∞
Ex =
∫ x(t)
2
∞
dt =
−∞
∫e
( −1− j 8π ) t
−∞
∞
u( t) dt = ∫ e( −1− j 8π ) t e( −1+ j 8π ) t dt 2
0
∞
∞
e −2 t 1 E x = ∫ e dt = = −2 0 2 0 −2 t
57. Find the average signal power of each of these signals: (a)
x( t) = 2 sin(200πt) Px =
1 T
This is a periodic function. Therefore
T 2
T 2
1
∫ [2 sin(200πt)] dt = T ∫ 2 − 2 cos(400πt)dt
−
2
T 2
4
−
1
T 2
T
2 sin( 400πt) 2 2 T sin(200πT ) T sin(−200πT ) = − Px = t − + + =2 400π − T T 2 400π 2 400π T 2
For any sinusoid, the average signal power is half the square of the amplitude. (b)
(c)
x( t) = comb( t) This is a periodic signal whose period, T, is 1. Between T/2 and +T/2, there is one impulse whose energy is infinite. Therefore the average power is the energy in one period, divided by the period, or infinite. x( t) = e j100πt This is a periodic function. Therefore
Solutions 2-47
M. J. Roberts - 7/12/03
Px =
1 1 2 x( t) dt = ∫ T0 T0 T0
T0 2
∫
−
2
e j100πt dt = 50
T0 2
1 100
∫
−
e j100πt e − j100πt dt
1 100
1 100
∫
Px = 50
dt = 1
1 − 100
58. Sketch these DT exponential and trigonometric functions. (a)
2πn g[ n ] = −4 cos 10
(b)
g[ n ] = −4 cos(2.2πn )
(c)
g[ n ] = −4 cos(1.8πn )
(d)
2πn 2πn g[ n ] = 2 cos − 3 sin 6 6
(e)
3 g[ n ] = 4
(f)
2πn n g[ n ] = 2(0.9) sin 4
n
(a)
(b)
g[n]
(c)
g[n]
4
g[n]
4
-5
20
n
-4
4
-5
20
n
-4
(d)
20
(f)
g[n]
4
g[n]
4
20
-4
n
n
-4
(e)
g[n]
-5
-5
4
-5
20
-4
n
-5
20
n
-4
59. Sketch these DT singularity functions. (a)
g[ n ] = 2 u[ n + 2]
(b)
g[ n ] = u[5 n ]
(c)
g[ n ] = −2 ramp[− n ]
(d)
n g[ n ] = 10 ramp 2
(e)
g[ n ] = 7δ [ n − 1]
(f)
g[ n ] = 7δ [2( n − 1)]
Solutions 2-48
M. J. Roberts - 7/12/03
(a)
(b)
g[n]
(c)
g[n]
2
g[n]
1
-5
20
n
-5
-5
20
(d)
n
20
-10
(e)
g[n]
(f)
g[n]
100
g[n]
7
-5
20
n
7
-5
20
n
-5
20
(g)
2 g[ n ] = −4δ n 3
(h)
2 g[ n ] = −4δ n − 1 3
(i)
g[ n ] = 8 comb 4 [ n ]
(j)
g[ n ] = 8 comb 4 [2 n ]
(k)
g[ n ] = rect 4 [ n ]
(l)
n g[ n ] = 2 rect 5 3
(g)
(h)
g[n] 20
n
-4
g[n]
-5
20
n
-4
(k)
(m)
n g[ n ] = tri 5
(o)
n + 1 g[ n ] = sinc 4
n
n
g[n]
1
20
20
(l)
g[n]
8
-5
8
-5
(j) g[n]
n
(i)
g[n]
-5
n
2
-5
20
(n)
n
-20
n g[ n ] = − sinc 4
Solutions 2-49
20
n
M. J. Roberts - 7/12/03
(m)
(n)
(o)
g[n]
g[n]
g[n]
1
1
1
-20
-10
(p)
10
20
n
n
-20
20
-1
n
-1
n g[ n ] = drcl , 9 10 (p) g[n] 1
-20
20
n
-1
60. Sketch these combinations of DT functions. (a)
g[ n ] = u[ n ] + u[− n ]
(b)
g[ n ] = u[ n ] − u[− n ]
(c)
2πn g[ n ] = cos comb 3 [ n ] 12
(d)
2πn n g[ n ] = cos comb 3 12 2
2πn sin u[ n ] 8
(f)
2πn g[ n ] = sin u[ n ] 4
(e)
g[ n ] = 5e
−
n 16
(a)
(b)
g[n]
g[n]
2
(c) g[n]
1
1
-10 10
-10
10
n
(g)
20
(e)
(f) g[n]
5
20
n
-5
n
-1
g[n]
1
-1
-5
-1
(d) g[n]
-5
n
1
n 30
20 -5
-5
-1
2π ( n + 1) 2πn g[ n ] = cos u[ n ] u[ n + 1] − cos 12 12
Solutions 2-50
n
M. J. Roberts - 7/12/03
(h)
(i)
(j)
(k)
g[ n ] = g[ n ] = g[ n ] =
2πm u[ m] 12
n
∑ cos
m =0 n
∑ (comb [m] − comb [m − 2]) 4
m =0
4
n
∑ (comb [m] + comb [m]) rect [m] 4
m =−∞
3
4
g[ n ] = comb 2 [ n + 1] − comb 2 [ n ] (g)
(h)
g[n]
20
n
20
(j)
-5
20
(k)
g[n]
1
1
-5
10
n
n
(l)
g[n]
6
∑ δ [ m]
m =−∞
n
-2
g[n]
n
1
-5
-1
∑
m =−∞
δ [ m] −
g[n]
2
-5
n +1
(i)
g[n]
1
-10
g[ n ] =
(l)
20 -1
n
-5
10
n
61. Sketch the magnitude and phase of each function versus k. πk
(a)
2πk − j 4 G[ k ] = 20 sin e 8
(c)
G[ k ] = (δ [ k + 8] − 2δ [ k + 4 ] + δ [ k ] − 2δ [ k − 4 ] + δ [ k − 8])e
(b)
2πk k G[ k ] = 20 cos sinc 8 40
Solutions 2-51
j
πk 8
M. J. Roberts - 7/12/03
(a)
(b)
(c)
|G[k]|
|G[k]|
|G[k]|
20
20
-16
16
k
Phase of G[k]
-16
16
k
-16
Phase of G[k]
π -16
2
16
Phase of G[k]
π 16
k
-16
-π
k
π
16
k
-16
-π
16
k
-π
62. Given the function definitions on the left, find the function values on the right. (a) g[ n ] =
3n + 6 −2 n e 10
g[ 3] =
3( 3) + 6 −2( 6) e = 0.0000092 10
(b) 1 + j n g[ n ] = Re 2 (c) 2 g[ n ] = ( j 2πn ) + j10πn − 4
1 + j 5 1 g[5] = Re =− 2 2 g[ 4 ] = ( j 2π ( 4 )) + j10π ( 4 ) − 4 = −635.7 + j125.7 2
63. Using MATLAB, for each function below plot the original function and the transformed function. (a)
5 , n ≤ 0 5 − 3n , 0 < n ≤ 4 g[ n ] = 2 −23 + n , 4 < n ≤ 8 41 , n > 8
g[ 3n ]
Solutions 2-52
vs. n
M. J. Roberts - 7/12/03
(a) g[n] 50
-10
20
-10
n
g[3n] 50
-10
(b)
20
-10
2πn 2πn g[ n ] = 10 cos cos 20 4
n
4 g[2( n + 1)]
vs. n
(b) g[n] 10
40
n
-10
4g[2(n+1)] 40
40
n
-40
(c) g[ n ] = 8e
j
2πn 16
u[ n ]
n g 2
Solutions 2-53
vs. n
M. J. Roberts - 7/12/03
(c) g[n] 10
-10
30
n
g[n/2] 10
-10
30
n
64. Given the graphical definition of a function, g[n], graph the indicated function(s), h[n]. (a) g[n] 6
h[ n ] = g[2 n − 4 ]
4 2 -8
-6
-4
-2
2
4
6
n
8
-2 -4 -6
g[ n ] = 0 , n > 8 g[2n - 4] 6 4 2 -8
-6
-4
-2
2
4
6
8
n
-2 -4 -6
(b) g[n] 6 4 2 -8
-6
-4
-2
2
4
6
8
n
n h[ n ] = g 2
-2 -4 -6
g[ n ] = 0 , n > 8
Solutions 2-54
M. J. Roberts - 7/12/03
g[
n 2
]
6 4 2 -8
-6
-4
-2
2
4
6
8
n
-2 -4 -6
(c) g[n]
n h[ n ] = g 2
6 4 2 -8
-6
-4
-2
2
4
6
n
8
-2 -4 -6
g[ n ] is periodic n
g[ 2 ] 6 4 2 -8
-6
-4
-2
2
4
6
8
n
-2 -4 -6
65. Sketch the accumulation from negative infinity to n of each of these DT functions. (a)
g[ n ] = cos(2πn ) u[ n ]
(b)
g[ n ] = cos( 4πn ) u[ n ]
(a) Accumulation of g[n] 20
-5
20
n
(b) Accumulation of g[n] 20
-5
20
n
66. Find and sketch the magnitude and phase of the even and odd parts of each of this “discrete-k” function.
Solutions 2-55
M. J. Roberts - 7/12/03
G[ k ] =
10 1− j4k
10 10 + 10 10 1− j4k 1 + j4k G e [k ] = = = 2 (1 − j 4 k )(1 + j 4 k ) 1 + 16k 2 10 10 − j 40 k j 40 k 1− j4k 1 + j4k G o[k ] = = = 2 (1 − j 4 k )(1 + j 4 k ) 1 + 16k 2 |Ge[k]|
|Go[k]|
10
10
-10
10
k
-10
10
Phase of Ge[k]
k
Phase of Go[k]
π
π
-10
10
k
-10
10
-π
k
-π
67. Find and sketch the even and odd parts of the DT function below. g[n] 6 4 2 -8
-6
-4
-2
2
4
6
8
n
-2 -4 -6
ge[n]
go[n]
6
6
4
4
2 -8
-6
-4
-2
2 2
4
6
8
n
-8
-6
-4
-2
2
-2
-2
-4
-4
-6
-6
4
6
8
n
68. Using MATLAB, plot each of these DT functions. If a function is periodic, find the period analytically and verify the period from the plot.
Solutions 2-56
M. J. Roberts - 7/12/03
(a)
3πn 6πn g[ n ] = sin = sin 2 4
(b)
2πn 10πn g[ n ] = sin + cos 3 3
Period is 4
2πn 6πn 4πn 2πn 4πn g[ n ] = sin + + cos = sin + cos 3 3 3 14234 3 142343 Period = 3
Period = 3
Period is 3 (c)
2πn 2πn g[ n ] = 5cos + 3sin 14284 3 14254 3 Period of 8
Period of 5
LCM of the periods is 40, therefore the period is 40. g[n]
(a)
1 60
n
-1
g[n]
(b)
2 60
n
-2
g[n]
(c)
8 60
n
-8
(d)
n g[ n ] = 10 cos 4
(e)
2πn 2πn g[ n ] = −3 cos sin (A trigonometric identity will be useful here.) 7 6
Not periodic.
3 2πn 2πn 2πn 2πn g[ n ] = − sin − + + sin 6 2 6 7 7 3 12πn 3 40πn 12πn 40πn g[ n ] = − sin − + sin = − sin − sin 42 42 2 42 2 42 The period is 42.
Solutions 2-57
M. J. Roberts - 7/12/03
(d)
g[n] 10
n
60 -10
(e)
g[n] 4
n
60 -4
69. Sketch the following DT functions. g[n]
(a)
g[ n ] = 5δ [ n − 2] + 3δ [ n + 1]
5 3 n -1
2 g[n]
(b)
g[ n ] = 5δ [2 n ] + 3δ [ 4 ( n − 2)]
5 3 n 2 g[n]
(c)
g[ n ] = 5(u[ n − 1] − u[ 4 − n ])
5 n -5
2
4
6
g[n]
(d)
g[ n ] = 8 rect 4 [ n + 1]
8 n -6
-4
-2
2
4
6
g[n] 10
(e)
2πn g[ n ] = 8 cos 7
5
n 5
10
-5
-10
Solutions 2-58
15
20
M. J. Roberts - 7/12/03
g[n]
(f)
n 4
n
g[ n ] = −10e u[ n ]
-2
2
4
6
8
-20
-40
-60
-80
g[n]
(g)
n
g[ n ] = −10(1.284 ) u[ n ] n
-2
2
4
6
8
-20
-40
-60
-80
g[n] 1 0.8
n
(h)
j g[ n ] = u[ n ] 4
0.6 0.4 0.2
-4
(i)
-2
2
4
6
8
n
g[ n ] = ramp[ n + 2] − 2 ramp[ n ] + ramp[ n − 2] g[n] 2
1.5
1
0.5
-4
-2
2
4
6
8
n
g[n] 1 0.8 0.6
(j)
g[ n ] = rect 2 [ n ] comb 2 [ n ]
0.4 0.2
-8
(k)
-6
-4
-2
g[ n ] = rect 2 [ n ] comb 2 [ n + 1]
Solutions 2-59
2
4
6
8
n
M. J. Roberts - 7/12/03
g[n] 1 0.8 0.6 0.4 0.2
-8
-6
-4
-2
2
4
6
8
5
10
15
n
g[n] 3 2
(l)
1
2πn g[ n ] = 3 sin rect 4 [ n ] 3
n -15
-10
-5 -1 -2 -3
g[n] 5
(m)
2πn n -5 g[ n ] = 5 cos u 8 2
n 5
10
15
20
25
30
-5
70. Graph versus k , in the range, −10 < k < 10 , the magnitude and phase of |X[k]| 1
(a)
k X[ k ] = sinc 2
-20
20
k
Phase of X[k] π
-20
20 -π
Solutions 2-60
k
M. J. Roberts - 7/12/03
|X[k]| 1
(b)
k − j X[ k ] = sinc e 2
2πk 4
-20
20
k
Phase of X[k] π
-20
20
k
-π
|X[k]| 1
(c)
X[ k ] = rect 3 [ k ]e
−j
2πk 3
-15
15
k
Phase of X[k] π
-15
15 -π
|X[k]| 1
(d)
X[ k ] =
1 k 1+ j 2
-15
15
k
Phase of X[k] π
-15
15 -π
Solutions 2-61
k
k
M. J. Roberts - 7/12/03
|X[k]| 2
(e)
X[ k ] =
jk -15
k 1+ j 2
15
k
Phase of X[k] π
-15
15
k
-π
|X[k]| 1
(f)
X[ k ] = comb 2 [ k ]e
−j
2πk 4
-15
15
k
Phase of X[k] π
-15
15 -π
71. Sketch the even and odd parts of these signals. All plots below. (a)
x[ n ] = rect 5 [ n + 2]
Solutions 2-62
k
M. J. Roberts - 7/12/03
xe[n] 1
-10
10
n
-1
xo[n] 1
-10
10
n
-1
(b)
x[ n ] = comb 3 [ n − 1] xe[n] 1
-10
10
n
-1
xo[n] 1
-10
10 -1
(c)
2πn π x[ n ] = 15 cos + 9 4 xe[n] 20
-10
10
n
-20
xo[n] 20
-10
10 -20
Solutions 2-63
n
n
M. J. Roberts - 7/12/03
(d)
2πn x[ n ] = sin rect 5 [ n − 1] 4 xe[n] 1
-10
10
n
-1
xo[n] 1
-10
10
n
-1
72. What is the numerical value of each of the following accumulations? 10
(a)
∑ ramp[n] = 0 + 1 + 2 + L + 10 = 55 n =0 6
(b)
(c)
1 1 +L + 6 . 2 2 n =0 N , α = 1 N −1 n Using ∑ α = 1 − α N , otherwise n =0 1− α 7 1 1 1− 1 − 6 1 2 128 = 127 = ∑ n = 1 1 256 n =0 2 1− 2 2 ∞ u[ n ] ∑ n n =−∞ 2 ∞ 1 , α <1 Using ∑ α n = 1− α n =0 ∞ u[ n ] 1 =2 ∑ n = 1 n =0 2 1− 2 1
∑2
n
=1+
10
(d)
∑ comb [n] = 7
n =−10 10
(e)
3
∑ comb [2n] = 7
n =−10
3
Solutions 2-64
M. J. Roberts - 7/12/03 ∞
(f)
∑ sinc[n] = 1
n =−∞
73. Find the signal energy of each of these signals: (a)
x[ n ] = 5 rect 4 [ n ]
Ex =
(b) x[ n ] = 2δ [ n ] + 5δ [ n − 3] E x = u[ n ] n
∞
∑
n =−∞
∞
∑ x[n]
2
n =−∞
∞
2
n =−∞
n =−4
∞
∑ 4 δ[n]
2δ [ n ] + 5δ [ n − 3] = 2
(c)
x[ n ] =
(d)
1 x[ n ] = − u[ n ] 3
(e)
πn x[ n ] = cos (u[ n ] − u[ n − 6]) 3
4
= 25 ∑ rect 4 [ n ] = 25 ∑ (1) = 225
2
n =−∞
+ 25 δ [ n − 3] = 29 2
The energy is infinite. ∞
2
∞ 1 1 E x = ∑ − u[ n ] = ∑ = 3 n =−∞ n =0 9
n
n
n
2
1 1 1− 9
πn πn E x = ∑ cos (u[ n ] − u[ n − 6]) = ∑ cos2 3 3 n =−∞ n =0 ∞
2
5
2
2
2
1 1 1 1 2 E x = 1 + + − + (−1) + − + = 3 2 2 2 2 2
74. Find the average signal power of each of these signals: (a) x[ n ] = u[ n ] 1 Px = lim N →∞ 2 N
N −1
∑ x[n]
n =− N
(b) x[ n ] = (−1)
n
2
1 = lim N →∞ 2 N
N −1
∑ u[n]
2
n =− N
1 Px = lim N →∞ 2 N
N −1
∑ (−1)
1 = lim N →∞ 2 N n 2
n =− N
N −1
1
∑ (1) = 2 n =0
1 = lim N →∞ 2 N
N −1
∑ (1) = 1
n =− N
(c) x[ n ] = A cos(2πF0 n + θ ) 1 N →∞ 2 N
Px = lim
N −1
∑
n =− N
A2 N →∞ 2 N
A cos(2πF0 n + θ ) = lim 2
N −1
∑ cos (2πF n + θ ) 2
n =− N
Using the trigonometric identity,
Solutions 2-65
0
=
9 8
M. J. Roberts - 7/12/03
cos( x ) cos( y ) = A2 Px = lim N →∞ 2 N
N −1
[
]
1 cos( x − y ) + cos( x + y ) , 2
[
]
A2 A2 1 F n 1 4 2 ∑ + cos( π 0 + θ ) = 2 + Nlim →∞ 4 N n =− N 2
N −1
∑ cos(4πF n + 2θ ) 0
n =− N
A2 , plus another term, So the average power is a constant, 2 A2 N →∞ 4 N
N −1
lim
∑ cos(4πF n + 2θ ) , 0
n =− N
whose value depends on the parameters, F0 and θ . Case 1. 2 F0 , an integer. A2 A2 lim Px = + 2 N →∞ 4 N
N −1
∑
n =− N
cos(2θ ) =
A2 [1 + cos(2θ )] 2
Case 2. 2 F0 , not an integer. Subcase 1. The signal is periodic. If it is periodic with period, N 0 = The square of the function is periodic with period, N 0 =
1 . F0
1 . For a periodic 2 F0
1 k + N −1 2 signal, Px = x[ n ] where “N” is any integer number of periods of the ∑ N n =k signal and where “k” is any integer. Then the average power is Px =
A2 A2 + 2 4N
k + N −1
∑ cos(2θ ) n =k
A 2 k + N −1 ∑ cos(2θ ) is zero because the samples are taken at 4 N n =k equal angular intervals over exactly an integer number of periods. Therefore A2 . the average power is Px = 2 Subcase 2. The signal is not periodic. This is the hardest case to examine. We cannot use the periodic formula so we must use
The summation,
Px =
A2 A2 + lim 2 N →∞ 4 N
N −1
∑ cos(4πF n + 2θ )
n =− N
0
The summation can be geometrically visualized as the real part of a sum of vectors in the complex plane, all with unit length and separated from their
Solutions 2-66
M. J. Roberts - 7/12/03
nearest neighbors by the angle, 4πF0 . Since this angle is not any integer multiple of π radians (that case has already been considered), the summation is of an infinity of vectors at angles which uniformly fill up the full 2π radians of a full circle. We don’t know what that sum is in the limit, but we do know it cannot grow to infinity because of the cancellation of the vectors arrayed in the circle and the sum is being divided by N which is going to A2 , just as in the infinity. Therefore, in the limit, the average power is Px = 2 periodic case. , ,17,18,19,L A , n = L, 0,1, 2, 3, 8, 9,10,1116 (d) x[ n ] = 0 , n = L, 4, 5, 6, 7,12,13,14,15, 20, 21, 22, 23,L This is a discrete-time rectangular wave with 50% duty cycle. Therefore the average A2 power is the average of the square of the signal’s values or Px = . 2 (e)
x[ n ] = e
−j
πn 2
This is a periodic function. 2
1 1 3 1 3 − j π2n 1 3 2 2 Px = x n = x n = e = [ ] [ ] ∑ ∑ ∑ ∑1 = 1 4 n =0 4 n =0 4 n =0 N 0 n = N0
Solutions 2-67
M. J. Roberts - 7/12/03
Chapter 3 - Mathematical Description and Analysis of Systems Solutions 1. Show that a system with excitation, x( t) , and response, y( t) , described by y( t) = u( x( t)) is non-linear, time invariant, stable and non-invertible. Homogeneity: Let x1 ( t) = g( t) . Then y1 ( t) = u(g( t)) . Let x 2 ( t) = K g( t) . Then y 2 ( t) = u(K g( t)) ≠ K y1 ( t) = K u(g( t)) . Not homogeneous Additivity: Let x1 ( t) = g( t) . Then y1 ( t) = u(g( t)) . Let x 2 ( t) = h( t) . Then y 2 ( t) = u(h( t)) . Let x 3 ( t) = g( t) + h( t) . Then y 3 ( t) = u(g( t) + h( t)) ≠ y1 ( t) + y 2 ( t) = u(g( t)) + u(h( t)) Not additive Since it is not homogeneous and not additive, it is not linear. It is also not incrementally linear. It is statically non-linear because it is non-linear without memory (lack of memory proven below). Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) = u(g( t)) . Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = u(g( t − t0 )) = y1 ( t − t0 ) . Time Invariant Stability: The unit step function can only have the values, zero or one. Stable Causality: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any future values. Causal
Solutions 3-1
M. J. Roberts - 7/12/03
Memory: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any past values. System has no memory. Invertibility: There are many value of the excitation that all cause a response of zero and there are many values of the excitation that all cause a response of one. Therefore the system is not invertible. 2. Show that a system with excitation, x( t) , and response, y( t) , described by y( t) = x( t − 5) − x( 3 − t) is linear but not causal and not invertible. Homogeneity: Let x1 ( t) = g( t) . Then y1 ( t) = g( t − 5) − g( 3 − t) . Let x 2 ( t) = K g( t) . Then y 2 ( t) = K g( t − 5) − K g( 3 − t) = K y1 ( t) . Homogeneous Additivity: Let x1 ( t) = g( t) . Then y1 ( t) = g( t − 5) − g( 3 − t) . Let x 2 ( t) = h( t) . Then y 2 ( t) = h( t − 5) − h( 3 − t) . Let x 3 ( t) = g( t) + h( t) . Then y 3 ( t) = g( t − 5) + h( t − 5) − g( 3 − t) − h( 3 − t) = y1 ( t) + y 2 ( t) Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) = g( t − 5) − g( 3 − t) . Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = g( t − t0 − 5) − g( 3 − t + t0 ) = y1 ( t − t0 ) . Time Invariant Stability: If x( t) is bounded then x( t − 5) and x( 3 − t) are bounded and so is y( t) = x( t − 5) − x( 3 − t) . Stable Causality: At time, t = 0, y(0) = x(−5) − x( 3) . Therefore the response at time, t = 0, depends on the excitation at a later time, t = 3. Not Causal Memory: At time, t = 0, y(0) = x(−5) − x( 3) . Therefore the response at time, t = 0, depends on the excitation at a previous time, t = −5.
Solutions 3-2
M. J. Roberts - 7/12/03
System has memory. Invertibility: A counterexample will demonstrate that the system is not invertible. Let the excitation be a constant, K. Then the response is y( t) = K − K = 0. This is the response, no matter what K is. Therefore when the output is a constant zero, the input cannot be determined. Not Invertible. 3. Show that a system with excitation, x( t) , and response, y( t) , described by t y( t) = x 2 is linear, time variant and non-causal. Homogeneity: t Let x1 ( t) = g( t) . Then y1 ( t) = g . 2 t Let x 2 ( t) = K g( t) . Then y 2 ( t) = K g = K y1 ( t) . 2 Homogeneous Additivity: t Let x1 ( t) = g( t) . Then y1 ( t) = g . 2 t Let x 2 ( t) = h( t) . Then y 2 ( t) = h . 2 t t Let x 3 ( t) = g( t) + h( t) . Then y 3 ( t) = g + h = y1 ( t) + y 2 ( t) 2 2 Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: t Let x1 ( t) = g( t) . Then y1 ( t) = g . 2 t − t0 t Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = g − t0 ≠ y1 ( t − t0 ) = g . 2 2 Time Variant Stability: If x( t) is bounded then y( t) is bounded. Stable
Solutions 3-3
M. J. Roberts - 7/12/03
Causality: At time, t = −2, y(−2) = x(−1) . Therefore the response at time, t = −2, depends on the excitation at a later time, t = −1. Not Causal Memory: At time, t = 2, y(2) = x(1) . Therefore the response at time, t = 2, depends on the excitation at a previous time, t = 1. System has memory. Invertibility: The system excitation at any arbitrary time, t = t0 , is uniquely determined by the system response at time, t = 2 t0 . Invertible. 4. Show that a system with excitation, x( t) , and response, y( t) , described by y( t) = cos(2πt) x( t) is time variant, BIBO stable, static and non-invertible. Homogeneity: Let x1 ( t) = g( t) . Then y1 ( t) = cos(2πt) g( t) . Let x 2 ( t) = K g( t) . Then y 2 ( t) = cos(2πt)K g( t) = K y1 ( t) . Homogeneous Additivity: Let x1 ( t) = g( t) . Then y1 ( t) = cos(2πt) g( t) . Let x 2 ( t) = h( t) . Then y 2 ( t) = cos(2πt) h( t) . Let x 3 ( t) = g( t) + h( t) . Then y 3 ( t) = cos(2πt)[g( t) + h( t)] = y1 ( t) + y 2 ( t) Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) = cos(2πt) g( t) . Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = cos(2πt) g( t − t0 ) ≠ y1 ( t − t0 ) = cos(2π ( t − t0 )) g( t − t0 ) . Time Variant Stability: If x( t) is bounded then y( t) is bounded because it is multiplied by a cosine which is bounded. Stable Causality:
Solutions 3-4
M. J. Roberts - 7/12/03
The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any later time. Causal Memory: The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any earlier time. System has no memory (static). Invertibility: Solve the system equation for the excitation as a function of the response, x( t) =
y( t) cos(2πt)
This system is not invertible because when the cosine function is zero the unique relationship between x and y is lost; any x produces the same y, zero. Not Invertible. 5. Show that a system whose response is the magnitude of its excitation is non-linear, BIBO stable, causal and non-invertible. y( t) = x( t) Homogeneity: Let x1 ( t) = g( t) . Then y1 ( t) = g( t) . Let x 2 ( t) = K g( t) . Then y 2 ( t) = K g( t) = K y1 ( t) . If K is negative, K y1 ( t) ≠ K y1 ( t) . Not Homogeneous. Additivity: Let x1 ( t) = g( t) . Then y1 ( t) = g( t) . Let x 2 ( t) = h( t) . Then y 2 ( t) = h( t) . Let x 3 ( t) = g( t) + h( t) . Then y 3 ( t) = g( t) + h( t) ≠ y1 ( t) + y 2 ( t) = g( t) + h( t) Not Additive Since it is neither homogeneous nor additive, it is also nonlinear. It is not incrementally linear because it is not linear except for the addition of a constant. It is statically non-linear because its excitation-response relationship is not a straight line. y(t)
x(t)
Solutions 3-5
M. J. Roberts - 7/12/03
Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) = g( t) . Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = g( t − t0 ) = y1 ( t − t0 ) . Time Invariant Stability: If x( t) is bounded then y( t) is bounded. Stable Causality: The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any later time. Causal Memory: The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any earlier time. System has no memory (static). Invertibility: Any response, y, can be caused by either x or –x. Not Invertible. 6. Show that the system in Figure E6 is linear, time invariant, BIBO unstable, and dynamic.
x(t)
0.1
∫
∫
∫
y(t)
1.4 -0.7 2.5 Figure E6 A CT system The differential equation of the system is 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y( t) = x( t) . Homogeneity: Let x1 ( t) = g( t) . Then 10 y1′′′ ( t) − 14 y1′′ ( t) + 7 y1′ ( t) − 25 y1 ( t) = g( t) . Let x 2 ( t) = K g( t) . Then 10 y′′′ 2 ( t) − 14 y′′ 2 ( t) + 7 y′2 ( t) − 25 y 2 ( t) = K g( t) . If we multiply the first equation by K, we get 10K y1′′′ ( t) − 14 K y1′′ ( t) + 7K y1′ ( t) − 25K y1 ( t) = K g( t) Therefore 10K y1′′′ ( t) − 14 K y1′′ ( t) + 7K y1′ ( t) − 25K y1 ( t) = 10 y′′′ 2 ( t) − 14 y′′ 2 ( t) + 7 y′2 ( t) − 25 y 2 ( t) This can only be true for all time if y 2 ( t) = K y1 ( t) . Homogeneous Solutions 3-6
M. J. Roberts - 7/12/03
Additivity: Let x1 ( t) = g( t) . Then 10 y1′′′ ( t) − 14 y1′′ ( t) + 7 y1′ ( t) − 25 y1 ( t) = g( t) . Let x 2 ( t) = h( t) . Then 10 y′′′ 2 ( t) − 14 y′′ 2 ( t) + 7 y′2 ( t) − 25 y 2 ( t) = h( t) . Let x 3 ( t) = g( t) + h( t) . Then 10 y′′′ 3 ( t) − 14 y′′ 3 ( t) + 7 y′3 ( t) − 25 y 3 ( t) = g( t) + h( t) Adding the first two equations, 10[ y1′′′ ( t) + y′′′ 2 ( t)] − 14[ y1′′ ( t) + y′′ 2 ( t)] + 7[ y1′ ( t) + y′2 ( t)] − 25[ y1 ( t) | + y 2 ( t)] = g( t) + h( t) Therefore 10[ y1′′′ ( t) + y′′′ 2 ( t)] − 14[ y1′′ ( t) + y′′ 2 ( t)] + 7[ y1′ ( t) + y′2 ( t)] − 25[ y1 ( t) | + y 2 ( t)] = 10 y′′′ 3 ( t) − 14 y′′ 3 ( t) + 7 y′3 ( t) − 25 y 3 ( t)
10[ y1 ( t) | + y 2 ( t)]′′′ − 14[ y1 ( t) | + y 2 ( t)]′′ + 7[ y1 ( t) | + y 2 ( t)]′ − 25[ y1 ( t) | + y 2 ( t)] = 10 y′′′ 3 ( t) − 14 y′′ 3 ( t) + 7 y′3 ( t) − 25 y 3 ( t)
This can only be true for all time if y 3 ( t) = y1 ( t) + y 2 ( t) . Additive Since it is homogeneous and additive, it is also linear. It is also incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 ( t) = g( t) . Then 10 y1′′′( t) − 14 y1′′( t) + 7 y1′ ( t) − 25 y1 ( t) = g( t) . Let x 2 ( t) = g( t − t0 ) . Then 10 y′′′ 2 ( t) − 14 y′′ 2 ( t) + 7 y′2 ( t) − 25 y 2 ( t) = g( t − t0 ) . The first equation can be written as 10 y1′′′( t − t0 ) − 14 y1′′( t − t0 ) + 7 y1′ ( t − t0 ) − 25 y1 ( t − t0 ) = g( t − t0 ) Therefore
10 y1′′′( t − t0 ) − 14 y1′′( t − t0 ) + 7 y1′ ( t − t0 ) − 25 y1 ( t − t0 ) = 10 y′′′ 2 ( t) − 14 y′′ 2 ( t) + 7 y′2 ( t) − 25 y 2 ( t)
This can only be true for all time if y 2 ( t) = y1 ( t − t0 ) . Time Invariant Stability: The characteristic equation is 10λ3 − 14 λ2 + 7λ − 25 = 0 . The eigenvalues are
Solutions 3-7
M. J. Roberts - 7/12/03
λ1 = 1.7895 λ 2 = -0.1948 + j1.1658 λ 3 = -0.1948 - j1.1658 So the homogeneous solution is of the form, y( t) = K1e1.7895t + K 2e( -0.1948 + j1.1658) t + K 3e( -0.1948 - j1.1658) t . If there is no excitation, but the zero-excitation response is not zero, the response will grow without bound as time increases. Unstable Causality: The system equation can be rewritten as t λ3 λ2 t λ3 λ2 ∫ ∫ ∫ x(λ1 ) dλ1dλ 2 dλ 3 + 25 ∫ ∫ ∫ y(λ1 ) dλ1dλ 2 dλ 3 1 −∞ −∞ −∞ −∞ −∞ −∞ y( t) = λ2 t t 10 − 7 ∫ ∫ y(λ1 ) dλ1dλ 2 + 14 ∫ y(λ1 ) dλ1 −∞ −∞ −∞
So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends on the excitation at times, t < t0 . System has memory. Invertibility: The system equation,
10 y ′′′ ( t) − 14 y ′′ ( t) + 7 y ′ ( t) − 25 y ( t) = x( t)
expresses the excitation in terms of the response and its derivatives. Therefore the excitation is uniquely determined by the response. Invertible. 7. Show that the system of Figure E7 is non-linear, BIBO stable, static and non-invertible. (The response of an analog multiplier is the product of its two excitations.)
Analog Multiplier x[n]
2
y[n]
Figure E7 A DT system Homogeneity:
Solutions 3-8
M. J. Roberts - 7/12/03
Let x1[ n ] = g[ n ] . Then y1[ n ] = 2 g 2 [ n ] 2 Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] = 2(K g[ n ]) = 2K 2 g 2 [ n ]. Multiplying the first equation by K, Then
K y1[ n ] = 2K g 2 [ n ] . K y1[ n ] ≠ y 2 [ n ] .
Not homogeneous.
Let x1[ n ] = g[ n ] . Then y1[ n ] = 2 g 2 [ n ] Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] = 2 h 2 [ n ] 2 2 2 Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] = 2(g[ n ] + h[ n ]) = 2 g[ n ] + h[ n ] + 2 g[ n ] h[ n ] . Adding the first two equations,
(
Not additive.
)
y1[ n ] + y 2 [ n ] = 2 g 2 [ n ] + 2 h 2 [ n ] ≠ y 3 [ n ] .
Since the system is not homogeneous and not additive it is also not linear. The system is also not incrementally linear. The system is statically non-linear because the non-linearity is inherent in the squared relationship between excitation and response. Time Invariance: Let x1[ n ] = g[ n ] . Then y1[ n ] = 2 g 2 [ n ] . Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] = 2 g 2 [ n − n 0 ] . The first equation can be rewritten as
Time Invariant
y1[ n − n 0 ] = 2 g 2 [ n − n o ] = y 2 [ n ]
Stability: If the excitation is bounded, the response is bounded. Stable Causality: At any discrete time, n = n 0 , the response depends only on the excitation at that same time. Causal. Memory: At any discrete time, n = n 0 , the response depends only on the excitation at that same time. System has no memory. Invertibility: Inverting the functional relationship,
Solutions 3-9
M. J. Roberts - 7/12/03
1
y[ n ] 2 x[ n ] = 2 The square root is multiple valued. Two different excitations can cause the same response. Not Invertible. 8. Show that a system with excitation, x[ n ] , and response, y[ n ] , described by y[ n ] = n x[ n ] , is linear, time variant and static. Homogeneity: Let x1[ n ] = g[ n ] . Then y1[ n ] = n g[ n ] Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] = nK g[ n ] . Multiplying the first equation by K, Then Homogeneous.
K y1[ n ] = nK g[ n ] . K y1[ n ] = y 2 [ n ] .
Let x1[ n ] = g[ n ] . Then y1[ n ] = n g[ n ] Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] = n h[ n ] Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] = n (g[ n ] + h[ n ]) = n g[ n ] + n h[ n ] = y1[ n ] + y 2 [ n ] . Additive. Since the system is homogeneous and additive it is also linear. The system is also incrementally linear. The system is not statically non-linear because it is linear. Time Invariance: Let x1[ n ] = g[ n ] . Then y1[ n ] = n g[ n ] . Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] = n g[ n − n 0 ] . The first equation can be rewritten as
Time Variant
y1[ n − n 0 ] = ( n − n 0 ) g[ n − n 0 ] ≠ y 2 [ n ]
Stability: If the excitation is bounded, the response is bounded. Stable Causality: At any discrete time, n = n 0 , the response depends only on the excitation at that same time. Causal. Solutions 3-10
M. J. Roberts - 7/12/03
Memory: At any discrete time, n = n 0 , the response depends only on the excitation at that same time. System has no memory. Invertibility: Inverting the functional relationship, x[ n ] = Invertible.
y[ n ] . n
9. Show that the system of Figure E9 is linear, time-invariant, BIBO unstable and dynamic. x[n] y[n]
D Figure E9 A DT system y[ n ] = x[ n ] + y[ n − 1] y[ n − 1] = x[ n − 1] + y[ n − 2] Then, by induction,
y[ n ] = x[ n ] + x[ n − 1] + y[ n − 2] ∞
y[ n ] = x[ n ] + x[ n − 1] + L + x[ n − k ] + L = ∑ x[ n − k ] k =0
Let m = n − k . Then y[ n ] =
−∞
∑ x[m] =
m =n
n
∑ x[m]
m =−∞
Homogeneity: Let x1[ n ] = g[ n ] . Then y1[ n ] =
n
∑ g[m]
m =−∞
Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] =
n
n
m =−∞
m =−∞
∑ K g[m] = K ∑ g[m] = K y [n] . 1
Homogeneous. Let x1[ n ] = g[ n ] . Then y1[ n ] =
n
Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] = Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] =
∑ g[m]
m =−∞ n
∑ h[m]
m =−∞
n
n
n
m =−∞
m =−∞
m =−∞
∑ (g[m] + h[m]) = ∑ g[m] + ∑ h[m] = y1[n] + y2[n] .
Additive.
Solutions 3-11
M. J. Roberts - 7/12/03
The excitation is uniquely determined by the response. Invertible. 10. Show that a system with excitation, x[ n ] , and response, y[ n ] , described by y[ n ] = rect ( x[ n ]) , is non-linear, time invariant and non-invertible. Homogeneity: Let x1[ n ] = g[ n ] . Then y1[ n ] = rect (g[ n ]) Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] = rect (K g[ n ]) ≠ K y1[ n ] . Not homogeneous. Let x1[ n ] = g[ n ] . Then y1[ n ] = rect (g[ n ]) Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] = rect (h[ n ]) Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] = rect (g[ n ] + h[ n ]) ≠ y1[ n ] + y 2 [ n ] . Not additive. Since the system is not homogeneous and not additive it is also not linear. The system is also not incrementally linear. The system is statically non-linear because the non-linearity arises from the non-linearity of the rectangle relation between excitation and response. Time Invariance: Let x1[ n ] = g[ n ] . Then y1[ n ] = rect (g[ n ]) . Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] = rect g[ n − n 0 ] . The first equation can be rewritten as
(
)
(
)
y1[ n − n o ] = rect g[ n − n o ] = y 2 [ n ] Time invariant
Stability: No matter what values the excitation may have the response can only have the values, zero or one. Stable Causality: At any discrete time, n = n 0 , the response depends only on the excitation at that discrete time and not on any future excitation. Causal. Memory: At any discrete time, n = n 0 , the response depends only on the excitation at that discrete time and not on any past excitation. Solutions 3-13
M. J. Roberts - 7/12/03
System has no memory. Invertibility: Many different excitations can cause the same response. Not invertible. 11. Show that the system of Figure E11 is non-linear, time-invariant, static and invertible.
5 x[n]
y[n]
10 Figure E11 A DT system y[ n ] = 10 x[ n ] − 5 ,
Homogeneity: Let x1[ n ] = g[ n ] . Then y1[ n ] = 10 g[ n ] − 5 Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] = 10K g[ n ] − 5 ≠ K y1[ n ]. Not homogeneous. Let x1[ n ] = g[ n ] . Then y1[ n ] = 10 g[ n ] − 5 Let x 2 [ n ] = h[ n ] . Then y1[ n ] = 10 h[ n ] − 5 Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] = 10(g[ n ] + h[ n ]) − 5 ≠ y1[ n ] + y 2 [ n ] . Not additive. Since the system is not homogeneous and not additive it is also not linear. The system is incrementally linear because the only deviation from linearity is caused by the presence of the non-zero, zero-excitation response. The system is not statically non-linear because it is incrementally linear. Time Invariance: Let x1[ n ] = g[ n ] . Then y1[ n ] = 10 g[ n ] − 5 . Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] = 10 g[ n − n 0 ] − 5 . The first equation can be rewritten as
Time invariant
y1[ n − n 0 ] = 10 g[ n − n 0 ] − 5 = y 2 [ n ]
Stability: If the excitation is bounded, the response is bounded. Stable Causality:
Solutions 3-14
M. J. Roberts - 7/12/03
At any discrete time, n = n 0 , the response depends only on the excitation at that discrete time and not on any future excitation. Causal. Memory: At any discrete time, n = n 0 , the response depends only on the excitation at that discrete time and not on any past excitation. System has no memory. Invertibility: Solving the system equation for the excitation as a function of the response, x[ n ] = Invertible.
y[ n ] + 5 10
12. Show that the system of Figure E12 is time-invariant, BIBO stable, and causal.
x[n]
1 4
y[n]
D
D
y[n-2]
1 4 1 2 Figure E12 A DT system Homogeneity: Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ] Let x 2 [ n ] = K g[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = K g[ n ] Multiply the first equation by K. 4 K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2] = K g[ n ] Then, equating results, 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = 4 K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2] If this equation is to be satisfied for all n, y 2 [ n ] = K y1[ n ] . Homogeneous. Additivity: Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ] Let x 2 [ n ] = h[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = h[ n ] Let x 3 [ n ] = g[ n ] + h[ n ] . Then 4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2] = g[ n ] + h[ n ] Add the two first two equations. 4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2]) = g[ n ] + h[ n ] Solutions 3-15
M. J. Roberts - 7/12/03
Then, equating results, 4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2]
= 4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2])
If this equation is to be satisfied for all n, y 3 [ n ] = y1[ n ] + y 2 [ n ].
Additive.
Since the system is both homogeneous and additive, it is linear. Since the system is linear it is also incrementally linear. Since the system is linear, it is not statically non-linear. Time Invariance: Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ] Let x 2 [ n ] = g[ n − n 0 ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = g[ n − n 0 ] We can re-write the first equation as 4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = g[ n − n 0 ] Then, equating results, 4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] If this equation is to be satisfied for all n, then y 2 [ n ] = y1[ n − n 0 ] . Time Invariant. Stability: The eigenvalues of the system homogeneous solution are found from the characteristic equation, 4α 2 − α + 2 = 0 . They are
α1,2 = 0.125 ± j 0.696 or α1,2 = 0.7071e ± j1.3931
Therefore the homogeneous solution is of the form, y h [ n ] = K h1 0.7071e + j1.3931 + K h 2e − j1.3931 and, as n approaches infinity the homogeneous solution approaches zero and the total solution approaches the particular solution. The particular solution is bounded because it
Solutions 3-16
M. J. Roberts - 7/12/03
consists of functions of the same form as the excitation and all its unique differences and the excitation is bounded in the BIBO stability test. Therefore if x is bounded, so is y. Stable. Causality: We can rearrange the system equation into y[ n ] =
1 (x[n] + y[n − 1] + 2 y[n − 2]) 4
showing that the response at time, n, depends on the excitation at time, n, and the response at previous times. It does not depend on any future values of the excitation. Causal. Memory: The response depends on past values of the response. The system has memory. Invertibility: The original system equation, 4 y[ n ] − y[ n − 1] + 2 y[ n − 2] = x[ n ], expresses the excitation in terms of the response. Invertible. 13. Find the impulse responses of these systems. (a)
y[ n ] = x[ n ] − x[ n − 1] The impulse response is very easily found by direct iteration to be h[ n ] = δ [ n ] − δ [ n − 1] .
(b)
25 y[ n ] + 6 y[ n − 1] + y[ n − 2] = x[ n ] The homogeneous solution is −3 − j 4 −3 + j 4 y h [ n ] = K1 + K2 25 25 n
n
and, after discrete-time, n = 0, this is the total solution because the excitation is zero. The first two values of the impulse response are (by direct iteration), y[0] =
1 6 and y[1] = − . 25 625
Solving for the constants, 1 4 + j3 = K1 + K 2 K1 = 25 200 ⇒ 6 4 − j3 −3 − j 4 −3 + j 4 − = K1 K2 = + K2 25 25 625 200
Solutions 3-17
M. J. Roberts - 7/12/03
Then the impulse response is 4 + j 3 −3 + j 4 4 − j 3 −3 − j 4 h[ n ] = + 200 25 200 25 n
h[ n ]
n
n n 4 + j 3)(−3 + j 4 ) + ( 4 − j 3)(−3 − j 4 ) ( =
h[ n ] = h[ n ] =
200(25)
n
(4 + j 3)5 n e j 2.214 n + (4 − j 3)5 n e − j 2.214 n 200(25)
n
=
4 (e j 2.214 n + e − j 2.214 n ) + j 3(e j 2.214 n − e − j 2.214 n ) 200(5)
n
4 cos(2.214 n ) − 3 sin(2.214 n ) n 100(5)
Then, using B A 2 + B 2 cos x − tan −1 A
A cos( x ) + B sin( x ) =
cos(2.214 n + 0.644 ) n 20(5) 4 y[ n ] − 5 y[ n − 1] + y[ n − 2] = x[ n ]
h[ n ] = (c)
The homogeneous solution is 1 y h [ n ] = K h1 + K h 2 4 By iteration, y[0] =
n
5 1 1 1 , y[1] = and K h1 = , K h 2 = − . 16 4 3 12
1 1 1 n So the impulse response is h[ n ] = − u[ n ] . 3 12 4 2 y[ n ] + 6 y[ n − 1] = x[ n ] − x[ n − 2]
(d) 1 K h1 + K h 2 4
n
The impulse response is the sum of the response, h1[ n ] to a unit impulse at time, n = 0, and the response, h 2 [ n ], to a negative unit impulse at time, n = 2. h1[ n ] =
1 (−3) n u[n] 2
and
Solutions 3-18
h2[n] = −
1 (−3)( n − 2) u[n − 2] 2
M. J. Roberts - 7/12/03
So the overall impulse response is h[ n ] =
1 1 1 1 (−3) n u[n] − (−3)( n − 2) u[n − 2] = (−3) n u[n] − u[n − 2] 9 2 2 2
14. Sketch g[ n ]. To the extent possible find analytical solutions. Where possible, compare analytical solutions with the results of using the MATLAB command, conv, to do the convolution. (a)
g[ n ] = u[ n ] ∗ u[ n ] =
∞
∑
m =−∞
u[ m] u[ n − m] =
∞
∑ u[n − m] = ramp[n + 1]
m =0
g[n] 8 4
...
(b)
g[ n ] = u[ n + 2] ∗ rect 3 [ n ] =
...
n
∞
∞
∑ u[m + 2]rect [n − m] = ∑ rect [n − m] 3
m =−∞
m =−2
3
g[n] 8 6 4
...
-2
... -8
-6
-4
-2
2
4
6
8
n
-2
(c)
∞
2
−∞
−2
g[ n ] = rect 2 [ n ] ∗ rect 2 [ n ] = ∑ rect 2 [ m] rect 2 [ n − m] = ∑ rect 2 [ n − m] g[n] 5
-10
(d)
10
n
g[ n ] = rect 2 [ n ] ∗ rect 4 [ n ] g[n] 5
-10
10
n
Solutions 3-19
M. J. Roberts - 7/12/03
n
(e)
3 3 g[ n ] = 3δ [ n − 4 ] ∗ u[ n ] = 3 4 4
n −4
u[ n − 4 ]
g[n] 3
2
4
6
8
10
12
14
∞ 7 7 (f) g[ n ] = 2 rect 4 [ n ] ∗ u[ n ] = ∑ 2 rect 4 [ m] 8 8 m =−∞ n
n −m
16
n
18
7 u[ n − m] = 2 ∑ m =−4 8 4
n −m
u[ n − m]
g[n] 12
-15
(g)
-10
-5
0
5
10
g[ n ] = rect 3 [ n ] ∗ comb14 [ n ] = rect 3 [ n ] ∗
15
20
n
25
∞
∞
m =−∞
m =−∞
∑ δ[n − 14 m] = ∑ rect [n − 14 m] 3
g[n] 1
-30
-20
-10
10
20
30
n
15. Given the excitations, x[ n ] , and the impulse responses, h[ n ], find closed-form expressions for and plot the system responses, y[ n ] . (a)
x[ n ] = e
j
2πn 32
,
h[ n ] = (0.95) u( n )
y[ n ] = h[ n ] ∗ x[ n ] =
n
∞
∑e
m =−∞
j
2πm 32
(0.95) n − m u[n − m]
Solutions 3-20
M. J. Roberts - 7/12/03
y[ n ] =
n
∑e
2πm j 32
m =−∞
(0.95) n − m
Making the change of variable, q = − m j 232π e n y[ n ] = (0.95) ∑ − q =−∞ 0.95 ∞
∑ rn = n =k
2π −j = (0.95) ∑ 0.95e 32 q =− n n
∞
q
rk , r <1 1− r
y[ n ] = (0.95)
2π −j 32 0 . 95 e n
1 − 0.95e
−j
−n
2π 32
e
=
j
2π n 32
1 − 0.95e
Real Excitation
= 5.0632e
1
-5
n
40
-5
-1
40
n
-1
Real Impulse Response h[n]
Imaginary Impulse Response h[n]
1
1
-5
n
40
-5
-1
40
n
-1
Real Response
Imaginary Response
y[n]
y[n]
5
5
n
40
-5
-5
2πn x[ n ] = sin 32
2π 32
x[n]
1
-5
−j
2π j n −1.218 32
Imaginary Excitation
x[n]
(b)
m
−q
−n
Using
j 232π e n = (0.95) ∑ m =−∞ 0.95 n
40
n
-5
h[ n ] = (0.95) u[ n ] n
,
From part (c), the response to x[ n ] = e
j
2πn 32
is y[ n ] = 5.0632e
2πn e sin = 32
j
2πn 32
−e j2
−j
2π j n −1.218 32
. Since
2πn 32
,
2πn by applying linearity and superposition, the response to x[ n ] = sin is 32
Solutions 3-21
M. J. Roberts - 7/12/03
y[ n ] = or
5.0632e
2π j n −1.218 32
− 5.0632e j2
2π − j n −1.218 32
2π y[ n ] = 5.0632 sin n − 1.218 32
Excitation
x[n] 1 -5
40
n
-1
Impulse Response
h[n] 1 -5
40
n
-1
Response
y[n] 5 -5
40
n
-5
16. Given the excitations, x[ n ] , and the impulse responses, h[ n ], use MATLAB to plot the system responses, y[ n ] . 2πn , h[ n ] = sin (a) x[ n ] = u[ n ] − u[ n − 8] (u[ n ] − u[ n − 8]) 8 y[ n ] = h[ n ] ∗ x[ n ] = y[ n ] =
∞
2πm (u[ m] − u[ m − 8])(u[ n − m] − u[ n − m − 8]) 8
∞
∑ sin
m =−∞
2πm u[ m] u[ n − m] − u[ m] u[ n − m − 8] 8 − u[ m − 8] u[ n − m] + u[ m − 8] u[ n − m − 8]
∑ sin
m =−∞
2πm 2πm 2πm 2πm y[ n ] = ∑ sin − ∑ sin − ∑ sin + ∑ sin 8 8 8 8 m =0 m =0 m =8 m =8 n −8
n
n
n −8
For n < 0 all the summations are zero and y[ n ] = 0. For n > 15,
2πm 2πm y[ n ] = ∑ sin − ∑ sin =0 . 8 m =n −7 8 m =n −7 n
n
So the response is only non-zero for 0 ≤ 7 < 16 (and can be zero at some points within that range).
Solutions 3-22
M. J. Roberts - 7/12/03
x[n]
Excitation
1
-5
30
h[n]
n
Impulse Response
1 -5
30
n
-1
y[n]
Response
3 -5
30
n
-3
(b)
2πn x[ n ] = sin (u[ n ] − u[ n − 8]) 8
y[ n ] = h[ n ] ∗ x[ n ] =
∞
2πn h[ n ] = − sin (u[ n ] − u[ n − 8]) 8
,
2π ( n − m) 2πm (u[ m] − u[ m − 8]) sin (u[ n − m] − u[ n − m − 8]) 8 8
∑ − sin
m =−∞
n n 2πm 2π ( n − m) 2πm 2π ( n − m) sin sin sin − sin ∑ ∑ m =8 8 8 8 m = 0 8 y[ n ] = − n − 8 n −8 − sin 2πm sin 2π ( n − m) + sin 2πm sin 2π ( n − m) ∑ m∑ 8 m =8 8 8 8 =0
x[n]
Excitation
1 -5
30
n
-1
h[n]
Impulse Response
1 -5
30
n
-1
y[n]
Response
3 -5
30 -3
17. Which of these systems are BIBO stable? (a)
Solutions 3-23
n
M. J. Roberts - 7/12/03
x[n]
y[n]
D
-0.9
The system equation is y[ n ] = x[ n ] − 0.9 y[ n − 1] The eigenvalue is
α = −0.9 . Its magnitude is less than one therefore the system is stable. (b) x[n]
y[n] 1.1
D
The system equation is y[ n ] = x[ n ] + 1.1 y[ n − 1] The eigenvalue is
α = 1.1 . Its magnitude is greater than one therefore the system is unstable. (c) x[n]
y[n] 1 2
D
1 2
D
The system equation is y[ n ] = x[ n ] − The eigenvalues are
1 (y[n − 1] + y[n − 2]) 2
α1,2 = −0.25 ± j 0.6614 .
Its magnitudes are less than one therefore the system is stable. (d)
Solutions 3-24
M. J. Roberts - 7/12/03
x[n]
The system equation is
y[n] 1.5
D
0.4
D
y[ n ] = x[ n ] − (1.5 y[ n − 1] + 0.4 y[ n − 2])
The eigenvalues are
α1,2 = −1.153, −0.3469 .
One magnitude is greater than one therefore the system is unstable. 18. Find and plot the unit-sequence responses of these systems. (a) x[n]
y[n]
D
0.7
D
-0.5
h[ n ] = h1[ n ] ∗ h 2 [ n ] h1[ n ] = (0.7) u[ n ] n
∞
h[ n ] =
m
m =−∞
∑ (0.7) (−0.5) m
m =0
h[ n ] = (−0.5)
n
∑ (0.7) u[m](−0.5)
h[ n ] = n
h 2 [ n ] = (−0.5) u[ n ]
and
n −m
n −m
u[ n − m] m
n 0.7 n m = (−0.5) ∑ − = (−0.5) ∑ (−1.4 ) 0.5 m =0 m =0 n
n
1 − (−1.4 ) n 1 − (−1.4 ) = (−0.5) 2.4 1 − (−1.4 ) n +1
n
1 − (−1.4 ) y[ n ] = h[ n ] ∗ u[ n ] = ∑ (−0.5) 2.4 m =−∞ ∞
y[ n ] =
1 − (−1.4 ) (−0.5) ∑ 2.4 m =0 n
m
m +1
m
m +1
n +1
u[ n ]
u[ m] u[ n − m]
[
1 n = (−0.5) m 1 − (−1.4) m +1 ∑ 2.4 m = 0
n n m m m y[ n ] = 0.4167 ∑ (−0.5) − ∑ (−1.4 )(−0.5) (−1.4 ) m = 0 m =0
Solutions 3-25
]
M. J. Roberts - 7/12/03 n n m m y[ n ] = 0.4167 ∑ (−0.5) + 1.4 ∑ (0.7) m = 0 m =0 n +1 1 − (−0.5) n +1 1 − (0.7) + 1.4 y[ n ] = 0.4167 u[ n ] 1 − (0.7) 1 + 0.5
{
(
y[ n ] = 0.4167 0.6667 1 − (−0.5) h- 1[n]
n +1
) + 4.6667(1 − (0.7) )} u[n] n +1
Unit-Sequence Response
3
-5
n
20
(b)
D
-0.8
x[n]
y[n]
D D
0.6
h[ n ] = h1[ n ] + h 2 [ n ] On the bottom path, y 2 [ n + 1] = x[ n ] + 0.6 y 2 [ n − 1]
or
y 2 [ n ] = x[ n − 1] + 0.6 y 2 [ n − 2] ⇒ α1,2 = ± 0.6 h 2 [ n ] = K h1
(
0.6
)
n
(
)
n + K h 2 − 0.6 u[ n ]
Fitting initial conditions, h 2 [0] = 0 and h 2 [1] = 1 K h1 = Therefore h1[ n ] = (−0.8) u[ n ] n
and
1 = 0.6455 = −K h 2 2 0.6
(
h 2 [ n ] = 0.6455
0.6
)
n
Solutions 3-26
(
)
n − 0.6455 − 0.6 u[ n ]
M. J. Roberts - 7/12/03
(
n h[ n ] = (−0.8) + 0.6455
0.6
(
n y[ n ] = h[ n ] ∗ u[ n ] = (−0.8) + 0.6455
y[ n ] =
∞
∑ (−0.8)
m
(
+ 0.6455
m =−∞
y[ n ] =
n
∑ (−0.8)
m
0.6
)
m =0
(
n
(
m
)
n − 0.6455 − 0.6 u[ n ]
0.6
(
+ 0.6455
)
)
n
(
(
− 0.6455 − 0.6
0.6
)
)
m
)
n − 0.6455 − 0.6 u[ n ] ∗ u[ n ]
)
m
u[ m] u[ n − m]
(
− 0.6455 − 0.6
(
)
m
)
n +1 n +1 n +1 1 − − 0.6 1 − 0.6 − − . 1 ( 0 8 ) u[ n ] y[ n ] = + 0.6455 − 0.6455 1 − (−0.8) 1 − − 0.6 1 − 0.6
(
(
)
)
(
(
n +1 n +1 1 − − 0.6 1 − 0.6 − − . 1 ( 0 8 ) y[ n ] = + 0.6455 − 0.6455 1.7746 0.2254 1.8
)
)
n +1
u[ n ]
Unit-Sequence Response
h- 1[n] 3
-5
20
n
19. Find the impulse responses of these systems: (a)
y′ ( t) + 5 y( t) = x( t) Following the example in the text, h( t) = e −5 t u( t)
(b)
y′′ ( t) + 6 y′ ( t) + 4 y( t) = x( t) For t < 0, h( t) = 0 . For t > 0, h h ( t) = K1e −5.23 t + K 2e −0.76 t Since the highest derivative of “x” is two less than the highest derivative of “y”, the general solution is of the form, h( t) = (K1e −5.23 t + K 2e −0.76 t ) u( t) Integrating the equation once from t = 0 − to t = 0 + ,
Solutions 3-27
M. J. Roberts - 7/12/03
[
h′ (0 ) − h′ (0 ) + 6 h(0 ) − h(0 +
−
+
−
0+
0+
0−
0−
)] + 4 ∫ h(t)dt = ∫ δ (t)dt = 1
We know that the impulse response cannot contain an impulse because its second derivative would be a triplet and there is no triplet excitation. We also know that the impulse response cannot be discontinuous at time, t = 0, because if it were the second derivative would be a doublet and there is no doublet excitation. Therefore, h′ (0 + ) − h′ (0 − ) = 1 ⇒ h′ (0 + ) = 1 This requirement, along with the requirement that the solution be continuous at time, t = 0, leads to the two equations,
[
h′ (0 + ) = 1 = −5.23K1e −5.23 t − 0.76K 2e −0.76 t and
]
t = 0+
= −5.23K1 − 0.76K 2
h(0 + ) = 0 = K1 + K 2 .
Solving, K1 = −0.2237 and K 2 = 0.2237 Then the total impulse response is h( t) = 0.2237(e −0.76 t − e −5.23 t ) u( t) . (c)
2 y′ ( t) + 3 y( t) = x′ ( t)
The homogeneous solution is y h ( t) = K h e h( t) = K h e
3 − t 2
3 − t 2
. The impulse response is of the form,
u( t) + K iδ ( t) .
Integrating from 0 − to 0 + ,
[
2 h(0 ) − h(0 +
or
0+
−
)] + 3 ∫ h(t) = x(0 ) − x(0 ) = 0 +
−
0−
2 h(0 + ) + 3K i = 2K h + 3K i = 0
Integrating a second time from 0 − to 0 + , 0+
0+
0−
0−
2 ∫ h( t) = ∫ δ ( t) = 1 ⇒ 2K i = 1 ⇒ K i =
Solutions 3-28
1 . 2
M. J. Roberts - 7/12/03
3 Then K h = − . Therefore 4 3 − 32 t 1 h( t) = − e u( t) + δ ( t) 4 2 (d)
4 y′ ( t) + 9 y( t) = 2 x( t) + x′ ( t)
The homogeneous solution is y h ( t) = K h e h( t) = K h e
3 − t 2
9 − t 4
. The impulse response is of the form,
u( t) + K iδ ( t) .
Integrating once from 0 − to 0 + ,
[
4 h(0 ) − h(0 +
0+
−
0+
)] + 9 ∫ h(t) = 2 ∫ δ (t) + x(0 ) − x(0 ) = 2 +
0−
or
0−
4 K h + 9K i = 2
Integrating a second time from 0 − to 0 + , 0+
4 ∫ h( t) = 1 ⇒ 4 K i = 1 ⇒ K i = 0−
Then K h = −
1 . 4
1 . Therefore 16 1 − 49 t 1 h( t) = − e u( t) + δ ( t) 16 4
20. Sketch g( t) . g(t) 1
(a) g( t) = rect ( t) ∗ rect ( t)
−1
t
1
g(t) 1
t (b) g( t) = rect ( t) ∗ rect 2
3 2
1 1 2 2
3 2
t
Solutions 3-29
−
M. J. Roberts - 7/12/03
g(t) 1
t
(c)
t g( t) = rect ( t − 1) ∗ rect 2
(d)
g( t) = [rect ( t − 5) + rect ( t + 5)] ∗ [rect ( t − 4 ) + rect ( t + 4 )]
1 3 2 2
1 2
5 2
g( t) = rect ( t − 5) * rect ( t − 4 ) + rect ( t + 5) * rect ( t − 4 )
+ rect ( t − 5) * rect ( t + 4 ) + rect ( t + 5) * rect ( t + 4 )
rect(t-5)∗rect(t-4)
rect(t-5)∗rect(t+4)
1
1 8
9
10
t
rect(t+5)∗rect(t-4)
1
t
2
rect(t+5)∗rect(t+4)
-1
1
1
1 -2
g(t)
t
-10
-9
t
-8
-9
-1 1
t
9
21. Sketch these functions. g(t) 1
(a)
g( t) = rect ( 4 t)
-1 8
t
1 8
g(t) 4
(b)
g( t) = rect(4 t) ∗ 4δ ( t)
-1 8
t
1 8
g(t) 4
(c)
2- 81
g( t) = rect ( 4 t) ∗ 4δ ( t − 2)
2+ 81
2
g(t) 2
(d)
g( t) = rect ( 4 t) ∗ 4δ (2 t)
-1 8
Solutions 3-30
1 8
t
t
M. J. Roberts - 7/12/03
g(t) 1
...
(e)
-2
g( t) = rect ( 4 t) ∗ comb( t)
-1
...
-1 1 88
t
1
g(t) 1
...
(f)
-2
g( t) = rect ( 4 t) ∗ comb( t − 1)
-1
...
-1 1 88
t
1
g(t) 1 2
...
(g)
-2
g( t) = rect ( 4 t) ∗ comb(2 t)
-1
...
-1 1 88
t
1
g(t) 1
...
(h)
g( t) = rect ( t) ∗ comb(2 t)
-2
-1
...
t
1
22. Plot these convolutions. (a)
t + 1 t + 2 t g( t) = rect ∗ [δ ( t + 2) − δ ( t + 1)] = rect − rect 2 2 2 g(t) 1
-4
1 -1
(b)
g( t) = rect ( t) ∗ tri( t) =
∞
1 2
∫ rect (τ ) tri(t − τ )dτ = ∫ tri(t − τ )dτ
−∞
−
1 2
Solutions 3-31
t
M. J. Roberts - 7/12/03
t < -3/2
-3/2 < t < -1/2
1
1
rect(τ) and tri(t- τ)
-4
τ
4
-1/2 < t < 1/2
rect(τ) and tri(t- τ)
rect(τ) and tri(t- τ) 1
-4
τ
4
1/2 < t < 3/2
-4
4
τ
3/2 < t
rect(τ) and tri(t- τ)
rect(τ) and tri(t- τ)
1
1
-4
4
τ
-4
4
τ
3 , g( t) = 0. 2 t +1 t +1 t +1 τ2 3 1 If − < t < − , g( t) = ∫ 1 − τ{ − t dτ = ∫ (1 − (τ − t)) dτ = τ − + τt 2 2 2 >0 − 1 1 1 − −
If t < −
2
2
2
1 − 2 (t + 1) 1 2 1 g( t) = t + 1 − + ( t + 1) t − − + − − t 2 2 2 2 2 t 3t 9 g( t) = + + 2 2 8 2
1 1 , g( t) = If − < t < 2 2
t
1 2
− t dτ + ∫ 1 − τ{ − t dτ ∫ 1 − τ{
−
1 2
<0
t
>0
1 2
1
t
τ2 2 τ2 g( t) = ∫ (1 − ( t − τ )) dτ + ∫ (1 − (τ − t)) dτ = τ − τt + + τ − + τt 2 − 1 2 t 1 t − t
2
2
t2 1 t 1 1 1 t t2 g( t) = t − t 2 + + − − + − + − t + − t 2 2 2 2 8 2 8 2 2 g( t) =
3 2 −t 4
By symmetry, g( t) = g(− t) and
Solutions 3-32
M. J. Roberts - 7/12/03
3 0 , t > 2 2 3t 9 1 3 t g( t) = − + , < t< 2 2 2 2 8 1 3 2 t< 4 − t , 2 g(t) 1
-2
(c)
g( t) = e − t u( t) ∗ e − t u( t) =
2
∞
( ∫ e u(τ )e
− t −τ )
−τ
t
u( t − τ ) dτ
−∞
For t < 0, g( t) = 0. For t > 0, t
g( t) = ∫ e e −τ
0
Therefore
−( t −τ )
t
dτ = ∫ e − t dτ = te − t 0
g( t) = te − t u( t) . g(t) 0.5
-1
(d)
5
t
1 1 1 t g( t) = tri 2 t + − tri 2 t − ∗ comb 2 2 2 2 1 1 1 1 t t g( t) = tri 2 t + ∗ comb − tri 2 t − ∗ comb 2 2 2 2 2 2 g( t) =
∞
1
1
∑ tri 2 t − 2n + 2 − tri 2 t − 2n − 2
n =−∞
Solutions 3-33
M. J. Roberts - 7/12/03
g(t) 1
-3
t
3 -1
(e)
1 1 g( t) = tri 2 t + − tri 2 t − ∗ comb( t) 2 2 1 1 g( t) = tri 2 t + ∗ comb( t) − tri 2 t − ∗ comb( t) 2 2 g( t) =
∞
1
1
∑ tri 2 t + n + 2 − tri 2 t + n − 2 = 0
n =−∞
g(t) 1
-3
3
t
-1
23. A system has an impulse response, h( t) = 4 e −4 t u( t) . Find and plot the response of the 1 system to the excitation, x( t) = rect 2 t − . 4 1 1 y( t) = x( t) ∗ h( t) = rect 2 t − ∗ 4 e −4 t u( t) = 4 u( t) − u t − ∗ e −4 t u( t) 2 4 u( t) ∗ e
−4 t
u( t) =
∞
∫ u(τ )e
−4 ( t −τ )
−∞
t
u( t − τ ) dτ = ∫ e 0
−4 ( t −τ )
1 −4 t (1 − e ) , t > 0 1 −4 t dτ = 4 = (1 − e ) u( t) 4 0 , t < 0
Invoking linearity and time-invariance, 1 −4 t − 1 1 y( t) = 4 u( t) − u t − ∗ e −4 t u( t) = (1 − e −4 t ) u( t) − 1 − e 2 u t − 2 2
y(t) 1
1.5 -1
Solutions 3-34
t
M. J. Roberts - 7/12/03
24. Change the system impulse response in Exercise 23 to h( t) = δ ( t) − 4 e −4 t u( t) and find 1 and plot the response to the same excitation, x( t) = rect 2 t − . 4 Using linearity, this response is the excitation convolved with a unit impulse at time zero minus the response calculated in Exercise 23. y( t) = rect 2 t −
1 −4 t − 1 1 2 −4 t 1 1 u t − − ( − e ) u( t) − − e 2 4
y(t) 1
1.5
t
-1
25. Find the impulse responses of the two systems in Figure E25. Are these systems BIBO stable?
∫
x(t) x(t)
∫
y(t)
y(t) (a) (b) Figure E25 Two single-integrator systems
(a)
y′ ( t) = x( t) ⇒ h( t) = u( t)
Impulse response is not absolutely integrable. BIBO unstable. (b)
y′ ( t) = x( t) − y( t) ⇒ h( t) = e − t u( t)
BIBO Stable.
Impulse response is absolutely integrable. BIBO stable. 26. Find the impulse response of the system in Figure E26. Is this system BIBO stable?
x(t)
∫
∫
Figure E26 A double-integrator system y′′ ( t) = x( t) − y( t) ⇒ h( t) = sin( t) u( t) Impulse response is not absolutely integrable. BIBO unstable.
Solutions 3-35
y(t)
M. J. Roberts - 7/12/03
27. In the circuit of Figure E27 the excitation is v i ( t) and the response is v o ( t) . (a)
Find the impulse response in terms of R and L.
(b)
If R = 10 kΩ and L = 100 µH graph the unit step response.
R
+ vi (t)
+ L
vo(t)
-
Figure E27 An RL circuit v i ( t) = R i( t) + L i′ ( t)
δ ( t) = R h( t) + L h′ ( t) h( t) =
1 − RL t e u( t) , Current impulse response L
R − RL t d v o ( t) = L (i( t)) = δ ( t) − e u( t) , Response-voltage impulse response L dt h −1 ( t) = v o ( t) = L
R − t 1 L − 1 e u( t) , Current step response R
R − t d i( t)) = e L u( t) , Response-voltage step response ( dt
vo(t) 1
-0.01
0.04
t (µs)
28. Find the impulse response of the system in Figure E28 and evaluate its BIBO stability.
Solutions 3-36
M. J. Roberts - 7/12/03
∫
x(t)
∫
y(t)
1 10 1 20 Figure E28 A two-integrator system y′′ ( t) = x( t) +
1 1 y′ ( t) − y( t) 10 20
The homogenoeus solution is
implying that
y h ( t) = e 0.05 t [K c cos(0.2179 t) + K s sin(0.2179 t)] h( t) = e 0.05 t [K c cos(0.2179 t) + K s sin(0.2179 t)] , t > 0 .
The response cannot have a discontinuity at zero. Integrating the differential equation once from 0 − to 0 + , h′ (0 + ) − h′ (0 − ) = 1 ⇒ h′ (0 + ) = 1 . Integrating the differential equation twice from 0 − to 0 + , h(0 + ) − h(0 − ) = 0 ⇒ h(0 + ) = 0 Therefore
h(0 + ) = K c = 0
Solving,
and
h′ (0 + ) = 0.05K c + 0.2179K s = 1 .
K c = 0 and K s = 4.589
and
h( t) = 4.589e 0.05 t sin(0.2179 t) u( t) .
29. Find the impulse response of the system in Figure E29 and evaluate its BIBO stability.
x(t)
∫
∫
2 3 1 8 Figure E29 A two-integrator system
Solutions 3-37
y(t)
M. J. Roberts - 7/12/03
2 1 y′′ ( t) = x( t) − y′ ( t) − y( t) 3 8 The homogenoeus solution is y h ( t) = e
−
implying that h( t) = e
−
t 3
[K
c
t 3
[K
c
cos(0.1179 t) + K s sin(0.1179 t)]
cos(0.1179 t) + K s sin(0.1179 t)] , t > 0 .
The response cannot have a discontinuity at zero. Integrating the differential equation once from 0 − to 0 + , h′ (0 + ) − h′ (0 − ) = 1 ⇒ h′ (0 + ) = 1 . Integrating the differential equation twice from 0 − to 0 + , h(0 + ) − h(0 − ) = 0 ⇒ h(0 + ) = 0 Therefore h(0 + ) = K c = 0 Solving,
and
1 h′ (0 + ) = − K c + 0.1179K s = 1 . 3
K c = 0 and K s = 8.482
and −
t 3
h( t) = 8.482e sin(0.1179 t) u( t) . 30. Plot the amplitudes of the responses of the systems of Exercise 19 to the excitation, e jωt , as a function of radian frequency, ω . (a)
y′ ( t) + 5 y( t) = x( t) y p ( t) = Ke jωt K=
1 jω + 5 |K| 0.2
-10 π
(b)
10π
y′′ ( t) + 6 y′ ( t) + 4 y( t) = x( t) K=
1 4 − ω + j 6ω 2
Solutions 3-38
ω
M. J. Roberts - 7/12/03
|K| 0.25
-5 π
(c)
ω
5π
2 y′ ( t) + 3 y( t) = x′ ( t) K=
jω j 2ω + 3 |K| 0.5
-5 π
(d)
4 y′ ( t) + 9 y( t) = 2 x( t) + x′ ( t) K=
ω
5π
2 + jω j 4ω + 9 |K| 0.25
-5 π
ω
5π
31. Plot the responses of the systems of Exercise 19 to a unit-step excitation. (a)
h( t) = e −5 t u( t) h −1 ( t) =
t
∫
h(λ ) dλ =
−∞
t
∫
−∞
t
e −5λ u(λ ) dλ = ∫ e −5λ dλ = − 0
[ ]
1 −5λ e 5
t 0
=
1 1 − e −5 t ) , t > 0 ( 5
h −1 ( t)0 , t < 0 h −1 ( t) =
1 (1 − e −5t ) u(t) 5 h (t) -1
0.2
1
Solutions 3-39
t
M. J. Roberts - 7/12/03
(b)
h( t) = 0.2237(e −0.76 t − e −5.23 t ) u( t) h −1 ( t) =
t
t
∫ h(λ )dλ = 0.2237 ∫ (e
−∞
t
−0.76λ
−∞
{[
] [
−e
−5.23λ
) u(λ )dλ = 0.2237∫ (e
− e −5.23λ ) dλ , t > 0
0
h −1 ( t) = 0.2237 −1.316e −0.76 t 0 − −0.1912e −5.23 t t
−0.76λ
]} t
, t>0
0
[
]
[
]
h −1 ( t) = 0.2237 1.316(1 − e −0.76 t ) − 0.1912(1 − e −5.23 t ) , t > 0 h −1 ( t) = 0.2237 1.316(1 − e −0.76 t ) − 0.1912(1 − e −5.23 t ) u( t) h (t) -1
0.25
5
(c)
t
3 −3t 1 h( t) = − e 2 u( t) + δ ( t) 4 2 t 3 − 32 λ 1 1 3 − 32 λ h −1 ( t) = ∫ h(λ ) dλ = ∫ − e u(λ ) + δ (λ ) dλ = − ∫ e dλ , t > 0 4 2 2 04 −∞ −∞ t
t
t
1 3 2 − 32 λ 1 1 − 32 t h −1 ( t) = + e = + e − 1 , t > 0 2 4 3 0 2 2 1 − 32 t h −1 ( t) = e u( t) 2 h (t) -1
0.5
5
(d)
t
1 1 − 49 t h( t) = δ ( t) − e u( t) 4 16 h −1 ( t) =
t
∫
−∞
h(λ ) dλ =
t 1 − 49 λ 1 1 1 − 49 λ e d e dλ , t > 0 − u ( λ ) + δ ( λ ) λ = − ∫ 16 4 4 ∫0 16 −∞ t
Solutions 3-40
M. J. Roberts - 7/12/03
t
1 1 4 − 49 λ 1 1 − 49 t e h −1 ( t) = + = + e − 1 , t > 0 4 16 9 0 4 36 1 1 −9t h −1 ( t) = + e 4 − 1 u( t) 4 36 h- 1(t) 0.25
5
t
32. A CT system is described by the block diagram in Figure E32.
x(t)
1 4
∫
∫
y(t)
1 4 3 4 Figure E32 A CT system Classify the system as to homogeneity, additivity, linearity, time-invariance, stability, causality, memory, and invertibility. 4 y′′ ( t) + y′ ( t) + 3 y( t) = x( t) Homogeneity: Let x1 ( t) = g( t) . Then 4 y1′′( t) + y1′ ( t) + 3 y1 ( t) = g( t) . Let x 2 ( t) = K g( t) . Then 4 y′′2 ( t) + y′2 ( t) + 3 y 2 ( t) = K g( t) . If we multiply the first equation by K, we get 4 K y1′′( t) + K y1′ ( t) + 3K y1 ( t) = K g( t) Therefore 4 K y1′′( t) + K y1′ ( t) + 3K y1 ( t) = 4 y′′2 ( t) + y′2 ( t) + 3 y 2 ( t) This can only be true for all time if y 2 ( t) = K y1 ( t) . Homogeneous Additivity: Let x1 ( t) = g( t) . Then 4 y1′′( t) + y1′ ( t) + 3 y1 ( t) = g( t) . Solutions 3-41
M. J. Roberts - 7/12/03
Let x 2 ( t) = h( t) . Then 4 y′′2 ( t) + y′2 ( t) + 3 y 2 ( t) = h( t) . Let x 3 ( t) = g( t) + h( t) . Then 4 y′′3 ( t) + y′3 ( t) + 3 y 3 ( t) = g( t) + h( t) Adding the first two equations, 4[ y1′′( t) + y′′2 ( t)] + [ y1′ ( t) + y′2 ( t)] + 3[ y1 ( t) + y 2 ( t)] = g( t) + h( t)
Therefore 4 y′′3 ( t) + y′3 ( t) + 3 y 3 ( t) = 4[ y1′′( t) + y′′2 ( t)] + [ y1′ ( t) + y′2 ( t)] + 3[ y1 ( t) + y 2 ( t)] 4 y′′3 ( t) + y′3 ( t) + 3 y 3 ( t) = 4[ y1 ( t) + y 2 ( t)]′′ + [ y1 ( t) + y 2 ( t)]′ + 3[ y1 ( t) + y 2 ( t)] This can only be true for all time if y 3 ( t) = y1 ( t) + y 2 ( t) . Additive Since it is homogeneous and additive, it is also linear. It is also incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 ( t) = g( t) . Then 4 y1′′( t) + y1′ ( t) + 3 y1 ( t) = g( t) . Let x 2 ( t) = g( t − t0 ) . Then 4 y′′2 ( t) + y′2 ( t) + 3 y 2 ( t) = g( t − t0 ) . The first equation can be written as 4 y1′′( t − t0 ) + y1′ ( t − t0 ) + 3 y1 ( t − t0 ) = g( t − t0 ) Therefore
4 y1′′( t − t0 ) + y1′ ( t − t0 ) + 3 y1 ( t − t0 ) = 4 y′′2 ( t) + y′2 ( t) + 3 y 2 ( t)
This can only be true for all time if y 2 ( t) = y1 ( t − t0 ) . Time Invariant Stability: The eigenvalues are
λ1 = -0.1250 + j 0.4841 λ 2 = -0.1250 - j 0.4841
Therefore the homogeneous solution is of the form, y( t) = K1e( -0.1250 + j 0.4841) t + K 2e( -0.1250 - j 0.4841) t . If there is no excitation, but the zero-excitation response is not zero, the response will decay to zero as time increases. Since the particular solution has the same form as the excitation plus all its unique derivatives, the response to any bounded input will be a bounded output. Stable Solutions 3-42
M. J. Roberts - 7/12/03
Causality: The system equation can be rewritten as t λ t t λ2 1 2 y( t) = ∫ ∫ x(λ1 ) dλ1dλ 2 − ∫ y(λ1 ) dλ1 − 3 ∫ ∫ y(λ1 ) dλ1dλ 2 4 −∞ −∞ −∞ −∞ −∞
So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends on the excitation at times, t < t0 . System has memory. Invertibility: The system equation,
4 y′′ ( t) + y′ ( t) + 3 y( t) = x( t)
expresses the excitation in terms of the response and its derivatives. Therefore the excitation is uniquely determined by the response. Invertible. 33. A system has a response that is the cube of its excitation. Classify the system as to homogeneity, additivity, linearity, time-invariance, stability, causality, memory, and invertibility. y( t) = x 3 ( t) Homogeneity: Let x1 ( t) = g( t) . Then y1 ( t) = g 3 ( t) .
Let x 2 ( t) = K g( t) . Then y 2 ( t) = [K g( t)] = K 3 g 3 ( t) ≠ Ky1 ( t) = K g 3 ( t) . Not homogeneous 3
Additivity: Let x1 ( t) = g( t) . Then y1 ( t) = g 3 ( t) . Let x 2 ( t) = h( t) . Then y 2 ( t) = h 3 ( t) . Let x 3 ( t) = g( t) + h( t) .
Then y 3 ( t) = [g( t) + h( t)] = g 3 ( t) + h 3 ( t) + 3 g 2 ( t) h( t) + 3 g( t) h 2 ( t) ≠ y1 ( t) + y 2 ( t) Not additive 3
Since it is not homogeneous and not additive, it is also linear. It is also not incrementally linear. It is statically non-linear because it is non-linear without memory (lack of memory proven below).
Solutions 3-43
M. J. Roberts - 7/12/03
Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) = g 3 ( t) . Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = g 3 ( t − t0 ) = y1 ( t − t0 ) . Time Invariant Stability: If x( t) is bounded then y( t) = x 3 ( t) is also bounded. Stable Causality: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any past values. System has no memory. Invertibility: 1
Solve y( t) = x 3 ( t) for x( t) . x( t) = y 3 ( t) . Therefore the system is not invertible.
The cube root operation is multiple valued.
34. A CT system is described by the differential equation, t y′ ( t) − 8 y( t) = x( t) . Classify the system as to linearity, time-invariance and stability. Homogeneity: Let x1 ( t) = g( t) . Then t y1′ ( t) − 8 y1 ( t) = g( t) . Let x 2 ( t) = K g( t) . Then t y′2 ( t) − 8 y 2 ( t) = K g( t) . If we multiply the first equation by K, we get tK y1′ ( t) − 8K y1 ( t) = K g( t) Therefore tK y1′ ( t) − 8K y1 ( t) = t y′2 ( t) − 8 y 2 ( t) This can only be true for all time if y 2 ( t) = K y1 ( t) . Homogeneous Additivity: Let x1 ( t) = g( t) . Then t y1′ ( t) − 8 y1 ( t) = g( t) . Let x 2 ( t) = h( t) . Then t y′2 ( t) − 8 y 2 ( t) = h( t) . Let x 3 ( t) = g( t) + h( t) . Then t y′3 ( t) − 8 y 3 ( t) = g( t) + h( t) Adding the first two equations,
Solutions 3-44
M. J. Roberts - 7/12/03
Therefore
t[ y1′ ( t) + y′2 ( t)] − 8[ y1 ( t) + y 2 ( t)] = g( t) + h( t) t[ y1′ ( t) + y′2 ( t)] − 8[ y1 ( t) + y 2 ( t)] = t y′3 ( t) − 8 y 3 ( t) t[ y1 ( t) + y 2 ( t)]′ − 8[ y1 ( t) + y 2 ( t)] = t y′3 ( t) − 8 y 3 ( t)
This can only be true for all time if y 3 ( t) = y1 ( t) + y 2 ( t) . Additive Since it is homogeneous and additive, it is also linear. It is also incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 ( t) = g( t) . Then t y1′ ( t) − 8 y1 ( t) = g( t) . Let x 2 ( t) = g( t − t0 ) . Then t y′2 ( t) − 8 y 2 ( t) = g( t − t0 ) . The first equation can be written as
(t − t0 ) y1′(t − t0 ) − 8 y1(t − t0 ) = g(t − t0 ) Therefore
(t − t0 ) y1′(t − t0 ) − 8 y1(t − t0 ) = t y′2 (t) − 8 y2 (t)
This equation is not satisfied if y 2 ( t) = y1 ( t − t0 ) therefore y 2 ( t) ≠ y1 ( t − t0 ) . Time Variant Stability: The homogeneous solution to the differential equation is of the form, ty ′ ( t) = 8 y ( t) To satisfy this equation the derivative of “y” times “t” must be of the same functional form as “y” itself. This is satisfied by a homogeneous solution of the form, y( t) = Kt 8 If there is no excitation, but the zero-excitation response is not zero, the response will increase without bound as time increases. Unstable Causality: The system equation can be rewritten as Solutions 3-45
M. J. Roberts - 7/12/03
x(λ ) y(λ ) y( t) = ∫ dλ + 8 ∫ dλ λ λ −∞ −∞ t
t
So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends on the excitation at times, t < t0 . System has memory. Invertibility: The system equation,
t y′ ( t) − 8 y( t) = x( t)
expresses the excitation in terms of the response and its derivatives. Therefore the excitation is uniquely determined by the response. Invertible. 35. A CT system is described by the equation, y( t) =
t 3
∫ x(λ )dλ
.
−∞
Classify the system as to time-invariance, stability and invertibility. Homogeneity: t 3
Let x1 ( t) = g( t) . Then y1 ( t) =
∫ g(λ )dλ .
−∞
Let x 2 ( t) = K g( t) . Then y 2 ( t) =
t 3
t 3
−∞
−∞
∫ K g(λ )dλ = K ∫ g(λ )dλ = K y (t) . 1
Homogeneous Additivity: t 3
Let x1 ( t) = g( t) . Then y1 ( t) =
∫ g(λ )dλ .
−∞ t 3
Let x 2 ( t) = h( t) . Then y 2 ( t) = Let x 3 ( t) = g( t) + h( t) .
∫ h(λ )dλ .
−∞
Solutions 3-46
M. J. Roberts - 7/12/03
Then y 3 ( t) =
t 3
t 3
t 3
−∞
−∞
−∞
∫ [g(λ ) + h(λ )]dλ = ∫ g(λ )dλ + ∫ h(λ )dλ = y (t) + y (t) 1
2
Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) =
∫ g(λ )dλ .
−∞
Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) =
t 3
t 3
t − t0 3
t −t 3 0
∫ g(λ − t )dλ = ∫ g(u)du ≠ y (t − t ) = ∫ g(λ )dλ . 0
−∞
1
0
−∞
−∞
Time Variant Stability: If x( t) is a constant, K, then y( t) =
t 3
t 3
−∞
−∞
∫ Kdλ = K ∫ dλ and, as t → ∞, y(t) increases without
bound. Unstable Causality: The response at time, t = −3, depends partially on the excitation at time t = −1 which is in the future. Not causal Memory: The response at any time, t = t0 , depends partially on excitations in the past, t < t0 . System has memory. Invertibility: Differentiate both sides of y( t) = that x( t) = y′ ( 3t) . Invertible.
t 3
∫ x(λ )dλ
−∞
t w.r.t. t yielding y′ ( t) = x . Then it follows 3
36. A CT system is described by the equation,
Solutions 3-47
M. J. Roberts - 7/12/03
y( t) =
t +3
∫ x(λ )dλ
.
−∞
Classify the system as to linearity, causality and invertibility. Homogeneity: Let x1 ( t) = g( t) . Then y1 ( t) =
t +3
∫ g(λ )dλ .
−∞
Let x 2 ( t) = K g( t) . Then y 2 ( t) =
t +3
∫
K g(λ ) dλ = K
−∞
t +3
∫ g(λ )dλ = K y (t) . 1
−∞
Homogeneous Additivity: Let x1 ( t) = g( t) . Then y1 ( t) =
t +3
∫ g(λ )dλ .
Let x 2 ( t) = h( t) . Then y 2 ( t) = t +3
∫
−∞
∫ h(λ )dλ .
−∞
Let x 3 ( t) = g( t) + h( t) . Then y 3 ( t) =
−∞ t +3
[g(λ ) + h(λ )]dλ =
t +3
∫
g(λ ) dλ +
−∞
t +3
∫ h(λ )dλ = y (t) + y (t) 1
2
−∞
Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) =
∫ g(λ )dλ .
−∞
Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) =
t +3
t − t0 + 3
t +3
∫ g(λ − t )dλ = ∫ g(u)du = y (t − t ) . 0
−∞
1
0
−∞
Time Invariant Stability: If x( t) is a constant, K, then y( t) =
t +3
t +3
−∞
−∞
∫ Kdλ = K ∫ dλ
and, as t → ∞, y( t) increases
without bound. Unstable Causality: The response at any time, t = t0 , depends partially on the excitation at times t0 < t < t0 + 3 which are in the future.
Solutions 3-48
M. J. Roberts - 7/12/03
Not causal Memory: The response at any time, t = t0 , depends partially on excitations in the past, t < t0 . System has memory. Invertibility: Differentiate both sides of y( t) = follows that x( t) = y′ ( t − 3) . Invertible.
t +3
∫ x(λ )dλ
w.r.t. t yielding y′ ( t) = x( t + 3) .
Then it
−∞
37. Show that the system described by y( t) = Re( x( t)) is additive but not homogeneous. (Remember, if the excitation is multiplied by any complex constant and the system is homogeneous, the response must be multiplied by that same complex constant.) y( t) = Re( x( t)) Homogeneity: Let x1 ( t) = g( t) + j h( t) , where g( t) and h( t) are both real-valued functions. Then y1 ( t) = Re(g( t) + j h( t)) = g( t) . Let x 2 ( t) = (K r + jK i )[g( t) + j h( t)], where K r and K i are both real constants. Then y 2 ( t) = Re (K r + jK i )[g( t) + j h( t)]
(
)
y 2 ( t) = Re(K r g( t) − K i h( t) + jK i g( t) + jK r h( t)) = K r g( t) − K i h( t) If we multiply the first equation by K r + jK i , we get
(Kr + jKi ) y1(t) = (Kr + jKi ) Re(g(t) + j h(t)) (Kr + jKi ) y1(t) = (Kr + jKi ) g(t) Therefore
y 2 ( t) ≠ (K r + jK i ) y1 ( t)
unless K i = 0. Not homogeneous Additivity: Let x1 ( t) = g1 ( t) + j h1 ( t) , where g1 ( t) and h1 ( t) are both real-valued functions. Then y1 ( t) = Re(g1 ( t) + j h1 ( t)) = g1 ( t) . Let x 2 ( t) = g 2 ( t) + j h 2 ( t) , where g 2 ( t) and h 2 ( t) are both real-valued functions. Then y 2 ( t) = Re(g 2 ( t) + j h 2 ( t)) = g 2 ( t) .
Solutions 3-49
M. J. Roberts - 7/12/03
Let x 3 ( t) = g1 ( t) + j h1 ( t) + g 2 ( t) + j h 2 ( t) . Then y 3 ( t) = Re(g1 ( t) + j h1 ( t) + g 2 ( t) + j h 2 ( t)) = g1 ( t) + g 2 ( t) = y1 ( t) + y 2 ( t) . Additive 38. Graph the magnitude and phase of the complex-sinusoidal response of the system described by y′ ( t) + 2 y( t) = e − j 2πft as a function of cyclic frequency, f. This is a complex exponential excitation which has been applied for all time. The response is the particular solution of the form, y( t) = Ke − j 2πft where K is a complex constant. Substituting the solution form into the equation, − j 2πfKe − j 2πft + 2Ke − j 2πft = e − j 2πft or K= Therefore y( t) =
1 1 . 2 1 − jπf
1 1 e − j 2πft 2 1 − jπf Exercise 38
0.5 0.4
|K|
0.3 0.2 0.1 0 -4
-3
-2
-1
-3
-2
-1
0
1
2
3
4
0
1
2
3
4
1.5
Phase of K
1 0.5 0 -0.5 -1 -1.5 -4
Frequency, f (Hz)
39. A DT system is described by y[ n ] =
n +1
∑ x(m) .
m =−∞
Classify this system as to time invariance, BIBO stability and invertibility.
Solutions 3-50
M. J. Roberts - 7/12/03
Homogeneity: Let x1[ n ] = g[ n ] . Then y1[ n ] =
n +1
∑ g[m]
m =−∞ n +1
Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] =
n +1
∑ K g[m] = K ∑ g[m] = K y [n] .
m =−∞
1
m =−∞
Homogeneous. Let x1[ n ] = g[ n ] . Then y1[ n ] =
n +1
∑ g[m]
m =−∞ n +1
Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] = Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] =
∑ h[m]
m =−∞
n +1
n +1
n +1
m =−∞
m =−∞
m =−∞
∑ (g[m] + h[m]) = ∑ g[m] + ∑ h[m] = y1[n] + y2[n] .
Additive. Since the system is homogeneous and additive it is also linear. The system is also incrementally linear because it is linear. The system is not statically non-linear because it is linear. Time Invariance: Let x1[ n ] = g[ n ] . Then y1[ n ] =
n +1
∑ g[m] .
m =−∞
Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] =
n +1
∑ g[m − n ] . 0
m =−∞
The first equation can be rewritten as y1[ n − n 0 ] =
n − n 0 +1
∑
m =−∞
g[ m] =
n +1
∑ g[q − n ] = y [n]
q =−∞
0
2
Time invariant Stability: If the excitation is a constant, the response increases without bound. Unstable Causality: At any discrete time, n = n 0 , the response depends on the excitation at the next discrete time in the future. Not causal. Memory: At any discrete time, n = n 0 , the response depends on the excitation at that discrete time and previous discrete times. System has memory.
Solutions 3-51
M. J. Roberts - 7/12/03
Invertibility: Inverting the functional relationship, y[ n ] =
n +1
∑ x[m] .
m =−∞
Invertible. Taking the first backward difference of both sides of the original system equation, y[ n ] − y[ n − 1] =
n +1
n +1−1
m =−∞
m =−∞
∑ x[m] − ∑ x[m]
x[ n + 1] = y[ n ] − y[ n − 1] The excitation is uniquely determined by the response. Invertible. 40. A DT system is described by n y[ n ] − 8 y[ n − 1] = x[ n ] . Classify this system as to time invariance, BIBO stability and invertibility. Homogeneity: Let x1[ n ] = g[ n ] . Then n y1[ n ] − 8 y1[ n − 1] = g[ n ] Let x 2 [ n ] = K g[ n ] . Then n y 2 [ n ] − 8 y 2 [ n − 1] = K g[ n ] Multiply the first equation by K. nK y1[ n ] − 8K y1[ n − 1] = K g[ n ] Then, equating results, nK y1[ n ] − 8K y1[ n − 1] = n y 2 [ n ] − 8 y 2 [ n − 1] If this equation is to be satisfied for all n, y 2 [ n ] = K y1[ n ] . Homogeneous. Additivity: Let x1[ n ] = g[ n ] . Then n y1[ n ] − 8 y1[ n − 1] = g[ n ] Let x 2 [ n ] = h[ n ] . Then n y 2 [ n ] − 8 y 2 [ n − 1] = h[ n ] Let x 3 [ n ] = g[ n ] + h[ n ] . Then n y 3 [ n ] − 8 y 3 [ n − 1] = g[ n ] + h[ n ] Add the two first two equations. n ( y1[ n ] + y 2 [ n ]) − 8( y1[ n − 1] + y 2 [ n − 1]) = g[ n ] + h[ n ] Then, equating results, n ( y1[ n ] + y 2 [ n ]) − 8( y1[ n − 1] + y 2 [ n − 1]) = n y 3 [ n ] − 8 y 3 [ n − 1] If this equation is to be satisfied for all n, Solutions 3-52
M. J. Roberts - 7/12/03
y 3 [ n ] = y1[ n ] + y 2 [ n ].
Additive.
Since the system is both homogeneous and additive, it is linear. Since the system is linear it is also incrementally linear. Since the system is linear, it is not statically non-linear. Time Invariance: Let x1[ n ] = g[ n ] . Then n y1[ n ] − 8 y1[ n − 1] = g[ n ] Let x 2 [ n ] = g[ n − n 0 ] . Then n y 2 [ n ] − 8 y 2 [ n − 1] = g[ n − n 0 ] We can re-write the first equation as
(n − n0 ) y1[n − n0 ] − 8 y1[n − n0 − 1] = g[n − n0 ] Then, equating results,
(n − n0 ) y1[n − n0 ] − 8 y1[n − n0 − 1] = n y2[n] − 8 y2[n − 1] This equation cannot be satisfied for all n, therefore y 2 [ n ] ≠ y1[ n − n 0 ] . Time Variant. Stability: If x is bounded, so is y. Stable. Causality: We can rearrange the system equation into y1[ n ] =
g[ n ] + 8 y1[ n − 1] n
showing that the response at time, n, depends on the excitation at time, n, and the response at a previous time. Causal. Memory: The response depends on past values of the response. The system has memory. Invertibility: The original system equation, n y[ n ] − 8 y[ n − 1] = x[ n ], expresses the excitation in terms of the response. Invertible.
Solutions 3-53
M. J. Roberts - 7/12/03
41. A DT system is described by
y[ n ] = x[ n ] .
Classify this system as to linearity, BIBO stability, memory and invertibility. Homogeneity: Let x1[ n ] = g[ n ] . Then y1[ n ] = g[ n ]
Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] = K g[ n ] = K g[ n ] . Multiplying the first equation by K, K y1[ n ] = K g[ n ] ≠ y 2 [ n ] . Not homogeneous. Let x1[ n ] = g[ n ] . Then y1[ n ] = g[ n ]
Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] = h[ n ]
Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] = g[ n ] + h[ n ] ≠ y1[ n ] + y 2 [ n ] . Not additive. Since the system is not homogeneous and not additive it is also not linear. The system is also not incrementally linear. The system is statically non-linear because of the square-root relationship between excitation and response. Time Invariance: Let x1[ n ] = g[ n ] . Then y1[ n ] = g[ n ] .
Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] = g[ n − n 0 ] . The first equation can be rewritten as y1[ n − n 0 ] = g[ n − n 0 ] = y 2 [ n ] Time invariant Stability: If the excitation is bounded, the response is bounded. Stable Causality: At any discrete time, n = n 0 , the response depends only on the excitation at that same time. Causal. Memory: At any discrete time, n = n 0 , the response depends only on the excitation at that same time. System has no memory. Invertibility:
Solutions 3-54
M. J. Roberts - 7/12/03
Inverting the functional relationship, Invertible.
x[ n ] = y 2 [ n ] .
42. Graph the magnitude and phase of the complex-sinusoidal response of the system described by 1 y[ n ] + y[ n − 1] = e − jΩn 2 as a function of Ω. The equation can be written as n 1 y[ n ] + y[ n − 1] = (e − jΩ ) = α n 2
where
α = e − jΩ
The particular solution has the form,
y[ n ] = Kα n .
Substituting the solution form into the equation, 1 Kα n + Kα n −1 = α n 2
or
1 Kα + K = α . 2 Solving for K, K=
Therefore y[ n ] =
e − jΩ e − jΩ
1 + 2
α 1 α+ 2
e − jΩ
= e
− jΩ
1 + 2
.
αn .
Solutions 3-55
M. J. Roberts - 7/12/03
|K| 2
-2
2
F
Phase of K π
-2
2
F
-π
43. Find the impulse response, h[ n ], of the system in Figure E43.
x[n]
y[n]
2
D
0.9
Figure E43 DT system block diagram or
y[ n ] = 2 x[ n ] + 0.9 y[ n − 1] y[ n ] − 0.9 y[ n − 1] = 2 x[ n ]
The homogeneous solution (for n ≥ 0) is of the form, y[ n ] = K hα n therefore the characteristic equation is K hα n − 0.9K hα n −1 = 0 . and the eigenvalue is α = 0.9 and, therefore, y[ n ] = K h (0.9)
n
We can find an initial condition to evaluate the constant, K h , by directly solving the difference equation for n = 0. y[0] = 2 x[0] + 0.9 y[−1] = 2 . Therefore
2 = K h (0.9) ⇒ K h = 2 . 0
Therefore the total solution is y[ n ] = 2(0.9)
n
Solutions 3-56
M. J. Roberts - 7/12/03
which is the impulse response. 44. Find the impulse responses of these systems. (a)
3 y[ n ] + 4 y[ n − 1] + y[ n − 2] = x[ n ] + x[ n − 1] This can be written as y[ n ] =
1 (x[n] + x[n − 1] − 4 y[n − 1] − y[n − 2]) . 3
The homogeneous solution is y h [ n ] = K1hα1n + K 2 hα 2n where 2 1 α1,2 = − ± . 3 3 The impulse response is the response to a single unit impulse at n = 0 plus the response to another single unit impulse at n = 1. The response to a single unit impulse at n = 0 is 1 n n n h 0 [ ] = K1h − + K 2 h (−1) u[ n ] 3 subject to the initial conditions, h 0 [0] = relation). Solving for the constants, K1h = − Therefore
1 4 and h 0 [1] = − (found from the recursion 3 9 1 1 , K2h = . 6 2
1 1 n 1 n h 0 [ n ] = − − + (−1) u[ n ] 2 6 3
and
1 1 n 1 1 1 n −1 1 n n −1 h[ n ] = − − + (−1) u[ n ] + − − + (−1) u[ n − 1] 2 2 6 3 6 3
or
n
1 1 1 n h[ n ] = − − (u[ n ] − 3 u[ n − 1]) + (−1) (u[ n ] − u[ n − 1]) 2 6 3 (b)
5 y[ n ] + 6 y[ n − 1] + 10 y[ n − 2] = x[ n ] 2 The homogeneous solution is
Solutions 3-57
M. J. Roberts - 7/12/03
y h [ n ] = K1hα1n + K 2 hα 2n
where
α1,2 = −1.2 ± j1.6 .
(
)
h[ n ] = K1h (−1.2 + j1.6) + K 2 h (−1.2 + j1.6) u[ n ] n
h[0] =
2 5
n
and
h[1] = −
24 25
Solving for the constants, K1h = 0.2 + j 0.15 , K 2 h = 0.2 − j 0.15
(
)
h[ n ] = (0.2 + j 0.15)(−1.2 + j1.6) + (0.2 − j 0.15)(−1.2 − j1.6) u[ n ] n
n
(
)
h[ n ] = (0.2 + j 0.15)e( 0.693 + j 2.214 ) n + (0.2 − j 0.15)e( 0.693 − j 2.214 ) n u[ n ] h[ n ] = e 0.693 n (0.2e j 2.214 n + 0.2e − j 2.214 n + j 0.15e j 2.214 n − j 0.15e − j 2.214 n ) u[ n ] h[ n ] = 2 n (0.4 cos(2.2143n ) − 0.3 sin(2.2143n )) u[ n ] 45. Plot g[ n ]. Use the MATLAB conv function if needed. g[n] 3
... -10
... 10
(a)
2πn g[ n ] = rect1[ n ] ∗ sin 9
n
-3
j 1 1 G( F ) = 3 drcl( F , 3) comb F + − comb F − 2 9 9 1 1 3 G( F ) = j drcl( F , 3) comb F + − drcl( F , 3) comb F − 9 9 2 1 1 3 1 1 G( F ) = j drcl − , 3 comb F + − drcl , 3 comb F − 9 9 9 9 2 Since the Dirichlet function is even, 1 1 3 1 1 G( F ) = j drcl , 3 comb F + − drcl , 3 comb F − 9 9 9 2 9
Solutions 3-58
M. J. Roberts - 7/12/03
1 2πn 2πn g[ n ] = 3 drcl , 3 sin = 2.5321sin 9 9 9 g[n] 3
... -10
... 10
(b)
2πn g[ n ] = rect 2 [ n ] ∗ sin 9
n
-3
j 1 1 G( F ) = 5 drcl( F , 5) comb F + − comb F − 2 9 9 1 1 5 G( F ) = j drcl( F , 5) comb F + − drcl( F , 5) comb F − 9 9 2 1 1 5 1 1 G( F ) = j drcl − , 5 comb F + − drcl , 5 comb F − 9 9 9 9 2 Since the Dirichlet function is even, 1 1 5 1 1 G( F ) = j drcl , 5 comb F + − drcl , 5 comb F − 9 9 9 9 2 1 2πn 2πn g[ n ] = 5 drcl , 5 sin = 2.8794 sin 9 9 9 g[n] 3
-10
(c)
2πn g[ n ] = rect 4 [ n ] ∗ sin 9
10
n
-3
j 1 1 G( F ) = 9 drcl( F , 9) comb F + − comb F − 2 9 9 1 1 9 G( F ) = j drcl( F , 9) comb F + − drcl( F , 9) comb F − 9 9 2 1 1 9 1 1 G( F ) = j drcl − , 9 comb F + − drcl , 9 comb F − 9 9 9 9 2 Since the Dirichlet function is even,
Solutions 3-59
M. J. Roberts - 7/12/03
1 1 9 1 1 G( F ) = j drcl , 9 comb F + − drcl , 9 comb F − 9 9 9 2 9 1 2πn g[ n ] = 9 drcl , 9 sin =0 9 9 (d)
g[ n ] = rect 3 [ n ] ∗ rect 3 [ n ] ∗ comb14 [ n ] g[n] 7
...
... -30
rect 3 [ n ] ∗ rect 3 [ n ] =
-20
-10
0
10
20
n
30
∞
n
3
∑ rect [m]rect [n − m] = ∑ rect [n − m] = 7 tri 7 3
m =−∞
3
m =−3
3
n g[ n ] = rect 3 [ n ] ∗ rect 3 [ n ] ∗ comb14 [ n ] = 7 tri ∗ comb14 [ n ] 7 ∞
∞
n − 14 m n g[ n ] = 7 tri ∗ ∑ δ [ n − 14 m] = 7 ∑ tri 7 m =−∞ 7 m =−∞ (e)
g[ n ] = rect 3 [ n ] ∗ rect 3 [ n ] ∗ comb 7 [ n ] g[n] 7
...
... n
rect 3 [ n ] ∗ rect 3 [ n ] =
∞
n
3
∑ rect [m]rect [n − m] = ∑ rect [n − m] = 7 tri 7
m =−∞
3
3
m =−3
3
n g[ n ] = rect 3 [ n ] ∗ rect 3 [ n ] ∗ comb 7 [ n ] = 7 tri ∗ comb 7 [ n ] 7
Solutions 3-60
M. J. Roberts - 7/12/03 ∞
∞
n − 7m n g[ n ] = 7 tri ∗ ∑ δ [ n − 7 m] = 7 ∑ tri =7 7 7 m =−∞ m =−∞ 2π 7 g[ n ] = 2 cos n ∗ u[ n ] 7 8 n
(f)
g[n] 0.1
0 10
15
20
25
30
35
n
-0.1
(g)
n n sinc sinc 4 4 g[ n ] = ∗ 2 2 2 2 g[n] 0.5
-40
-30
-20
-10
10
20
30
40
n
46. Find the impulse responses of the subsystems in Figure E 46 and then convolve them to find the impulse response of the cascade connection of the two subsystems. You may find this formula for the summation of a finite series useful, , α =1 N N . α = 1 − α ∑ , α ≠1 n =0 1− α N −1
n
x[n]
y1 [n]
D
y2 [n]
D
4 5
Figure E 46 Two cascaded subsystems y1[ n ] = x[ n ] − y1[ n − 1]
and
y 2 [ n ] = y1[ n ] −
Solutions 3-61
4 y [ n − 1] 5 2
M. J. Roberts - 7/12/03
n
h1[ n ] = (−1) u[ n ] n
4 h 2 [ n ] = − u[ n ] 5
and
∞ 4 4 m h[ n ] = h1[ n ] ∗ h 2 [ n ] = (−1) u[ n ] ∗ − u[ n ] = ∑ (−1) u[ m] − 5 5 m =−∞ n
n
∞
4 h[ n ] = ∑ (−1) − 5 m =0
n −m
m
4 h[ n ] = − 5 5 n 1− 4 4 h[ n ] = − 5 5 1− 4
n n
m 4 u[ n − m] = ∑ (−1) − 5 m =0 n
4 (−1) − ∑ 5 m =0 m
−m
4 = − 5
n n
5 ∑ m =0 4
n −m
u[ n − m]
n −m
m
n +1
4 = −4 − 5
n
n n 4 h[ n ] = −4 (−1) − 5
4 n 5 5 n +1 n 4 1 − = − − − (−1) 4 4 5 n +1 5 4 n 5 1 1 ( ) − = − 5 4
47. For the system of Exercise 43, let the excitation, x[ n ] , be a unit-amplitude complex sinusoid of DT cyclic frequency, F. Plot the amplitude of the response complex sinusoid versus F over the range, −1 < F < 1. Since we only want the steady-state sinusoidal response we can assume a particular solution of the form, y[ n ] = Ke − j 2πFn . Then
Ke − j 2πFn − 0.9Ke − j 2πF ( n −1) = 2e − j 2πFn K=
and
2e − j 2πF e − j 2πF − 0.9
2e − j 2πF e − j 2πFn . y[ n ] = − j 2πF e − 0.9
Solutions 3-62
M. J. Roberts - 7/12/03
|K| 20
15
10
5
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0.4
0.6
0.8
1
F
Phase of K 1.5 1 0.5 0.2 -1
-0.8
-0.6
-0.4
-0.2
0
F
-0.5 -1 -1.5
48. In the second-order DT system below what is the relationship between a, b and c that ensures that the system is stable? 1 x[n] y[n] a
y[ n ] =
b
D
c
D
x[ n ] − (b y[ n − 1] + c y[ n − 2]) a
Stability is determined by the eigenvalues of the homogeneous solution. a y[ n ] + b y[ n − 1] + c y[ n − 2] = 0 The eigenvalues are
α1,2 =
−b ± b 2 − 4 ac 2a
For stability the magnitudes of all the eigenvalues must be less than one. Therefore 2
b c b − + − <1 2a 2a a −
b 1 b 2 − 4 ac < 1 + 2a 2a
2
and
b c b − − − <1 2a 2a a
and
−
b 1 b 2 − 4 ac < 1 − 2a 2a
Solutions 3-63
M. J. Roberts - 7/12/03
−b + b 2 − 4 ac < 2 a
and
−b − b 2 − 4 ac < 2 a
−b + j 4 ac − b 2 < 2 a
and
−b − j 4 ac − b 2 < 2 a
If b 2 − 4 ac < 0 ,
In either case b 2 + 4 ac − b 2 < 4 a 2
or
ac < a 2 From the requirement, b 2 − 4 ac < 0 we know that ac must be positive. Then we can divide both sides by the positive number, ac, yielding a >1 . c
If b 2 − 4 ac ≥ 0 ,
(
−b + b 2 − 4 ac
)
2
< 4 a2
and
b 2 − 2b b 2 − 4 ac + b 2 − 4 ac < 4 a 2 and −2b b 2 − 4 ac < 4 a 2 − 2b 2 + 4 ac −b b 2 − 4 ac < 2 a 2 − b 2 + 2 ac
and
(
−b − b 2 − 4 ac
)
2
< 4 a2
b 2 + 2b b 2 − 4 ac + b 2 − 4 ac < 4 a 2 2b b 2 − 4 ac < 4 a 2 − 2b 2 + 4 ac
and
b b 2 − 4 ac < 2 a 2 − b 2 + 2 ac
Taken together, these two requirements lead to 2 a 2 − b 2 + 2 ac > b b 2 − 4 ac ≥ 0 2 a( a + c ) ≥ b 2 and
(2a
2
− b 2 + 2 ac ) > b 2 (b 2 − 4 ac ) 2
4 a 4 + b 4 + 4 a 2c 2 − 4 a 2b 2 − 4 ab 2c + 8 a 3c > b 4 − 4 ab 2c 4 a 2 ( a 2 + c 2 − b 2 + 2 ac ) > 0 a 2 + c 2 − b 2 + 2 ac > 0 a 2 − 2 ac + c 2 > b 2 − 4 ac
(a − c ) 2 > b 2 − 4 ac
Solutions 3-64
M. J. Roberts - 7/12/03
49. Given the excitations, x[ n ] , and the impulse responses, h[ n ], find closed-form expressions for and plot the system responses, y[ n ] . (a)
n
x[ n ] = u[ n ]
7 h[ n ] = n u[ n ] 8
,
1 − r N , r ≠1 with respect to r.) (Hint: Differentiate ∑ r n = 1 − r n =0 N , r =1 N −1
y[ n ] = h[ n ] ∗ x[ n ] =
∞
m
n 7 7 m u m u n − m = m [ ] [ ] ∑ ∑ 8 8 m =−∞ m =0
m
1 − r N , r ≠1 with respect to r, Differentiating ∑ r = 1 − r n =0 N , r =1 N −1
n
N −1
∑ nr
n −1
n =0
(1 − r)(− Nr N −1) − (1 − r N )(−1) , r ≠1 = (1 − r) 2
N −1
r ∑ nr n −1 = r n =0
− Nr N −1 + Nr N + 1 − r N , r ≠1 (1 − r) 2
Nr N −1 ( r − 1) + 1 − r N nr = r , r ≠1 ∑ (1 − r) 2 n =0 N −1
n
7 (n + 1) 7 8 y[ n ] = 8
n
7 7 − 1 + 1 − 8 8 7 1 − 8
n +1
2
u[ n ]
n n +1 7 1 7 1 − + − y[ n ] = 56 ( n + 1) u[ n ] 8 8 8
7 n n y[ n ] = 56 1 − + 1 u[ n ] 8 8
Solutions 3-65
M. J. Roberts - 7/12/03
Excitation
x[n] 1
-5
60
n
Impulse Response
h[n] 3
-5
60
y[n]
n
Response
50
-5
(b)
60 n
x[ n ] = u[ n ] y[ n ] = h[ n ] ∗ x[ n ] = y[ n ] =
n
4 3 h[ n ] = δ [ n ] − − u[ n ] 4 7
,
m m n 4 4 3 3 − δ m − − u m u n m = δ m − − [ ] [ ] [ ] [ ] ∑ ∑ 4 4 m =−∞ 7 m =0 7 ∞
m
n 4 4 3 3 u[ n ] − ∑ − = − 4 7 7 4 m =0
x[n]
n +1
u[ n ]
Excitation
1
-5
40
h[n]
n
Impulse Response
1 -5
40
n
-1
y[n]
Response
1 -5
40
n
-1
50. A CT function is non-zero over a range of its argument from 0 to 4. It is convolved with a function which is non-zero over a range of its argument from -3 to -1. What is the non-zero range of the convolution of the two? -3 to 3 51. What function convolved with −2cos( t) would produce 6sin( t) ?
Solutions 3-66
M. J. Roberts - 7/12/03
π π −2 cos( t) = −2 sin t + = 2 sin t − 2 2 Therefore an impulse of some weight and position can produce the desired function. π 2 sin t − ∗ Aδ ( t − t0 ) = 6 sin( t) 2 Then π π π A = 3 and t − t0 − = t or t0 = − and the desired function is 3δ t + . 2 2 2 This answer is not unique. 52. Sketch these functions. (a) 1 1 g( t) = 3 cos(10πt) ∗ 4δ t + = 12 cos10π t + = 12 cos(10πt + π ) = −12 cos(10πt) 10 10 g(t) 12
-0.5
0.5
t
-12
(b) g( t) = tri(2 t) ∗ comb( t) = tri(2 t) ∗
∞
∑ δ (t − n) =
n =−∞
∞
∑
n =−∞
tri(2 t) ∗ δ ( t − n ) =
∞
∑ tri(2(t − n))
n =−∞
g(t) 1
-2
2
t
∞ t t (c) g( t) = [tri(2 t) − rect ( t − 1)] ∗ comb = [tri(2 t) − rect ( t − 1)] ∗ ∑ δ − n 2 2 n =−∞ ∞
∞
n =−∞
n =−∞
[
]
g( t) = 2 ∑ [tri(2 t) − rect ( t − 1)] ∗ δ ( t − 2 n ) = 2 ∑ tri(2( t − 2 n )) − rect (( t − 2 n ) − 1) g(t) 2
-2
2 -2
Solutions 3-67
t
M. J. Roberts - 7/12/03 ∞ ∞ t t t t (d) g( t) = tri comb( t) ∗ comb = tri ∑ δ ( t − m) ∗ ∑ δ − n 8 4 m =−∞ 4 n =−∞ 8 ∞ ∞ ∞ t t m t g( t) = ∑ ∑ tri δ ( t − m) ∗ δ − n = ∑ ∑ tri δ ( t − m) ∗ δ − n 4 8 4 8 n =−∞ m =−∞ n =−∞ m =−∞ ∞
∞
∞ 3 m m tri δ ( t − m ) ∗ δ ( t − 8 n ) = 8 tri δ ( t − 8 n − m) ∑ ∑ ∑ 4 4 n =−∞ m =−3 n =−∞ m =−3
g( t) = 8 ∑
3
g(t) 8
-10
10
t
∞ 1 t (e) g( t) = sinc( 4 t) ∗ comb = ∑ sinc( 4 ( t − 2 n )) 2 n =−∞ 2
g(t) 1
-3
3
t
-1
(f) g( t) = e −2 t u( t) ∗ g( t) =
∞
∑e
n =−∞
−2( t − 4 n )
1 t − 2 t comb − comb 4 4 4
u( t − 4 n ) − e −2( t − 2 − 4 n ) u( t − 2 − 4 n ) g(t) 1
-6
6
t
-1
∞ t − 2n t 1 t ∗ = (g) g( t) = sinc( t) rect comb ∑ sinc( t − 2 n ) rect 2 2 2 2 n =−∞
Solutions 3-68
M. J. Roberts - 7/12/03
g(t) 1
-6
t
6
∞ 1 t t t (h) g( t) = sinc(2 t) ∗ comb rect = rect ∑ sinc(2( t − 2 n )) 2 4 4 n =−∞ 2
g(t) 1
-3
t
3 -1
53. Find the signal power of these signals. t x( t) = rect ( t) ∗ comb 4
(a)
∞ t rect ( t) ∗ comb = 4 ∑ rect ( t − 4 n ) 4 n =−∞ This is a periodic signal whose period, T, is 4. Between -T/2 and +T/2, there is one rectangle whose height is 4 and whose width is 1. Therefore, between -T/2 and +T/2, the square of the signal is
[4 rect (t)]
2
1 = 16 rect ( t) and P = T 2
T 2
−
t x( t) = tri( t) ∗ comb 4
(b)
2
16 2 ∫T 16 rect (t)dt = 4 −∫ rect (t)dt = 4 2 2
2
∞ t tri( t) ∗ comb = 4 ∑ tri( t − 4 n ) 4 n =−∞
This is a periodic signal whose period, T, is 4. Between -T/2 and +T/2, there is triangle whose height is 4 and whose width is 2. Therefore, between -T/2 and T/2, the square of the signal is [4 Λ(t)]2 = 16Λ2 (t) and 1 P= T
T 2
2
1
1
1
16 2 2 2 2 ∫T 16 tri (t)dt = 4 −∫ tri (t)dt = 4 −∫ tri (t)dt = 8∫ tri (t)dt = 8∫ (1 − 2t + t )dt 2 1 0 0 2
−
2
Solutions 3-69
M. J. Roberts - 7/12/03
1
t3 8 P = 8 t − t 2 + = 3 0 3 54. A rectangular voltage pulse which begins at t = 0, is 2 seconds wide and has a height of 0.5 V drives an RC lowpass filter in which R = 10 kΩ and C = 100 µF . (a) (b) (c) (d)
Sketch the voltage across the capacitor versus time. Change the pulse duration to 0.2 s and the pulse height to 5 V and repeat. Change the pulse duration to 2 ms and the pulse height to 500 V and repeat. Change the pulse duration to 2 µs and the pulse height to 500 kV and repeat. Pulse Width = 2 s, Height = 0.5V
Pulse Width = 0.2 s, Height = 5V
v(t)
v(t)
1
1
-1
5
t
-1
Pulse Width = 0.002 s, Height = 500V
5
Pulse Width = 2e-06 s, Height = 500000V
v(t)
v(t)
1
1
-1
5
t
t
-1
5
t
Based on these results what do you think would happen if you let the input voltage be a unit impulse? v out ( t) = e − t u( t) 55. Write the differential equation for the voltage, vC ( t) , in the circuit below for time, t > 0, then find an expression for the current, i( t) , for time, t > 0. R1 = 2 Ω
C=3F iC (t) - + i(t) vC(t)
i s(t)
t=0
Vs = 10 V
i( t) = is ( t) + iC ( t)
,
vC ( t) + iC ( t) R2 = 0
i s ( t) =
,
Vs R1
R2 = 6 Ω
iC ( t) = C
,
vC ( t) + R2C
d (v (t)) dt C
d (v (t)) = 0 dt C
Solutions 3-70
M. J. Roberts - 7/12/03
vC ( t) = Ke
−αt
vC (0) = −10 = Ke
−
1 α= R2C
, t R 2C
vC ( t) = Ke
,
= K ⇒ K = −10
−
t R 2C
vC ( t) = −10e
, t
−
t R 2C
t
− d 10 10 − Ce R 2C = e R 2C iC ( t) = C ( vC ( t)) = dt R2C R2 t
V 10 − 5 −t i( t) = s + e R 2C = 5 + e 18 R1 R2 3 56. The water tank in Figure E56 is filled by an inflow, x( t) , and is emptied by an outflow, y( t) . The outflow is controlled by a valve which offers resistance, R, to the flow of water out of the tank. The water depth in the tank is d( t) and the surface area of the water is A, independent of depth (cylindrical tank). The outflow is related to the water depth (head) by d( t) y( t) = . R The tank is 1.5 m high with a 1m diameter and the valve resistance is 10 (a) (b)
(c) (d)
s . m2
Write the differential equation for the water depth in terms of the tank dimensions and valve resistance. m3 If the inflow is 0.05 , at what water depth will the inflow and outflow rates be s equal, making the water depth constant? Find an expression for the depth of water versus time after 1 m3 of water is dumped into an empty tank. m3 after If the tank is initially empty at time, t = 0, and the inflow is a constant 0.2 s time, t = 0, at what time will the tank start to overflow?
Surface area, A
Inflow, x(t)
d(t)
R
Valve
Outflow, y(t) Solutions 3-71
M. J. Roberts - 7/12/03
Figure E56 Water tank with inflow and outflow (a) y( t) =
d( t) R
d A d( t) = x( t) − y( t) 1 dt 23 volume
A d′ ( t) = x( t) − A d′ ( t) + (b)
d( t) R
d( t) = x( t) R
For the water height to be constant, d′ ( t) = 0. Then s m3 d( t) = R x( t) = 10 2 × 0.05 = 0.5 m m s
(c) Dumping 1 m3 of water into an empty tank is exciting this system with a unit impulse of water inflow. The impulse response, h( t) , of the system is the solution of A h′ ( t) + The solution is
h( t) = δ ( t) R
h( t) = Ke
−
t AR
u( t) .
We can find K by finding the initial water depth in response to 1 m3 being suddenly dumped in. The surface area is 0.7854 m2 . Therefore the initial depth is 1.273 m and h( t) = 1.273e
−
t AR
u( t) .
(d) The response to a step of flow is the convolution of the impulse response with the step excitation. d( t) = h( t) ∗ x( t) = 1.273e ∞
d( t) = 0.2546 ∫ e For t < 0, d( t) = 0
−∞
−
τ AR
−
t AR
u( t) ∗ 0.2 u( t) ∞
u(τ ) u( t − τ ) dτ = 0.2546 ∫ e 0
For t > 0,
Solutions 3-72
−
τ AR
u( t − τ ) dτ
M. J. Roberts - 7/12/03
t
d( t) = 0.2546 ∫ e
−
τ AR
0
t
t − −τ dτ = −0.2546 AR e AR = −0.2546 AR1 − e AR 0
For all time, t − d( t) = 21 − e 7.854 u( t) .
Solving for a depth of 1.5 m, t t − − t 1.5 = 21 − e 7.854 ⇒ 0.25 = e 7.854 ⇒ −1.386 = − ⇒ 10.886 s 7.854
57. The suspension of a car can be modeled by the mass-spring-dashpot system of Figure N E57 Let the mass, m, of the car be 1500 kg, let the spring constant, K s, be 75000 m N⋅s and let the shock absorber (dashpot) viscosity coefficient, K d , be 20000 . m At a certain length, d0 , of the spring, it is unstretched and uncompressed and exerts no force. Let that length be 0.6 m. (a)
What is the distance, y( t) − x( t) , when the car is at rest?
(b) Define a new variable z( t) = y( t) − x( t) − constant such that, when the system is at rest, z( t) = 0 and write a describing equation in z and x which describes an LTI system. Then find the impulse response. (c) The effect of the car striking a curb can be modeled by letting the road surface height change discontinuously by the height of the curb, hc . Let hc = 0.15 m. Graph z( t) versus time after the car strikes a curb. Automobile Chassis
Shock Absorber
Spring
y(t) x(t)
Figure E57 Car suspension model Using the basic principle, F = ma , we can write K s[ y( t) − x( t) − d0 ] + K d
d [y(t) − x(t)] + mg = − m y′′(t) dt Solutions 3-73
M. J. Roberts - 7/12/03
or (a)
m y′′ ( t) + K d y′ ( t) + K s y( t) = K d x′ ( t) + K s x( t) + K sd0 − mg . At rest all the derivatives are zero and K s ( y( t) − x( t) − d0 ) + mg = 0 .
Solving, y( t) − x( t) = (b)
K sd0 − mg 75000 × 0.6 − 1500 × 9.8 = = 0.404 m Ks 75000
The describing equation is m y′′ ( t) + K d y′ ( t) + K s y( t) = K d x′ ( t) + K s x( t) + K sd0 − mg .
which can be rewritten as
or
m y′′ ( t) + K d [ y′ ( t) − x′ ( t)] + K s[ y( t) − x( t)] − K sd0 + mg = 0 mg =0 m y′′ ( t) + K d [ y′ ( t) − x′ ( t)] + K s y( t) − x( t) − d0 + K s
Let z( t) = y( t) − x( t) − d0 +
or
mg . Then y′′ ( t) = z′′ ( t) + x′′ ( t) and Ks m[z′′ ( t) + x′′ ( t)] + K d z′ ( t) + K s z( t) = 0 m z′′ ( t) + K d z′ ( t) + K s z( t) = − m x′′ ( t)
This equation is in a form which describes an LTI system. We can find its impulse response. After time, t = 0, the impulse response is the homogenous solution. The eigenvalues are −K d ± K d2 − 4 mK s K λ1,2 = =− d ± 2m 2m
K d2 K s − = −6.667 ± j 2.357 . 4m2 m
The homogeneous solution is h( t) = K h1eλ 1 t + K h 2eλ 2 t = K h1e( −6.667 + j 2.357) t + K h 2e( −6.667 − j 2.357) t . Since the system is underdamped another (equivalent) form of homogeneous solution will be more convenient, h( t) = e −6.667 t [K h1 cos(2.357 t) + K h 2 sin(2.357 t)] . The impulse response can have a discontinuity at t = 0 and an impulse but no higher-order singularity there. Therefore the general form of the impulse response is
Solutions 3-74
M. J. Roberts - 7/12/03
h( t) = Kδ ( t) + e −6.667 t [K h1 cos(2.357 t) + K h 2 sin(2.357 t)] u( t) Integrating both sides of the describing equation,
(
m h′ (0 ) − h′ (0 +
0+
−
)) + K (h(0 ) − h(0 )) + K ∫ h(t)dt = 0 . +
−
d
s
0−
(The integral of the doublet, which is the derivative of the impulse excitation, is zero.) Since the impulse response and all its derivatives are zero before time, t = 0, it follows then that 0+
m h′ (0 ) + K d h(0 ) + K s ∫ h( t) dt = 0 +
+
0−
and
m(−6.667K h1 + 2.357K h 2 ) + K d K h1 + K sK = 0 .
Integrating the describing equation a second time, 0+
m h(0 ) + K d ∫ h( t) dt = 0 +
0−
or
mK h1 + K d K = 0 .
Integrating the describing equation a third time, 0+
m ∫ h( t) dt = − m 0−
or
mK = − m ⇒ K = −1 .
Solving for the other two constants, K h1 =
Kd and m
K K m −6.667 d + 2.357K h 2 + K d d − K s = 0 m m or K s K d2 K − 2 + 6.667 d m Kh 2 = m m 2.357 Kd m Therefore
(c)
h( t) = −δ ( t) + e −6.667 t [13.333 cos(2.357 t) − 16.497 sin(2.357 t)] u( t)
The response to a step of size 0.15 is then the convolution,
Solutions 3-75
M. J. Roberts - 7/12/03
z( t) = 0.15 u( t) ∗ h( t) or
∞
{
}
z( t) = 0.15 ∫ −δ (τ ) + e −6.667τ [13.333 cos(2.357τ ) − 16.497 sin(2.357τ )] u(τ ) u( t − τ ) dτ −∞
∞
{
}
z( t) = 0.15 ∫ −δ (τ ) + e −6.667τ [13.333 cos(2.357τ ) − 16.497 sin(2.357τ )] u( t − τ ) dτ 0−
For t < 0, z( t) = 0. For t > 0, using ax ∫ e sin(bx )dx =
e ax [a sin(bx ) − b cos(bx )] a2 + b2
e ax ∫ e cos(bx )dx = a2 + b2 [a cos(bx ) + b sin(bx )] ax
we get
t
e −6.667τ 13.333 50 [−6.667 cos(2.357τ ) + 2.357 sin(2.357τ )] z( t) = −0.15 u( t) + 0.15 e −6.667τ −16.497 50 [−6.667 sin(2.357τ ) − 2.357 cos(2.357τ )] − 0 or e −6.667 t 13 . 333 −6.667 cos(2.357 t) + 2.357 sin(2.357 t)] [ 50 e −6.667 t z( t) = −0.15 u( t) + 0.15 −16.497 −6.667 sin(2.357 t) − 2.357 cos(2.357 t)] [ 50 6 − −13.333 .667 + 16.497 −2.357 50 50
{
}
z( t) = −0.15 u( t) + 0.15 e −3.333 t [2.812 sin(2.357 t) − cos(2.357 t)] + 1 u( t) or
z( t) = 0.15e −3.333 t [2.812 sin(2.357 t) − cos(2.357 t)] u( t) z(t) 0.1 2
t
-0.2
58. As derived in the text, a simple pendulum is approximately described for small angles, θ , by the differential equation, Solutions 3-76
M. J. Roberts - 7/12/03
mLθ ′′ ( t) + mgθ ( t) ≅ x( t) where m is the mass of the pendulum, L is the length of the massless rigid rod supporting the mass and θ is the angular deviation of the pendulum from vertical. (a)
Find the general form of the impulse response of this system. mL h′′ ( t) + mg h( t) ≅ δ ( t)
The form of the homogeneous solution is j θ h ( t) = K h1e
g t L
+ K h 2e
−j
g t L
u( t)
or, more conveniently, g g t u( t) . t + K h 2 sin θ h ( t) = K h1 cos L L There can be no discontinuity or impulse in the impulse response therefore this is also the impulse response. Integrate the differential equation once through zero. g Kh 2 ≅ 1 ⇒ Kh 2 = mL L
L 1 1 1 = g mL m gL
Now integrate again through zero. K h1 = 0 . Therefore h( t) = (b)
g 1 1 sin t u( t) m gL L
If the mass is 2 kg and the rod length is 0.5 m, at what cyclic frequency will the pendulum oscillate? The cyclic frequency is
1 2π
9.8 = 0.704 . The mass is irrelevant. 1 2
59. Pharmacokinetics is the study of how drugs are absorbed into, distributed through, metabolized by and excreted from the human body. Some drug processes can be approximately modeled by a “one compartment” model of the body in which V is the volume of the compartment, C( t) is the drug concentration in that compartment, ke is a rate constant for excretion of the drug from the compartment and k0 is the infusion rate at which the drug enters the compartment.
Solutions 3-77
M. J. Roberts - 7/12/03
(a) Write a differential equation in which the infusion rate is the excitation and the drug concentration is the response. mg (where “l” is the (b) Let the parameter values be ke = 0.4 hr −1, V = 20 l and k0 = 200 hr mg symbol for “liter”). If the initial drug concentration is C(0) = 10 , plot the drug l concentration as a function of time (in hours) for the first 10 hours of infusion. Find the solution as the sum of the zero-excitation response and the zero-state response. (a)
The differential equation is V
or
d (C(t)) = k0 − Vke C(t) dt k d C( t)) + ke C( t) = 0 ( V dt
(b)
The eigenvalue is –0.4 and the zero-excitation response is C( t) = 10e −0.4 t
mg (“t” in hours). l
e −0.4 t mg . The u( t) 20 l mg mg is then C( t) = 25(1 − e −0.4 t ) step response to an infusion rate of k0 = 200 . The sum hr l of the two responses is mg mg . C( t) = 10e −0.4 t + 25(1 − e −0.4 t ) = (25 − 15e −0.4 t ) l l
The impulse response (to a unit impulse of “infusion rate”) is h( t) =
(
)
C (t) 25
10
t (hours)
60. At the beginning of the year 2000, the country, Freedonia, had a population, p, of 100 million people. The birth rate is 4% per annum and the death rate is 2% per annum, compounded daily. That is, the births and deaths occur every day at a uniform fraction of the current population and the next day the number of births and deaths changes because the population changed the previous day. For example, every day the number 0.02 of people who die is the fraction, , of the total population at the end of the previous 365 day (neglect leap-year effects). Every day 275 immigrants enter Freedonia. (a)
Write a difference equation for the population at the beginning of the nth day after January 1, 2000 with the immigration rate as the excitation of the system.
Solutions 3-78
M. J. Roberts - 7/12/03
(b)
By finding the zero-exctiation and zero-state responses of the system determine the population of Freedonia be at the beginning of the year 2050.
(a)
The difference equation is p[ n + 1] = p[ n ] +
0.04 0.02 p[ n ] − p[ n ] + 275 365 365
p[ n + 1] − (1 + 5.48 × 10 −5 ) p[ n ] = 275 The eigenvalue is 1 + 5.48 × 10 −5 = 1.0000548 and the zero-excitation response is p[ n ] = 10 8 (1.0000548)
n
The impulse response is h[ n ] = (1.0000548) u[ n ] . The response to the immigration rate is the convolution of the impulse response with the immigration rate, 275u[ n ], or n
n −1
p[ n ] = (1.0000548) u[ n ] ∗ 273.9 u[ n ] = 273.9 ∑ (1.0000548) n
m
m =0
Using the summation formular for a geometric series, p[ n ] = 275
1.0000548 n − 1 1 − 1.0000548 n u[ n ] = 5018248.2(1.0000548 n − 1) u[ n ] u[ n ] = 275 0.0000548 1 − 1.0000548
The total solution is n p[ n ] = 10 8 (1.0000548) + 5018248.2(1.0000548 n − 1) u[ n ] , n ≥ 0 p[ n ] = 1.05 × 10 8 (1.0000548) − 5018248.2 , n ≥ 0 n
(b)
The beginning of the year 2050 is the 18250th day. p[18250] = 1.05 × 10 8 (1.0000548)
18250
− 5018248.2 = 280, 420, 000
61. A car rolling on a hill can be modeled as shown in Figure E61. The excitation is the force, f ( t) , for which a positive value represents accelerating the car forward with the motor and a negative value represents slowing the car by braking action. As it rolls, the car experiences drag due to various frictional phenomena which can be approximately modeled by a coefficient, k f , which multiplies the car’s velocity to produce a force which tends to slow the car when it moves in either direction. The mass of the car is m and gravity acts on it at all times tending to make it roll down the hill in the absence of other forces. Let the mass, m, of the car be 1000 kg, let the friction coefficient, k f , be N⋅s π . 5 and let the angle, θ , be m 12
Solutions 3-79
M. J. Roberts - 7/12/03
(a) Write a differential equation for this system with the force, f ( t) , as the excitation and the position of the car, y( t) , as the response. (b) If the nose of the car is initially at position, y(0) = 0 , with an initial velocity, [y′(t)]t = 0 = 10 ms , and no applied acceleration or braking force, graph the velocity of the car, y′ ( t) , for positive time. (c) If a constant force, f ( t) , of 200 N is applied to the car what is its terminal velocity ?
f(t) y(t)
θ
(θ)
sin
mg
Figure E61 Car on an inclined plane (a)
Summing forces,
or
(b)
f ( t) − mg sin(θ ) − k f y′ ( t) = m y′′ ( t) m y′′ ( t) + k f y′ ( t) + mg sin(θ ) = f ( t)
The zero-excitation response can be found by setting the force, f ( t) , to zero yielding m y′′ ( t) + k f y′ ( t) = − mg sin(θ ) −
kf
The homogeneous solution is y h ( t) = K h1 + K h 2e m . The particular solution must be in the form of a linear function of t, to satisfy the differential equation. Choosing the form, t
y p ( t) = K p t and solving, K p = −
mg sin(θ ) . Then the total zero-excitation response is kf y( t) = K h1 + K h 2e
Using the initial conditions,
−
kf m
t
−
mg sin(θ ) t kf
y(0) = 0 = K h1 + K h 2
and y′ (0) = 10 = −
kf mg Kh 2 − sin(θ ) . kf m
Solving,
Solutions 3-80
M. J. Roberts - 7/12/03
2
m m K h 2 = − g sin(θ ) − 10 = −1.0146 × 10 5 − 2000 = −1.0346 × 10 5 kf kf and K h1 = 1.0346 × 10 5 t − y( t) = 1.0346 × 10 5 1 − e 200 − 507.28 t t t t − − 200 1.0346 × 10 5 − 200 200 y′ ( t) = − 507.28 = 517.28 e − 1 + 10 e − 507.28 = 517.28e 200
y’(t) 1000
t
-550
(c)
The differential equation is m y′′ ( t) + k f y′ ( t) + mg sin(θ ) = f ( t)
We can re-write the equation as m y′′ ( t) + k f y′ ( t) = f ( t) − mg sin(θ ) treating the force due to gravity as part of the excitation. Then the impulse response is the solution of m h′′ ( t) + k f h′ ( t) = δ ( t) which is of the form, kf − t h( t) = K h1 + K h 2e m u( t) .
Integrating both sides of the differential equation through t = 0 we get kf m h′ (0 + ) + k f h(0 + ) = 1 = m − K h 2 + k f (K h1 + K h 2 ) m Integrating a second time yields, m h(0 + ) = 0 = m(K h1 + K h 2 ) . Solving,
we get
k f 1
0 K h1 1 = 1 K h 2 0
Solutions 3-81
M. J. Roberts - 7/12/03
K h1 =
1 1 , Kh 2 = − . kf kf
So the impulse response is −
1− e h( t) = kf
kf m
t
u( t) .
Now, if we say that the force, f ( t) , is a step of size, 200 N, the excitation of the system is x( t) = 200 u( t) − mg sin(θ ) . But this is going to cause a problem. The problem is that the term, − mgsin(θ ) , is a constant, therefore presumed to have acted on the system for all time before time, t = 0. The implication from that is that the position at time, t = 0, is at infinity. Since we are only interested in the final velocity, not position, we can assume that the car was held in place at y( t) = 0 until the force was applied and gravity was allowed to act on the car. That makes the excitation, x( t) = [200 − mg sin(θ )] u( t) and the response is −
1− e y( t) = x( t) ∗ h( t) = [200 − mg sin(θ )] u( t) ∗ kf
kf m
t
u( t)
or kf t − τ 200 − mg sin(θ ) y( t) = ∫0 1 − e m dτ kf
or k k m − mf t m m − mf τ 200 − mg sin(θ ) 200 − mg sin(θ ) y( t) = − = τ + e t + k e kf kf kf kf 0 f t
The terminal velocity is the derivative of position as time approaches infinity which, in this case is 200 − mg sin(θ ) 200 − 2536.43 m . y′ ( +∞) = = = −467.3 kf 5 s Obviously a force of 200 N is insufficient to move the car forward and its terminal velocity is negative indicating it is rolling backward down the hill. 62. A block of aluminum is heated to a temperature of 100 °C. It is then dropped into a flowing stream of water which is held at a constant temperature of 10°C. After 10 seconds the temperature of the ball is 60°C. (Aluminum is such a good heat conductor that its temperature is essentially uniform throughout its volume during the cooling process.) The rate of cooling is proportional to the temperature difference between the ball and the water. (a) Write a differential equation for this system with the temperature of the water as the excitation and the temperature of the block as the response.
Solutions 3-82
M. J. Roberts - 7/12/03
(b)
Compute the time constant of the system.
(c)
Find the impulse response of the system and, from it, the step response.
(d) If the same block is cooled to 0 °C and dropped into a flowing stream of water at 80 °C, at time, t = 0, at what time will the temperature of the block reach 75°C? (a)
The controlling differential equation is d T ( t) = K ( Tw − Ta ( t)) dt a
or
1 d T ( t) + Ta ( t) = Tw K dt a where Ta is the temperature of the aluminum ball and Tw is the temperature of the water. The solution is Ta ( t) = 90e − Kt + 10 (b)
We can find the constant, K, by using the temperature after 10 seconds, Ta (10) = 60 = 90e −10 K + 10 ⇒ K = 0.0588 .
(c)
The impulse response is the solution of the equation,
The form of the solution is
1 d h( t) + h( t) = δ ( t) . K dt h( t) = K h e − Kt u( t)
Integrating both sides of the differential equation through t = 0,
Therefore
1 K h(0 + ) = 1 = h ⇒ K h = K K K h( t) = Ke − Kt u( t) = 0.0588e −0.0588 t u( t) .
The unit step response is the integral of the impulse response, h −1 ( t) = (1 − e −0.0588 t ) u( t) . (d)
The response is the response to a step of 80 °C . Ta ( t) = 80 h −1 ( t) = 80(1 − e −0.0588 t ) u( t) .
To find the time at which the temperature is 75 °C, t75 , solve Ta ( t75 ) = 75 = 80(1 − e −0.0588 t 75 ) u( t75 ) . Solutions 3-83
M. J. Roberts - 7/12/03
Solving,
t75 = 47.153 .
63. A well-stirred vat has been fed for a long time by two streams of liquid, fresh water at 0.2 cubic meters per second and concentrated blue dye at 0.1 cubic meters per second. The vat contains 10 cubic meters of this mixture and the mixture is being drawn from the vat at a rate of 0.3 cubic meters per second to maintain a constant volume. The blue dye is suddenly changed to red dye at the same flow rate. At what time after the switch does the mixture drawn from the vat contain a ratio of red to blue dye of 99:1? Let the concentration of red dye be denoted by Cr ( t) and the concentration of blue 2 dye be denoted by Cb ( t) . The concentration of water is constant throughout at . The 3 rates of change of the dye concentrations are governed by d (VCb (t)) = −Cb (t) f draw dt d (VCr (t)) = f r − Cr (t) f draw dt where V is the constant volume, 10 cubic meters, f draw is the flow rate of the draw from the vat and f r is the flow rate of red dye into the tank. Solving the two differential equations, 1 − f draw t Cb ( t) = e V 3
and
f − draw t 1 Cr ( t) = 1 − e V . 3
Then the ratio of red to blue dye concentration is Cr ( t) = Cb ( t)
f − draw t 1 V 1 − e 3 t 1 − f draw e V 3
=
1− e e
−
−
f draw t V
f draw t V
=e
f draw t V
−1 .
Setting that ratio to 99 and solving for t99 , 99 = e
0.3 t 10 99
− 1 ⇒ t99 = 153.5 seconds
64. Some large auditoriums have a noticeable echo or reverberation. While a little reverberation is desirable, too much is undesirable. Let the response of an auditorium to an acoustic impulse of sound be ∞ n h( t) = ∑ e − nδ t − . 5 n =0
Solutions 3-84
M. J. Roberts - 7/12/03
We would like to design a signal processing system that will remove the effects of reverberation. In later chapters on transform theory we will be able to show that the compensating system that can remove the reverberations has an impulse response of the form, ∞ n h c ( t) = ∑ g[ n ]δ t − . 5 n =0 Find the function, g[ n ]. Removal of the reverberation is equivalent to making the overall impulse response, h 0 ( t) , an impulse. That means that ∞ −n h o ( t) = h( t) ∗ h c ( t) = ∑ e δ t − n =0 ∞
∞
∑∑e
n m δ t − ∗ g[ m]δ t − = Kδ ( t) 5 5
−n
n =0 m =0 ∞
n ∞ m ∗ ∑ g[ m]δ t − = Kδ ( t) 5 m = 0 5
∞
n + m ∑ ∑ e g[m]δ t − 5 = Kδ (t) −n
n =0 m =0 ∞
∞ n + m g m [ ]∑ e − nδ t − = Kδ ( t) ∑ 5 m =0 n =0
1 2 −1 −2 g[0]δ ( t) + e δ t − + e δ t − + L 5 5 2 3 1 + g[1]δ t − + e −1δ t − + e −2δ t − + L 5 5 5 = Kδ ( t) + g[2]δ t − 2 + e −1δ t − 3 + e −2δ t − 4 + L 5 5 5 M g[0] = K
g[1] + e −1 g[0] = 0 ⇒ g[1] = −Ke −1
g[2] + e −1 g[1] + e −2 g[0] = 0 ⇒ g[2] = Ke −2 − Ke −2 = 0
g[ 3] + e −1 g[2] + e −2 g[1] + e −3 g[0] = 0 ⇒ g[ 3] = Ke −3 − Ke −3 = 0 M So the compensating impulse response is 1 h c ( t) = Kδ ( t) − Ke −1δ t − 5 and the function, g, is
Solutions 3-85
M. J. Roberts - 7/12/03
g[ n ] = Kδ [ n ] − Ke −1δ [ n − 1] . 65. Show that the area property and the scaling property of the convolution integral are in agreement by finding the area of x( at) ∗ h( at) and comparing it with the area of x( t) ∗ h( t) . x( t) ∗ h( t) = y( t) x( at) ∗ h( at) =
1 y( at) a
The areas of x( at) and h( at) are the areas of x( t) and h( t) divided by the magnitude of a because the scaling factor, a, does not change their heights but compresses them in time by a. Therefore the product of the areas of x( at) and h( at) is smaller by a factor of a 2 than the product of the areas of x( t) and h( t) . The area of y( at) is smaller by a factor of a than the 1 y( at) is smaller by a factor of a 2 than the area of area of y( t) . Therefore the area of a y( t) . QED. 66. The convolution of a function, g( t) , with a doublet can be written as ∞
g( t) ∗ u1 ( t) =
∫ g(τ ) u (t − τ )dτ 1
.
−∞
Integrate by parts to show that g( t) ∗ u1 ( t) = g′ ( t) . Let u = g(τ ) and let dv = u1 ( t − τ ) dτ . Then du = g′ (τ ) dτ and v = u 0 ( t − τ ) = −δ ( t − τ ) d(t − τ ) d d δ ( t − τ )) = δ (t − τ ) (because = u1 ( t − τ )(−1) = − u1 ( t − τ ) ). ( dτ d(t − τ ) dτ and
∞
g( t) ∗ u1 ( t) = g(τ )δ ( t − τ ) −∞ − ∫ (−δ ( t − τ )) g′ (τ ) dτ = g′ ( t) 144244 3 −∞ ∞
=0
67. Derive the “sampling” property for a unit triplet. That is, find an expression for the integral, ∞
∫ g(t) u (t)dt 2
−∞
which is analogous to the sampling property of the unit doublet, − g′ ( t) =
∞
∫ g(t) u (t)dt . 1
−∞
In − g′ ( t) =
∞
∫ g(t) u (t)dt , let u = g(t) and let dv = u (t)dt . 1
2
Then du = g′ ( t) dt and v = u1 ( t)
−∞
and Solutions 3-86
M. J. Roberts - 7/12/03 ∞
∞
∞
) u (t) − ∫ u (t) g′ (t)dt = − ∫ u (t) g′ (t)dt ∫ g(t) u (t)dt = g1(t4 243 ∞
2
1
−∞
Then, applying − g′ ( t) =
−∞
1
∞
1
−∞
=0
−∞
∫ g(t) u (t)dt , we get 1
−∞
∞
∫ g(t) u (t)dt = g′′(t) . 2
−∞
68. Sketch block diagrams of the systems described by these equations. For the differential equation use only integrators in the block diagrams. (a)
y′′ ( t) + 3 y′ ( t) + 2 y( t) = x( t) x(t)
+
∫
+
∫
3
+
(b)
y(t)
2
6 y[ n ] + 4 y[ n − 1] − 2 y[ n − 2] + y[ n − 3] = x[ n ]
x[n]
+
1 6
+ +
y[n]
-4
D
2
D
-1
D
+ + +
Solutions 3-87
M. J. Roberts - 7/12/03
Chapter 4 - The Fourier Series Solutions (In this solution manual, the symbol, ⊗, is used for periodic convolution because the preferred symbol which appears in the text is not in the font selection of the word processor used to create this manual.) 1. Using MATLAB plot each sum of complex sinusoids over the time period indicated. (a)
x( t) =
1 30 k sinc e j 200πkt , −15ms < t < 15ms ∑ 10 10 k =−30 x(t) 1
-15
(b)
x( t) =
t (ms)
15
j 9 k − 2 j10πkt k + 2 sinc e − sinc ∑ 2 2 4 k =−9
,
−200ms < t < 200ms
x(t) 1
-0.2
0.2
t
-1
2. Show by direct analytical integration that the integral of the function,
g( t) = A sin(2πt) B sin( 4πt) 1 1 is zero over the interval, − < t < . 2 2
Solutions 4-1
M. J. Roberts - 7/12/03 1 2
1 2
∫ g(t)dt = AB ∫ sin(2πt) sin(4πt)dt −
Using sin( x ) sin( y ) =
1 2
∫ g(t)dt =
1 − 2
1 2
AB ∫1 g(t)dt = 2 2
1 2
∫ −
1 2
1 [cos( x − y ) − cos( x + y )] , 2
1 2
−
−
∫ [cos(2πt − 4πt) − cos(2πt + 4πt)]dt −
1 2
1 2
1
AB sin(−2πt) sin(6πt) 2 ∫1 [cos(−2πt) − cos(6πt)]dt = 2 −2π − 6π 1 −
−
g( t) dt =
1 2
AB 2
1 2
2
2
AB sin(−π ) sin( 3π ) sin(π ) sin(−3π ) − − − = 0 −2π 2 −2π 6π 6π
3. Convert the function, g( t) = (1 + j )e j 4 πt + (1 − j )e − j 4 πt , to an equivalent form in which “j” does not appear.
g( t) = e j 4 πt + je j 4 πt + e − j 4 πt − je − j 4 πt = e j 4 πt + e − j 4 πt + je j 4 πt − je − j 4 πt g( t) = 2 cos( 4πt) + j (e j 4 πt − e − j 4 πt ) = 2 cos( 4πt) − 2 sin( 4πt)
4. Using MATLAB plot these products over the time range indicated and observe in each case that the net area under the product is zero. (a)
x( t) = −3 sin(16πt) × 2 cos(24πt) , 0 < t <
1 4
x(t) 6
0.25 -6
(b)
x( t) = −3 sin(16πt) × 2 cos(24πt) , 0 < t < 1
Solutions 4-2
t
M. J. Roberts - 7/12/03
x(t) 6
1
t
-6
(c)
x( t) = −3 sin(16πt) × 2 cos(24πt) , −
1 3
x(t) 6
-0.0625
0.1875
t
-6
(d)
x( t) = x1 ( t) x 2 ( t) where
x1 ( t) is an even, 50%-duty-cycle square wave with a fundamental period
of 4 seconds and an average value of zero and
x 2 ( t) is an odd, 50%-duty-cycle square wave with a fundamental period of 4 seconds and an average value of zero
x(t) 6
4
t
-6
(e)
x( t) = x1 ( t) x 2 ( t)
1 1 t 1 where x1 ( t) = rect (2 t) ∗ comb( t) and x 2 ( t) = rect 4 t − ∗ comb − 2 8 8 2
Solutions 4-3
M. J. Roberts - 7/12/03 x(t) 1
2
t
-1
5. A sine function can be written as
sin(2πf 0 t) =
e j 2πf 0 t − e − j 2πf 0 t . j2
This is a very simple complex CTFS in which the harmonic function is only non-zero at two harmonic numbers, +1 and –1. Verify that we can write the harmonic function directly as j X[ k ] = (δ [ k + 1] − δ [ k − 1]) . 2 Write the equivalent expressions for sin(2π (− f 0 ) t) and show that the harmonic function is the complex conjugate of the previous one for sin(2πf 0 t) .
If the harmonic function is correct this equality should hold:
sin(2πf 0 t) =
∞ ∞ j e j 2πf 0 t − e − j 2πf 0 t = ∑ X[ k ]e j 2π ( kf 0 ) t = ∑ (δ [ k + 1] − δ [ k − 1])e j 2π ( kf 0 ) t j2 k =−∞ k =−∞ 2
e j 2πf 0 t − e − j 2πf 0 t j − j 2π ( kf 0 ) t e j 2πf 0 t − e − j 2πf 0 t = e − e j 2π ( kf 0 ) t = j2 j2 2
(
In
)
Check.
e − j 2πf 0 t − e j 2πf 0 t sin(−2πf 0 t) = j2
the signs of the terms are changed and * j j j X[ k ] = (δ [ k − 1] − δ [ k + 1]) = − (δ [ k + 1] − δ [ k − 1]) = (δ [ k + 1] − δ [ k − 1]) 2 2 2
Solutions 4-4
M. J. Roberts - 7/12/03
6. For each signal, find a complex CTFS which is valid for all time, plot the magnitude and phase of the harmonic function versus harmonic number, k, then convert the answers to the trigonometric form of the harmonic function. (a)
x( t) = 4 rect ( 4 t) ∗ comb( t)
X[ k ] =
1 2
1 8
1 x( t)e − j 2π ( kf 0 ) t dt = 4 ∫ rect ( 4 t)e − j 2πkt dt = 4 ∫ e − j 2πkt dt ∫ T T0 0 1 1 −
2
−
8
πk 1 +j − j πk e − j 2πkt 8 4 e 4 − e 4 4 k πk = sin = sinc X[ k ] = 4 = 4 πk 4 − j2 − j 2πk − 1 πk 8
k X c [ k ] = X[ k ] + X* [ k ] = 2 sinc , X s[ k ] = X[ k ] − X* [ k ] = 0 4
Using the table entry, TF = T0
t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 TF = T0
t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 t FS → wf 0 sinc( wkf 0 ) rect ∗ f 0 comb( f 0 t) ← w from Appendix E we get 1 k FS rect ( 4 t) ∗ comb( t) ← → sinc 4 4 k FS 4 rect ( 4 t) ∗ comb( t) ← → sinc 4
Check with CTFT: ∞ ∞ f f k X( f ) = sinc comb( f ) = sinc ∑ δ ( f − k ) = ∑ sinc δ ( f − k ) 4 4 k =−∞ 4 k =−∞
Solutions 4-5
M. J. Roberts - 7/12/03 |X[k]| 1
-18
18
k
Phase of X[k] π
-18
18
k
-π
(b)
1 t x( t) = 4 rect ( 4 t) ∗ comb 4 4 2
1 4 X[ k ] = ∫ x( t)e − j 2π ( kf 0 ) t dt = ∫ rect ( 4 t)e T 4 −2 T0 0
2πk t −j 4
1 8
dt =
∫e
−j
πk t 2
dt
1 − 8
1
πk 8 sin 2 t πk πk πk X[ k ] = ∫ cos t dt − j ∫ sin t dt = 2 ∫ cos t dt = 2 π k 2 2 2 1 1 0 − − 8 8 2 0 1 8
1 8
1 8
πk πk sin sin 16 1 16 1 k X[ k ] = 2 = = sinc πk 16 4 4 πk 16 2 1 k X c [ k ] = X[ k ] + X* [ k ] = sinc , X s[ k ] = X[ k ] − X* [ k ] = 0 16 2
Using the table entry,
Solutions 4-6
M. J. Roberts - 7/12/03
TF = T0 t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 TF = T0
t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 t FS → wf 0 sinc( wkf 0 ) rect ∗ f 0 comb( f 0 t) ← w from Appendix E we get 1 t FS 4 k 1 k 4 rect ( 4 t) ∗ comb ← → sinc = sinc 4 16 4 16 4 16
Check with CTFT:
1 ∞ k f k X ( f ) = sinc comb( 4 f ) = ∑ sinc δ f − 4 4 k =−∞ 16 4
|X[k]| 0.25
-50
50
k
Phase of X[k] π
-50
50
k
-π
(c)
A periodic signal which is described over one period by sgn( t) , t < 1 x( t) = , 1< t < 2 0
X[ k ] =
πk πk −j t −j t 1 1 1 − j 2π ( kf 0 ) t 2 2 x ( t ) e dt = x ( t ) e dt = sgn ( t ) e dt 4 −∫2 4 −∫1 T0 ∫T0 2
Solutions 4-7
1
M. J. Roberts - 7/12/03
1 1 1 πk πk X[ k ] = ∫ sgn( t)cos t dt − j ∫ sgn( t)sin t dt = 123 2 4 −1 123 14224 −1 odd 1 3 424 3 odd odd 3 even 3 144244 144244 even odd 1 πk cos t 1 2 −j −j πk = − X[ k ] = sin t dt = ∫ πk 2 0 2 2 2 0
−j πk sgn( t) sin t dt ∫ 2 2 0 1
πk cos − 1 2 j πk
πk πk πk cos − 1 cos − 1 cos − 1 2 2 2 X c [k ] = j −j = 0 , X s[ k ] = X[ k ] − X* [ k ] = j 2 πk πk πk |X[k]| 0.5
-20
20
k
Phase of X[k] π
-20
20
k
-π
7 . Using the CTFS table of transforms and the CTFS properties, find the CTFS harmonic function of each of these periodic signals using the time interval, TF , indicated. (a)
x( t) = 10 sin(20πt) , TF = FS sin(2πf 0 t) ← →
1 10
j (δ[k + 1] − δ[k − 1]) , TF = T0 2
FS 10 sin(20πt) ← → j 5(δ [ k + 1] − δ [ k − 1]) , TF = T0 =
Solutions 4-8
1 10
M. J. Roberts - 7/12/03
(b)
x( t) = 2 cos(100π ( t − 0.005)) , TF = FS cos(2πf 0 t) ← →
1 50
1 (δ[k − 1] + δ[k + 1]) , TF = T0 2
FS 2 cos(100πt) ← → δ [ k − 1] + δ [ k + 1] , TF = T0 =
1 50
FS 2 cos(100π ( t − 0.005)) ← →(δ [ k − 1] + δ [ k + 1])e − j 2πkf 0 t 0 , TF = T0 =
FS 2 cos(100π ( t − 0.005)) ← →(δ [ k − 1] + δ [ k + 1])e − j 0.5πk , TF = T0 =
FS 2 cos(100π ( t − 0.005)) ← → j (δ [ k + 1] − δ [ k − 1]) , TF = T0 =
FS 2 sin(100πt) ← → j (δ [ k + 1] − δ [ k − 1]) , TF = T0 =
(c)
x( t) = −4 cos(500πt) , TF = f 0 = 250 ⇒ T0 =
1 50
1 50
1 ⇒ TF = 5T0 250
Using the table entry,
TF = mT0 1 FS cos(2πf 0 t) ← → (δ [ k − m] + δ [ k + m]) 2 1 FS cos(ω 0 t) ← → (δ [ k − m] + δ [ k + m]) 2 (m an integer) we get FS cos(500πt) ← →
1 (δ[k − 5] + δ[k + 5]) , TF = 5T0 2
FS −4 cos(500πt) ← → −2(δ [ k − 5] + δ [ k + 5]) , TF = 5T0
Solutions 4-9
1 50
1 50
1 50
M. J. Roberts - 7/12/03
(d)
x( t) =
d − j10πt 1 , TF = e ( ) dt 5
TF = T0 =
(e)
1 5 FS e j 2πf 0 t ← → δ [ k − 1]
,
TF = T0 =
1 5
FS e − j10πt ← → δ [ k + 1]
,
TF = T0 =
1 5
d − j10πt FS e ← → j 2πkf 0δ [ k + 1] ( ) dt
,
TF = T0 =
1 5
d − j10πt FS e ← → j10πkδ [ k + 1] ( ) dt
,
TF = T0 =
1 5
d − j10πt FS e ← → − j10πδ [ k + 1] ( ) dt
,
TF = T0 =
1 5
t x( t) = rect ( t) ∗ comb 4
, TF = 4
t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0
,
TF = T0 = 4
1 t FS 1 k rect ( t) ∗ comb ← → sinc 4 4 4 4 t FS k rect ( t) ∗ comb ← → sinc 4 4 (f)
x( t) = rect ( t) ∗ comb( t) , TF = 1
t FS w w t 1 Using rect ∗ comb ←→ sinc k w T0 T0 T0 T0
, with w = 1 and T
FS rect ( t) ∗ comb( t) ← → sinc( k ) = δ [ k ]
Solutions 4-10
0
= 1,
M. J. Roberts - 7/12/03
The only non-zero value of sinc( k ) (k an integer harmonic number) is at k = 0 . Therefore sinc( k ) = δ [ k ] . (g)
x( t) = tri( t) ∗ comb( t) , TF = 1
t FS w w t 1 Using tri ∗ comb ←→ sinc 2 k w T0 T0 T0 T0
, with w = 1 and T
0
= 1,
FS tri( t) ∗ comb( t) ← → sinc 2 ( k ) = δ [ k ]
The only non-zero value of sinc 2 ( k ) (k an integer harmonic number) is at k = 0 . Therefore sinc 2 ( k ) = δ [ k ] . 8. If a periodic signal, x( t) , has a fundamental period of 10 seconds and its harmonic function is k X[ k ] = 4 sinc , 20
with a representation period of 10 seconds, what is the harmonic function of z( t) = x( 4 t) using the same representation period, 10 seconds?
k k , an integer X Using the time-scaling property, Z[ k ] = a a , 0 , otherwise k k k k an integer 4 sinc , an integer X , 80 4 Z[ k ] = 4 4 = 0 , otherwise 0 , otherwise This can be written more compactly as k Z[ k ] = 4 sinc comb 4 [ k ] . 80
9. A periodic signal, x( t) , has a fundamental period of 4 ms and its harmonic function is
X[ k ] = 15(δ [ k − 1] + δ [ k + 1]) ,
Solutions 4-11
M. J. Roberts - 7/12/03
with a representation period of 4 ms. Find the integral of x( t) . FS cos(2πf 0 t) ← →
1 (δ[k − 1] + δ[k + 1]) 2
FS 30 cos(500πt) ← →15(δ [ k − 1] + δ [ k + 1])
,
TF = T0 = 4 ms
,
TF = T0 = 4 ms
Using the integration property, t
∫ 30 cos(500πλ )dλ ←→ FS
−∞
t
∫ 30 cos(500πλ )dλ ←→ FS
−∞
t
15(δ [ k − 1] + δ [ k + 1]) j 2πkf 0
,
TF = T0 = 4 ms
15(δ [ k − 1] + δ [ k + 1]) j 500πk
,
TF = T0 = 4 ms
δ [ k − 1] δ [ k + 1] − 500π 500π
∫ 30 cos(500πλ )dλ ←→ − j15
,
TF = T0 = 4 ms
15 ∫ 30 cos(500πλ )dλ ←→ j 500π (δ[k + 1] − δ[k − 1]) ,
TF = T0 = 4 ms
FS
−∞
t
FS
−∞
15 3 FS sin(500πt) ← →j (δ[k + 1] − δ[k − 1]) 500π 50π
,
TF = T0 = 4 ms
10. If X[ k ] is the harmonic function over one fundamental period of a unit-amplitude 50%-duty-cycle square wave with an average value of zero and a fundamental period of 1 µs , find an expression consisting of only real-valued functions for the signal whose harmonic function is X[ k − 10] + X[ k + 10] . x( t) = 2 rect (2 × 10 6 t) ∗ 10 6 comb(10 6 t) − 1
t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0
Solutions 4-12
,
TF = T0
M. J. Roberts - 7/12/03
t k FS 6 6 −7 rect × 10 6 sinc −7 ∗ 10 comb(10 t) ←→ 5 × 10 5 × 10 2 k FS 2 rect (2 × 10 6 t) ∗ 10 6 comb(10 6 t) ← → sinc 2 FS −1← → −δ [ k ]
, TF = T0 = 1 µs
, TF = T0 = 1 µs
, TF = T0 = 1 µs
k FS 2 rect (2 × 10 6 t) ∗ 10 6 comb(10 6 t) − 1← → sinc − δ [ k ] 2
[2 rect(2 × 10 t) ∗10 comb(10 t) − 1]e 6
6
6
[2 rect(2 × 10 t) ∗10 comb(10 t) − 1]e 6
6
6
[2 rect(2 × 10 t) ∗10 comb(10 t) − 1](e 6
6
k − 10 FS ← → sinc − δ [ k − 10] 2
− j 20 ×10 6 πt
j 2 ×10 7 πt
6
j 20 ×10 6 πt
k + 10 FS ← → sinc − δ [ k + 10] 2
+ e − j 2 ×10
7
πt
)
k − 10 sinc 2 − δ [ k − 10] FS ← → + sinc k + 10 − δ [ k + 10] 2
k − 10 sinc 2 − δ [ k − 10] FS 2 2 rect (2 × 10 6 t) ∗ 10 6 comb(10 6 t) − 1 cos(2 × 10 7 πt) ← → + sinc k + 10 − δ [ k + 10] 2
[
]
11. Find the harmonic function for a sine wave of the general form, A sin(2πf 0 t) . Then, using Parseval’s theorem, find its signal power and verify that it is the same as the signal power found directly from the function itself. FS A sin(2πf 0 t) ← →j
Px =
1 T0
T0 2
∫
T − 0 2
A sin(2πf 0 t) dt = 2
2
A T0
A (δ[k + 1] − δ[k − 1]) 2 T0 2
, TF = T0 T0 2
∫ sin (2πf t)dt = 2T ∫ [1 − cos(4πf t)]dt A
2
2
0
T − 0 2
Solutions 4-13
0
0
T − 0 2
M. J. Roberts - 7/12/03
Px =
A2 2
By Parseval’s theorem,
Px =
∞
∑
k =−∞
2
A A2 A2 j (δ [ k + 1] − δ [ k − 1]) = (1 + 1) = . Check. 2 4 2
12. Show for a cosine and a sine that the CTFS harmonic functions have the property,
X[ k ] = X* [− k ] . For a cosine,
X[ k ] = X[− k ] =
1 (δ[k − 1] + δ[k + 1]) 2
1 1 δ [− k − 1] + δ [− k + 1]) = (δ [ k + 1] + δ [ k − 1]) = X[ k ] = X* [ k ] . Check. ( 2 2
For a sine,
X[ k ] = X[− k ] =
j (δ[k + 1] − δ[k − 1]) 2
j j j δ [− k + 1] − δ [− k − 1]) = (δ [ k − 1] − δ [ k + 1]) = − (δ [ k + 1] − δ [ k − 1]) = X* [ k ] ( 2 2 2
1 3 . Find and sketch the time functions associated with these harmonic functions assuming TF = 1. (a)
X[ k ] = δ [ k − 2] + δ [ k ] + δ [ k + 2]
From the table,
TF = mT0 1 FS cos(2πf 0 t) ← → (δ [ k − m] + δ [ k + m]) 2 1 FS cos(ω 0 t) ← → (δ [ k − m] + δ [ k + m]) 2 (m an integer)
Solutions 4-14
M. J. Roberts - 7/12/03
TF = 2T0 ⇒ T0 = FS cos(2π (2) t) ← →
FS cos( 4πt) ← →
1 ⇒ f0 = 2 2
1 (δ[k − 2] + δ[k + 2]) 2
1 (δ[k − 2] + δ[k + 2]) 2
FS From the table, 1← → δ[k ]
FS 2 cos( 4πt) + 1← → δ [ k − 2] + δ [ k ] + δ [ k + 2]
(Plot below) (b)
k X[ k ] = 10 sinc 10
t FS w w t 1 From the table, rect ∗ comb ←→ sinc k w T0 T0 T0 T0 FS rect (10 t) ∗ comb( t) ← →
1 k sinc 10 10
k FS 100 rect (10 t) ∗ comb( t) ← →10 sinc 10 (a) x(t) 3
2
t
-1
(b) x(t) 100
-2.5
2.5
14. Find the even and odd parts, x e ( t) and x o ( t) , of
Solutions 4-15
t
M. J. Roberts - 7/12/03
π x( t) = 20 cos 40πt + . 6
Then find the harmonic functions, X e [ k ] and X o [ k ], corresponding to them. Then, using the time shifting property find the harmonic function, X[ k ] and compare it to the sum of the two harmonic functions, X e [ k ] and X o [ k ]. Using cos( x + y ) = cos( x ) cos( y ) − sin( x ) sin( y ) ,
π π x( t) = 20 cos( 40πt) cos − sin( 40πt) sin 6 6 π x e ( t) = 20 cos( 40πt) cos 6
and
π x o ( t) = −20 sin( 40πt) sin . 6
π π X e [ k ] = 10 cos (δ [ k − 1] + δ [ k + 1]) and X o [ k ] = − j10 sin (δ [ k + 1] − δ [ k − 1]) . 6 6 j 2πk 20 π FS 240 20 cos 40πt + ← →10(δ [ k − 1] + δ [ k + 1])e 6 1
πk
j π FS 20 cos 40πt + ← →10(δ [ k − 1] + δ [ k + 1])e 6 6
π π j −j π FS 20 cos 40πt + ← →10δ [ k − 1]e 6 + δ [ k + 1]e 6 6
π FS π π π π 20 cos 40πt + ← →10δ [ k − 1]cos + j sin + δ [ k + 1]cos − j sin 6 6 6 6 6 π π X e [ k ] + X o [ k ] = 10 cos (δ [ k − 1] + δ [ k + 1]) − j10 sin (δ [ k + 1] − δ [ k − 1]) . Check. 6 6
15. Using the direct summation formula find and sketch the DTFS harmonic function of comb N 0 [ n ] with N F = N 0 .
X[ k ] =
1 1 comb N 0 [ n ]e − j 2π ( kF0 ) n = ∑ N 0 n = N0 N0
N0 −1 2
∑
n =−
Solutions 4-16
N0 2
comb N 0 [ n ]e − j 2π ( kF0 ) n =
1 N0
M. J. Roberts - 7/12/03
|X[k]| 1 N0 -8
-6
-4
-2
2
4
6
8
2
4
6
8
k
X[k] -8
-6
-4
-2
k
16. Using the DTFS table of transforms and the DTFS properties, find the DTFS harmonic function of each of these periodic signals using the representation period, N F , indicated. (a)
2πn x[ n ] = 6 cos 32 2πn FS 1 Using cos ←→ comb N 0 [ k − 1] + comb N 0 [ k + 1] 2 N0
(
)
,
X[ k ] = 3(comb 32 [ k − 1] + comb 32 [ k + 1]) |X[k]| 3
-40
40
k
Phase of X[k] π
-40
40 -π
(b)
2π ( n − 2) x[ n ] = 10 sin 12
Solutions 4-17
k
N F = N 0 = 32
M. J. Roberts - 7/12/03
2πn FS j Using sin ←→ comb N 0 [ k + 1] − comb N 0 [ k − 1] 2 N0
(
)
,
N F = N 0 = 12
2πn FS 10 sin ←→ = j 5(comb12 [ k + 1] − comb12 [ k − 1]) 12 FS Then, using x[ n − n 0 ] ← → e − j 2π ( kF0 ) n 0 X[ k ]
X[ k ] = j 5(comb12 [ k + 1] − comb12 [ k − 1])e
−j
π k 3
|X[k]| 5
-40
40
k
Phase of X[k] π
-40
40
k
-π
We can demonstrate that this solution is correct by reconstituting the signal using x[ n ] =
∑ X[k ]e
j 2π
kn N0
=
k = N0
∑ j 5(comb [k + 1] − comb [k − 1])e 12
12
−j
π k 3
j 2π
e
kn N0
k = N0
Since the summation extends only over one period, N 0 = 12 , choose the simplest period,
−6 ≤ k < 6 . In that period the two comb functions are simply two impulses at k = ±1. x[ n ] =
5
∑ j 5(δ[k + 1] − δ[k − 1])e
n 1 j 2πk − 12 6
k =−6
n 1 n 1 j 2π n − 1 − j 2π 12n − 16 j 2π − − j 2π − 12 6 x[ n ] = j 5 e − e 12 6 = − j 5 e 12 6 − e
n 1 n − 2 x[ n ] = − j 5 j 2 sin 2π − = 10 sin 2π 12 6 12
Solutions 4-18
M. J. Roberts - 7/12/03
We could have chosen a different period, for example 4 ≤ k < 16 . Then n 1 j 2πk − 12 6
15
x[ n ] = ∑ j 5(δ [ k − 11] − δ [ k − 13])e k =4
n 1 13 n 13 j 2π 1112n −116 j 22π 12n − 16 j 26π − j 2π − 12 6 − e 12 6 x[ n ] = j 5 e −e = j 5 e
x[ n ] = j 5e
12 n 12 j 2π − 12 6
n 1 n 1 − j 2π 12n − 16 j 2π −12n + 16 j 2π − j 2π − 12 6 j 2πn − j 4 π − e 12 6 −e = j 5e{ e{ e e =1 =1
n 1 n − 2 x[ n ] = j 5 − j 2 sin 2π − = 10 sin 2π 12 6 12 which is the same answer as before (with somewhat more effort).
(c)
n n an integer , x1 , N F = 48 x[ n ] = 8 8 0 , otherwise 2πn where x1[ n ] = sin 6
Using the time-scaling property of the DTFS,
X[ k ] =
j (comb6[k + 1] − comb6[k − 1]) 2
,
NF = N0 = 6
j (comb6[k + 1] − comb6[k − 1]) 16
,
N F = N 0 = 48
X1[ k ] = X[ k ] =
1 X [k ] 8 1
Solutions 4-19
M. J. Roberts - 7/12/03 |X[k]| 0.0625
-48
48
k
Phase of X[k] π
-48
48
k
-π
(d)
x[ n ] = e j 2πn , N F = 6 FS e j 2πn = 1← → comb 6 [ k ]
|X[k]| 1
-40
40
k
Phase of X[k] π
-40
40
k
-π
(e)
2π ( n − 1) 2πn x[ n ] = cos − cos 16 16
,
N F = 16
2π ( n − 1) FS 1 comb16 [ k − 1] + comb16 [ k + 1] 2πn cos πk πk − cos ←→ 16 16 2 − comb [ k − 1]e − j 8 − comb [ k + 1]e − j 8 16 16 −j
2π ( n − 1) FS 1 − e 2πn cos − cos ←→ 16 16 2
π 8
comb16 [ k − 1] +
Solutions 4-20
1− e 2
j
π 8
comb16 [ k + 1]
M. J. Roberts - 7/12/03 |X[k]| 0.25
-40
40
k
Phase of X[k] π
-40
40
k
-π
(f)
33 × 2πn 33πn x[ n ] = − sin = − sin 64 32
2πmn FS j Using sin ←→ = comb N 0 [ k + m] − comb N 0 [ k − m] 2 N0
(
X[ k ] = −
)
j (comb64 [k + 33] − comb64 [k − 33]) 2 |X[k]| 0.5
-64
64
k
Phase of X[k] π
-64
64
k
-π
We can demonstrate that this solution is correct by reconstituting the signal using x[ n ] =
∑ X[k ]e
k = N0
j 2π
kn N0
=
∑
k = N0
kn
j 2π j − (comb 64 [ k + 33] − comb 64 [ k − 33])e N 0 2
Solutions 4-21
M. J. Roberts - 7/12/03
Since the summation extends only over one period, N 0 = 64 , choose the simplest period,
−32 ≤ k < 32 . kn
j 2π j x[ n ] = ∑ − (comb 64 [ k + 33] − comb 64 [ k − 33])e N 0 k =−32 2 31
We must now determine for which values of k, −32 ≤ k < 32 , the comb functions are not zero. Take the first comb function,
comb 64 [ k + 33] . Its impulses occur whenever k + 33 is an integer multiple of 64. The only value of k in the range, −32 ≤ k < 32 , for which that is true is k = 31. Similarly, for
comb 64 [ k − 33] the only value for which it is non-zero is k = −31 . Therefore we can write the summation as x[ n ] =
31
∑
k =−32
−31n j 2π j 2π j j j 2π 3164n N0 64 31 31 δ k − − δ k + e = − e − e [ ] [ ]) ( 2 2 kn
−
31n j 2π j j − j 2π 3164n 31n 31πn 31n x[ n ] = e − e 64 = − j 2 sin 2π = sin 2π = sin 32 64 64 2 2
But this can also be written as 64 n 33 n 33 n j j 2π 3364n j 2π − j 2π − j 2π j − j 2π 6464n j 2π 3364n 64 64 64 x[ n ] = e12 e − e e = e − e 4 4 3 123 2 =1 2 =1
j 33n 33πn x[ n ] = j 2 sin 2π = − sin 32 64 2 If these two results are to both be correct 31πn 33πn sin = − sin 32 32
Solutions 4-22
M. J. Roberts - 7/12/03
for any integer value of n. We can write 31πn − 64πn 33πn 33πn 31πn 31πn − 2πn = sin sin = sin − = − sin = sin 32 32 32 32 32
proving that the two expressions are indeed equivalent for integer values of n. (g)
x[ n ] = rect 5 [ n ] ∗ comb11[ n ] FS Using rect W [ n ] ∗ comb N 0 [ n ] ←→
k 2W + 1 drcl , 2W + 1 N0 N0
n and drcl , 2 m + 1 = (2 m + 1) comb 2 m +1[ n ] and the fact that 2W + 1 = N 0 2m + 1
X[ k ] = comb11[ k ] FS Agrees with 1← → comb N 0 [ k ] because x[ n ] = rect 5 [ n ] ∗ comb11[ n ] = 1
FS and in 1← → comb N 0 [ k ], N 0 can be arbitrarily chosen. The meaning of
the result is the same regardless of the choice of N 0 . |X[k]| 1
-40
40
k
Phase of X[k] π
-40
40
k
-π
This result was obtained for a period of N 0 = 11. But when the function is a constant, we can choose any period we like and get an equivalent result (because a constant repeats exactly in any “period” you choose). For example, if we let N 0 = 4 the FS transform pair, 1← → comb N 0 [ k ], yields X[ k ] = comb 4 [ k ] . Then, reconstituting the
signal from its DTFS,
Solutions 4-23
M. J. Roberts - 7/12/03
x[ n ] =
∑ X[k ]e
j 2π
kn N0
=
k = N0
∑
k = N0
comb 4 [ k ]e
j 2π
kn N0
Summing over the period, −2 ≤ k < 2 , yields
x[ n ] =
1
∑ comb [k ]e
j 2π
kn 4
4
k =−2
=e
j 2π
( 0) n 4
=1 .
If we chose the period, 2 ≤ k < 6 we would get 5
x[ n ] = ∑ comb 4 [ k ]e
j 2π
kn 4
=e
j 2π
( 4) n 4
=1 ,
k =2
which is exactly the same result. In general, for any choice of period and any range of k covering one period, x[ n ] =
k0 + N 0 −1
∑
k = k0
comb N 0 [ k ]e
j 2π
kn N0
=e
j 2π
qN 0 n N0
=1
where the integer, q, lies in the range, k0 ≤ q < k0 + N 0 . (h)
x[ n ] = rect 2 [ n ] ∗ comb 21[ n − 3]
N F = 21
,
FS Using rect N w [ n ] ∗ comb N 0 [ n ] ←→
k 2N w + 1 drcl , 2 N w + 1 N0 N0
FS and x[ n − n 0 ] ← → e − j 2π ( kF0 ) n 0 X[ k ] ,
X[ k ] =
2πk 5 k −j drcl , 5 e 7 21 21
Solutions 4-24
M. J. Roberts - 7/12/03 |X[k]| 0.25
-40
k
40
Phase of X[k] π
-40
k
40 -π
17. Find the DTFS harmonic function of
x[ n ] =
n
∑ comb [m] − comb [m − 1]
m =−∞
3
3
with N F = N 0 = 3 . 1 1 − j 2π 3k FS , NF = N0 = 3 comb 3 [ m] − comb 3 [ m − 1] ← → − e 3 3
X[ k ]
n
Using
∑ x[m]←→ 1 − e FS
m =−∞
− j 2π ( kF0 )
, k≠0
n
1 1− e
∑ comb [m] − comb [m − 1]←→ 3 FS
m =−∞
3
3
− j 2π
1− e
k 3
2πk −j 3
=
1 , NF = N0 = 3 3
This is correct because, since the representation period is 3, the corresponding DTdomain function is comb 3 [ m] which is the accumulation of the backward difference of itself,
x[ n ] =
n
∑ comb [m] − comb [m − 1] = comb [m] . 3
m =−∞
3
18. Find the average signal power of
x[ n ] = rect 4 [ n ] ∗ comb 20 [ n ] Solutions 4-25
3
M. J. Roberts - 7/12/03
directly in the DT domain and then find its harmonic function, X[ k ], and find the signal power in the “k” domain and show that they are the same. In the DT domain: 1 1 1 9 2 2 2 Px = x n = rect n ∗ comb n = rect 4 [ n ] ∗ comb 20 [ n ] [ ] [ ] [ ] ∑ ∑ ∑ 4 20 20 n = 20 20 n =−10 N 0 n = N0
Px =
9 1 4 1= ∑ 20 20 n =−4
In the “k” domain: FS rect 4 [ n ] ∗ comb 20 [ n ] ← →
Px =
∑ X[k ]
k = N0
2
=
∑
k = 20
9 k drcl , 9 20 20
2
9 k 9 drcl , 9 = 20 20 20
Doing the summation numerically in MATLAB, Px =
2
∑
k = 20
k drcl , 9 20
2
9 . Check. 20
19. Using the frequency shifting property of the DTFS find the DT-domain signal, x[ n ] , corresponding to the harmonic function,
X[ k ] =
7 k − 16 drcl , 7 . 32 32
FS rect 3 [ n ] ∗ comb 32 [ n ] ← →
(rect 3[n] ∗ comb32[n])e
j 2π
16 n 32
7 k drcl , 7 32 32
FS ← →
7 k − 16 drcl , 7 32 32
FS → drcl (rect 3[n] ∗ comb32[n])e jπn ← 32 7
k − 16 , 7 32
FS → drcl (rect 3[n] ∗ comb32[n])(−1)n ← 32 7
Solutions 4-26
k − 16 , 7 32
M. J. Roberts - 7/12/03
This frequency shifting causes the sign of the DT-domain function to alternate. 20. Find the DTFS harmonic function for
x[ n ] = rect 3 [ n ] ∗ comb 8 [ n ] with the representation period, N F = 8 .
Then, using MATLAB, plot the DTFS
representation, 7
x F [ n ] = ∑ X[ k ]e
j 2π
kn 8
k =0
over the DT range, −8 ≤ n < 8 . For comparison, plot the function,
xF 2[n] =
20
∑ X[k ]e
j 2π
kn 8
k =13
over the same range. The plots should be identical. 7 k FS rect 3 [ n ] ∗ comb 8 [ n ] ← → drcl , 7 8 8
k = 0:7 xF[n] 1
-8
8
n
k = 13:20 xF[n] 1
-8
8
n
21. A periodic signal, x(t), with a period of 4 seconds is described over one period by
x( t) = 3 − t , 0 < t < 4 .
Solutions 4-27
M. J. Roberts - 7/12/03
Plot the signal and find its trigonometric CTFS description. Then plot on the same scale approximations to the signal, x N ( t) , given by
x N ( t) = X c [0] + ∑ X c [ k ] cos(2π ( kf F ) t) + X s[ k ] sin(2π ( kf F ) t) N
k =1
for N = 1, 2 and 3 . (In each case the time scale of the plot should cover at least two periods of the original signal.) t 0 +T0
4 4 1 1 1 t2 ∫ x(t)dt = 4 ∫ x(t)dt = 4 ∫ (3 − t)dt = 4 3t − 2 = 1 , t0 0 0 0 t 0 +T0 4 2 1 πk X c [k ] = x( t) cos(2π ( kf 0 ) t) dt = ∫ x( t) cos t dt , ∫ 2 20 T0 t
1 X c [0] = T0
4
0
4 1 πk X c [ k ] = ∫ ( 3 − t) cos t dt , 2 2 0
1 1 πk πk X c [ k ] = ∫ 3 cos t dt − ∫ t cos t dt . 2 2 20 20 4
Using
∫x
n
4
cos( x ) dx = x n sin( x ) − n ∫ x n −1 sin( x ) dx ,
and making the change of variable
λ=
πk t 2
in the second integral, we get 1 2 X c [k ] = ∫ 3 cos(λ )dλ − πk 0 πk 2πk
X c [k ] =
2πk
∫ λ cos(λ )dλ 0
1 2 2 2πk 2πk 3 sin(λ ) 0 − [λ sin(λ )]0 + πk πk πk
2 X s[ k ] = T0
t 0 +T0
∫
t0
2πk
0
1 πk x( t) sin(2π ( kf 0 ) t) dt = ∫ x( t) sin t dt 2 20 4
1 1 πk πk X s[ k ] = ∫ 3 sin t dt − ∫ t sin t dt 2 2 20 20 4
Using
∫x
n
∫ sin(λ )dλ = 0
4
sin( x ) dx = − x n cos( x ) + n ∫ x n −1 cos( x ) dx ,
and making the change of variable,
Solutions 4-28
.
M. J. Roberts - 7/12/03
λ= we get
πk t, 2
2πk
2πk
1 1 2 λ sin(λ ) dλ X s[ k ] = 3 sin(λ ) dλ − ∫ ∫ πk 0 πk 0 πk 2πk 1 2 2 2πk 2πk X s[ k ] = − 3 cos(λ ) 0 + [λ cos(λ )]0 − sin(λ ) πk πk πk 0 4 X s[ k ] = πk x1(t) 3
8
-1
t
x2(t) 3
8
-1
t
x3(t) 3
8
-1
t
22. A periodic signal, x(t), with a period of 2 seconds is described over one period by
1 ( π ) < sin t , t 2 2 . x( t) = 1 0 , < t <1 2 Plot the signal and find its complex CTFS description. Then plot on the same scale approximations to the signal, x N ( t) , given by
x N ( t) =
N
∑ X[k ]e
j 2π ( kf F ) t
k =− N
for N = 1, 2 and 3 . (In each case the time scale of the plot should cover at least two periods of the original signal.)
Solutions 4-29
M. J. Roberts - 7/12/03
T0 = 2 , f 0 = 1 X[ k ] = T0
t 0 +T0
∫ x(t)e
− j 2π ( kf 0 ) t
t0
1 2 1
1 dt = ∫ x( t)e − jkπt dt 2 −1
Using the fact that x(t) is an odd function, 1 1 1 1 X[ k ] = ∫ x( t)[cos( kπt) − j sin( kπt)]dt = ∫ x( t) cos( kπt) − j x( t) sin( kπt) dt 14243 2 −1 2 −1 14243 odd even 1 1 j X[ k ] = − ∫ x( t) sin( kπt) dt = − j ∫ x( t) sin( kπt) dt 2 −1 0 1 2
1 2
1 [cos(2πt − kπt) − cos(2πt + kπt)]dt 2 0
X[ k ] = − j ∫ sin(2πt) sin( kπt) dt = − j ∫ 0
1
j sin[π (2 − k ) t] sin[π (2 + k ) t] 2 X[ k ] = − − π (2 + k ) 0 2 π (2 − k ) π π sin (2 − k ) sin (2 + k ) j 2 − 2 = − j sinc 2 − k − sinc 2 + k X[ k ] = − 2 2 2 π (2 − k ) 4 π (2 + k ) x1(t) 1 -1
3
t
-1
x (t) 2
1 -1
3
t
-1
x3(t) 1 -1
3
t
-1
23. Find and plot two periods of the complex CTFS description of cos(2πt)
and
(a)
Over the interval, 0 < t < 1,
(b)
Over the interval, 0 < t < 1.5 .
Solutions 4-30
M. J. Roberts - 7/12/03
(a)
The period of cos(2πt) is 1, therefore the complex Fourier series description is
simply the exponential description of the cosine function e j 2πt + e − j 2πt . cos(2πt) = 2 That is, 1 X[1] = X[−1] = and X[ k ] = 0 , k ≠ 1 . 2 The plot is simply the plot of two periods of the cosine because the complex Fourier series description is exact everywhere. (b) 3 2 TF = and f F = 2 3 1 X[ k ] = TF X[ k ] =
t 0 +TF
∫ x(t)e t0
− j 2π ( kf F ) t
3 2
j 4 kπt − 1 dt = ∫ x( t)e 3 dt 30 2
3 2
2 4 kπt 4 kπt cos(2πt) cos − j sin dt ∫ 3 3 30
3 32 2 2 4 kπt 4 kπt X[ k ] = ∫ cos(2πt) cos dt − j ∫ cos(2πt) sin dt 3 3 3 0 0 3 3 2 2 2k 2k ∫ cos2πt 3 − 1 dt + ∫ cos2πt 3 + 1 dt 1 0 0 X[ k ] = 3 3 3 2 2 − j sin 2πt 2 k − 1 dt − j sin 2πt 2 k + 1 dt ∫0 3 ∫0 3
3
2 2k 2k sin 2πt − 1 sin 2πt + 1 3 3 + 2k 2k 2π + 1 2π − 1 3 3 1 X[ k ] = 3 2k 2k cos2πt − 1 cos2πt + 1 3 3 +j + j 2k 2k 2π − 1 2π + 1 3 0 3
Solutions 4-31
M. J. Roberts - 7/12/03
2k 2k sin 3π − 1 sin 3π + 1 3 3 + 2k 2k 2π + 1 2π − 1 3 3 1 X[ k ] = 3 2k 2k cos3π − 1 − 1 cos3π + 1 − 1 3 3 +j + j 2k 2k 2π − 1 2π + 1 3 3 Fourier Series Fit to 1 1/2 Cycles of cos(2πt) 1
s(t)
0.5 0 -0.5 -1
0
0.5
1
1.5 Time, t (s)
2
2.5
3
24. Using MATLAB, plot the following signals over the time range, −3 < t < 3 . (a)
x 0 ( t) = 1
(b)
x1 ( t) = x 0 ( t) + 2 cos(2πt)
(c)
x 2 ( t) = x1 ( t) + 2 cos( 4πt)
(d)
x 20 ( t) = x19 ( t) + 2 cos( 40πt)
For each part, (a) through (d), numerically evaluate the area of the signal over the time 1 1 range, − < t < . 2 2 The “Nth” member of this sequence of functions can be represented by the general form
x N ( t) = x N −1 ( t) + 2 cos(2πNt) . Based on what you observed in parts (a) through (d) what is the limit as “N” approaches infinity? Find the trigonometric Fourier series expression for the unit comb function and compare it to this result.
Solutions 4-32
M. J. Roberts - 7/12/03 Successive Cumulative Sums of Higher Harmonic Cosines so(t), N=1
5 0
so(t), N=2
-5 -3 5
so(t), N=3
-1
0
1
2
3
-2
-1
0
1
2
3
-2
-1
0
1
2
3
-2
-1
0 Time, t (s)
1
2
3
0 -5 -3 10 0
-10 -3 50 so(t), N=20
-2
0 -50 -3
>>>Improve above plot
x( t) = comb( t) X c [0] =
X c [k ] =
2 T0
X s[ k ] =
2 T0
1 T0
T0 2
T0 2
T0 2
T0 2
∫ x(t)dt = ∫ comb(t)dt = ∫ δ (t)dt = 1
−
T0 2
−
T0 2
−
T0 2
T0 2
T0 2
2πkt 2πkt ∫ x(t) cos(2π (kf )t)dt = 2 ∫ comb(t) cos T dt = 2 ∫ δ (t) cos T dt = 2 0
−
T0 2 T0 2
T0 2 T0 2
0
T0 2
0
−
0
T0 2 T0 2
−
2πkt 2πkt ∫ x(t) sin(2π (kf )t)dt = 2 ∫ comb(t) sin T dt = 2 ∫ δ (t) sin T dt = 0 0
T − 0 2
−
−
T0 2
25. Using the CTFS table of transforms and the CTFS properties, find the CTFS harmonic function of each of these periodic signals using the time interval, TF , indicated. (a)
1 x( t) = 3 rect 2 t − ∗ comb( t) , TF = 1 4
t FS w w t 1 Using rect ∗ comb ←→ sinc k w T0 T0 T0 T0
, with w = 12 and T
1 k FS rect (2 t) ∗ comb( t) ← → sinc . 2 2
Solutions 4-33
0
= 1,
M. J. Roberts - 7/12/03
Then using the linearity property, 3 k FS 3 rect (2 t) ∗ comb( t) ← → sinc . 2 2 FS Then using the time-shifting property, x( t − t0 ) ← → e − j 2π ( kf 0 ) t 0 X[ k ],
kπ 3 1 k − j FS 3 rect 2 t − ∗ comb( t) ← → sinc e 2 2 4 2
(b)
1 t x( t) = 5[tri( t − 1) − tri( t + 1)] ∗ comb , TF = 4 4 4
t FS w w t 1 Using tri ∗ comb ←→ sinc 2 k w T0 T0 T0 T0
, with w = 1 and T
0
= 4,
1 t FS 1 k tri( t) ∗ comb ← → sinc 2 4 4 4 4 FS Then using the time-shifting property, x( t − t0 ) ← → e − j 2π ( kf 0 ) t 0 X[ k ],
πk 1 t FS 1 2 k − j 2 tri( t − 1) ∗ comb ←→ sinc e 4 4 4 4
and πk 1 t FS 1 k j tri( t + 1) ∗ comb ← → sinc 2 e 2 4 4 4 4
Then, using linearity, πk j 1 − j πk 1 k t FS 5[tri( t − 1) −tri( t + 1)] ∗ comb ← → 5 e 2 − e 2 sinc 2 4 4 4 4
or 1 5 πk t FS k 5[tri( t − 1) −tri( t + 1)] ∗ comb ← → − j sin sinc 2 4 4 4 2 2
Solutions 4-34
M. J. Roberts - 7/12/03
(c)
x( t) = 3 sin(6πt) + 4 cos(8πt) , TF = 1 FS 3 sin(6πt) + 4 cos(8πt) ← →j
(d)
3 (δ[k + 3] − δ[k − 3]) + 2(δ[k − 4] + δ[k + 4]) 2
x( t) = 2 cos(24πt) − 8 cos( 30πt) + 6 sin( 36πt) , TF = 2 δ [ k − 24 ] + δ [ k + 24 ] FS 2 cos(24πt) − 8 cos( 30πt) + 6 sin( 36πt) ← → −4 (δ [ k − 30] + δ [ k + 30]) + j 3 δ k + 36 − δ k − 36 ] [ ]) ([
(e)
x( t) =
t
1
∫ comb(λ ) − comb λ − 2 dλ
, TF = 1
−∞
FS comb( t) ← → comb1[ k ] = 1
1 FS comb( t) − comb t − ← →1 − e − jπk 2 j
πk
−j
πk −j e 2 −e 1 1 − e − jπk FS 2 e comb ( λ ) − comb λ − d λ ← → = ∫ j 2πk j 2πk 2 −∞ t
−j
πk 2
πk
e 2 1 k FS comb ( λ ) − comb λ − d λ ← → = sinc ∫−∞ 2 2 2 t
(f)
x( t) = 4 cos(100πt) sin(1000πt) , TF =
1 50
FS 4 cos(100πt) sin(1000πt) ← → 2(δ [ k − 1] + δ [ k + 1]) ∗
j (δ[k + 10] − δ[k − 10]) 2
FS 4 cos(100πt) sin(1000πt) ← → j (δ [ k + 9] + δ [ k + 11] − δ [ k − 9] − δ [ k − 11])
(g)
t t t t x( t) = 14 rect ∗ comb ⊗ 7 rect ∗ comb , TF = 24 8 12 5 8
Solutions 4-35
M. J. Roberts - 7/12/03
Let t t x1 ( t) = rect ∗ comb 8 12
and
t t x 2 ( t) = rect ∗ comb 5 8
Then, using
t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0
,
x( t) = 98 x1 ( t) ⊗ x 2 ( t) 2k X1[ k ] = 8 sinc 3
, TF = 12
k k 8 sinc , an integer 3 2 X1[ k ] = 0 , otherwise 5k X 2 [ k ] = 5 sinc 8
, TF = 24
, TF = 8
k 5k 5 sinc , an integer 24 3 X 2[k ] = 0 , otherwise
, TF = 24
k k 5k 40 sinc sinc , an integer 3 24 6 X[ k ] = 98 × 24 0 , otherwise
, TF = 24
All the values of this function are zero except when k = 0 . Then
X[ k ] = 94080δ [ k ]
, TF = 24 .
(This result indicates that the periodic convolution of the two signals is a constant, 94080. That can be confirmed by graphically periodically convolving the two signals. That is by finding the area under the product over one cycle and observing that, as one signal is shifted, the area does not change. Signal one is a periodic sequence of pulses of height 168, width 8 and period 12. Signal two is a periodic sequence of pulses of height 56, width 5 and period 8. The overall overlap width is a constant 10. That product is 168 × 56 × 10 = 94080 .)
Solutions 4-36
M. J. Roberts - 7/12/03
(h)
t t t t x( t) = 8 rect ∗ comb ⊗ −2 rect ∗ comb , TF = 20 2 5 6 20
Let t t x1 ( t) = rect ∗ comb 2 5
and
t t x 2 ( t) = rect ∗ comb 6 20
Then using
t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0
,
x( t) = −16 x1 ( t) ⊗ x 2 ( t) 2k X1[ k ] = 2 sinc 5
, TF = 5
k k an integer 2 sinc , 10 4 X1[ k ] = 0 , otherwise 3k X 2 [ k ] = 6 sinc 10
, TF = 20
, TF = 20
k k 3k an integer 12 sinc sinc , 10 20 4 X[ k ] = −16 0 , otherwise
, TF = 20
k k 3k an integer sinc sinc , 10 20 4 X[ k ] = −192 0 , otherwise
, TF = 20
26. A signal, x( t) , is described over one period by
T0 − A , − 2 < t < 0 . x( t) = A , 0 < t < T0 2
Solutions 4-37
M. J. Roberts - 7/12/03
Find its complex CTFS and then, using the integration property find the CTFS of its integral and plot the resulting CTFS representation of the integral.
1 1 X[ k ] = ∫ x( t)e − j 2π ( kf 0 ) t dt = ∫ x{ ( t) cos(2π ( kf 0 ) t) − j sin(2π ( kf 0 ) t) dt 4244 3 144244 3 T0 T0 T0 T0 odd 14 even odd function function function
X[ k ] = − j
1 T0
T0 2
T0 2
T0 2
2 2A x( t) sin(2π ( kf ) t) dt = − j ∫ x( t) sin(2π ( kf ) t) dt = − j sin(2π ( kf ) t) dt ∫1 442443 T ∫ T 0
−
T0 2
0
0
0 0
even function
0
T0
A 2 A cos(2π ( kf 0 ) t) 2 X[ k ] = j cos(2π ( kf 0 ) t) = j kπ T0 2π ( kf 0 ) 0
[
]
T0 2 0
= jA
0
cos( kπ ) − 1 kπ
The CTFS integration property is
X[ k ]
t
∫ x(λ )dλ ←→ j 2π (kf ) FS
−∞
if X[0] = 0 .
0
Therefore the CTFS of the integral of the function would be
X[ k ] =
AT0 cos( kπ ) − 1 . 2 (kπ ) 2
Using L’Hôpital’s rule,
X[0] =
cos( kπ ) − 1 AT0 − k sin( kπ ) AT0 − k 2 cos( kπ ) AT0 lim = lim = lim =0 . 2 k → 0 ( kπ ) 2 2 k → 0 2 kπ 2 2 k→ 0 2π 2
The CTFS representation of the signal is
x( t) =
∞
∑ X[k ]e
j 2π ( kf 0 ) t
k =−∞
Solutions 4-38
M. J. Roberts - 7/12/03 x(t) 0.3
1
-1
t
-0.3
27. In some types of communication systems binary data are transmitted using a technique called binary phase-shift keying (BPSK) in which a “1” is represented by a “burst” of a CT sine wave and a “0” is represented by a burst which is the exact negative of the burst that represents a “1”. Let the sine frequency be 1 MHz and let the burst width be 10 periods of the sine wave. Find and plot the CTFS harmonic function for a periodic binary signal consisting of alternating “1’s” and “0’s” using its fundamental period as the representation period.
[
]
x( t) = sin(2 × 10 6 πt) 2 rect (10 5 t) ∗ 5 × 10 4 comb(5 × 10 4 t) − 1 X[ k ] =
j k comb[ k + 20] − comb[ k − 20]) ⊗ 10 5 × 10 −5 sinc − δ [ k ] ( 2 2
X[ k ] =
j k δ [ k + 20] − δ [ k − 20]) ∗ 10 5 × 10 −5 sinc − δ [ k ] ( 2 2
k δ [ k + 20] ∗ sinc δ [ k + 20] ∗ δ [ k ] 2 j X[ k ] = − 2 k −δ [ k − 20] ∗ δ [ k ] −δ [ k − 20] ∗ sinc 2
X[ k ] =
j k − 20 k + 20 − (δ [ k + 20] − δ [ k − 20]) − sinc sinc 2 2 2
Solutions 4-39
M. J. Roberts - 7/12/03 |X[k]| 0.5
-30
30
k
Phase of X[k] π
-30
30
k
-π
28. Using the DTFS table of transforms and the DTFS properties, find the harmonic function of each of these periodic signals using the representation period, N F , indicated. (a)
x[ n ] = e
−j
2πn 16
⊗ comb 24 [ n ] , N F = 48
From the table, e
j
2πn 16
F ← → comb16 [ k − 1] , N F = 16
Therefore, using the change of period property,
e
−j
2πn 16
k k comb16 + 1 , an integer , N F = 48 ←→ 3 3 0 , otherwise F
Also, from the table,
N F = mN 0 j
e
2πn N0
FS ← → comb N F [ k − m]
48 = 3 × 16 e
j
2πn 16
FS ← → comb 48 [ k − 3]
Solutions 4-40
M. J. Roberts - 7/12/03
k The function, comb16 + 1 , is a sequence of impulses. The impulses occur wherever 3 k + 1 = 16 m , m any integer. Rearranging, the impulses occur where k + 3 = 48 m . This 3 describes a comb of the form, comb 48 [ k + 3]. Therefore the two answers agree.
From the table, F comb N 0 [ n ] ← →
1 , N F = 24 N0
Therefore, using the time scaling property, k 1 an integer , comb N 0 [ n ] ←→ 24 2 , N F = 48 0 , otherwise F
Then, since the two DTFS’s are both done with reference to the same representation period, using the multiplication-convolution duality property, k k 1 an integer an integer , comb 48 [ k + 3] , X[ k ] = 2 3 2 24 0 , otherwise 0 , otherwise
The non-zero impulses in the first harmonic function occur at values of k for which k + 3 is an integer multiple of 48. Therefore all these k’s must be odd. The values of k for which the second harmonic function is non-zero are all even. Therefore
X[ k ] = 0 , for all k. −j
2πn
This result implies that the original DT function, x[ n ] = e 16 ⊗ comb 24 [ n ] , is zero. A periodic convolution of two periodic signals using a common period (48 in this case) is equivalent to an aperiodic convolution either of the two functions with one period of the other function. Therefore
x[ n ] = e
−j
2πn 16
∗ (δ [ n ] + δ [ n − 24 ]) = e 144 42444 3 one period of the comb function using the common period of 48
Solutions 4-41
−j
2πn 16
+e
−j
2π ( n − 24 ) 16
M. J. Roberts - 7/12/03
x[ n ] = e
(b)
−j
2πn 16
+e
−j
2πn 16
e
j
48π 16
=e
2πn 16
+e
−j
2πn 16
j 3π
e{ = e
−j
2πn 16
−e
−j
2πn 16
=0
=−1
2πn x[ n ] = (rect 5 [ n ] ∗ comb 24 [ n ]) sin , N F = 24 6 X[ k ] =
11 k j drcl ,11 ⊗ (comb[ k + 4 ] − comb[ k − 4 ]) 24 2 24
X[ k ] =
11 k j drcl ,11 ∗ (δ [ k + 4 ] − δ [ k − 4 ]) 24 2 24
X[ k ] = j
(c)
−j
11 k + 4 k − 4 drcl ,11 − drcl ,11 24 48 24
n x[ n ] = x1[ n ] − x1[ n − 1] where x1[ n ] = tri ∗ comb 20 [ n ] , N F = 20 8 X1[ k ] =
8 2 8k 2k sinc 2 ∗ comb 20 [ k ] = sinc 2 ∗ comb 20 [ k ] 5 20 20 5
2 ∞ k − 20q X1[ k ] = ∑ sinc 2 5 5 q =−∞
(
)
FS Using x[ n ] − x[ n − 1] ← → 1 − e − j 2π ( kF0 ) X[ k ]
πk ∞ −j 2 k − 20q X[ k ] = 1 − e 10 ∑ sinc 2 5 5 q =−∞
29. Find the signal power of
14πn 26πn x[ n ] = 5 sin − 8 cos . 15 30
The fundamental period of this signal is 30. 2π × 13 × n 2π × 14 × n x[ n ] = 5 sin − 8 cos 30 30
Solutions 4-42
Check.
M. J. Roberts - 7/12/03
X[ k ] =
Px =
j5 (comb30[k + 14] − comb30[k − 14]) − 4(comb30[k − 13] + comb30[k + 13]) 2
∑ X[k ]
2
14
∑
=
k = 30
k =−15
j5 (comb30[k + 14] − comb30[k − 14]) − 4(comb30[k − 13] + comb30[k + 13]) 2 2
2
89 25 5 5 Px = + + 4 2 + 4 2 = + 32 = = 44.5 2 2 2 2 3 0F i .n d
the
DTFS
harmonic
x[ n ] = (rect1[ n − 1] − rect1[ n − 4 ]) ∗ comb 6 [ n ].
xN [n] =
N
∑ X[k ]e
k =− N
πnk j 3
X[ k ],
function, Then
plot
the
partial 5
, for N = 0,1, 2 and then the total sum, x[ n ] = ∑ X[ k ]e k =0
x[ n ] = rect1[ n − 1] ∗ comb 6 [ n ] − rect1[ n − 4 ] ∗ comb 6 [ n ] kπ sin (2 N W + 1) N0 1 FS Using rect NW [ n ] ∗ comb N 0 [ n ] ←→ N0 kπ sin N0
1 X[ k ] = 6
kπ kπ kπ sin sin sin 4 πk −j 2 − j π3k 1 2 − j 4 3πk 1 2 − j π3k 3 − − = e e e e 6 kπ 6 kπ kπ sin sin sin 6 6 6
j X[ k ] = e 3
5πk −j 6
kπ sin 2 2 kπ sin 6
kπ 2 kπ sin 2 πk 5 5πk πnk N sin 2 −j 6 j 3 2 j 3 n − 2 j j xN [n] = ∑ e e = ∑ e 3 k =− N 3 k =− N kπ kπ sin sin 6 6 N
Solutions 4-43
of πnk j 3
sum, .
2
M. J. Roberts - 7/12/03
N=0 x[n] 2 -12
12
n
12
n
12
n
12
n
-2
N=1 x[n] 2 -12 -2
N=2 x[n] 2 -12 -2
Total Sum from 0 to 5 x[n] 2 -12 -2
31. Find and sketch the magnitude and phase of the DTFS harmonic function of 2π x[ n ] = 4 cos 7
2π n + 3 sin 3
n
which is valid for all discrete-time. The least common period of these two signals is N 0 = 21. Using the tables and the change-of-period property of the DTFS, the DTFS harmonic function is
X[ k ] = 2(comb 21[ k − 3] + comb 21[ k + 3]) + j
3 (comb21[k + 7] − comb21[k − 7]) 2
|X[k]| 2
-21
21
k
Phase of X[k] π
-21
21 -π
Solutions 4-44
k
M. J. Roberts - 7/12/03
32. The sun shining on the earth is a system in which the radiant power from the sun is the excitation and the atmospheric temperature (among many other things) is a response. A simplified model of the radiant power falling on a typical mid-latitude location in North America is that it is periodic with a fundamental period of one year and that every day the radiant power of the sunlight rises linearly from the time the sun rises until the sun is at its zenith then falls linearly until the sun sets. The earth absorbs and stores the radiant energy and re-radiates some of the energy into space every night. To keep the model of the excitation as simple as possible assume that the energy loss every night can be modeled as a continuation of the daily linear radiant power pattern except negative at night. There is also a variation with the seasons caused by the tilt of the earth’s axis of rotation. This causes this linear riseand-fall pattern to rise and fall sinusoidally on a much longer time scale as illustrated below. Radiant Power
June 21 2
1.75
Day
-1
(a)
365
-0.75 2
December 21
Write a mathematical description of the radiant power from the sun. 1 1 2πt p( t) = 2 tri(2 t) ∗ comb( t) − + sin 2 4 365
(b) Assume that the earth is a first-order system with a time constant of 0.16 years. What day of the year should be the hottest according to this simplified model? The differential equation for the rate of heat flow into the earth is
K T( t) d K T( t)) = p( t) − ( dt 58.4
Solutions 4-45
M. J. Roberts - 7/12/03
where K is a proportionality constant relating the temperature of the earth to its stored energy. Since the radiant power striking the earth is periodic the temperature is also periodic and both can be expressed in a CTFS with a representation period of 365 days.
t FS w w t 1 Using tri ∗ comb ←→ sinc 2 k w T0 T0 T0 T0 k 1 2 k an integer j sinc − δ [ k ] , 2 2 365 P[ k ] = + (δ [ k + 1] − δ [ k − 1]) 0 8 , otherwise and
p( t) =
∞
∑ P[k ]e
j 2π ( kf 0 ) t
and
k =−∞
T( t) =
∞
∑ Τ[k ]e
j 2π ( kf 0 ) t
.
k =−∞
Substituting into the differential equation,
j 2π ( kf 0 )K
∞
∑ Τ[k ]e
j 2π ( kf 0 ) t
=
k =−∞
∞
∑ P[k ]e
j 2π ( kf 0 ) t
−
k =−∞
K ∞ Τ[ k ]e j 2π ( kf 0 ) t ∑ 58.4 k =−∞
or
j 2π ( kf 0 )KΤ[ k ] = P[ k ] − or Τ[ k ] =
K Τ[ k ] 58.4
P[ k ] j 2π ( kf 0 )K +
K 58.4
or 1 k 2 k an integer j sinc − δ [ k ] , 2 2 365 + (δ [ k + 1] − δ [ k − 1]) 8 , otherwise 1 0 Τ[ k ] = . 2 π k K + 0.01712 j 365
Notice that, except for the impulse at k = 0 , the first term in the numerator is zero up to the 365th harmonic. At that harmonic the denominator is much larger than it is at the first harmonic. Therefore the daily variation is neglible and the seasonal variation (the sinusoid) determines the hottest day. The harmonic function of the temperature is therefore approximately,
Solutions 4-46
M. J. Roberts - 7/12/03
1 1 j 1 − 2 δ [ k ] + 8 (δ [ k + 1] − δ [ k − 1]) 1 − 2 δ [ k ] j δ [ k + 1] δ [ k − 1] = − Τ[ k ] = + 2πk 2π 2π 0 01712 8 . K K j + 0.01712 + 0.01712 − j + 0.01712 j 365 365 365 .
1 2π 2π δ [ k + 1] − j + 0.01712 − δ [ k − 1] j + 0.01712 − δ[k ] 365 365 1 j Τ[ k ] = 2 + 2 K 0.01712 8 2π 2 + (0.01712) 365 2π −j 1 1 0 01712 1 1 + + − + . + − − δ δ δ δ k k k k [ ] [ ] [ ] [ ] ( ) ( ) 1 j δ[k ] 365 Τ[ k ] = − 2 0.03424 K 8 2π 2 + (0.01712) 365 2πt 2πt 0.0344 cos + 0.03424 sin 365 365 1 1 Τ ( t) = − 2 0.03424 K 8K 2π 2 + (0.01712) 365 Since the sine and cosine are about the same size, the maximum temperature occurs where they are both positive and equal which is where
2πt π = ⇒ t = 45.625 . 365 4 So the hottest day should occur about 46 days after the summer solstice or about August 6. 33. The speed and timing of computer computations are controlled by a clock. The clock is a periodic sequence of rectangular pulses, typically with 50% duty cycle. One problem in the design of computer circuit boards is that the clock signal can interfere with other signals on the board by being coupled into adjacent circuits through stray capacitance. Let the computer clock be modeled by a square-wave voltage source alternating between 0.4 and 1.6 V at a frequency of 2 GHz and let the coupling into an adjacent circuit be modeled by a series combination of a 0.1 pF capacitor and a 50
Solutions 4-47
M. J. Roberts - 7/12/03
Ω resistance. Find and plot over two fundamental periods the voltage across the 50 Ω resistance. The excitation can be described by v i ( t) = 1.2 rect ( 4 × 10 9 t) ∗ 2 × 10 9 comb(2 × 10 9 t) + 0.4 C = 0.1 pF + vi (t)
+ R = 50 Ω
-
vo (t) -
We can find the CTFS for v i ( t) = 1.2 rect ( 4 × 10 9 t) ∗ 2 × 10 9 comb(2 × 10 9 t) + 0.4 from the table entry,
t FS rect ∗ f 0 comb( f 0 t) ← → wf 0 sinc( wkf 0 ) . w The CTFS is k Vi [ k ] = 0.4δ [ k ] + 0.6 sinc 2
and
v i ( t) =
∞
∑ V [k ]e
k =−∞
i
+ j 2π ( kf 0 ) t
=
∞
k
∑ 0.4δ[k ] + 0.6 sinc 2 e
j 4 ×10 9 kπt
.
k =−∞
The differential equation relating the response to the excitation is
RC[ v′i ( t) − v′o ( t)] = v o ( t) or
v o ( t) + v′o ( t) = v′i ( t) . RC
Therefore we need an expression for the derivative of the input signal, ∞ 9 d ∞ k j 4 ×10 9 kπt k 9 v′i ( t) = ∑ 0.4δ [ k ] + 0.6 sinc e = j 4 × 10 ∑ kπ 0.4δ [ k ] + 0.6 sinc e j 4 ×10 kπt 2 2 dt k =−∞ k =−∞
Solutions 4-48
M. J. Roberts - 7/12/03
Since the impulse, δ [ k ], occurs at k = 0 , and k multiplies all the terms in the derivative, ∞ 9 d ∞ k j 4 ×10 9 kπt k 9 v′i ( t) = ∑ 0.4δ [ k ] + 0.6 sinc e = j 2.4 × 10 ∑ kπ sinc e j 4 ×10 kπt . 2 2 dt k =−∞ k =−∞
Let the output signal be expressed as
v o ( t) =
∞
∑ V [k ]e
+ j 2π ( kf 0 ) t
o
k =−∞
.
Then ∞
∑ V [k ]e
k =−∞
o
RC
j 4 ×10 9 kπt
+ j 4 × 10 9
∞
∑
k =−∞
kπ Vo [ k ]e j 4 ×10
9
kπt
= j 2.4 × 10 9
∞
k
∑ kπ sinc 2 e
j 4 ×10 9 kπt
k =−∞
or ∞
∞ 9 1 j 4 ×10 9 kπt k 9 9 Vo [ k ] + j 4 × 10 kπ e = j 2.4 × 10 ∑ kπ sinc e j 4 ×10 kπt . ∑ 2 RC k =−∞ k =−∞
Solving for the CTFS harmonic function of the output voltage, k j 2.4 × 10 9 kπ sinc 2 Vo [ k ] = 1 + j 4 × 10 9 kπ RC or substituting numbers for R and C, k k kπ sinc j 2.4 × 10 9 kπ sinc 2 2 . Vo [ k ] = = j 2.4 11 9 2 × 10 + j 4 × 10 kπ 200 + j 4 kπ
Therefore, k kπ sinc 2 j 4 ×10 9 kπt e v o ( t) = j 2.4 ∑ . k =−∞ 200 + j 4 kπ ∞
v (t) o
1
-6.25e-10
6.25e-10 -1
Solutions 4-49
t
M. J. Roberts - 7/12/03
34. Find and plot versus F the magnitude of the response, y[ n ] , to the periodic excitation, x[ n ] = cos(2πFn ) , in the system below.
x[n]
y[n] 0.9
D
y[ n ] = x[ n ] + 0.9 y[ n − 1] or
y[ n ] − 0.9 y[ n − 1] = x[ n ] .
Since we are seeking the response of the system to a periodic excitation, we want the particular solution. The excitation is e j 2πFn + e − j 2πFn x[ n ] = cos(2πFn ) = 2 which is the linear combination of two complex exponentials,
x1[ n ] = e j 2πFn and x[ n ] =
x 2 [ n ] = e − j 2πFn ,
1 (x [n] + x2[n]) . 2 1
Therefore the response is of the form,
y[ n ] = where,
y1[ n ] = K1e j 2πFn
1 (y [n] + y2[n]) , 2 1 and
y 2 [ n ] = K 2e − j 2πFn ,
Substituting solution form #1 into the equation,
K1e j 2πFn − 0.9K1e j 2πF ( n −1) = e j 2πFn
Solutions 4-50
M. J. Roberts - 7/12/03
[
]
K1 1 − 0.9e − j 2πF e j 2πFn = e j 2πFn
Solving,
K1 =
1 1 − 0.9e − j 2πF
Similarly,
K2 =
1 1 − 0.9e j 2πF
Therefore the total response is
e − j 2πFn 1 e j 2πFn y[ n ] = + . 2 1 − 0.9e − j 2πF 1 − 0.9e j 2πF This can be written as 1e y[ n ] = 2
j 2πFn
(1 − 0.9e ) + e (1 − 0.9e (1 − 0.9e )(1 − 0.9e ) j 2πF
− j 2πFn
− j 2πF
− j 2πF
j 2πF
)
e j 2πFn − 0.9e j 2πF e j 2πFn + e − j 2πFn − 0.9e − j 2πF e − j 2πFn 1 y[ n ] = 2 (1 − 0.9 cos(2πF ) − j 0.9 sin(2πF ))(1 − 0.9 cos(2πF ) + j 0.9 sin(2πF )) j 2πFn + e − j 2πFn ) − j 0.9 sin(2πF )(e j 2πFn − e − j 2πFn ) 1 (1 − 0.9 cos(2πF ))(e y[ n ] = 2 (1 − 0.9 cos(2πF ))2 + (0.9 sin(2πF ))2
y[ n ] =
(1 − 0.9 cos(2πF )) cos(2πFn) + 0.9 sin(2πF ) sin(2πFn) (1 − 0.9 cos(2πF ))2 + (0.9 sin(2πF ))2
Using A cos( x ) + B sin( x ) =
B A 2 + B 2 cos x − tan −1 , A
(1 − 0.9 cos(2πF ))2 + (0.9 sin(2πF ))2 sin(2πF ) −1 y[ n ] = 2 2 cos 2πFn − tan 1 − 0.9 cos(2πF ) (1 − 0.9 cos(2πF )) + (0.9 sin(2πF ))
y[ n ] =
Therefore the amplitude is
sin(2πF ) cos 2πFn − tan −1 1 − 0.9 cos(2πF )
(1 − 0.9 cos(2πF ))2 + (0.9 sin(2πF ))2 1
(1 − 0.9 cos(2πF )) + (0.9 sin(2πF )) 2
Solutions 4-51
2
=
1 . 1 − 0.9e − j 2πF
M. J. Roberts - 7/12/03 Chapter Problem 6-9 10
Amplitude
8 6 4 2 0 -1
-0.8
-0.6
-0.4
-0.2
0 F
0.2
0.4
0.6
0.8
1
The response to a sinusoid with a DT frequency, F, near zero is a sinusoid of amplitude 1 near 10. The response to a sinusoid with a DT frequency, F, near is a sinusoid of 2 amplitude near 0.526. This is a lowpass DT filter.
Solutions 4-52
M. J. Roberts - 7/12/03
Chapter 5 - The Fourier Transform Solutions (In this solution manual, the symbol, ⊗, is used for periodic convolution because the preferred symbol which appears in the text is not in the font selection of the word processor used to create this manual.)
1. The transition from the CTFS to the CTFT is illustrated by the signal,
t t 1 x( t) = rect ∗ comb w T0 T0 or
x( t) =
∞
t − nT0 . w
∑ rect
n =−∞
The complex CTFS for this signal is given by
X[ k ] =
Plot the “modified” CTFS,
kw Aw sinc . T0 T0
T0 X[ k ] = Aw sinc( w ( kf 0 )) ,
for w = 1 and f 0 = 0.5, 0.1 and 0.02 versus kf 0 for the range −8 < kf 0 < 8 .
5-1
M. J. Roberts - 7/12/03 T0|X[k]| 1
-8
8
kf
0
T |X[k]| 0
π
-8
8
kf0
-π
f 0 = 0.5 T |X[k]| 0
1
-8
8
kf
0
T0|X[k]| π
-8
8
kf
0
-π
f 0 = 0.1 T0|X[k]| 1
-8
8
kf
0
T0|X[k]| π
-8
8 -π
f 0 = 0.02
5-2
kf0
M. J. Roberts - 7/12/03
kg and is a function of spatial position, x, in m3 meters. Write the mathematical expression for its CTFT, M( y ) . What are the units of
2. Suppose a function, m( x ) , has units of
M and y?
M( y ) =
∞
∫ m( x )e
− j 2πyx
dx
−∞
The units of M are
kg and the units of y are m−1 . m2
3 . Using the integral definition of the Fourier transform, find the CTFT of these functions. (a) x( t) = tri( t)
X( f ) =
∞
∫
tri( t)e − j 2πft dt =
−∞
1
∫ (1 − t )(cos(2πft) + j sin(2πft))dt
−1
1 1 1 X( f ) = 2 ∫ (1 − t)(cos(2πft)) dt = 2 ∫ cos(2πft) dt − ∫ t cos(2πft) dt 0 0 0
1 sin(2πft) 1 1 sin(2πft) X( f ) = 2∫ cos(2πft) dt − t −∫ dt 2 π f 2 π f 0 0 0
sin(2πft) 1 1 sin(2πft) 1 X( f ) = 2 +∫ dt = 2πf 0 0 2πf πf
1
∫ sin(2πft)dt 0
1 − cos(2πft) 1 − cos(2πf ) X( f ) = = 2 πf 2πf 2(πf ) 0 1
Then, using sin( x ) sin( y ) =
X( f ) =
sin 2 (πf )
(πf )
2
1 1 − cos(2 x ) cos( x − y ) − cos( x + y )] ⇒ sin 2 ( x ) = [ 2 2
= sinc 2 ( f )
5-3
M. J. Roberts - 7/12/03
1 1 (b) x( t) = δ t + − δ t − 2 2
X( f ) =
∞
1
1
∫ δ t + 2 − δ t − 2 e
− j 2πft
dt = e jπf − e − jπf = j 2 sin(πf )
−∞
4. In Figure E4 there is one example each of a lowpass, highpass, bandpass and bandstop signal. Identify them. x(t)
x(t)
(a)
(b)
t
x(t)
t
x(t)
(c) t
(d) t
Figure E4 Signals with different frequency content (a) (b) (c) (d)
bandstop bandpass lowpass highpass
5 . Starting with the definition of the CTFT find the radian-frequency form of the generalized CTFT of a constant. Then verify that a change of variable, ω → 2πf , yields the correct result in cyclic-frequency form. Check your answer against the Fourier transform table in Appendix E. Let g( t) = A . Then
G( jω ) =
∞
∫ Ae
−∞
− jωt
∞
dt = lim+ A ∫ e −σ t (cos(ωt) − j sin(ωt)) dt σ →0
−∞
∞ ∞ −σ t −σ t G( jω ) = lim+ A ∫ e{ cos(ωt) dt − j ∫ e{ sin(ωt) dt 123 σ →0 −∞ even 123 −∞ even even odd
5-4
M. J. Roberts - 7/12/03
∞
G( jω ) = lim+ 2 A ∫ e −σt cos(ωt) dt σ →0
0
∞
G( jω ) = lim+ 2 A ∫ e −σt cos(ωt) dt σ →0
Using
az ∫ e cos(bz)dz =
0
e az (a cos(bz) + b sin(bz)) , a2 + b2 ∞
e −σt G( jω ) = lim+ 2 A 2 2 (ω sin(ωt) − σ cos(ωt)) σ →0 σ + ω 0
G( jω ) = lim+ 2 A σ →0
For ω ≠ 0, lim+ 2 A
σ = 0. σ + ω2
The area under 2 A
σ is σ + ω2
σ →0
2
2
∞
Area =
∫ 2A σ
−∞
Using
∫a
2
σ σ + ω2 2
dx 1 bx tan −1 2 2 = a +b x ab
2
σ dω . + ω2
∞
1 ω Area = 2 Aσ tan −1 = 2πA . σ −∞ σ Therefore the CTFT pair is
F A ← → 2πAδ (ω ) .
For the f form the pair is Letting f =
ω in the second pair, 2π
F A ← → Aδ ( f ) .
ω F A ← → Aδ = 2πAδ (ω ) . QED. 2π
5-5
M. J. Roberts - 7/12/03
6. Starting with the definition of the CTFT, find the generalized CTFT of a sine of the form, A sin(ω 0 t) and check your answer against the results given above. Check your answer against the Fourier transform table in Appendix E. Let g( t) = A sin(ω 0 t) . Then
G( jω ) =
∞
∫ A sin(ω t)e
− jωt
0
−∞
∞
dt = lim+ σ →0
∫ Ae
−σ t
−∞
sin(ω 0 t)(cos(ωt) − j sin(ωt)) dt
∞ ∞ −σ t −σ t G( jω ) = A lim+ ∫ e{ sin(ω 0 t)cos(ωt) dt − j ∫ e{ sin(ω 0 t)sin(ωt) dt 12 4 4 3 123 12 4 4 3 123 σ →0 −∞ even even odd −∞ even odd odd
∞
G( jω ) = − j 2 A lim+ ∫ e −σt sin(ω 0 t) sin(ωt) dt σ →0
∞
0
[
]
G( jω ) = − jA lim+ ∫ e −σt cos((ω 0 − ω ) t) − cos((ω 0 + ω ) t) dt σ →0
Using
az ∫ e cos(bz)dz =
0
e az (a cos(bz) + b sin(bz)) a2 + b2
e −σt 2 2 ω sin((ω 0 − ω ) t) − σ cos((ω 0 − ω ) t) σ + (ω 0 − ω ) G( jω ) = − jA lim+ σ →0 e −σt − 2 ω sin((ω 0 + ω ) t) − σ cos((ω 0 + ω ) t) σ 2 + (ω 0 + ω )
(
)
(
)
σ σ G( jω ) = − jA lim+ 2 − lim 2 2 + 2 σ → 0 σ + (ω 0 − ω ) σ → 0 σ + (ω 0 + ω ) By the same reasoning as in Exercise 5, when ω ≠ ω 0 ,
G( jω ) = lim+ σ →0
σ
and, when ω ≠ −ω 0
G( jω ) = lim+ σ →0
2
=0
2
= 0.
σ + (ω 0 − ω ) 2
σ
σ + (ω 0 + ω ) 2
5-6
∞
0
M. J. Roberts - 7/12/03
The areas under these functions are ∞
Area1 = − jA lim+ σ →0
∫σ
−∞
∞
σdω
2
+ (ω 0 − ω )
Area2 = − jA lim+
and
2
σ →0
∫σ
−∞
σdω
2
+ (ω 0 + ω )
2
Making the change of variable, ω − ω 0 = λ ⇒ dω = dλ , in the first integral, ∞
Area1 = − jA lim+ σ →0
Similarly,
∫σ
−∞
∞
Area2 = jA lim+ σ →0
∫σ
−∞
σdλ = − jπA 2 + λ2
σdλ = jπA . 2 + λ2
These areas are independent of the value of σ . Therefore
[
]
G( jω ) = jπA δ (ω + ω 0 ) − δ (ω − ω 0 ) . QED.
7. Find the CTFS and CTFT of each of these periodic signals and compare the results. After finding the transforms, formulate a general method of converting between the two forms for periodic signals. (a)
x( t) = A cos(2πf 0 t) The CTFS is simply two impulses, X[ k ] = The CTFT is X( f ) =
A (δ ( f − f 0 ) + δ ( f + f 0 )) = X[1]δ ( f − f 0 ) + X[−1]δ ( f + f 0 ) . 2
X( f ) =
(b)
A (δ[k − 1] + δ[k + 1]) . 2
∞
∑ X[k ]δ ( f − kf )
k =−∞
x( t) = comb( t) The CTFS is found from
5-7
0
M. J. Roberts - 7/12/03
X[ k ] =
1 T0
T0 2
∫ comb(t)e
− j 2π ( kf 0 ) t
T − 0 2
1 2
dt =
∫ δ (t)e
− j 2π ( kf 0 ) t
dt = 1 .
1 − 2
The CTFT is
X( f ) = comb( f ) =
∞
∞
k =−∞
k =−∞
∑ δ ( f − k ) = ∑ X[k ]δ ( f − kf ) 0
8. Let a signal be defined by
x( t) = 2 cos( 4πt) + 5 cos(15πt) . 1 1 Find the CTFT’s of x t − and x t + and identify the resultant phase shift of each 40 20 sinusoid in each case. Plot the phase of the CTFT and draw a straight line through the 4 phase points which result in each case. What is the general relationship between the slope of that line and the time delay?
1 1 1 x t − = 2 cos 4π t − + 5 cos15π t − 40 40 40 3π π 1 + 5 cos15πt − x t − = 2 cos 4πt − 40 8 10 { { phase phase shift shift πf −j 5 15 15 − j 3πf 1 F x t − ← →[δ ( f − 2) + δ ( f + 2)]e 20 + δ f − + δ f + e 8 40 2 2 2 π π −j +j 5 15 − j 4516π 5 15 + j 4516π 1 F 10 10 x t − ←→ δ ( f − 2)e + δ ( f + 2)e + δ f − e + δ f + e 40 2 2 2 2
1 1 1 x t + = 2 cos 4π t + + 5 cos15π t + 20 20 20
5-8
M. J. Roberts - 7/12/03
π 3π 1 + 5 cos15πt + x t + = 2 cos 4πt + 20 4 5 { { phase phase shift shift πf +j 5 15 15 + j 34πf 1 F 10 x t + ←→[δ ( f − 2) + δ ( f + 2)]e + δ f − + δ f + e 20 2 2 2 π π +j −j 5 15 + j 45π 5 15 − j 45π 1 F x t + ← → δ ( f − 2)e 5 + δ ( f + 2)e 5 + δ f − e 8 + δ f + e 8 20 2 2 2 2
X( f ) 3π 4 Slope = π 10 -8 -7 -6 -5 -4 -3 -2 -1
1 2 3 4 5 6 7 8 Slope =
f
π 20
The slope of the line is −2πf times the delay.
9. Using the frequency-shifting property, find and plot versus time the inverse CTFT of f + 20 f − 20 X( f ) = rect . + rect 2 2
x( t) = 2 sinc(2 t)e j 40πt + 2 sinc(2 t)e − j 40πt = 4 sinc(2 t) cos( 40πt) x(t) 4
-2
2
-4
5-9
t
M. J. Roberts - 7/12/03
10. Find the CTFT of
x( t) = sinc( t) .
Then make the transformation, t → 2 t , in x( t) and find the CTFT of the transformed signal. F sinc( t) ← → rect ( f ) 1 f F sinc(2 t) ← → rect 2 2 | 12 rect( 2f )|
|rect( f )| 1 1 2 1 2
1 2
f
1 2
f
-1
1
rect( f )
rect( f )
f
1 2
-1
1
f
11. Using the multiplication-convolution duality of the CTFT, find an expression for y( t) which does not use the convolution operator, ∗ , and plot y( t) . (a)
y( t) = rect ( t) ∗ cos(πt) 1 1 1 y( t) = F −1 sinc( f ) δ f − + δ f + 2 2 2
y( t) =
y( t) =
1 −1 1 1 1 1 F δ f − sinc + δ f + sinc − 2 2 2 2 2
1 −1 2 1 2 1 1 1 1 F δ f − + δ f + = F −1 δ f − + δ f + 2 2 π 2 π 2 2 π y( t) =
2 cos(πt) π
y(t) 1 -4
4 -1
5-10
t
M. J. Roberts - 7/12/03
(b)
y( t) = rect ( t) ∗ cos(2πt) 1 y( t) = F −1 sinc( f ) [δ ( f − 1) + δ ( f + 1)] 2
y( t) =
1 1 −1 1 F δ f − sinc(1) + δ f + sinc(−1) = 0 2 2 2 y(t) 1 -8
8
t
-1
(c)
t y( t) = sinc( t) ∗ sinc 2 t y( t) = F −1 {rect ( f ) × 2 rect (2 f )} = 2 F −1 {rect (2 f )} = sinc 2 y(t) 1
-8
8
t
-0.5
(d)
t y( t) = sinc( t) ∗ sinc 2 2
t y( t) = F −1 {rect ( f ) × 2 tri(2 f )} = 2 F −1 {rect ( f ) tri(2 f )} == 2 F −1 {tri(2 f )} = sinc 2 2 y(t) 1
-8
(e)
8
t
y( t) = e − t u( t) ∗ sin(2πt)
j δ ( f + 1) δ ( f − 1) 1 j δ ( f + 1) − δ ( f − 1)] = F −1 − y( t) = F −1 [ 2 1 2 1 + j 2πf + π j f 1 + j 2πf 2
5-11
M. J. Roberts - 7/12/03
j δ ( f + 1) δ ( f − 1) −1 j δ ( f + 1)(1 + j 2π ) − δ ( f − 1)(1 − j 2π ) − y( t) = F −1 = F (1 − j 2π )(1 + j 2π ) 2 2 1 − j 2π 1 + j 2π j δ ( f + 1) − δ ( f − 1) + j 2π [δ ( f + 1) + δ ( f − 1)] y( t) = F −1 2 1 + (2π ) 2
j δ ( f + 1) − δ ( f − 1) [δ ( f + 1) + δ ( f − 1)] y( t) = F −1 −π 2 2 1 + (2π ) 1 + (2π ) 2
y( t) = Then, using
sin(2πt) − 2π cos(2πt) 2 1 + (2π )
B A cos( x ) + B sin( x ) = A 2 + B 2 cos x − tan −1 A
we get y( t) =
cos(2πt + 0.158) 1 + (2π )
2
y(t) 0.2 -2
2
t
-0.2
12. Using the CTFT of the rectangle function and the differentiation property of the CTFT find the Fourier transform of
x( t) = δ ( t − 1) − δ ( t + 1) . Check your answer against the CTFT found using the table and the time-shifting property.
d t Let y( t) = − rect . Then x( t) = ( y( t)) . 2 dt t F − rect ← → −2 sinc(2 f ) 2
Using the differentiation property of the CTFT,
5-12
M. J. Roberts - 7/12/03
d t F − rect ←→ j 2πf [−2 sinc(2 f )] = − j 4πf sinc(2 f ) dt 2 sin(2πf ) d t F = − j 2 sin(2πf ) − rect ←→ − j 4πf dt 2 2πf Using the direct approach, F δ ( t − 1) − δ ( t + 1) ← → e − j 2πf − e + j 2πf = − j 2 sin(2πf ) . Check.
13. Find the CTFS and CTFT of these periodic functions and compare answers. (a)
1 t x( t) = rect ( t) ∗ comb 2 2
1 X[ k ] = T0
T0 2
1
1 1 1 t − j 2π kf t t − jπkt ∫T rect (t) ∗ 2 comb 2 e ( 0 ) dt = 2 ∫1rect (t) ∗ 2 comb 2 e dt − 0
−
2
X[ k ] =
1 2
1 2
∫e
− jπkt
dt =
1 − 2
1 2
1 2
∫ [cos(πkt) − j sin(πkt)]dt −
1 2
k sinπ sin(πkt) 1 2 1 k = = sinc X[ k ] = ∫ cos(πkt) dt = 2 2 πk 0 2 πk 0 1 2
1 2
∞ 1 k X( f ) = sinc( f ) comb(2 f ) = sinc( f ) ∑ δ f − 2 2 k =−∞
∞
∞ k 1 k X( f ) = ∑ sinc δ f − = ∑ X[ k ]δ ( f − kf 0 ) 2 2 k =−∞ k =−∞ 2
(b)
x( t) = tri(10 t) ∗ 4 comb( 4 t)
X[ k ] =
1 T0
T0 2
∫ tri(10t) ∗ 4 comb(4 t)e
−
T0 2
− j 2π ( kf 0 ) t
1 8
dt = 4 ∫ tri(10 t)e − j 8πkt dt −
5-13
1 8
M. J. Roberts - 7/12/03
X[ k ] = 4
1 10
1 10
1 10 1 10
0
∫ tri(10t)(cos(8πkt) − j sin(8πkt))dt = 8 ∫ tri(10t) cos(8πkt)dt
−
1 101 10 X[ k ] = 8 ∫ (1 − 10 t) cos(8πkt) dt = 8 ∫ cos(8πkt) dt − ∫ 10 t cos(8πkt) dt 0 0 0 1 8 1 1 πk 10 10 10 sin(8πkt) 10 sin(8πkt) 10 − 10 ∫ t cos(8πkt) dt = 8 − X[ k ] = 8 2 ∫ λ cos(λ ) dλ 8πk 0 (8πk ) 0 8πk 0 0 4 4 πk 5 4 sin 5 πk 10 πk 5 ( ) X[ k ] = 8 − sin − sin ( ) d λ λ λ λ ∫0 0 (8πk ) 2 8πk 4 4 sin 5 πk 10 4 πk 4 5 X[ k ] = 8 − πk sin πk − [cos(λ )]0 5 (8πk ) 2 5 8πk 4 sin 5 πk 10 4 4 4 − X[ k ] = 8 − k sin k cos π π πk + 1 5 5 (8πk ) 2 5 8πk
4 cos πk − 1 5 5 X[ k ] = 4 (πk ) 2
X( f ) =
∞ 1 f f 4 f sinc 2 comb = sinc 2 ∑ δ ( f − 4 k ) 10 4 10 10 k =−∞ 10
X( f ) =
2 ∞ 2k sinc 2 δ ( f − 4 k ) ∑ 5 5 k =−∞
2 2πk 4πk sin cos −1 ∞ 5 5 5 1 2 δ ( f − 4 k) = ∑ X( f ) = ∑ δ ( f − 4 k) 2 k =−∞ 2 5 k =−∞ 2πk 2 (πk ) 2 5 ∞
5-14
M. J. Roberts - 7/12/03
4πk cos −1 5 5 X( f ) = ∑ δ ( f − 4 k ) . Checks with CTFS. (πk ) 2 k =−∞ 4 ∞
14. Using Parseval’s theorem, find the signal energy of these signals. (a)
t x( t) = 4 sinc 5 ∞
Ex =
∫
x( t) dt = 2
−∞
∞
∫
X( f ) df = 2
−∞
∞
∫
−∞
∞
20 rect (5 f ) df = 400 ∫ rect (5 f ) df 2
−∞
1 10
E x = 400 ∫ df = 80 −
(b)
x( t) = 2 sinc 2 ( 3t) ∞
Ex =
∫
−∞
x( t) dt = 2
∞
∫
X( f ) df = 2
−∞
1 10
∞
∫
−∞
2
∞
2 f 4 f tri df = ∫ tri2 df 3 3 9 −∞ 3
3 3 3 f f 2 8 8 8 2f 2 f 1 1 tri df = − df = − + df 3 9 9 ∫0 3 9 ∫0 3 9 ∫0 2
Ex =
3
f 2 f 3 8 9 27 8 8 Ex = f − + = 3− + = 3 27 9 9 3 27 0 9
t − 8 15. What is the total area under the function, g( t) = 100 sinc ? 30 ∞
∫ g(t)dt = G(0)
−∞
G( f ) = 3000 rect ( 30 f )e − j16πf G(0) = 3000 =
∞
∫ g(t)dt
−∞
16. Using the integration property, find the CTFT of these functions and compare with the CTFT found using other properties.
5-15
M. J. Roberts - 7/12/03
(a)
1 , t <1 g( t) = 2 − t , 1 < t < 2 0 , elsewhere g(t) 1
t -3
-2
-1
1
2
3
1
2
3
g'(t) 1 -3
-2
t
-1 -1
3 3 g′ ( t) = rect t + − rect t − 2 2 3 3 F g′ ( t) = rect t + − rect t − ← → sinc( f )e j 3πf − sinc( f )e − j 3πf 2 2 F g( t) ← →
sinc( f ) j 3πf sinc( f ) sin( 3πf ) e − e − j 3πf ) = j 2 sin( 3πf ) = sinc( f ) ( πf j 2πf j 2πf F g( t) ← → 3 sinc( 3 f ) sinc( f )
Alternate Method:
t F g( t) = rect ∗ rect ( t) ← → 3 sinc( 3 f ) sinc( f ) Check. 3
(b)
t g( t) = 8 rect 3 3 3 F g′ ( t) = 8δ t + − 8δ t − ← → 8(e j 3πf − e − j 3πf ) = j16 sin( 3πf ) 2 2 F g( t) ← → j16
Alternate Method:
sin( 3πf ) sin( 3πf ) = 24 = 24 sinc( 3 f ) j 2πf 3πf
F g( t) ← → 24 sinc( 3 f ) Check.
5-16
M. J. Roberts - 7/12/03
17. Sketch the magnitudes and phases of the CTFT’s of these signals in the f form. (a)
F x( t) = δ (t − 2) ← → X( f ) = e − j 4 πf
|X( f )| 1
-1
1
f
Phase of X( f ) π
-1
1
f
-π
(b)
1 1 F x( t) = u( t) − u( t − 1) ← → X( f ) = + δ ( f ) (1 − e − j 2πf ) j 2πf 2
F x( t) = u( t) − u( t − 1) ← → X( f ) =
1 (1 − e − j 2πf ) , (impulses cancel) j 2πf
F x( t) = u( t) − u( t − 1) ← → X( f ) = e − jπf
e + jπf − e − jπf = e − jπf sinc( f ) j 2πf
|X( f )| 1
-5
5
f
Phase of X( f ) π
-5
5 -π
5-17
f
M. J. Roberts - 7/12/03
(c)
t + 2 F j 4 πf x( t) = 5 rect ←→ X( f ) = 20 sinc( 4 f )e 4 |X( f )| 20
-1
1
f
Phase of X( f ) π
-1
1
f
-π
(d)
f F x( t) = 25 sinc(10( t − 2)) ← → X( f ) = 2.5 rect e − j 4 πf 10 |X( f )| 3
-10
10
f
Phase of X( f ) π
-10
10
f
-π
(e)
F x( t) = 6 sin(200πt) ← → X( f ) = j 3[δ ( f + 100) − δ ( f − 100)]
5-18
M. J. Roberts - 7/12/03 |X( f )| 3
-100
100
f
Phase of X( f ) π
-100
100
f
-π
(f)
F 2e − t u( t) ← →
2 1 + j 2πf
F x( t) = 2e −3 t u( 3t) ← → X( f ) =
1 2 2 = 3 1 + j 2π f 31+ 3
1 2 j πf 3
|X( f )| 1
-2
2
f
Phase of X( f ) π 2
-2
2
f
π 2
(g)
F e −πt ← → e −πf 2
x( t) = 4 e
−3 t
2
2
= 4e
3 −π t π
2
π 3
π −π F ← → X( f ) = 4 e 3
5-19
f
2
π − π3 =4 e 3
2
f2
M. J. Roberts - 7/12/03 |X( f )| 5
-3
3
f
Phase of X( f ) π
-3
3
f
-π
18. Sketch the magnitudes and phases of the CTFT’s of these signals in the ω form. (a)
∞ 1 t F ω x( t) = comb ←→ X( jω ) = comb = π ∑ δ (ω − kπ ) 2 π 2 k =−∞
|X(jω)| π
4π
-4π
ω
Phase of X(jω) π
4π
-4π -π
(b)
F x( t) = sgn(2 t) ← → X( jω ) =
2 jω
5-20
ω
M. J. Roberts - 7/12/03
|X(jω)| 6
4π
-4π
ω
Phase of X(jω) π
4π
-4π
ω
-π
(c)
t − 4 F 2 10 − j 4ω x( t) = 10 tri ←→ X( jω ) = 200 sinc ω e 20 π |X(jω)| 200
-2
2
ω
2
ω
Phase of X(jω) π
-2 -π
(d)
t + 1 sinc 2 3 F 3 3 ←→ X( jω ) = tri ω e jω x( t) = 10 10 2π
5-21
M. J. Roberts - 7/12/03
|X(jω)| 0.3
-4
4
ω
4
ω
Phase of X(jω) π
-4 -π
(e)
π cos 200πt − ω 1 ω ω − j 800 4 F x( t) = ←→ X( jω ) = δ − 100 + δ + 100 e 2π 4 8 2π
π cos 200πt − ω −j π 4 F 800 x( t) = ←→ X( jω ) = [δ (ω − 200π ) + δ (ω + 200π )]e 4 4 π cos 200πt − π π −j +j π 4 F x( t) = ←→ X( jω ) = δ (ω − 200π )e 4 + δ (ω + 200π )e 4 4 4 |X(jω)| 1
-700
700
ω
700
ω
Phase of X(jω) π
-700 -π
(f)
F x( t) = 2e −3 t u( t) = 2e −3 t u( 3t) ← → X( jω ) =
5-22
2 1 31+ j ω 3
M. J. Roberts - 7/12/03
|X(jω)| 1
-10
10
ω
10
ω
Phase of X(jω) 1.5708
-10 -1.5708
(g)
F x( t) = 7e −5 t ← → X( jω ) = 7
2×5 70 2 = 5 +ω 25 + ω 2 2
|X(jω)| 3
-10
10
ω
10
ω
Phase of X(jω) π
-10 -π
19. Sketch the inverse CTFT’s of these functions. (a)
f F x( t) = −60 sinc( 4 t) ← → X( f ) = −15 rect 4 x(t) 20 -1
1
-60
5-23
t
M. J. Roberts - 7/12/03
(b)
t rect − 10 F sinc(−10 f ) x( t) = ←→ X( f ) = 300 30 x(t) 0.005
-10
(c)
F e − a t ← →
10
2a
a 2 + (2πf )
, f →
2
F x( t) = 6πe −3 2πt ← → X( f ) =
t
f , x( t) → 2π x(2πt) 2π
18 9+ f2
x(t) 20
-0.25
(d)
F e − at u( t) ← →
0.25
1 , a>0 a + j 2πf
F x( t) = 2πe −20πt u( t) ← → X( f ) =
t
, f → 1 10 + jf
x(t) 8
t
-0.016667
(e)
F cos(2πf 0 t) ← →
0.066667
[
]
1 δ ( f − f0 ) + δ ( f + f0 ) 2
5-24
f , x( t) → 2π x(2πt) 2π
M. J. Roberts - 7/12/03
δ ( f − 3) + δ ( f + 3) 1 F x( t) = cos(6πt) ← → X( f ) = 3 6 x(t) 0.5
-1
1
t
-0.5
(f)
8 8 F x( t) = ← → X( f ) = 8δ (5 f ) = δ ( f ) 5 5 x(t) 2
-1
(g)
F sgn( t) ← →
1
t
1 jπf
F x( t) = −3 sgn( t) ← → X( f ) = −
3 jπf
x(t) 3
-1
1
t
-3
20. Sketch the inverse CTFT’s of these functions.
(a)
e
−πt 2
←→ e F
−
ω2 4π
, ω → 4 πω , x( t) →
1 4 π
t x 4 π
t ω −16π − 2 1 − 16 F x( t) = e ← → X( jω ) = e −4ω = e 4 π = e 4 π 2
2
5-25
(4
πω )
4π
2
M. J. Roberts - 7/12/03 x(t) 0.2
-8
(b)
ω F tri( t) ← → sinc 2 2π x( t) =
8
t
, ω → 2ω , x( t) →
1 t x 2 2
7 t F ω tri ←→ X( jω ) = 7 sinc 2 π 2 2 x(t) 4
-4
(c)
4
[
t
]
F sin(ω 0 t) ← → jπ δ (ω + ω 0 ) − δ (ω − ω 0 )
F x( t) = sin(10πt) ← → X( jω ) = jπ [δ (ω + 10π ) − δ (ω − 10π )]
x(t) 1
-0.4
0.4
t
-1
(d)
ω F comb( t) ← → comb 2π
, ω → 8ω , x( t) →
4ω comb π 1 t F x( t) = comb ← → X( jω ) = 8 5 40 x(t) 0.2
-40
40
5-26
t
1 t x 8 8
M. J. Roberts - 7/12/03
(e)
F sgn( t) ← →
x( t) =
2 jω
F , 1← → 2πδ (ω )
5π 5π F sgn( t) + 5 ← → X( jω ) = + 10πδ (ω ) 2 jω x(t) 18
-1
(f)
F e − at u( t) ← →
1
-4
t
1 , a>0 a + jω
F x( t) = 6e −3 t u( t) ← → X( jω ) =
6 3 + jω
x(t) 6
-0.2
(g)
ω F sinc 2 ( t) ← → tri 2π x( t) =
1.5
t
, ω → 16πω , x( t) →
1 16π
t x 16π
5 t F sinc 2 ←→ X( jω ) = 20 tri(8ω ) 16π 4π x(t) 0.5
-200
200
t
21. Find the CTFT’s of these signals in either the f or ω form, whichever is more convenient. 5-27
M. J. Roberts - 7/12/03
(a)
x( t) = 3 cos(10 t) + 4 sin(10 t) X( f ) =
3 5 5 5 5 δ f − + δ f + + j 2 δ f + − δ f − 2 π π π π 5 3 + j4 5 3 − j4 X( f ) = δ f + δ f − + 2 π 2 π
5 5 5 5 X( f ) = e − j 0.927 δ f − + e j 0.927 δ f + 2 π 2 π X( jω ) = (5πe − j 0.927 )δ (ω − 10) + (5πe j 0.927 )δ (ω + 10)
(b)
t − 1 t x( t) = comb − comb 2 2 X( f ) = 2 comb(2 f ) − 2 comb(2 f )e − j 2πf = 2 comb(2 f )(1 − e − j 2πf ) X( f ) = e − jπf 2 comb(2 f )(e jπf − e − jπf ) = j 4 e − jπf comb(2 f ) sin(πf ) X( jω ) = j 4 e
−j
ω 2
ω ω comb sin π 2
(a)
|X(jω)|
x(t)
20
6
1
-20
t
20
ω
20
ω
4
f
4
f
Phase of X(jω) π
-6
-20
(b)
-π
|X( f )| 2
x(t) 2 -4 -4
4
Phase of X( f )
t
π -4
-2
(c)
x( t) = 4 sinc( 4 t) − 2 sinc 4 t −
-π
1 − 2 sinc 4 t + 4
5-28
1 4
M. J. Roberts - 7/12/03 πf πf 1 f 1 f −j f j X( f ) = rect − rect e 2 − rect e 2 4 2 4 4 2
πf πf j f 1 f −j 2 πf f f X( f ) = rect − rect e + e 2 = rect − rect cos 4 2 4 4 4 2
ω ω ω X( jω ) = rect − rect cos 8π 8π 4
(d)
[
]
x( t) = 2e( −1+ j 2π ) t + 2e( −1− j 2π ) t u( t) F → Using e − at u( t) ←
X( jω ) =
1 , a>0 a + jω
2 2 4 + j 4ω + = 1 − j 2π + jω 1 + j 2π + jω (1 − j 2π + jω )(1 + j 2π + jω )
X( jω ) = 4 X( f ) = 4
jω + 1
( jω + 1) 2 + (2π ) 2
( j 2πf
j 2πf + 1
+ 1) + (2π ) 2
2
Alternate Solution:
[
]
x( t) = 2e( −1+ j 2π ) t + 2e( −1− j 2π ) t u( t) = 4 e − t cos(2πt) u( t) F Using e − at cos(ω 0 t) u( t) ← →
X( jω ) = 4
a + jω
( jω + a) 2 + ω 02 jω + 1
( jω + 1) 2 + (2π ) 2
5-29
M. J. Roberts - 7/12/03
(c)
|X( f )|
x(t)
2
4
-2
2
-4
t
4
f
4
f
3
f
3
f
Phase of X( f ) π
-4
-4
-π
(d)
|X( f )| 2
x(t) 4 -3 -1
3
Phase of X( f )
t
π -3
-4
(e)
x( t) = 4 e
−
-π
t 16
F Using e − a t ← →
2a , Re( a) > 0 , a + ω2 2
e
4e
t − 16
−
t 16
F ← →
F ← →
1 8 2
1 2 +ω 16
1 2 2
1 2 +ω 16
(e)
=
128 1 + 256ω 2
|X(jω)|
x(t)
128
4 -0.25
0.25
ω
0.25
ω
Phase of X(jω) π -50
50
t
-0.25
-π
22. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFT’s of the functions also. 10 4 F x( t) = (10e −3 t − 4 e −5 t ) u( t) ← → X( jω ) = − (a) 3 + jω 5 + jω
5-30
M. J. Roberts - 7/12/03
|X(jω)|
x(t)
4
6 -20
20
ω
20
ω
Phase of X(jω) π -1
(b)
2
t
-20
-π
f + 1 f − 1 F x( t) = 16 rect (2 t) cos(2πt) ← → X( f ) = 4 sinc + sinc 2 2 |X( f )| 5
x(t) 16 -10 -1
1
t
f
10
f
π -10
-16
(c)
10
Phase of X( f )
π
F x( t) = 1.6 sinc 2 (8 t) sin( 4πt) ← → X( f ) =
j f + 2 f − 2 tri − tri 8 10 8 |X( f )|
x(t)
0.1
0.5
-0.5
0.5
-15
t
15
f
15
f
Phase of X( f ) π
-0.5
-15
-π
(d)
δ ( f + 1050) + δ ( f + 950) F x( t) = 2[cos(2100πt) + cos(1900πt)] ← → X( f ) = +δ ( f − 950) + δ ( f − 1050)
or
δ ( f + 1050) + δ ( f + 950) F x( t) = 4 cos(100πt) cos(2000πt) ← → X( f ) = +δ ( f − 950) + δ ( f − 1050)
5-31
M. J. Roberts - 7/12/03 |X( f )| 1
x(t) 4 -1200
t -4
-0.04
(e) or
1200
f
1200
f
Phase of X( f ) π
-1200 -π
0.04
δ ( f + 1050) + 2δ ( f + 1000) cos(2100πt) F → X( f ) = +δ ( f + 950) + δ ( f − 950) x( t) = 2 + cos(1900πt) ← 2δ +2 cos(2000πt) + ( f − 1000) + δ ( f − 1050) δ ( f + 1050) + 2δ ( f + 1000) cos(2000πt) cos(100πt) F x( t) = 4 ←→ X( f ) = +δ ( f + 950) + δ ( f − 950) + cos(2000πt) 2 f 1000 ) + δ ( f − 1050) + δ ( − |X( f )|
x(t)
2
8
-0.04
0.04
-1000
t
1000
f
1000
f
Phase of X( f ) π
-8
-1000
-π
23. Sketch these signals versus time. Sketch the magnitudes and phase of their CTFT’s in either the f or ω form, whichever is more convenient. (a)
1 x( t) = rect (2 t) ∗ comb( t) − rect (2 t) ∗ comb t − 2 1 f X( f ) = sinc comb( f )(1 − e − jπf ) 2 2 X( f ) = je
X( f ) =
−j
∞
πf 2
∑e
k =−∞
πf f sinc comb( f ) sin 2 2
π − j ( k −1) 2
k πk sinc sin δ ( f − k ) 2 2
5-32
M. J. Roberts - 7/12/03
Non-zero only for odd values of k. At those odd values, e
π − j ( k −1) 2
always evaluates to +1. Therefore
X( f ) =
∞
k
∑ sinc 2 δ ( f − k )
k =−∞ k ≠0
|X( f )| x(t)
1
1 -8 -3
3
t
f
8
f
π -8
-1
(b)
8
Phase of X( f )
-π
x( t) = −1 + 2 rect (2 t) ∗ comb( t) f X( f ) = sinc comb( f ) − δ ( f ) 2
X( f ) = −δ ( f ) +
∞
∞ k k sinc δ f − k = sinc δ ( f − k ) ( ) ∑ ∑ 2 2 k =−∞ k =−∞ k ≠0
Same as answer in part (a). |X( f )| x(t)
1
1 -8 -3
3
t
x( t) = e
−
X( jω ) =
t 4
f
8
f
π -8
-1
(c)
8
Phase of X( f )
-π
u( t) ∗ sin(2πt) 1 jπ [δ (ω + 2π ) − δ (ω − 2π )] 1 + jω 4
δ (ω + 2π ) δ (ω − 2π ) − X( jω ) = jπ 1 1 − j 2π + j 2π 4 4 5-33
πk sin , 2
M. J. Roberts - 7/12/03
Rationalizing the denominator,
jπ jπ − 2π 2 + 2π 2 4 4 X( jω ) = δ (ω + 2π ) − δ (ω − 2π ) 1 1 2 2 + (2π ) + (2π ) 16 16
X( jω ) =
x( t) =
x( t) =
jπ 4
2π 2
[δ (ω + 2π ) − δ (ω − 2π )] − 1 [δ (ω + 2π ) + δ (ω − 2π )] 1 2 2 + (2π ) + (2π ) 16 16 1 4
1 2 + (2π ) 16
sin(2πt) −
2π 1 2 + (2π ) 16
cos(2πt)
4 sin(2πt) − 32π cos(2πt) 1 + 64π 2 |X(jω)|
x(t)
1
0.2
-2
2
-8
t
8
ω
8
ω
Phase of X(jω) π
-0.2
-8 -π
(d)
x( t) = e −πt ∗ [rect (2 t) ∗ comb( t)] 2
X( f ) = e −πf
X( f ) = x( t) =
2
1 f sinc comb( f ) 2 2
1 ∞ −πf 2 1 ∞ −πk 2 f k e sinc δ f − k = e sinc δ ( f − k ) ) ( ∑ ∑ 2 2 2 k =−∞ 2 k =−∞
2 1 ∞ −πk 2 1 ∞ k k e sinc e j 2πkt = + ∑ e −πk sinc cos(2πkt) ∑ 2 2 2 k =−∞ 2 k =1
5-34
M. J. Roberts - 7/12/03 |X( f )| 1
x(t) 1 -8
8
f
8
f
Phase of X( f ) π
-2
(e)
2
-8
t
-π
x( t) = rect ( t) ∗ [tri(2 t) ∗ comb( t)] 1 f X( f ) = sinc( f ) sinc 2 comb( f ) 2 2
All the comb impulses have zero weight except the one at zero.
1 X( f ) = δ ( f ) 2 x( t) =
1 2 |X( f ) |
x(t)
1
1 -8
8
f
8
f
Phase of X( f ) π -2
(f)
2
t
-8 -π
x( t) = sinc(2.01t) ∗ comb( t) X( f ) =
x( t) =
1 1 f rect comb( f ) = [δ ( f + 1) + δ ( f ) + δ ( f − 1)] 2.01 2.01 2.01
1 + 2 cos(2πt) 2.01
5-35
M. J. Roberts - 7/12/03 |X( f )| 1
x(t) 2 -1
1
f
1
f
2
f
2
f
1
f
1
f
Phase of X( f ) -2
2
π
t -1
-1
(g)
-π
x( t) = sinc(1.99 t) ∗ comb( t) X( f ) =
x( t) =
1 1 f rect δ( f ) comb( f ) = 1.99 1.99 1.99
1 = 0.5025 1.99 |X( f ) |
x(t)
1
1 -2
Phase of X( f ) π -2
(h)
2
x( t) = e − t ∗ e − t 2
X ( f ) = π e −π
t
-2 -π
2
2
π e −π
f2
π −π x( t) = e 2π
2π t
2
2
= πe −2π
f2
2
f2
π − t2 = e 2
2
|X( f )| 4
x(t) 2 -1
Phase of X( f ) π
-4
4
-1
t
-π
24. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFT’s of the functions also. 5-36
M. J. Roberts - 7/12/03
(a)
f X( f ) = sinc ∗ δ ( f − 1000) + δ ( f + 1000)] 100 [ x( t) = 100 rect (100 t)(e j 2000πt + e − j 2000πt ) = 200 rect (100 t) cos(2000πt) |X( f )|
x(t)
1
-1000
200
f
1000
Phase of X( f )
-0.01
π
-1000
(b)
-π
0.01
f
1000
t
-200
X( f ) = sinc(10 f ) ∗ comb( f )
x( t) =
1 1 ∞ t n rect comb( t) = rect δ ( t − n ) ∑ 10 10 10 10 n =−∞ |X( f )| 1
x(t) 0.1
-2
2
f
2
f
Phase of X( f ) π
-2
-π
-10
10
t
25. Sketch these signals versus time. Sketch the magnitudes and phases of the CTFT’s of these signals in either the f or ω form, whichever is more convenient. In some cases the time sketch may be conveniently done first. In other cases it may be more convenient to do the time sketch after the CTFT has been found, by finding the inverse CTFT. (a)
x( t) = e −πt sin(20πt) 2
j e −π ( f +10) − e −π ( f −10) ∗ [δ ( f + 10) − δ ( f − 10)] = j 2 2 2
X( f ) = e
−πf 2
5-37
2
M. J. Roberts - 7/12/03 |X( f )|
x(t)
0.5
1
-2
2
-12
t
f
12
f
π
-1
(b)
12
Phase of X( f ) -12
x( t) = cos( 400πt) comb(100 t) =
-π
1 ∞ n cos( 4πn )δ t − ∑ 100 100 n −−∞
1 1 f δ ( f − 200) + δ ( f + 200)] ∗ comb [ 100 100 2 1 f + 200 f − 200 X( f ) = + comb comb 100 200 144 100 2443 1442443 f f = comb = comb 100 100
X( f ) =
X( f ) =
1 f comb 100 100 |X( f )| 0.01
x(t) 0.01 -1000
1000
f
1000
f
Phase of X( f ) π
-0.1
(c)
0.1
-1000
t
-π
x( t) = [1 + cos( 400πt)] cos( 4000πt)
1 1 X( f ) = δ ( f ) + [δ ( f − 200) + δ ( f + 200)] ∗ [δ ( f − 2000) + δ ( f + 2000)] 2 2
δ ( f ) ∗ [δ ( f − 2000) + δ ( f + 2000)] 1 1 X( f ) = + δ ( f − 200) ∗ [δ ( f − 2000) + δ ( f + 2000)] 2 2 1 + 2 δ ( f + 200) ∗ [δ ( f − 2000) + δ ( f + 2000)]
5-38
M. J. Roberts - 7/12/03
δ ( f − 2000) + δ ( f + 2000) 1 X( f ) = 1 2 + [δ ( f − 2200) + δ ( f + 1800) + δ ( f − 1800) + δ ( f + 2200)] 2 |X( f )|
x(t)
0.5
2 -2500
t
2500
f
2500
f
Phase of X( f ) π
-2
-0.01
(d)
-2500
0.01
-π
x( t) = [1 + rect (100 t) ∗ 50 comb(50 t)] cos(500πt)
1 f f 1 X( f ) = δ ( f ) + sinc comb ∗ [δ ( f − 250) + δ ( f + 250)] 50 2 100 100 1 1 ∞ k X( f ) = δ ( f ) + ∑ sinc δ ( f − 50 k ) ∗ [δ ( f − 250) + δ ( f + 250)] 2 2 k =−∞ 2 1 2 [δ ( f − 250) + δ ( f + 250)] X( f ) = ∞ k + 1 sinc [δ ( f − 250 − 50 k ) + δ ( f + 250 − 50 k )] 4 k∑ 2 =−∞ |X( f )| 1
x(t) 0.08 -500
(e)
-0.08
f
500
f
Phase of X( f )
t -0.04
500 π
-500
0.04
-π
t x( t) = rect comb( t) 7
X( f ) = 7 sinc( 7 f ) ∗ comb( f ) = 7 sinc( 7 f ) ∗ ∞
X( f ) = 7 ∑ sinc( 7( f − k )) k =−∞
5-39
∞
∑ δ ( f − k)
k =−∞
M. J. Roberts - 7/12/03 |X( f )|
x(t)
7
1 -4
4
f
4
f
Phase of X( f ) π
-6
6
t
-4
-π
26. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFT’s of the functions also. f (a) X( f ) = sinc comb( f ) 4
x( t) = 4 rect ( 4 t) ∗ comb( t) = 4 rect ( 4 t) ∗ ∞
∞
∑ δ (t − n)
n =−∞
x( t) = 4 ∑ rect ( 4 ( t − n )) n =−∞
|X( f )|
x(t)
1
4 -16
16
f
16
f
Phase of X( f ) π -16
-π
-2
5-40
2
t
M. J. Roberts - 7/12/03
(b)
f + 1 f − 1 X( f ) = sinc comb( f ) + sinc 4 4 x( t) = 4 rect ( 4 t)(e
j 2πt
+e
j 2πt
∞
) ∗ comb(t) = 8 rect(4 t) cos(2πt) ∗ ∑ δ (t − n) n =−∞
∞
x( t) = 8 ∑ rect ( 4 ( t − n )) cos(2π ( t − n )) n =−∞
|X( f )| 2
x(t) 1
-8
8
f
8
f
Phase of X( f ) π -8
(c)
-π
-2
2
t
X( f ) = sinc( f ) sinc(2 f ) 1 t 1 t x( t) = rect ( t) ∗ rect = rect ( t) ∗ rect 2 2 2 2
The result of the graphical convolution can be expressed in the form,
x( t) =
1 2t 3 tri − tri(2 t) 4 3 |X( f )|
x(t)
1 0.5 -2
2
f
2
f
Phase of X( f ) π -2
-π
-2
2
t
27. Sketch these signals versus time and the magnitudes and phases of their CTFT’s. (a)
x( t) =
d sin(πt) tπ cos(πt) − sin(πt) d = sinc( t)] = [ dt πt dt πt 2
X( f ) = j 2πf rect ( f )
5-41
M. J. Roberts - 7/12/03
(a)
|X( f )|
x(t)
π
2
-8
8
-1
t
x( t) =
f
1
f
π
-2
(b)
1
Phase of X( f ) -1
-π
d t 4 rect 6 dt
X( f ) = j 2πf × 24 sinc(6 f ) = j 2πf × 24
sin(6πf ) = j 8 sin(6πf ) 6πf
Alternate Solution:
d t F 4 rect = 4[δ ( f + 3) − δ ( f − 3)] ← → 4 (e j 6πf − e − j 6πf ) 6 dt F 4[δ ( f + 3) − δ ( f − 3)] ← → j 8 sin(6πf )
|X( f )| 8
x(t) 4 -0.5 3 -3
t
x( t) =
-0.5
d tri(2 t) ∗ comb( t)] = 2rect 2 t + [ dt
X( f ) = j 2πf
f
0.5
f
π
-4
(c)
0.5
Phase of X( f )
-π
1 − rect 2 t − 4
1 ∗ comb( t) 4
∞ 1 f k sinc 2 comb( f ) = jπ ∑ k sinc 2 δ ( f − k ) 2 2 2 k =−∞
5-42
M. J. Roberts - 7/12/03 |X( f )|
x(t)
2
2
-2
-8
t
2
8
f
8
f
Phase of X( f ) π
-2
-8
-π
28. Sketch these signals versus time and the magnitudes and phases of their CTFT’s. (a)
x( t) =
t
∫ sin(2πλ )dλ
−∞
X( f ) =
[
]
1 1 j × [δ ( f + 1) − δ ( f − 1)] + F (sin(2πt)) 2 j 2πf 2 X( f ) = −
f =0
δ( f ) =
δ ( f + 1) − δ ( f − 1) 4πf
δ ( f + 1) δ ( f − 1) 1 1 − =− × [δ ( f + 1) + δ ( f − 1)] 4π 4π 2π 2 |X( f )|
x(t)
0.1
0.2
-2
-1
t
2
1
f
1
f
Phase of X( f ) π -1
-0.2 -π
(b)
1 , t<− 0 2 t 1 1 x( t) = ∫ rect (λ ) dλ = t + , t < 2 −∞ 2 1 , t> 1 2 sinc( f ) 1 sinc( f ) 1 X( f ) = + F (rect ( t)) f = 0 δ ( f ) = + δ( f ) 2 2 j 2πf j 2πf
[
]
|X( f )| 1
x(t) 1 -2
2
f
2
f
Phase of X( f ) π -2 -1
1
t
-π
5-43
M. J. Roberts - 7/12/03
(c)
x( t) =
t
∫ 3sinc(2λ )dλ
−∞
3 Let u = 2πλ . Then x( t) = 2π
2πt
3 u ∫−∞ sinc π du = 2π
sin( u) du u −∞
2πt
∫
For t ≤ 0 : −∞ 0 0 2πt sin( u) 3 sin( u) sin( u) 3 sin( u) du du + ∫ du = du − ∫ x( t) = ∫ − ∫ u u 2π −∞ u 2π 0 u 2πt 0
∞ 2πt sin( u) 3 π 3 sin(λ ) 3 π + Si(2πt) = du = x( t) = dλ + ∫ − Si(−2πt) ∫ u 2π 0 λ 2π 2 2π 2 0 4243 14243 1 =Si( 2πt ) =Si( ∞ ) = π 2 For t ≥ 0 : x( t) =
−∞ 0 2πt 2πt sin( u) 3 sin( u) sin( u) 3 sin( u) du du du du = − + + ∫ ∫ ∫ ∫ u u 2π −∞ u 2π 0 u 0 0
x( t) =
3 π + Si(2πt) 2π 2
Therefore, for any t, x( t) = X( f ) =
3 π + Si(2πt) . 2π 2
3 1 3 f 1 f 3 rect + F[ 3 sinc(2 t)] f = 0 δ ( f ) = rect + δ ( f ) 2 4 2 2 j 2πf 2 j 4πf |X( f )|
x(t)
1
2 -2 -4
4
t
-1
f
2
f
π -2
29. From the definition, find the DTFT of
x[ n ] = 10 rect 4 [ n ] . 5-44
2
Phase of X( f )
-π
M. J. Roberts - 7/12/03
and compare with the Fourier transform table in Appendix E.
X( F ) =
∞
∑ x[n]e
− j 2πFn
∞
=
n =−∞
∑ 10 rect [n]e
− j 2πFn
4
n =−∞
8
8
m =0
m =0
4
= 10 ∑ e − j 2πFn n =−4
X( F ) = 10 ∑ e − j 2πF ( m − 4 ) = 10e j 8πF ∑ e − j 2πFm = 10e j 8πF X( F ) = 10e
j 8πF
1 − e − j18πF 1 − e − j 2πF
e − j 9πF e j 9πF − e − j 9πF sin(9πF ) = 90 drcl( F , 9) − jπF − jπF − jπF = 10 e e −e sin(πF )
From the table, F rect N w [ n ] ← →(2 N w + 1) drcl( F , 2 N w + 1)
F rect 4 [ n ] ← → 9 drcl( F , 9)
F 10 rect 4 [ n ] ← → 90 drcl( F , 9)
Check
30. From the definition, derive a general expression for the F and Ω forms of the DTFT of functions of the form, x[ n ] = A sin(2πF0 n ) = A sin(Ω0 n ) . (It should remind you of the CTFT of x( t) = A sin(2πf 0 t) = A sin(ω 0 t) .) Compare with the Fourier transform table in Appendix E.
X( F ) =
∞
∑ x[n]e
n =−∞
− j 2πFn
=
∞
∑ A sin(2πF n)e
X( F ) = Then, using
0
n =−∞
− j 2πFn
e j 2πF0 n − e − j 2πF0 n − j 2πFn e j2 n =−∞ ∞
=A∑
A ∞ ∑ e j 2π ( F0 − F ) n − e − j 2π ( F0 + F ) n j 2 n =−∞
[
∞
∑e
n =−∞
j 2πxn
= comb( x )
5-45
]
M. J. Roberts - 7/12/03
we get X( F ) =
[
]
[
]
j A comb( F0 − F ) − comb(− F0 − F ) = A − comb( F0 − F ) + comb(− F0 − F ) j2 2 j X( F ) = A comb( F + F0 ) − comb( F − F0 ) 2
[
]
The Ω form can be found by the transformation, F →
Ω . 2π
j Ω Ω0 Ω Ω0 X( jΩ) = A comb + − comb − 2π 2π 2π 2π 2
[
]
X( jΩ) = jπA comb(Ω + Ω0 ) − comb(Ω − Ω0 )
31. A DT signal is defined by
n x[ n ] = sinc . 8
Sketch the magnitude and phase of the DTFT of x[ n − 2] . From the table of transform pairs, n F → wrect ( wF ) ∗ comb( F ) sinc ← w n F sinc ← → 8rect (8 F ) ∗ comb( F ) 8 n − 2 F − j 4 πF sinc ←→[8rect (8 F ) ∗ comb( F )]e 8 ∞
n − 2 F − j 4 πF sinc ←→ 8e ∑ rect (8(F − k )) 8 k =−∞
5-46
M. J. Roberts - 7/12/03
|X( F )|
x[n]
8
1 -1
n -32
F
1
F
π
32 -1
32. A DT signal is defined by
1
Phase of X( F )|
-π
πn x[ n ] = sin . 6
Sketch the magnitude and phase of the DTFT of x[ n − 3] and x[ n + 12]. From the table, F sin(2πF0 n ) ← →
[
]
j comb( F + F0 ) − comb( F − F0 ) 2
1 1 πn F j sin ← → comb F + − comb F − 6 2 12 12 π ( n − 3) F j 1 1 − j 6πF sin ←→ comb F + − comb F − e 6 2 12 12 π ( n − 3) F j ∞ − j 6πF 1 1 sin δ F + − k − e − j 6πFδ F − − k ←→ ∑ e 6 2 k =−∞ 12 12 1 1 − j 6π k + π ( n − 3) F j ∞ − j 6π k −12 1 1 12 sin ← → e δ F + − k − e δ F − − k ∑ 43 243 6 2 k =−∞ 142 12 14=− 12 =j j
π ( n − 3) F 1 ∞ 1 1 sin ← → − δ F + − k + δ F − − k ∑ 6 2 k =−∞ 12 12 π ( n − 3) F 1 1 1 sin ←→ − comb F + + comb F − 6 2 2 2 Similarly, 1 1 + j 24 π k + π ( n + 12) F j ∞ + j 24 π k −12 1 1 12 sin ← → e δ F + − k − e δ F − − k ∑ 6 2 k =−∞ 12 12
5-47
M. J. Roberts - 7/12/03
π ( n + 12) F j ∞ + j ( 24 πk − 2π ) 1 1 sin δ F + − k − e + j ( 24 πk + 2π )δ F − − k ←→ ∑ e 6 2 k =−∞ 12 12 π ( n + 12) F j ∞ 1 1 sin ←→ ∑ δ F + − k − δ F − − k 6 2 k =−∞ 12 12 π ( n + 12) F j 1 1 sin ←→ comb F + − comb F − 6 2 2 2 This is the same as the transform of the unshifted function because a shift of 12 is a shift over exactly one period. |X( F )|
x[n]
0.5
1 -12
12
-1
n
1
F
1
F
1
F
1
F
Phase of X( F )| π
-1
-1
-π
|X( F )| 0.5
x[n] 1 -1 -12
12
Phase of X( F )|
n
π -1
-1
-π
33. The DTFT of a DT signal is defined by
Sketch x[ n ] .
π π 2 2 Ω X( jΩ) = 4 rect Ω − + rect Ω + ∗ comb . 2π π π 2 2
From the table,
n F wΩ Ω → wrect sinc ← ∗ comb w 2π 2π n F 2 Ω → 4rect Ω ∗ comb sinc ← 4 π 2π
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M. J. Roberts - 7/12/03
e
e
j
π n 2
−j
π n 2
2 π n F Ω sinc ← → 4rect Ω − ∗ comb 4 2π π 2
π 2 n F Ω sinc ← → 4rect Ω + ∗ comb 4 2π π 2
π π − j n π π 2 2 n j 2 n Ω F + e 2 ← → 4 rect Ω − + rect Ω + ∗ comb sinc e 4 2π π π 2 2
π π 2 2 n π F Ω 2 sinc cos n ← → 4 rect Ω − + rect Ω + ∗ comb 4 2 2π π π 2 2 Therefore
n π x[ n ] = 2 sinc cos n 4 2 x[n] 2
-16
16
-2
34. Sketch the magnitude and phase of the DTFT of
Then sketch x[ n ] .
2πn x[ n ] = rect 4 [ n ] ∗ cos . 6
From the table, F rect N w [ n ] ← →(2 N w + 1) drcl( F , 2 N w + 1)
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n
M. J. Roberts - 7/12/03
F rect N w [ n ] ← →
and F cos(2πF0 n ) ← →
X( F ) =
The function, Then
sin(πF (2 N w + 1)) sin(πF )
[
]
1 comb( F − F0 ) + comb( F + F0 ) 2
sin(9πF ) 1 1 1 comb F − + comb F + sin(πF ) 2 6 6
sin(9πF ) is periodic with period, one. To prove that, let k be any integer. sin(πF )
sin(9π ( F + k )) sin(9πF + 9πk ) sin(9πF ) cos(9πk ) + sin(9πk ) cos(9πF ) = = sin(π ( F + k )) sin(πF + πk ) sin(πF ) cos(πk ) + sin(πk ) cos(πF ) sin(9π ( F + k )) − sin(9πF ) sin(9πF ) , k odd = = sin(π ( F + k )) sin(πF ) − sin(πF ) sin(9π ( F + k )) sin(9πF ) , k even = sin(π ( F + k )) sin(πF )
X( F ) =
1 sin(9πF ) ∞ 1 1 δ F − − k + δ F + − k ∑ 2 sin(πF ) k =−∞ 6 6
1 1 sin 9π k − sin 9π k + 6 1 6 1 1 X( F ) = ∑ δ F − − k + δ F + − k 1 6 1 6 2 k =−∞ sinπ k − sinπ k + 6 6 ∞
1 1 sin 9π k − sin 9π k + sin(9πF ) 6 6 Since is periodic with period, one, so are and . sin(πF ) 1 1 sinπ k − sinπ k + 6 6
Therefore, for any k, 1 1 3π 3π sin 9π k − sin − sin 9π k + sin 2 2 −1 6 1 6 = = = −2 and = = = −2 1 1 π 1 1 π − sin − sin sinπ k − sinπ k + 6 6 2 2 6 6
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M. J. Roberts - 7/12/03
and therefore
1 1 1 1 1 X( F ) = −2 comb F − − 2 comb F + = −comb F − + comb F + . 6 6 6 6 2 Then, using F cos(2πF0 n ) ← →
[
]
1 comb( F − F0 ) + comb( F + F0 ) 2
1 1 2πn F −2 cos ←→ − comb F − + comb F + 6 6 6 and, therefore,
2πn x[ n ] = −2 cos 6 |X( F )|
x[n]
1
2 -12
12
-1
n
1
F
1
F
Phase of X( F )| π
-2
-1
-π
35. Sketch the inverse DTFT of
X( F ) = [rect ( 4 F ) ∗ comb( F )] ⊗ comb(2 F ) . From the table, n F → wrect ( wF ) ∗ comb( F ) sinc ← w
and
F comb N 0 [ n ] ← → comb( N 0 F )
Therefore using multiplication-convolution duality, x[ n ] =
1 n sinc comb 2 [ n ] . 4 4
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M. J. Roberts - 7/12/03 x[n] 0.25
-16
16
n
-0.1
36. Using the differencing property of the DTFT and the transform pair, n F tri ← →1 + cos(2πF ) , 2
1 (δ[n + 1] + δ[n] − δ[n − 1] − δ (n − 2)) . Compare it with Fourier 2 transform found using the table in Appendix E.
find the DTFT of
1 n The first backward difference of tri is (δ [ n + 1] + δ [ n ] − δ [ n − 1] − δ ( n − 2)) . 2 2 Applying the differencing property, n − 1 F n − j 2πF tri − tri ←→(1 − e )(1 + cos(2πF )) 2 2
1 F δ [ n + 1] + δ [ n ] − δ [ n − 1] − δ [ n − 2]) ← →(1 − e − j 2πF )(1 + cos(2πF )) ( 2 j 2πF e j 2πF + e − j 2πF + e − j 2πF 1 F − j 2πF − j 2πF e 1 1 2 1 δ n + + δ n − δ n − − δ n − ← → − e + − e [ ] [ ] [ ] [ ] ( ) 2 2 2
e j 2πF e − j 2πF 1 e − j 4 πF 1 F δ [ n + 1] + δ [ n ] − δ [ n − 1] − δ [ n − 2]) ← →1 − e − j 2πF + + − − ( 2 2 2 2 2
1 1 F δ [ n + 1] + δ [ n ] − δ [ n − 1] − δ [ n − 2]) ← → (e j 2πF + 1 − e − j 2πF − e − j 4 πF ) ( 2 2 Other route to the DTFT:
1 1 F δ [ n + 1] + δ [ n ] − δ [ n − 1] − δ [ n − 2]) ← → (e j 2πF + 1 − e − j 2πF − e − j 4 πF ) ( 2 2 Check. 37. Using Parseval’s theorem, find the signal energy of
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M. J. Roberts - 7/12/03
n 2πn x[ n ] = sinc sin . 10 4
Ex =
From the table,
∞
∑ x[n]
n =−∞
2
= ∫ X( F ) dF 2
1
n F → wrect ( wF ) ∗ comb( F ) sinc ← w
and F sin(2πF0 n ) ← →
[
]
j comb( F + F0 ) − comb( F − F0 ) 2
Using the multiplication-convolution duality of the DTFT,
j 1 n 2πn F sinc sin ←→10rect (10 F ) ∗ comb( F ) ⊗ comb F + − comb F − 10 4 2 4
1 4
Periodic convolution is the same as aperiodic convolution with one period.
1 1 n 2πn F sinc sin ←→ j 5rect (10 F ) ∗ comb( F ) ∗ δ F + − δ F − 10 4 4 4 1 n 2πn F sinc sin ←→ j 5rect (10 F ) ∗ comb F + − comb F − 10 4 4
1 4
1 1 n 2πn F sinc sin ←→ j 5 rect (10 F ) ∗ comb F + − rect (10 F ) ∗ comb F − 10 4 4 4 2
Ex = ∫
1
1 1 j 5 rect (10 F ) ∗ comb F + − rect (10 F ) ∗ comb F − dF 4 4
Since we are integrating only over a range of one, only one impulse in each comb is significant. 2 1 1 E x = 25 ∫ rect (10 F ) ∗ δ F + − rect (10 F ) ∗ δ F − dF 1 4 4
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M. J. Roberts - 7/12/03
1 E x = 25 ∫ rect 10 F + − rect 10 F − 1 4
2
1 dF 4
The square of the sum equals the sum of the squares because there is no cross product; the two rectangles do not overlap. 1 1 E x = 25∫ rect 10 F + dF + ∫ rect 10 F − dF 1 4 4 1 1 1 + − 14 + 201 4 20 1 1 E x = 25 ∫ dF + ∫ dF = 25 + = 5 10 10 1 1 − 1 − 1 − 4 20 4 20
38. Sketch the magnitude and phase of the CTFT of
x1 ( t) = rect ( t) and of the CTFS of
1 t x 2 ( t) = rect ( t) ∗ comb . 8 8
For comparison purposes, sketch X1 ( f ) versus f and T0 X 2 [ k ] versus kf 0 on the same set 1 of axes. ( T0 is the period of x 2 ( t) and T0 = .) f0
X1 ( f ) = sinc( f ) 1 k X 2 [ k ] = sinc 8 8
5-54
M. J. Roberts - 7/12/03 |T0 X2[ k]|
|X1( f )| 1
1
-4
4
f
-4
4
kf0
4
kf0
Phase of T0 X2[ k]|
Phase of X1( f ) π
π
-4
4
-4
f
-π
-π
39. Sketch the magnitude and phase of the CTFT of
x1 ( t) = 4 cos( 4πt) and of the DTFT of
x 2 [ n ] = x1 ( nTs )
1 . For comparison purposes sketch X1 ( f ) and Ts X 2 (Ts f ) versus f on the 16 same set of axes. X1 ( f ) = 2[δ ( f − 2) + δ ( f + 2)]
where Ts =
πn x 2 [ n ] = 4 cos( 4πnTs ) = 4 cos 4
1 1 X 2 ( F ) = 2 comb F − + comb F + 8 8 ∞ 1 1 X 2 ( F ) = 2 ∑ δ F − − k + δ F + − k 8 8 k =−∞ ∞ f 1 f 1 X 2 (Ts f ) = 2 ∑ δ − − k + δ + − k 16 8 16 8 k =−∞ ∞
X 2 (Ts f ) = 32 ∑ [δ ( f − 2 − 16 k ) + δ ( f + 2 − 16 k )] k =−∞ ∞
Ts X 2 (Ts f ) = 2 ∑ [δ ( f − 2 − 16 k ) + δ ( f + 2 − 16 k )] k =−∞
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M. J. Roberts - 7/12/03
|X ( f )|
|T X (T f )|
1
s 2
2
s
2
... -16
16
f
...
-16
Phase of X1( f )
16
Tsf
16
... Tsf
Phase of TsX2(Tsf )
π
π
... -16
16
f
-16
-π
-π
40. Sketch the magnitude and phase of the DTFT of
n sinc 16 x1[ n ] = 4 and of the DTFS of
n sinc 16 x2[n] = ∗ comb 32 [ n ] . 4 For comparison purposes sketch X1 ( F ) versus F and N 0 X 2 [ k ] versus kF0 on the same set of axes. From the table, n F → wrect ( wF ) ∗ comb( F ) sinc ← w
n sinc 16 F ←→ 4rect (16 F ) ∗ comb( F ) 4
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M. J. Roberts - 7/12/03
therefore ∞
X1 ( F ) = 4rect (16 F ) ∗ comb( F ) = 4 ∑ rect (16( F − q)) . q =−∞
The fundamental frequency of the periodic signal is the reciprocal of its period, 1 1 = . N 0 32
F0 =
Using the results of the analysis of periodic extensions of aperiodic DT signals, X 2[k ] =
1 X ( kF ) N0 1 0
∞ ∞ 1 k k 1 1 k X 2[k ] = X1 = ∑ rect 16 − q = ∑ rect − 16q 2 32 8 q =−∞ 32 32 8 q =−∞
|X (F )| 1
4
-1
1
F
Phase of X1(F ) π
-1
1
F
-π
N0|X2[k]| 4
-1
1
kF0
Phase of X2[k] π
-32
32 -π
5-57
kF
0
M. J. Roberts - 7/12/03
41. A system is excited by a signal, t x( t) = 4 rect 2
y( t) = 4 (1 − e −( t +1) ) u( t + 1) − 4 (1 − e −( t −1) ) u( t − 1) and its response is
[
]
y( t) = 10 (1 − e −( t +1) ) u( t + 1) − (1 − e −( t −1) ) u( t − 1) .
What is its impulse response? 1 1 jω 1 1 − jω Y( jω ) = 10 + πδ (ω ) − + πδ (ω ) − e − e 1 + jω 1 + jω jω jω
1 1 1 1 jω − jω Y( jω ) = 10 − − (e − e ) = j 20 sin(ω ) jω 1 + jω jω 1 + jω 1 1 1 1 − j 20 − sin(ω ) sin(ω ) jω 1 + jω 5 1 1 5 jω 1 + jω = j H( jω ) = = jω − sin(ω ) 2 jω 1 + jω 2 ω 8 sinc π ω
5 5 2 H( jω ) = jω = 2 jω (1 + jω ) 1 + jω
5 h( t) = e − t u( t) 2 42. Sketch the magnitudes and phases of the CTFT’s of the following functions. (a)
g( t) = 5δ ( 4 t) F 5δ ( 4 t) ← →
(b)
t − 3 t + 1 g( t) = comb − comb 4 4
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5 4
M. J. Roberts - 7/12/03
t − 3 F t + 1 j 2πf comb − 4 comb( 4 f )e − j 6πf ←→ 4 comb( 4 f )e − comb 4 4 t − 3 F t + 1 − j 2πf comb ←→ 4 comb( 4 f )e − comb (e j 4πf − e − j 4πf ) 4 4 t − 3 F t + 1 − j 2πf comb sin( 4πf ) comb( 4 f ) = 0 ←→ j 8e − comb 4 4 t − 3 t + 1 Also, g( t) = comb =0 − comb 4 4 (a)
(b)
|G( f )|
|G( f )|
2
2
-1
1
f
-1
Phase of G( f )
π
-1
1
f
-1
-π
1
f
-π
g( t) = u(2 t) + u( t − 1) F u(2 t) + u( t − 1) ← →
1 2
1 1 1 − j 2πf 1 + δ( f ) + e + δ( f ) f 2 j 2πf 2 j 2π 2
F u(2 t) + u( t − 1) ← →
(d)
f
Phase of G( f )
π
(c)
1
1 1 + e − j 2πf ) + δ ( f ) ( jπf
g( t) = sgn( t) − sgn(− t) F sgn( t) − sgn(− t) ← →
5-59
1 1 1 2 − = jπf −1 jπ (− f ) jπf
M. J. Roberts - 7/12/03
(c)
(d)
|G( f )|
|G( f )|
4
8
-2
2
f
-2
Phase of G( f )
f
Phase of G( f )
π
π
-2
2
f
-1
-π
(e)
2
1
f
-π
t − 1 t + 1 g( t) = rect + rect 2 2 t − 1 F t + 1 j 2πf rect + 2 sinc(2 f )e − j 2πf ←→ 2 sinc(2 f )e + rect 2 2 t − 1 F t + 1 rect ←→ 4 sinc(2 f ) cos(2πf ) + rect 2 2
(f)
t g( t) = rect 4 t F rect ← → 4 sinc( 4 f ) 4 (e)
(f)
|G( f )|
|G( f )|
4
-1
4
1
f
-1
Phase of G( f )
f
Phase of G( f )
π
-1
1
π
1
f
-π
-1
1 -π
5-60
f
M. J. Roberts - 7/12/03
(g)
t t g( t) = 5 tri − 2 tri 5 2 t t F 5 tri − 2 tri ← → 25 sinc 2 (5 f ) − 4 sinc 2 (2 f ) 5 2
(h)
3 t t g( t) = rect ∗ rect 8 2 2 3 t t F rect ∗ rect ← → 24 sinc(8 f ) sinc(2 f ) 8 2 2 (g)
(h)
|G( f )|
|G( f )|
25
-0.4
24
0.4
f
-0.5
Phase of G( f )
0.5
Phase of G( f )
π
-0.4
f
π
0.4
f
-0.5
-π
0.5
f
-π
43. Sketch the magnitudes and phases of the CTFT’s of the following functions. (a) rect ( 4t)
1 f F rect ( 4 t) ← → sinc 4 4 1 ω F rect ( 4 t) ← → sinc 8π 4
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M. J. Roberts - 7/12/03
| 41 sinc( 4f ) | 1 4 -4
f
4
f 1 4 sinc 4
( )
rect(4t)
π
1 -1 8
(b) rect(4 t) ∗ 4δ ( t)
-4
t
1 8
f
4 −π
1 f f F rect( 4 t) ∗ 4δ ( t) ← → sinc × 4 = sinc 4 4 4 ω F rect( 4 t) ∗ 4δ ( t) ← → sinc 8π f
| sinc( 4 ) | 1 -4
4 f sinc 4
f
( )
rect(4t)∗4δ(t)
π
4 -1 8
1 8
t
-4
4
f
−π
(c) rect(4 t) ∗ 4δ ( t − 2)
1 f f F rect( 4 t) ∗ 4δ ( t − 2) ← → sinc × 4 e − j 4 πf = sinc e − j 4 πf 4 4 4 ω F rect( 4 t) ∗ 4δ ( t − 2) ← → sinc e − j 2ω 8π
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M. J. Roberts - 7/12/03
| sinc( 4f ) e
-j4πf
|
1 -4
f
4
f sinc 4
( )e
-j4πf
rect(4t)∗4δ(t-2) 16π
4
π
2- 81
f
t
2+ 81
2
4
-4
(d) rect ( 4 t) ∗ 4δ (2 t)
1 1 f f F rect ( 4 t) ∗ 4δ (2 t) ← → sinc × 2 = sinc 4 2 4 4 1 ω F rect ( 4 t) ∗ 4δ (2 t) ← → = sinc 8π 2
|
( 4f ) |
1 sinc 2 1 2
-4
4
f
( 4f )
1 sinc 2
rect(4t)∗4δ(2t)
π
2 -1 8
1 8
t
-4
4
f
−π
(e) rect ( 4t) ∗ comb( t)
∞ 1 1 f f F rect ( 4 t) ∗ comb( t) ← → sinc comb( f ) = sinc ∑ δ ( f − k ) 4 4 k =−∞ 4 4
F rect ( 4 t) ∗ comb( t) ← →
F rect ( 4 t) ∗ comb( t) ← →
1 ∞ k sinc δ ( f − k ) ∑ 4 4 k =−∞
∞ 1 ∞ k ω π k k sinc δ − = sinc δ (ω − 2πk ) ∑ ∑ 4 2π 2 k =−∞ 4 4 k =−∞
5-63
M. J. Roberts - 7/12/03 f
| 41 sinc( 4 ) comb( f )| 1 4 -4
f
4
f 1 4 sinc 4
( ) comb( f )
rect(4t)∗comb(t) 1
... -2
-1
-1 1 88
π ...
1
f
4
-4
t
−π
(f) rect ( 4 t) ∗ comb( t − 1)
∞ 1 1 f f F rect ( 4 t) ∗ comb( t − 1) ← → sinc comb( f )e − j 2πf = sinc e − j 2πf ∑ δ ( f − k ) 4 4 4 4 k =−∞
F rect ( 4 t) ∗ comb( t − 1) ← →
1 ∞ 1 ∞ k − j 2πk k sinc e δ f − k = sinc δ ( f − k ) ( ) ∑ ∑ 1 2 3 4 =1 4 4 k =−∞ 4 k =−∞
π ∞ k rect ( 4 t) ∗ comb( t) ←→ ∑ sinc δ (ω − 2πk ) 4 2 k =−∞ F
Same as part (e). f
| 41 sinc( 4 ) comb( f )| 1 4 -4
f
4
f 1 4 sinc 4
( ) comb( f )
rect(4t)∗comb(t-1) 1
... -2
-1
-1 1 88
π ...
1
-4
t
4
f
−π
(g) rect ( 4 t) ∗ comb(2 t)
∞ 1 f1 f 1 f f F rect ( 4 t) ∗ comb(2 t) ← → sinc comb = sinc ∑ δ − k 4 4 2 2 8 4 k =−∞ 2
∞ 1 ∞ 1 f k F rect ( 4 t) ∗ comb(2 t) ← → sinc ∑ δ ( f − 2 k ) = ∑ sinc δ ( f − 2 k ) 4 k =−∞ 2 4 k =−∞ 4
5-64
M. J. Roberts - 7/12/03
∞ 1 ∞ k ω π k rect ( 4 t) ∗ comb(2 t) ←→ ∑ sinc δ − 2 k = ∑ sinc δ (ω − 4πk ) 2 2π 2 k =−∞ 2 4 k =−∞ F
f
| 81 sinc( 4 ) comb( 2f )| 1 4 -4
f
4
f 1 8 sinc 4
( ) comb( 2f )
rect(4t)∗comb(2t) 1 2
... -2
-1
π
...
-1 1 88
1
f
4
-4
t −π
(h) rect ( t) ∗ comb(2 t)
∞ 1 f 1 f F rect ( t) ∗ comb(2 t) ← → sinc( f ) × comb = sinc( f ) ∑ δ − k 2 2 2 2 k =−∞ ∞
F rect ( t) ∗ comb(2 t) ← → sinc( f ) ∑ δ ( f − 2 k ) = k =−∞
∞
∑ sinc(2k )δ ( f − 2k ) = δ ( f )
k =−∞
ω F rect ( t) ∗ comb(2 t) ← → δ = 2πδ (ω ) 2π |δ( f )| 1 f rect(t)∗comb(2t) 1
... -2
-1
δ( f ) ... t
1
f
44. Plot these signals over two periods centered at t = 0 . (a)
x( t) = 2 cos(20πt) + 4 sin(10πt) + 3 cos(−20πt) − 3 sin(−10πt)
(b)
x( t) = 5 cos(20πt) + 7 sin(10πt)
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M. J. Roberts - 7/12/03
(a) x(t) 10
1
t
-10
(b) x(t) 10
1 -10
Compare the results of parts (a) and (b).
5-66
t
M. J. Roberts - 7/12/03
45. A periodic signal has a period of four seconds. (a) zero?
What is the lowest positive frequency at which its CTFT could be non-
1 Hz 4 (b)
What is the next-lowest positive frequency at which its CTFT could be
nonzero?
1 Hz 2 46. Sketch the magnitude and phase of the CTFT of each of the following signals (ω form): (a)
t x( t) = 0.1rect 40
X( f ) = 4 sinc( 40 f ) 20ω X( jω ) = 4 sinc π π )| | 4sinc( 20ω 4 2π 40
2π 40
ω
4sinc(40 f ) π
x(t) 0.1
(b)
t + 5 x( t) = 3 rect 10
−20
t
20
X( f ) = 30 sinc(10 f )e + j10πf
5-67
2π 40
2π 40 -π
ω
M. J. Roberts - 7/12/03
5ω X( jω ) = 30 sinc e + j 5ω π
| 30sinc( 5ωπ ) e | j5ω
30 2π 10
ω
2π 10
j5ω 30sinc( 5ω π )e
2π π
x(t) 3
t
−10
(c)
x( t) =
ω
7 t comb 5 5
X( f ) = 7 comb(5 f ) ∞ 5ω 5ω 14π X( jω ) = 7 comb = 7 ∑ δ − k = 2π 2π 5 k =−∞
∞
∑ δ ω −
k =−∞
2πk 5
|X(jω)|
14π 5 ...
...
4π 2π 5 5
... −10 −5
(d)
x( t) =
ω
X(jω)
x(t) 7
2π 4π 5 5
...
... 5 10
...
4π 2π 5 5
t
2π 4π 5 5
ω
7 t + 3 t − 2 7 comb = comb 5 5 5 5
X( f ) = 7 comb(5 f )e − j 4 πf = 7 comb(5 f )e + j 6πf ∞ 14π 5ω − j 2ω 5ω X( jω ) = 7 comb e = 7 ∑ δ − k e − j 2ω = 2π 2π 5 k =−∞
5-68
2πk − j ∑ δ ω − 5 e k =−∞ ∞
4 πk 5
M. J. Roberts - 7/12/03
∞ 14π 5ω + j 3ω 5ω X( jω ) = 7 comb e = 7 ∑ δ − k e + j 3ω = 2π 2π 5 k =−∞
2πk + j δ ω − e ∑ 5 k =−∞ ∞
|X(jω)|
14π 5 ...
...
4π 2π 5 5 x(t) ... −13 −8
−3
X(jω) ...
...
7
ω
2π 4π 5 5
... 2 7
t
6π 5 4π 5
...
ω
...
47. Sketch the inverse CTFT’s of the following functions: (a)
f X( f ) = 20 rect 8
x( t) = 160 sinc(8 t)
|X( f )| 20 -4
4
f
160sinc(8t) 160
X( f )
4
-4
(b)
f 5 8
π f j8f X( f ) = 20 rect e 8 |X( f )| 20
-4
4
4 8
3 8
2 8
1 8
1 8
π 2
4
3 8
4 8
t
5 8
1 x( t) = 160 sinc 8 t + 16
1 160sinc[8(t + 16 )]
f
160
X( f ) π 2 -4
2 8
f
5 8
4 8
3 8
2 8
1 8
1 8 1 16
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2 8
3 8
4 8
5 8
t
6πk 5
M. J. Roberts - 7/12/03
(c)
X( f ) = 2[δ ( f + 5) + δ ( f − 5)]e
j
πf 5
[
]
X( f ) = 2 δ ( f + 5)e − jπ + δ ( f − 5)e jπ = −2[δ ( f + 5) + δ ( f − 5)]
x( t) = −4 cos(10πt) | X( f ) | 2
−5
f
5
-4cos(10πt)
X( f )
4 4 5
π −5
(d)
-π
3 5
2 5
1 5
1 5
2 5
3 5
4 5
f
5
t
X( f ) = 8δ ( f ) + 5δ ( f − 5) + 5δ ( f + 5)
x( t) = 8 + 10 cos(10πt) | X( f ) | 8 5
8+10cos(10πt) −5
f
5
18
X( f ) −5
2 5
f
5
1 5
-2
1 5
2 5
t
48. Find the inverse CTFT of this real, frequency-domain function (Figure E48) and sketch it. (Let A = 1, f1 = 95 kHz and f 2 = 105 kHz .) f − f0 f + f0 X( f ) = A rect + rect ∆f ∆f
[
x( t) = A ∆f sinc( ∆f t)e + j 2πf 0 t + ∆f sinc( ∆f t)e − j 2πf 0 t
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]
M. J. Roberts - 7/12/03
[
]
x( t) = A∆f sinc( ∆f t) e + j 2πf 0 t + e − j 2πf 0 t = 2 A∆f sinc( ∆f t) cos(2πf 0 t) x( t) = 20, 000 sinc(10, 000 t) cos(2 × 10 5 πt) X( f ) Α -f2
-f 1
f
f1 f2
Figure E48 A real frequency-domain function x(t) x 10
4
2 1 0 -5
-4
-3
-2
-1
0
1
2
3
4
-1
5 -4 x 10
t
-2
49. Find the CTFT (either form) of this signal (Figure E49) and sketch its magnitude and phase versus frequency on separate graphs. (Let A = − B = 1 and let t1 = 1 and t2 = 2 .) Hint: Express this signal as the sum of two functions and use the linearity property.
t t x( t) = A rect − ( A − B) rect 2 t2 2 t1
X( f ) = 2 t2 A sinc(2 t2 f ) − 2 t1 ( A − B) sinc(2 t1 f ) X( f ) = 4 sinc( 4 f ) − 4 sinc(2 f )
x(t) A
- t2 - t1
t1 t2 B
Figure E49 A CT function
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t
M. J. Roberts - 7/12/03 |X( f )| 3
-4
4
f
Phase of X( f ) π
-4
4
f
-π
50. In many communication systems a device called a “mixer” is used. In its simplest form a mixer is simply an analog multiplier. That is, its response signal, y(t), is the product of its two excitation signals. If the two excitation signals are
x1 ( t) = 10 sinc(20 t)
x 2 ( t) = 5 cos(2000πt)
and
plot the magnitude of the CTFT of y(t), Y( f ) , and compare it to the magnitude of the CTFT of x1 ( t) . In simple terms what does a mixer do?
y( t) = x1 ( t) x 2 ( t) = 50 sinc(20 t) cos(2000πt) Y( f ) = X1 ( f ) ∗ X 2 ( f ) =
Y( f ) =
5 f rect ∗ [δ ( f − 1000) + δ ( f + 1000)] 20 4
5 f + 1000 f − 1000 rect + rect 20 4 20
|X1( f )|
|Y( f )| 5 4
1 2 -10 10
f
-1010 -990
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990 1010
f
M. J. Roberts - 7/12/03
51. Sketch a graph of the convolution of the two functions in each case: rect(t) * rect(t) 1 −1
(a) rect ( t) ∗ rect ( t)
t
1
rect(t- 12 ) * rect(t+ 12 ) 1 −1
1 1 (b) rect t − ∗ rect t + 2 2
(c)
t
1
tri( t) ∗ tri( t − 1)
1
tri(t) -1
tri(t-1) 1
1
tri(t-τ) t-1 -1 t
t+1 1
2
t
tri(τ-1) 2
τ
For t < -1, the non-zero portions of the two functions do not overlap and the convolution is zero. For t > 3, the non-zero portions of the two functions do not overlap and the convolution is zero. For -1 < t < 0: The non-zero portions overlap for 0 < τ < t+1 and, in that range of τ,
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M. J. Roberts - 7/12/03
tri( t − τ ) = t + 1 − τ and tri(τ − 1) = τ Therefore, for -1 < t < 0,
tri( t) ∗ tri( t − 1) =
t +1
∫ (t + 1 − τ )τdτ = 0
t +1
t +1
∫ [(t + 1)τ − τ ]dτ 2
0
τ2 τ3 (t + 1) (t + 1) (t + 1) tri( t) ∗ tri( t − 1) = ( t + 1) − = − = 2 3 0 2 3 6 3
3
3
For 0 < t < 1:
tri(t-τ) 1
τ
t 1 t+1 2
-1 t-1 tri( t) ∗ tri( t − 1) =
tri(τ-1)
∞
∫ tri(t − τ ) tri(τ )dτ
−∞
The non-zero portions overlap for 0 < τ < t+2 and, in that range of τ, there are three cases to consider, 0 < τ < t, t < τ < 1 and 1 < τ < t+1. Therefore t
1
t +1
0
t
1
tri( t) ∗ tri( t − 1) = ∫ tri( t − τ ) tri(τ ) dτ + ∫ tri( t − τ ) tri(τ ) dτ + Case 1:0 < τ < t
Case 2:t < τ < 1
Case 3:1 < τ < t+1
∫ tri(t − τ ) tri(τ )dτ
tri( t − τ ) = 1 − t + τ and tri(τ − 1) = τ
tri( t − τ ) = 1 + t − τ and tri(τ − 1) = τ
tri( t − τ ) = 1 + t − τ and tri(τ − 1) = 2 − τ
Therefore
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M. J. Roberts - 7/12/03 t
1
t +1
0
t
1
tri( t) ∗ tri( t − 1) = ∫ (1 − t + τ )τdτ + ∫ (1 + t − τ )τdτ + t
[
1
]
[
]
tri( t) ∗ tri( t − 1) = ∫ (1 − t)τ + τ dτ + ∫ (1 + t)τ − τ dτ + 2
0
2
t
t
∫ (1 + t − τ )(2 − τ )dτ
t +1
∫ [2(1 + t) − 2τ − (1 + t)τ + τ ]dτ 2
1
t +1
1
τ2 τ3 τ2 τ3 τ2 τ3 tri( t) ∗ tri( t − 1) = (1 − t) + + (1 + t) − + 2(1 + t)τ − τ 2 − (1 + t) + 2 3 0 2 3 t 2 3 1
t2 t3 t2 t3 1 1 tri( t) ∗ tri( t − 1) = (1 − t) + + (1 + t) − − (1 + t) + 2 3 2 3 2 3 1 1 (t + 1) 2 (t + 1) 3 2 2 + − 2(1 + t) + 1 + (1 + t) − + 2(1 + t) − ( t + 1) − (1 + t) 2 3 2 3
t2 1 t 2 2t 3 (t + 1) 2 tri( t) ∗ tri( t − 1) = (1 − t) + − (1 + t)1 + + + (1 + t) − 6 2 3 2 3
3
t3 t2 t 1 tri( t) ∗ tri( t − 1) = − + + + 2 2 2 6 For the remaining regions of t, the convolution simply repeats with even symmetry about the point, t = 1. The analytical solutions can be found by the following successive changes of variable: t → t + 1 , t → −t , t → t − 1
These three successive changes of variable can be condensed into one,
t → −t + 2 Then, for 1 < t < 2,
(2 − t) 3 (2 − t) 2 (2 − t) 1 t3 t2 t 1 tri( t) ∗ tri( t − 1) = − + + + = − + + + 2 2 2 6 2 2 2 6 t →− t + 2
and, for 2 < t < 3,
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M. J. Roberts - 7/12/03
( t + 1) 3 [(− t + 2) + 1] tri( t) ∗ tri( t − 1) = = 6 6 t →− t + 2
3
tri(t) * tri(t -1) 0.8
-5
5
t
3δ(t)∗10cos(t)
(d) 3δ ( t) ∗ 10 cos( t)
30 2π
t
(e) 10comb( t) ∗ rect ( t)
The constant, 10.
(f) 5comb( t) ∗ tri( t)
The constant, 10.
52. In electronics, one of the first circuits studied is the rectifier. There are two forms, the half-wave rectifier and the full-wave rectifier. The half-wave rectifier cuts off half of an excitation sinusoid and leaves the other half intact. The full-wave rectifier reverses the polarity of half of the excitation sinusoid and leaves the other half intact. Let the excitation sinusoid be a typical household voltage, 120 Vrms at 60 Hz, and let both types of rectifiers alter the negative half of the sinusoid while leaving the positive half unchanged. Find and plot the magnitudes of the CTFT’s of the responses of both types of rectifiers (either form). Half-Wave Case:
x( t) = 120 2 cos(120πt)[rect (120 t) ∗ 60 comb(60 t)]
1 f f X( f ) = 60 2 [δ ( f − 60) + δ ( f + 60)] ∗ sinc comb 120 60 120 X( f ) =
2 f f δ ( f − 60) + δ ( f + 60)] ∗ sinc comb [ 120 60 2
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M. J. Roberts - 7/12/03
∞ 2 k f X( f ) = δ ( f − 60) + δ ( f + 60)] ∗ ∑ sinc δ − k [ 2 60 2 k =−∞
X( f ) = 30 2
∞
k =−∞
X( f ) = 30 2
Full-Wave Case:
k
∑ sinc 2 [δ ( f − 60) + δ ( f + 60)] ∗ δ ( f − 60k ) ∞
k
∑ sinc 2 [δ ( f − 60 − 60k ) + δ ( f + 60 − 60k )]
k =−∞
x( t) = 120 2 cos(120πt)[2 rect (120 t) ∗ 60 comb(60 t) − 1]
1 f f X( f ) = 60 2 [δ ( f − 60) + δ ( f + 60)] ∗ sinc comb − δ ( f ) 120 60 60 X( f ) =
2 f f δ ( f − 60) + δ ( f + 60)] ∗ sinc comb − δ ( f ) [ 120 60 2
∞ k X( f ) = 60 2 −δ ( f − 60) − δ ( f + 60) + ∑ sinc [δ ( f − 60 − 60 k ) + δ ( f + 60 − 60 k )] 2 k =−∞
|X( f )|
|X( f )| 30 2
60 2
f
f
60
120
53. Find the DTFT of each of these signals: n
(a)
1 x[ n ] = u[ n − 1] 3 X( jΩ) =
∞
∑
n =−∞
x[ n ]e − jΩn =
∞
∞ 1 1 − jΩn 1 u n − e = [ ] ∑ 3 ∑1 3 e − jΩn n =−∞ n= n
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n
M. J. Roberts - 7/12/03 ∞
1 X( jΩ) = ∑ m =0 3 e − jΩ X( jΩ) = 3
m +1
e
− jΩ( m +1)
m
e − jΩ = ∑ m =0 3
e − jΩ e − jΩ ∑ = 3 m =0 3 ∞
∞
m +1
e − jΩ 1 = e − jΩ 3 − e − jΩ 1− 3
Alternate Solution: n
n
1 1 x[ n ] = u[ n − 1] = u[ n ] − δ [ n ] 3 3 Using F α n u[ n ] ← →
1 1 − αe − jΩ
and
n
1 1 x[ n ] = u[ n ] − δ [ n ] = 3 3
F δ [ n ] ← →1
1 −1 e − jΩ 1− 3
e − jΩ e − jΩ 1 − 1 − 3 e − jΩ 3 x[ n ] = = = e − jΩ e − jΩ 3 − e − jΩ 1− 1− 3 3 Second Alternate Solution:
1 1 x[ n ] = 3 3 X( jΩ) =
n −1
u[ n − 1]
e − jΩ 1 1 e − jΩ = 3 − e − jΩ 3 1 − 1 e − jΩ 3
π 1 x[ n ] = sin n u[ n − 2] 4 4 n
(b)
x[ n ] =
e
j
π n 4
−e j2
−j
π n 4
j π n − j π n 1 e 4 e 4 1 − u[ n − 2] u[ n − 2] = 4 j 2 4 4 n
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j π 2 j π n − 2 − j π 2 − j π n − 2 1 e 4 e 4 e 4 e 4 − x[ n ] = u[ n − 2] 4 4 j2 4 4 n
j π4 F e u[ n ] ← → 4
j π4 e 4
1 1−
e
j
π 4
4
e − jΩ
n −2
e − j 2Ω
F u[ n − 2] ← →
1−
e
j
π 4
4
e − jΩ
2 π 2 − j j π4 − j 2Ω 4 1 e e − j 2Ω e e − X( jΩ) = π π −j j 4 j2 4 e 4 − jΩ e 4 − jΩ 1− 1− e e 4 4 π π 2 π jπ 2 −j −j j 4 4 4 4 e e e 1 − e − jΩ − j Ω 1 − − e e 4 4 4 4 e − j 2Ω X( jΩ) = π π j −j j2 4 4 e e 1 − e − jΩ 1 − e − jΩ 4 4
2
2
2
2
− j π4 − j π4 j π4 j π4 j π4 − j π4 e e e e − e e + e − jΩ − e − jΩ 4 4 4 4 4 4 e − j 2Ω X( jΩ) = π π −j j4 j2 4 e e e − jΩ + e − j 2Ω + 1− 4 4
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M. J. Roberts - 7/12/03
j
π 2
−j
π 2
j
π 4
−j
π 4
e e e e − jΩ − jΩ e − j 2Ω 16 − 16 − 64 e + 64 e X( jΩ) = 1 π j2 1 − cos e − jΩ + e − j 2Ω 2 4 π −j j π4 j e − e 4 − jΩ − e 8 64 − j 2Ω e X( jΩ) = 1 π j2 1 − cos e − jΩ + e − j 2Ω 2 4
1 π − jΩ sin e e 4 4 X( jΩ) = 1 π 16 1 − cos e − jΩ + e − j 2Ω 2 4 1−
− j 2Ω
(c)
2π ( n − 4 ) 2πn x[ n ] = sinc ∗ sinc 8 8 Using n F → wrect ( wF ) ∗ comb( F ) sinc ← w
8 8 X( F ) = rect F ∗ comb( F ) rect F ∗ comb( F ) e − j 8πF 2π 2π 8 X( F ) = rect F ∗ comb( F ) e − j 8πF 2π 2π ( n − 4 ) x[ n ] = sinc 8 (d)
Using
2πn x[ n ] = sinc 2 8
n F → wrect ( wF ) ∗ comb( F ) sinc ← w
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M. J. Roberts - 7/12/03
8 8 8 8 X( F ) = rect F ∗ comb( F ) ⊗ rect F ∗ comb( F ) 2π 2π 2π 2π 8 X( F ) = 2π
2
8 rect 2π
8 F ∗ comb( F ) ⊗ rect 2π
F ∗ comb( F )
2
8 8 8 X( F ) = rect F ∗ rect F ∗ comb( F ) 2π 2π 2π 8 8 8 2π 8 X( F ) = tri F ∗ comb( F ) = tri F ∗ comb( F ) 2π 8 2π 2π 2π 2
54. Sketch the magnitudes and phases of the DTFT’s of the following functions: (a) rect 2 [ n ] F Using rect N w [ n ] ← →(2 N w + 1) drcl( F , 2 N w + 1) ,
F rect 2 [ n ] ← → 5 drcl( F , 5)
|X( F )| 5
-1
1
F
Phase of X( F ) π
-1
1
F
-π
(b) rect 2 [ n ] ∗ (−5δ [ n ]) F F Using δ [ n ] ← →1 and x[ n ] ∗ y[ n ] ← → X( F ) Y( F ) F rect 2 [ n ] ∗ (−5δ [ n ]) ← → 5 drcl( F , 5) × (−5) = −25 drcl( F , 5)
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M. J. Roberts - 7/12/03
|X( F )| 25
-1
1
F
Phase of X( F ) π
-1
1 -π
(c) rect 2 [ n ] ∗ 3δ [ n + 3] F Using x[ n − n 0 ] ← → e − j 2πFn 0 X( F )
F rect 2 [ n ] ∗ 3δ [ n + 3] ← →15 drcl( F , 5)e j 6πF
|X( F )| 15
-1
1
F
Phase of X( F ) π
-1
1 -π
(d) rect 2 [ n ] ∗ (−5δ [ 4 n ]) = rect 2 [ n ] ∗ (−5δ [ n ]) F rect 2 [ n ] ∗ (−5δ [ 4 n ]) ← → −25 drcl( F , 5)
5-82
F
F
M. J. Roberts - 7/12/03 |X( F )| 25
-1
1
F
Phase of X( F ) π
-1
1
F
-π
(e) rect 2 [ n ] ∗ comb 8 [ n ] F Using comb N 0 [ n ] ← → comb( N 0 F ) ,
F rect 2 [ n ] ∗ comb 8 [ n ] ← → 5 drcl( F , 5) comb(8 F )
F rect 2 [ n ] ∗ comb 8 [ n ] ← → 5 drcl( F , 5)
∞
∑ δ (8F − m)
m =−∞
m 5 ∞ rect 2 [ n ] ∗ comb 8 [ n ] ←→ ∑ drcl( F , 5)δ F − 8 m =−∞ 8 F
F rect 2 [ n ] ∗ comb 8 [ n ] ← →
m 5 ∞ m drcl , 5 δ F − ∑ 8 8 m =−∞ 8
|X( F )| 0.625
-1
1
F
Phase of X( F ) π
-1
1 -π
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M. J. Roberts - 7/12/03
(f) rect 2 [ n ] ∗ comb 8 [ n − 3] F Using the result of (e) and x[ n − n 0 ] ← → e − j 2πFn 0 X( F ) ,
∞ m 5 m F rect 2 [ n ] ∗ comb 8 [ n − 3] ← → e − j 6πF ∑ drcl , 5 δ F − 8 8 8 m =−∞
|X( F )| 0.625
-1
1
F
Phase of X( F ) π
-1
1
F
-π
(g) rect 2 [ n ] ∗ comb 8 [2 n ] = rect 2 [ n ] ∗
rect 2 [ n ] ∗ comb 8 [2 n ] = rect 2 [ n ] ∗
∞
∑
m =−∞
δ [2 n − 8 m] = rect 2 [ n ] ∗
∞
∑ δ[2(n − 4 m)]
m =−∞
∞
∑ δ[n − 4 m] = rect [n] ∗ comb [n] 2
m =−∞
Therefore F rect 2 [ n ] ∗ comb 8 [2 n ] ← → 5 drcl( F , 5) comb( 4 F )
F rect 2 [ n ] ∗ comb 8 [2 n ] ← → 5 drcl( F , 5)
∞
∑ δ ( 4 F − m)
m =−∞
m 5 ∞ rect 2 [ n ] ∗ comb 8 [2 n ] ←→ ∑ drcl( F , 5)δ F − 4 m =−∞ 4 F
F rect 2 [ n ] ∗ comb 8 [2 n ] ← →
m 5 ∞ m drcl , 5 δ F − ∑ 4 4 m =−∞ 4
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M. J. Roberts - 7/12/03
|X( F )| 1.25
-1
1
F
Phase of X( F ) π
-1
1
F
-π
(h) rect 2 [ n ] ∗ comb 5 [ n ] F rect 2 [ n ] ∗ comb 5 [ n ] ← → 5 drcl( F , 5) comb(5 F )
F rect 2 [ n ] ∗ comb 5 [ n ] ← → 5 drcl( F , 5)
∞
∑ δ (5F − m)
m =−∞
m 5 ∞ rect 2 [ n ] ∗ comb 5 [ n ] ←→ ∑ drcl( F , 5)δ F − 5 m =−∞ 5 F
F rect 2 [ n ] ∗ comb 5 [ n ] ← →
∞
m
m
∑ drcl 5 ,5 δ F − 5
m =−∞
n Then, using drcl , 2 m + 1 = comb 2 m +1[ n ] 2m + 1
rect 2 [ n ] ∗ comb 5 [ n ] ←→ F
∞
m
∑ comb [m]δ F − 5
m =−∞
5
F rect 2 [ n ] ∗ comb 5 [ n ] ← → comb( F ) F and, since 1← → comb( F ) , that implies that rect 2 [ n ] ∗ comb 5 [ n ] = 1.
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M. J. Roberts - 7/12/03 |X( F )| 1
-1
1
F
Phase of X( F ) π
-1
1
F
-π
55. Sketch the inverse DTFT’s of these functions. 1 (a) X( F ) = comb( F ) − comb F − 2 F F Using 1← → X( F − F0 ) → comb( F ) and e j 2πF0 n x[ n ] ←
1 F → comb( F ) − comb F − 1 − e jπn ← 2
e
j
πn 2
πn +j − j π2n 1 F 2 e − e ←→ comb( F ) − comb F − 2
− j 2e
−2e
j
πn 2
1 πn F sin ← → comb( F ) − comb F − 2 2
π j ( n +1) 2
1 πn F sin ← → comb( F ) − comb F − 2 2
π 1 π πn F → comb( F ) − comb F − −2 cos ( n + 1) + j sin ( n + 1) sin ← 2 2 2 2 π 1 πn π πn F −2 cos ( n + 1) sin + j sin ( n + 1) sin ←→ comb( F ) − comb F − 2 2 2 2 2 144 244 3 1444 424444 3 =0 πn − sin 2 5-86
M. J. Roberts - 7/12/03
1 πn F 2 sin 2 ← → comb( F ) − comb F − 2 2 x[n] 2
-12
12
n
1 1 (b) X( F ) = j comb F + − j comb F − 8 8 F F Using 1← → X( F − F0 ) → comb( F ) and e j 2πF0 n x[ n ] ←
je
−j
πn 4
− je
j
πn 4
1 1 F ← → j comb F + − j comb F − 8 8
1 1 πn F − j j 2 sin ← → j comb F + − j comb F − 4 8 8 1 1 πn F 2 sin ← → j comb F + − j comb F − 4 8 8 x[n] 2
-12
12
n
-2
1 1 (c) X( F ) = sinc10 F − + sinc10 F + ∗ comb( F ) 4 4 Using
T πt T t sinc ∗ f 0 comb( f 0 t) = wf 0 cos + 0 − 1 drcl f 0 t, 0 − 1 w w w w
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from Appendix A (because
X( F ) =
T0 is an integer), 2w
1 1 1 1 1 cos10π F − + 9 drcl F − , 9 + cos10π F + + 9 drcl F + , 9 4 4 4 4 10
∫ cos(10πF )e
j 2πFn
1
F dF ← → cos(10πF )
1 F e j10πF + e − j10πF )e j 2πFn dF ← → cos(10πF ) ( ∫ 2 1 1 2
[
[∫ e
j 2πF ( n + 5)
1
]
F dF + ∫ e j 2πF ( n − 5) dF ← → cos(10πF ) 1
]
1 ∫1 cos(2πF ( n + 5)) + j sin(2πF ( n + 5)) dF F ←→ cos(10πF ) 2 + cos(2πF ( n − 5)) + j sin(2πF ( n − 5)) dF ∫1
[
]
These integrals are zero unless n = ±5 . Therefore
1 F δ [ n + 5] + δ [ n − 5]) ← → cos(10πF ) . ( 2 F Then using rect N w [ n ] ← →(2 N w + 1) drcl( F , 2 N w + 1) ,
F rect 4 [ n ] ← → 9 drcl( F , 9)
Combining inverse transforms,
1 F δ [ n + 5] + δ [ n − 5]) + rect 4 [ n ] ← → cos(10πF ) + 9 drcl( F , 9) . ( 2 F Then, using e j 2πF0 n x[ n ] ← → X( F − F0 )
e and
e
−j
j
πn 2
πn 2
1 F (δ [ n + 5] + δ [ n − 5]) + rect 4 [ n ] ←→ cos10π F − 2
1 1 + 9 drcl F − , 9 4 4
1 1 F 1 (δ [ n + 5] + δ [ n − 5]) + rect 4 [ n ] ←→ cos10π F + + 9 drcl F + , 9 . 4 4 2
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M. J. Roberts - 7/12/03
Then, finally 1 1 j π2n 1 − + − 9 10 , 9 cos π F drcl F 5 5 e δ n + + δ n − + rect n [ ] [ ] [ ] ( ) 4 4 4 1 F 1 2 ←→ 10 10 − j π2n 1 1 1 + cos10π F + + 9 drcl F + , 9 (δ [ n + 5] + δ [ n − 5]) + rect 4 [ n ] +e 4 4 2
The impulses on the left side cancel and we get 1 1 cos10π F − + 9 drcl F − , 9 4 4 rect 4 [ n ] πn F 1 cos ←→ 2 5 10 1 1 + cos10π F + + 9 drcl F + , 9 4 4 x[n] 0.2
-12
12
n
-0.2
1 3 5 (d) X( F ) = δ F − + δ F − + δ F − ∗ comb(2 F ) 4 16 16 X( F ) = δ F −
1 3 5 1 1 + δ F − + δ F − ∗ comb( F ) + comb F − 4 16 16 2 2
1 3 5 δ F − + δ F − + δ F − ∗ comb( F ) 4 16 16 1 X( F ) = 2 1 3 5 1 + δ F − + δ F − + δ F − ∗ comb F − 4 16 16 2 1 3 5 comb F − + comb F − + comb F − 4 16 16 1 X( F ) = 2 1 1 3 1 5 1 + comb F − − + comb F − − + comb F − − 4 2 16 2 16 2
5-89
M. J. Roberts - 7/12/03 F F Then, using 1← → X( F − F0 ) we get → comb( F ) and e j 2πF0 n x[ n ] ←
1 3 5 3πn 5πn j j j π2n comb F − + comb F − + comb F − 8 8 +e 4 16 16 1 e + e 1 F ← → π π π n n n 3 11 13 2 j 2 2 j j 1 1 3 1 5 1 + comb F − − + comb F − − + comb F − − +e 8 +e 8 +e 4 2 16 2 16 2 or 1 3 5 3πn 5πn j j j π2n comb F − + comb F − + comb F − 8 8 +e 4 16 16 1 e + e 1 F ← → 5πn 3πn 2 − j π2n 2 −j −j 13 3 11 8 8 + comb F − + comb F − + comb F − +e +e +e 16 4 16 or 1 3 5 comb F − + comb F − + comb F − 4 16 16 πn 3πn 5πn F 1 cos + cos + cos ←→ 2 8 8 2 3 11 13 + comb F − + comb F − + comb F − 4 16 16 x[n] 3
-16
16
n
-3
56. Using the relationship between the CTFT of a signal and the CTFS of a periodic extension of that signal, find the CTFS of
t t 1 x( t) = rect ∗ comb w T0 T0 and compare it with the table entry.
X( f ) = w sinc( wf )
X[ k ] = f 0 X( kf 0 ) = f 0 w sinc( wkf 0 ) =
w w sinc k T0 T0
57. Using the relationship between the DTFT of a signal and the DTFS of a periodic extension of that signal, find the DTFS of
5-90
M. J. Roberts - 7/12/03
rect N w [ n ] ∗ comb N 0 [ n ] and compare it with the table entry.
X( F ) = (2 N w + 1) drcl( F , 2 N w + 1) From the text,
X p [k ] =
( )
1 X kFp . Np
Therefore,
X p [k ] =
k 2N w + 1 drcl , 2 N w + 1 . N0 N0
5-91
M. J. Roberts - 7/12/03
Chapter 6 - Fourier Transform Analysis of Signals and Systems Solutions (In this solution manual, the symbol, ⊗, is used for periodic convolution because the preferred symbol which appears in the text is not in the font selection of the word processor used to create this manual.) 1. A system has an impulse response, h LP ( t) = 3e −10 t u( t) , and another system has an impulse response, h HP ( t) = δ ( t) − 3e −10 t u( t) . (a) Sketch the magnitude and phase of the transfer function of these two systems in a parallel connection. 3 3 H LP ( jω ) = , H HP ( jω ) = 1 − jω + 10 jω + 10 H P ( jω ) =
3 3 + 1− =1 jω + 10 jω + 10 |HP(jω)| 1
-40
40
ω
40
ω
Phase of HP(jω) π
-40 -π
(b) Sketch the magnitude and phase of the transfer function of these two systems in a cascade connection. 3( jω + 10) − 9 3 3 9 3 1− HC ( jω ) = = − 2 = jω + 10 jω + 10 jω + 10 ( jω + 10) ( jω + 10) 2
Solutions 6-1
M. J. Roberts - 7/12/03
HC ( jω ) =
j 3ω + 21
( jω + 10)
2
=3
jω + 7
( jω + 10) 2
|HC(jω)| 0.25
-40
40
ω
40
ω
Phase of HC(jω) π
-40 -π
2. Below are some pairs of signals, x( t) and y( t) . In each case decide whether or not y( t) is a distorted version of x( t) . (a)
(b)
x(t)
x(t)
2
2
2
t
-2
2
t
-2
y(t)
y(t)
2
2
2
t
-2
(a)
Inverted and amplified, undistorted
(b)
Time shifted, undistorted
2 -2
Solutions 6-2
t
M. J. Roberts - 7/12/03
(c)
(d)
x(t)
x(t)
2
1
2
t
-2
2
t
-1
y(t)
y(t)
2
1
2
t
-2
2
t
-1
(c)
Clipped at a negative value, distorted
(d)
Time compressed, distorted (e)
(f)
x(t)
x(t)
2
2
2
t
-2
2 -2
y(t)
y(t)
2
2
2 -2
(e)
Constant added, distorted
(f)
Log-amplified, distorted
t
t
2
t
-2
3. Classify each of these transfer functions as having a lowpass, highpass, bandpass or bandstop frequency response.
Solutions 6-3
M. J. Roberts - 7/12/03
(a)
(b)
|H( f )|
|H(F)|
1
1
-10
10
f
-2
(c)
(d)
|H(jΩ)|
|H(jω)|
1
(a)
Lowpass
-4 π
4π
(b)
Bandpass
Ω
-100
(c)
(f)
100
ω
Lowpass
(e)
(f)
|H(F)|
|H(jΩ)|
(d)
Bandpass
1
-2
Highpass
F
1
1
(e)
2
2
F
-4 π
4π
Ω
Bandstop
4. Classify each of these transfer functions as having a lowpass, highpass, bandpass or bandstop frequency response. (a)
f − 100 H( f ) = 1 − rect 10
Bandstop
(b)
H( F ) = rect (10 F ) ∗ comb( F )
Lowpass
(c)
π π Ω H( jΩ) = rect 20π Ω − + rect 20π Ω + ∗ comb 2π 4 4 Bandpass
5. A system has an impulse response,
t − 0.01 h( t) = 10 rect . 0.02
What is its null bandwidth? The transfer function is
H( f ) = 0.2 sinc(0.02 f )e − j 0.02πf .
Solutions 6-4
M. J. Roberts - 7/12/03
Its first positive-frequency null occurs at f = 50. The null bandwidth is therefore 50. 6. A system has an impulse response, n
7 h[ n ] = u[ n ] . 8 What is its half-power DT-frequency bandwidth? Using F α n u[ n ] ← →
1 1 − αe − jΩ
the transfer function is H( jΩ) = The low-frequency gain is
8 1 . = 7 − jΩ 8 − 7e − jΩ 1− e 8 H(0) = 8
The -3 dB point occurs where H( jΩ−3 dB ) Solving, H( jΩ) = 2
1 7 − jΩ 1− e 1− 8
H( jΩ−3 dB ) = 2
2
82 = = 32 . 2
1 1 = 2 7 jΩ 7 − jΩ 7 jΩ e 1 − (e + e ) + 8 8 8 1
7 7 1 − cos(Ω−3 dB ) + 8 4
2
= 32
2
1 7 64 49 2 1+ − + − 8 32 64 64 64 111 × 4 = 0.991 cos(Ω−3 dB ) = = = 7 7 64 × 7 4 4 Ωhp = 0.1337 ± 2 nπ So the -3 dB DT-frequency bandwidth in radians is 0.1337. In cycles it is 0.0213. 7. Determine whether or not the CT systems with these transfer functions are causal. (a)
H( f ) = sinc( f )
h( t) = rect ( t)
Solutions 6-5
Not Causal
M. J. Roberts - 7/12/03
(b)
H( f ) = sinc( f )e − jπf
1 h( t) = rect t − 2
(c)
H( jω ) = rect (ω )
h( t) =
1 t sinc 2π 2π
(d)
H( jω ) = rect (ω )e − jω
h( t) =
1 t − 1 sinc Not Causal 2π 2π
(e)
H( f ) = A
h( t) = Aδ ( t)
Causal
(f)
H( f ) = Ae j 2πf
h( t) = Aδ ( t + 1)
Not Causal
Causal
Not Causal
8. Determine whether or not the DT systems with these transfer functions are causal. (a)
H( F ) =
sin( 7πF ) sin(πF )
h[ n ] = rect 3 [ n ]
Not Causal
(b)
H( F ) =
sin( 7πF ) − j 2πF e sin(πF )
h[ n ] = rect 3 [ n − 1]
Not Causal
(c)
H( F ) =
sin( 3πF ) − j 2πF e sin(πF )
h[ n ] = rect 2 [ n − 1]
Causal
(d)
H( F ) = rect (10 F ) ∗ comb( F )
h[ n ] =
1 n sinc 10 10
Not Causal
9. Find and sketch the frequency response of each of these circuits given the indicated excitation and response. (a)
Excitation, v i ( t) - Response, v L ( t) R = 10 Ω C = 1 µF + vi (t)
+ L = 1 mH
H( jω ) =
vL(t) -
VL ( jω ) −ω 2 LC jωL = = Vi ( jω ) jωL + 1 + R 1 − ω 2 LC + jωRC jωC
Solutions 6-6
M. J. Roberts - 7/12/03
|H( jω )| 3
ω
-150000
150000
Phase of H( jω ) π
ω
-150000
150000 -π
(b)
Excitation, v i ( t) - Response, iC ( t) R = 1 kΩ + iC(t) C = 1 µF
vi (t) H( jω ) =
IC ( jω ) 1 jωC = = 1 Vi ( jω ) R + jωRC + 1 jωC |H( jω )| 0.001
-1500
1500
ω
1500
ω
Phase of H( jω ) π -1500
-π
(c)
Excitation, v i ( t) - Response, v R ( t)
Solutions 6-7
M. J. Roberts - 7/12/03
+
-
vR(t) R = 1 kΩ
+ vi (t)
C = 1 µF
L = 1 mH
H( jω ) =
VR ( jω ) = Vi ( jω ) R+
R jωL jωC jωL +
=
R(1 − ω 2 LC ) R = jωL R(1 − ω 2 LC ) + jωL R+ 1 − ω 2 LC
1 jωC |H( jω )| 1
ω
-50000
50000
Phase of H( jω ) π
ω
-50000
50000 -π
(d)
Excitation, ii ( t) - Response, v R ( t)
+
ii (t) R = 100 Ω
vR(t)
L = 1 mH C = 1 µF
H( jω ) =
VR ( jω ) I ( jω ) jωL ω 2 LC = R RC =R = −R 1 I i ( jω ) I i ( jω ) 1 − ω 2 LC + jωRC jωL + R + jωC Solutions 6-8
M. J. Roberts - 7/12/03
|H( jω )| 100
ω
-1000000
1000000
Phase of H( jω ) π
ω
-1000000
1000000 -π
10. Classify each of these transfer functions as having a lowpass, highpass, bandpass or bandstop frequency response. (a)
H( f ) =
1 1 + jf
Lowpass
(b)
H( f ) =
jf 1 + jf
Highpass
(c)
H( jω ) = −
(d)
H( F ) =
(e)
H( jΩ) = j[sin(Ω) + sin(2Ω)]
j10ω 100 − ω 2 + j10ω
sin( 3πF ) sin(πF )
Bandpass
Lowpass Bandpass
11. Plot the magnitude frequency responses, both on a linear-magnitude and on a logmagnitude scale, of the systems with these transfer functions, over the frequency range specified. 20 H( f ) = , −100 < f < 100 (a) 2 2 20 − 4π f + j 42πf (b)
H( jω ) =
2 × 10 5 , −500 < ω < 500 (100 + jω )(1700 − ω 2 + j 20ω )
Solutions 6-9
M. J. Roberts - 7/12/03
(a)
(b)
|H( f )|
|H( jω )|
1
2
-100
100
f
-500
ω
500
ω
ln(|H( jω )|)
ln(|H( f )|) -100
500
100
f
-500
-10
-10
12. Draw asymptotic and exact magnitude and phase Bode diagrams for the frequency responses of the following circuits and systems. (a)
An RC lowpass filter with R = 1 MΩ and C = 0.1 µF . 1 1 1 1 jωC H( jω ) = = = = 6 −7 1 + R jωRC + 1 jω10 10 + 1 j 0.1ω + 1 jωC 0
|H(jω)|dB
-10 -20 -30 -40 -50 -1 10
0
10
1
10
ω
2
10
3
10
Phase of H(jω)
0
-0.5
-1
-1.5
-2 -1 10
0
10
1
10
ω
2
10
(b)
Solutions 6-10
3
10
M. J. Roberts - 7/12/03
R = 10 Ω C = 1 µF +
+
vi (t)
vL(t)
L = 1 mH
-
-
From Exercise 9 (a) VL ( jω ) −ω 2 LC jωL H( jω ) = = = Vi ( jω ) jωL + 1 + R 1 − ω 2 LC + jωRC jωC H( jω ) =
−10 −9 ω 2 1 − 10 −9 ω 2 + jω10 −5
20
|H(jω)|dB
0 -20 -40 -60 -80 -100 2 10
3
10
4
10
5
ω
10
6
10
7
10
Phase of H(jω)
3.5 3 2.5 2 1.5 1 0.5 0 2 10
3
10
4
10
5
ω
10
6
10
7
10
Vo ( f ) , of these active filters and identify them as Vi ( f ) lowpass, highpass, bandpass or bandstop.
13. Find the transfer functions, H( f ) =
(a)
Solutions 6-11
M. J. Roberts - 7/12/03
R1
C1
C2
K
+
vi (t)
vx(t)
+
vo(t) R2
-
-
1 Vx ( f ) j 2πfC1 + + G1 − Vi ( f ) j 2πfC1 − Vo ( f )G1 = 0 1 + R2 j 2πfC2 Vx ( f )
R2 1 + R2 j 2πfC2
K = Vo ( f )
j 2πfC2 Vx ( f ) j 2πfC1 + + G1 − Vi ( f ) j 2πfC1 − Vo ( f )G1 = 0 1 + j 2πfR2C2 Vx ( f )
j 2πfR2C2 K = Vo ( f ) 1 + j 2πfR2C2
j 2πfC2 + G1 j 2πfC1 + 1 + j 2πfR2C2 j 2πfKR2C2
Vx ( f ) j 2πfC1 V ( f ) = 0 Vi ( f ) −(1 + j 2πfR2C2 ) o −G1
j 2πfC2 ∆ = − j 2πfC1 + + G1 (1 + j 2πfR2C2 ) + j 2πfKG1R2C2 1 + j 2πfR2C2 ∆ = (2πf ) R2C1C2 − j 2πf [C1 + C2 + G1R2C2 (1 − K )] − G1 2
j 2πfC2 1 j 2πfC1 + + G1 Vo ( f ) = 1 + j 2πfR2C2 ∆ j 2πfKR2C2
j 2πfC1
Vi ( f )
0
V (f) (2πf ) KR2C1C2 H( f ) = o = 2 Vi ( f ) (2πf ) R2C1C2 − j 2πf [C1 + C2 + G1R2C2 (1 − K )] − G1 2
Solutions 6-12
M. J. Roberts - 7/12/03
H( f ) =
(2πf ) 2 KR1R2C1C2
(2πf ) 2 R1R2C1C2 − j 2πf (R1(C1 + C2 ) + R2C2 (1 − K )) − 1 Highpass
(b) C1
R1
R2
K
+
vx(t)
vi (t)
+
vo(t) C2
-
-
1 Vx ( f ) G1 + + j 2πfC1 − Vi ( f )G1 − Vo ( f ) j 2πfC1 = 0 1 R2 + j 2πfC2 1 j 2πfC2 Vx ( f ) K = Vo ( f ) 1 + R2 j 2πfC2 j 2πfC2 Vx ( f ) G1 + + j 2πfC1 − Vi ( f )G1 − Vo ( f ) j 2πfC1 = 0 j 2πfR2C2 + 1 Vx ( f )K − (1 + j 2πfC2 R2 ) Vo ( f ) = 0 j 2πfC2 + j 2πfC1 G1 + j 2πfR2C2 + 1 K
Vx ( f ) G1 V ( f ) = 0 Vi ( f ) −(1 + j 2πfC2 R2 ) o − j 2πfC1
j 2πfC2 ∆ = −(1 + j 2πfC2 R2 ) G1 + + j 2πfC1 + jK 2πfC1 j 2πfR2C2 + 1 ∆ = (2πf ) R2C1C2 − j 2πf (G1R2C2 − C1 (K − 1) + C2 ) − G1 2
j 2πfC2 1 G1 + + j 2πfC1 Vo ( f ) = j 2πfR2C2 + 1 ∆ K
Solutions 6-13
G1 0
Vi ( f )
M. J. Roberts - 7/12/03
H( f ) = H( f ) = −
(2πf ) (2πf )
−KG1
2
R2C1C2 − j 2πf (G1R2C2 − C1 (K − 1) + C2 ) − G1 K
2
R1R2C1C2 − j 2πf ( R1C2 + R2C2 + R1C1 (1 − K )) − 1 Lowpass
14. Show that this system has a highpass frequency response. y(t) x(t)
∫ Writing the differential equation,
[
]
[
]
d −( y( t) − x( t)) = x( t) − −( y( t) − x( t)) dt
or
− or
d d y( t) + x( t) = x( t) + y( t) − x( t) dt dt d d y( t) + y( t) = x( t) dt dt
Fourier transforming both sides, jω Y( jω ) + Y( jω ) = jω X( jω ) or H( jω ) =
Y( jω ) jω = X( jω ) jω + 1
Highpass 15. Draw the block diagram of a system with a bandpass frequency response using two integrators as functional blocks. Then find its transfer function and verify that it has a bandpass frequency response.
∫
x(t)
∫ From Exercise 14, the first stage is characterized by the transfer function, Solutions 6-14
y(t)
M. J. Roberts - 7/12/03
H1 ( jω ) =
jω . jω + 1
The second stage differential equation is y′ ( t) = x( t) − y( t) . Then, Fourier transforming both sides, jω Y( jω ) = X( jω ) − Y( jω ) and H 2 ( jω ) =
1 jω + 1
Therefore the overall system transfer function is H( jω ) =
jω jω 1 = . jω + 1 jω + 1 ( jω + 1) 2 Bandpass
Y( jΩ) , and sketch the frequency response of each X( jΩ) of these DT filters over the range, −4π < Ω < 4π .
16. Find the transfer function, H( jΩ) =
(a) x[n]
y[n]
D y[ n ] = x[ n ] − x[ n − 1] Y( jΩ) = X( jΩ) − e − jΩ X( jΩ) H( jΩ) =
Y( jΩ) = 1 − e − jΩ X( jΩ)
Solutions 6-15
M. J. Roberts - 7/12/03
|H(jΩ)| 2
-2 π
2π
Ω
Phase of H(jΩ) π
-2 π
2π
Ω
-π
(b) x[n]
y[n]
D y[ n ] = x[ n ] − y[ n − 1] Y( jΩ)(1 + e − jΩ ) = X( jΩ) H( jΩ) =
Y( jΩ) 1 = X( jΩ) 1 + e − jΩ |H(jΩ)| 5
-2 π
2π
Ω
Phase of H(jΩ)| π
-2 π
2π
Ω
-π
(c)
Solutions 6-16
M. J. Roberts - 7/12/03
y[n]
x[n]
D
D
y[ n ] = x[ n ] − ( x[ n − 1] + y[ n − 1]) H( jΩ) =
1 − e − jΩ 1 + e − jΩ
|H(jΩ)| 5
-2 π
2π
Ω
Phase of H(jΩ) π
-2 π
2π
Ω
-π
(d)
x[n]
D
y[n]
D y[ n ] = x[ n ] − x[ n − 1] + z[ n ]
z[ n ] = x[ n ] − z[ n − 1]
Y( jΩ) = X( jΩ) − e − jΩ X( jΩ) + Z( jΩ) Z( jΩ) + e − jΩ Z( jΩ) = X( jΩ)
Y( jΩ) = (1 − e − jΩ ) X( jΩ) + Z( jΩ) Z( jΩ) =
X( jΩ) 1 + e − jΩ
Solutions 6-17
M. J. Roberts - 7/12/03
Y( jΩ) = (1 − e − jΩ ) X( jΩ) + H( jΩ) =
X( jΩ) 1 + e − jΩ
Y( jΩ) 2 − e − j 2Ω = X( jΩ) 1 + e − jΩ |H(jΩ)| 5
-2 π
2π
Ω
Phase of H(jΩ)| π
-2 π
2π
Ω
-π
17. Find the minimum stop band attenuation of a moving-average filter with N = 3. Define 1 the stop band as the frequency region, Fc < F < , where Fc is the DT frequency of the 2 first null in the frequency response. H( F ) =
e − jπNF sin(π ( N + 1) F ) N +1 sin(πF )
The first null in the frequency response occurs at
π ( N + 1) F = π ⇒ F =
1 1 = . N +1 4
The biggest magnitude response after that frequency is at the next maximum of H( F ) which occurs at 3π 3 3 π ( N + 1) F = ⇒F= = . 2 2( N + 1) 8 At that DT frequency, − jπ
3 e H = 8 4
9 8
3π 9 − jπ sin 2 e 8 (−1) = = 0.271 or − 11.35 dB 4 0.924 3π sin 8
Solutions 6-18
M. J. Roberts - 7/12/03
18. In the system below, x t ( t) = sinc( t) , f c = 10 and the cutoff frequency of the lowpass filter is 1 Hz. Plot the signals, x t ( t) , y t ( t) , y d ( t) and y f ( t) and the magnitudes and phases of their CTFT’s. yd (t)
yt (t) = x r(t)
x t(t)
yf (t)
LPF
cos(2πfct)
cos(2πfct)
X t ( f ) = rect ( f )
x t ( t) = sinc( t)
Modulation |Xt( f )|
x (t) t
1
1
-2
2
f
Phase of Xt( f ) -4
4
π
t -2
-0.5
y t ( t) = sinc( t) cos(20πt)
2
f
-π
Yt ( f ) = rect ( f ) ∗ Yt ( f ) =
1 [δ ( f − 10) + δ ( f + 10)] 2
1 [rect( f − 10) + rect( f + 10)] 2
Solutions 6-19
M. J. Roberts - 7/12/03
Modulated Carrier |Yt( f )|
yt(t)
0.5
1
-10
-4
4
10
f
Phase of Yt( f )
t
π
-10
10
-1
f
-π
y d ( t) = sinc( t) cos2 (20πt)
Yd ( f ) =
1 rect ( f − 10) 1 δ ( f − 10) ∗ 2 + rect ( f + 10) 2 +δ ( f + 10)
Yd ( f ) =
1 [rect( f − 20) + 2 rect( f ) + rect( f + 20)] 4
Demodulated Carrier |Yd( f )|
yd(t)
0.5
1
-20
20
f
Phase of Yd( f ) -4
4
π
t -20
-0.5
1 y f ( t) = sinc( t) 2
20 -π
Yf ( f ) =
2 rect ( f ) 4
Solutions 6-20
f
M. J. Roberts - 7/12/03
Demodulated and Filtered Carrier |Xf( f )|
xf(t)
0.5
0.5
-2
2
f
Phase of Xf( f ) -4
4
π
t -2
2
-0.25
f
-π
19. In the system below, x t ( t) = sinc(10 t) ∗ comb( t) , m = 1, f c = 100 and the cutoff frequency of the lowpass filter is 10 Hz. Plot the signals, x t ( t) , y t ( t) , y d ( t) and y f ( t) and the magnitudes and phases of their CTFT’s. x t(t)
yd (t)
yt (t) = x r(t)
m
cos(2πfct)
cos(2πfct)
1
x t ( t) = sinc(10 t) ∗ comb( t) = sinc(10 t) ∗
LPF
∞
∞
n =−∞
n =−∞
∑ δ (t − n) = ∑ sinc(10(t − n))
1 1 ∞ 1 5 f k X t ( f ) = rect comb( f ) = δ ( f − k) ∑ rect 10 δ ( f − k ) = 10 ∑ 10 10 10 k =−∞ k =−5
Modulation |Xt( f )|
xt(t)
0.1
1
-5
5
f
Phase of Xt( f ) -1
1
π
t -5
-0.5
5 -π
Solutions 6-21
f
yf (t)
M. J. Roberts - 7/12/03 ∞ y t ( t) = [1 + sinc(10 t) ∗ comb( t)] cos(200πt) = cos(200πt) 1 + ∑ sinc(10( t − n )) n =−∞
1 1 δ ( f − 100) f Yt ( f ) = δ ( f ) + rect comb( f ) ∗ 10 10 2 +δ ( f + 100) f − 100 rect comb( f − 100) δ ( f − 100) 10 1 1 Yt ( f ) = + 2 +δ ( f + 100) 10 f + 100 + rect comb( f + 100) 10
Modulated Carrier |Yt( f )|
yt(t)
0.5
2
-100
-1
1
100
f
Phase of Yt( f )
t
π
-100 -2
100
f
-π
∞ y d ( t) = [1 + sinc(10 t) ∗ comb( t)] cos2 (200πt) = cos2 (200πt) 1 + ∑ sinc(10( t − n )) n =−∞
f − 100 rect comb( f − 100) δ ( f − 100) δ ( f − 100) 10 1 1 ∗ 1 Yd ( f ) = + 2 +δ ( f + 100) 10 2 +δ ( f + 100) f + 100 + rect 10 comb( f + 100) f − 200 rect 10 comb( f − 200) 1 δ ( f − 200) + 2δ ( f ) 1 f Yd ( f ) = + +2 rect comb( f ) 10 4 +δ ( f + 200) 10 f + 200 200 + rect comb + f ( ) 10
Solutions 6-22
M. J. Roberts - 7/12/03
Demodulated Carrier |Yd( f )|
yd(t)
0.5
2
-200
200
f
Phase of Yd( f ) π
-200 -1
Yf ( f ) = y f ( t) =
1
200
t
f
-π
1 1 f δ ( f ) + rect comb( f ) 2 10 10
1 [1 + sinc(10t) ∗ comb(t)] 2
Demodulated and Filtered Carrier |Yf( f )|
yf(t)
0.5
1
-10
10
f
Phase of Yf( f ) π
-10 -1
1
t
10
f
-π
20. An RC lowpass filter with a time constant of 16 ms is excited by a DSBSC signal, x( t) = sin(2πt) cos(20πt) . Find the phase and group delays at a the carrier frequency. The transfer function of the RC lowpass filter is
Solutions 6-23
M. J. Roberts - 7/12/03
H( jω ) =
A A = . 1 + jωτ 1 + j 0.016ω
The phase of the transfer function is
φ ( jω ) = − tan −1 (ωτ ) = − tan −1 (0.016ω ) . The carrier frequency is 10 Hz. Therefore
φ ( j20π ) = − tan −1 (0.016 × 20π ) = 0.788 and the phase delay is −
φ ( j 20π ) 0.788 = = 0.01254 or 12.54 ms . The derivative of the 20π ωc
phase shift function is
τ d φ ( jω )) = − ( 2 . dω 1 + (ωτ ) Evaluating this derivative at the carrier frequency we get
τ 0.016 d φ ( jω )) − = ( 2 = 2 = 7.95 ms . 1 + (0.016 × 20π ) dω ω =ω c 1 + (ω cτ ) 21. A pulse train, is modulated by a signal,
p( t) = rect (100 t) ∗ 10 comb(10 t) x( t) = sin( 4πt) .
Plot the response of the modulator, y( t) , and the CTFT’s of the excitation and response. X( f ) =
j [δ ( f + 2) − δ ( f − 2)] 2
Solutions 6-24
M. J. Roberts - 7/12/03
PAM Modulator Excitation |X( f )| 1
-2
2
f
Phase of X( f ) π
-2
2
f
-π
y( t) = [rect (100 t) ∗ 10 comb(10 t)] sin( 4πt) ∞ n y( t) = rect (100 t) ∗ ∑ δ t − sin( 4πt) 10 n =−∞ ∞ n y( t) = sin( 4πt) ∑ rect 100 t − 10 n =−∞
1 f f j Y( f ) = sinc comb ∗ [δ ( f + 2) − δ ( f − 2)] 100 10 2 100 Y( f ) = Y( f ) =
j f − 2 f − 2 f + 2 f + 2 sinc comb comb − sinc 10 200 100 10 100
∞ ∞ j f + 2 f − 2 + − 2 10 sinc δ f k − sinc δ ( f − 2 − 10 k ) ( ) ∑ ∑ 100 k =−∞ 100 k =−∞ 20
∞ j ∞ k k Y( f ) = sinc δ ( f + 2 − 10 k ) − ∑ sinc δ ( f − 2 − 10 k ) ∑ 10 10 20 k =−∞ k =−∞
Solutions 6-25
M. J. Roberts - 7/12/03
PAM Modulator Response
|Y( f )|
x(t) and y(t)
1 20
1
-200
-1
1
t
200
f
Phase of Y( f ) π
-200
200
f
-1 -π
22. In the system below, let the excitation be x( t) = rect (1000 t) ∗ 250 comb(250 t) and let the filter be ideal, with unity passband gain. Plot the signal power of the response, y( t) , of this system versus the sweep frequency, f c , over the range, 0 < f c < 2000 for a LPF bandwidth of
and
(a)
5 Hz
(b)
50 Hz
(c)
500 Hz. x(t)
Multiplier xsh(t)
LPF
y(t)
cos(2πfc t) x sh ( t) = [rect (1000 t) ∗ 250 comb(250 t)] cos(2πf c t) ∞ n x sh ( t) = cos(2πf c t) ∑ rect 1000 t − 250 n =−∞
1 f 1 f X sh ( f ) = sinc comb ∗ δ ( f − fc ) + δ ( f + fc ) 250 2 1000 1000
[
X sh ( f ) =
]
∞ 1 f sinc δ ( f − 250 k ) ∗ δ ( f − f c ) + δ ( f + f c ) ∑ 1000 k =−∞ 8
[
Solutions 6-26
]
M. J. Roberts - 7/12/03
1 ∞ k X sh ( f ) = ∑ sinc δ ( f − f c − 250 k ) + δ ( f + f c − 250 k ) 4 8 k =−∞
[
]
1 ∞ k f Y( f ) = ∑ sinc δ ( f − f c − 250 k ) + δ ( f + f c − 250 k ) rect 4 2B 8 k =−∞
[
]
where “B” is the bandwidth of the LPF. ∞ f k rect 2 B ∑ sinc 4 δ ( f − f c − 250 k ) 1 k =−∞ Y( f ) = ∞ 8 f k + rect ∑ sinc δ ( f + f c − 250 k ) 2 B k =−∞ 4
1 k Y( f ) = ∑ sinc δ ( f − f c − 250 k ) + 4 8 f c + 250 k < B Py =
1 2 k ∑ sinc + 64 f c + 250 k < B 4
k sinc δ ( f + f c − 250 k ) 4 f c − 250 k < B
∑
k sinc 2 4 f c − 250 k < B
∑
Signal Power 0.1
2000
f
c
Signal Power 0.1
2000
fc
2000
fc
Signal Power 0.1
23. A signal, x(t) is described by x( t) = 500 rect (1000 t) ∗ comb(500 t) (a)
If x(t), is the excitation of an ideal lowpass filter with a cutoff frequency of 3 kHz, plot the excitation , x(t) and the response, y(t) on the same scale and compare.
Solutions 6-27
M. J. Roberts - 7/12/03
f f sinc comb ∞ 1000 500 1 f X( f ) = = sinc ∑ δ ( f − 500 n ) 1000 n =−∞ 1000 2 ∞ 1 n X( f ) = ∑ sinc δ ( f − 500 n ) 2 2 n =−∞ ∞ 1 f n Y( f ) = rect ∑ sinc δ ( f − 500 n ) 6000 n =−∞ 2 2 ∞ 1 n n Y( f ) = ∑ rect sinc δ ( f − 500 n ) 12 2 2 n =−∞ 6 1 n Y( f ) = ∑ sinc δ ( f − 500 n ) 2 2 n =−6 3 1 5 sinc δ ( f + 2500) + sinc δ ( f + 1500) + sinc δ ( f + 500) 2 2 2 1 Y( f ) = 2 5 3 1 +δ ( f ) + sinc 2 δ ( f − 500) + sinc 2 δ ( f − 1500) + sinc 2 δ ( f − 2500) 1 δ ( f ) + sinc [δ ( f − 500) + δ ( f + 500)] 2 1 Y( f ) = 2 3 5 + sinc [δ ( f − 1500) + δ ( f + 1500)] + sinc [δ ( f − 2500) + δ ( f + 2500)] 2 2 1 1 3 5 y( t) = 1 + sinc 2 cos(1000πt) + sinc 2 cos( 3000πt) + sinc 2 cos(5000πt) 2 2 2 2 1
-2 ms
(b)
-2 ms
t
If x(t), is the excitation of an ideal bandpass filter with a low cutoff frequency of 1 kHz and a high cutoff frequency of 5 kHz, plot the excitation x(t) and the response, y(t) on the same scale and compare.
1 ∞ n X( f ) = ∑ sinc δ ( f − 500 n ) 2 2 n =−∞ ∞ 1 f n f Y( f ) = rect − rect ∑ sinc δ ( f − 500 n ) 10000 2000 n =−∞ 2 2 ∞ 1 n n n Y( f ) = ∑ rect − rect sinc δ ( f − 500 n ) 20 4 2 2 n =−∞ −1 10 1 n n Y( f ) = ∑ sinc δ ( f − 500 n ) + ∑ sinc δ ( f − 500 n ) 2 2 2 n =−10 n =1
Solutions 6-28
M. J. Roberts - 7/12/03
3 sinc 2 [δ ( f − 1500 n ) + δ ( f + 1500 n )] 5 + sinc 2 [δ ( f − 2500 n ) + δ ( f + 2500 n )] 1 Y( f ) = 2 7 + sinc 2 [δ ( f − 3500 n ) + δ ( f + 3500 n )] 9 + sinc [δ ( f − 4500 n ) + δ ( f + 4500 n )] 2
5 3 sinc 2 cos( 3000πt) + sinc 2 cos(5000πt) y( t) = 7 9 + sinc 2 cos( 7000πt) + sinc 2 cos(9000πt) x(t) 1
-2 ms
-2 ms
t
24. Determine whether or not the CT systems with these transfer functions are causal. (a)
H( jω ) =
2 jω
h( t) = sgn( t)
Not Causal
(b)
H( jω ) =
10 6 + j 4ω
5 −3t h( t) = e 2 u( t) 2
Causal
(c)
H( jω ) =
4 25 − ω + j 6ω 2
=
4
( jω + 3) 2 + 16
F Using e − at sin(ω 0 t) u( t) ← →
ω0 ( jω + a) 2 + ω 02
h( t) = e −3 t sin( 4 t) u( t) (d)
H( jω ) =
4 25 − ω + j 6ω 2
Causal e jω =
4
( jω + 3)
2
+ 16
e jω
h( t) = e −3( t +1) sin( 4 ( t + 1)) u( t + 1)
Solutions 6-29
Not Causal
M. J. Roberts - 7/12/03
(e)
H( jω ) =
4 25 − ω + j 6ω 2
e − jω =
4
( jω + 3)
2
+ 16
e − jω
h( t) = e −3( t −1) sin( 4 ( t − 1)) u( t − 1) (f)
H( jω ) =
Causal
jω + 9 jω + 3 6 = + 2 2 45 − ω + j 6ω ( jω + 3) + 36 ( jω + 3) 2 + 36
h( t) = e −3 t [cos(6 t) + sin(6 t)] u( t) (g)
H( jω ) =
49 49 + ω 2
F Using e − a t ← →
h( t) =
Causal
2a , Re( a) > 0 , a + ω2 2
49 −7 t e 14
Not Causal
25. Determine whether or not the DT systems with these transfer functions are causal. (a)
H( F ) = [rect (10 F ) ∗ comb( F )]e − j 20πF h[ n ] =
(b)
H( F ) = j sin(2πF ) = h[ n ] =
1 n − 10 sinc Not Causal 10 10
e j 2πF − e − j 2πF 2
1 (δ[n + 1] − δ[n − 1]) 2
Not Causal
(c)
H( F ) = 1 − e − j 4 πF
h[ n ] = δ [ n ] − δ [ n − 2] Causal
(d)
e jΩ 8e jΩ H( jΩ) = = 8 − 5e − jΩ 1 − 5 e − jΩ 8
5 h[ n ] = 8
n +1
u[ n + 1] Not Causal
26. Find and sketch the frequency response of each of these circuits given the indicated excitation and response. (a)
Excitation, v i ( t) - Response, vC 2 ( t)
Solutions 6-30
M. J. Roberts - 7/12/03
R1 = 1 kΩ
R 2 = 10 kΩ
+
+
vi (t)
C1 = 1 µF
C2 = 0.1 µF
vC2(t)
-
-
1 V ( jω ) Zπ ( jω ) Zπ ( jω ) 1 jωC2 = H( jω ) = C 2 = 1 Vi ( jω ) R1 + Zπ ( jω ) R + R1 + Zπ ( jω ) jωR2C2 + 1 2 jωC2 1 1 1 R2 + R2 + jωC1 jωC2 jωR2C2 + 1 jωC2 Zπ ( jω ) = = = 1 1 C jω (C1 + C2 ) − ω 2 R2C1C2 1 + jωC1R2 + 1 + R2 + jωC1 jωC2 C2 jωR2C2 + 1 jω (C1 + C2 ) − ω 2 R2C1C2 1 H( jω ) = jωR2C2 + 1 jωR2C2 + 1 R1 + 2 jω (C1 + C2 ) − ω R2C1C2 H( jω ) =
1 ( jω (C1 + C2 ) − ω R2C1C2 )R1 + jωR2C2 + 1
H( jω ) =
1 1 − ω R1R2C1C2 + jω (C1 + C2 ) R1 + R2C2
2
[
2
|H( jω )| 1
ω
-15000
15000
Phase of H( jω ) π
ω
-15000
15000 -π
(b)
Excitation, v i ( t) - Response, iC1 ( t)
Solutions 6-31
]
M. J. Roberts - 7/12/03
R1 = 1 kΩ
R 2 = 10 kΩ
+ iC1(t) C1 = 1 µF
vi (t)
C2 = 0.1 µF
-
H( jω ) =
IC1 ( jω ) I R1 ( jω ) = Vi ( jω ) Vi ( jω )
R2 +
1 jωC2
1 1 + R2 + jωC1 jωC2
Zi ( jω ) = R1 + Zπ ( jω ) = R1 +
=
1 jωR2C2 + 1 Zi ( jω ) jωR C + 1 + C2 2 2 C1
jωR2C2 + 1 jω (C1 + C2 ) − ω 2 R2C1C2
[
]
R1 jω (C1 + C2 ) − ω 2 R2C1C2 + jωR2C2 + 1 Zi ( jω ) = jω (C1 + C2 ) − ω 2 R2C1C2 jω (C1 + C2 ) − ω 2 R2C1C2 jωR2C2 + 1 H( jω ) = 2 R1 jω (C1 + C2 ) − ω R2C1C2 + jωR2C2 + 1 jωR C + 1 + C2 2 2 C1
[
]
C jωC1 1 + 2 + jωR2C2 C1 jωR2C2 + 1 H( jω ) = 2 R1 jω (C1 + C2 ) − ω R2C1C2 + jωR2C2 + 1 jωR C + 1 + C2 2 2 C1
[
]
H( jω ) =
jωC1 ( jωR2C2 + 1) R1 jω (C1 + C2 ) − ω 2 R2C1C2 + jωR2C2 + 1
H( jω ) =
jωC1 ( jωR2C2 + 1) 1 − ω R1R2C1C2 + jω R1 (C1 + C2 ) + R2C2
[
]
2
[
Solutions 6-32
]
M. J. Roberts - 7/12/03
|H( jω )| 0.001
ω
-15000
15000
Phase of H( jω ) π
ω
-15000
15000 -π
(c)
Excitation, v i ( t) - Response, v R 2 ( t) C1 = 1 µF
C2 = 1 µF
+
vi (t)
+
R 1 = 10 kΩ
R 2 = 10 kΩ
-
H( jω ) =
vR2(t)
-
VR 2 ( jω ) = Vi ( jω )
jωC1Zπ ( jω ) Zπ ( jω ) R2 jωR2C2 = 1 1 + Zπ ( jω ) + R2 1 + jωC1Zπ ( jω ) 1 + jωR2C2 jωC1 jωC2
1 R1 + R2 jωC2 1 + jωR2C2 Zπ ( jω ) = = R1 1 jωC2 ( R1 + R2 ) + 1 R1 + + R2 jωC2 1 + jωR2C2 jωC2 ( R1 + R2 ) + 1 jωR2C2 H( jω ) = 1 + jωR2C2 1 + jωR2C2 1 + jωC1R1 jωC2 ( R1 + R2 ) + 1 jωC1R1
ω 2 R1R2C1C2 − jωC2 ( R1 + R2 ) + 1 H( jω ) = 1 + jωR2C2 1 + jωC1R1 jωC2 ( R1 + R2 ) + 1 Solutions 6-33
M. J. Roberts - 7/12/03
ω 2 R1R2C1C2 H( jω ) = − jωC2 ( R1 + R2 ) + 1 + jωC1R1 (1 + jωR2C2 ) H( jω ) = −
ω 2 R1R2C1C2 1 − ω 2 R1R2C1C2 + jω (C2 ( R1 + R2 ) + C1R1 ) |H( jω )| 1
-2000
2000
ω
2000
ω
Phase of H( jω ) π
-2000 -π
(d)
Excitation, ii ( t) - Response, v R1 ( t) C1 = 1 µF i i(t)
R 2 = 10 kΩ + C2 = 1 µF
vR1(t)
R 1 = 10 kΩ -
1 V ( jω ) I ( jω ) jωC2 H( jω ) = R1 = R1 R1 = R1 1 I i ( jω ) I i ( jω ) R1 + R2 + jωC2 R2 +
H( jω ) = R1
jωR2C2 + 1 jω ( R1 + R2 )C2 + 1
Solutions 6-34
M. J. Roberts - 7/12/03
|H( jω )| 10000
-500
500
ω
500
ω
Phase of H( jω ) π
-500 -π
(e)
Excitation, v i ( t) - Response, v RL ( t) R 2 = 10 kΩ
+
+ C1 = 1 µF
vi (t)
C2 = 1 µF
R 1 = 10 kΩ
vRL(t)
RL = 1 kΩ
-
-
VR1 ( jω )[ jωC1 + jωC2 + G1 ] − Vi ( jω ) jωC1 − VRL ( jω ) jωC2 = 0 VRL ( jω )[ jωC2 + GL + G2 ] − Vi ( jω )G2 − VR1 ( jω ) jωC2 = 0 jω (C1 + C2 ) + G1 − jωC2
VR1 ( jω ) jωC1 − jωC2 V ( jω ) = jωC2 + (GL + G2 ) VRL ( jω ) G2 i
[
][
]
∆ = jω (C1 + C2 ) + G1 jωC2 + (GL + G2 ) + ω 2C22
[
]
∆ = −ω 2C1C2 + jω (C1 + C2 )(GL + G2 ) + G1C2 + G1 (GL + G2 ) VRL ( jω ) =
1 jω (C1 + C2 ) + G1 ∆ − jωC2
jωC1 Vi ( jω ) G2
Solutions 6-35
M. J. Roberts - 7/12/03
−ω 2C1C2 + jω (C1 + C2 )G2 + G1G2 VRL ( jω ) H( jω ) = = Vi ( jω ) −ω 2C1C2 + jω (C1 + C2 )(GL + G2 ) + G1C2 + G1 (GL + G2 )
[
]
C1 + C2 1 + R2 R1R2 H( jω ) = 1 1 C 1 1 1 + + 2+ + −ω 2C1C2 + jω (C1 + C2 ) RL R2 R1 R1 RL R2 −ω 2C1C2 + jω
H( jω ) =
−ω 2 R1R2C1C2 + jωR1 (C1 + C2 ) + 1 R + R2 R −ω 2 R1R2C1C2 + jω (C1 + C2 )1 + 2 R1 + R2C2 + L RL RL |H( jω )| 1
ω
-25000
25000
Phase of H( jω ) π
ω
-25000
25000 -π
27. Find and sketch versus frequency the magnitude and phase of the input impedance, V(f) V (f) Z in ( f ) = i and transfer function, H( f ) = o , for each of these filters. Ii ( f ) Vi ( f ) +
i i(t) 1 µF
vi (t) -
(a)
1 kΩ
vo(t) -
1 1 1 + R , Z in ( f ) = +R= + 1000 sC j 2πfC j 2π × 10 −6 f
Z ins ( s) = H s ( s) =
+
R R+
1 sC
=
j 2πfCR j 2π × 10 −3 f sCR = , H( f ) = j 2πfCR + 1 j 2π × 10 −3 f + 1 sCR + 1
Solutions 6-36
M. J. Roberts - 7/12/03
|Z ( f )|
|H( f )|
in
50000
1
-800
f
800
-800
800
Phase of Zin( f )
Phase of H( f ) π
π
-800
f
800
-800
800
i i(t) 100 Ω 50 mH
vi (t) (b)
f
-π
-π
+
f
+
vo(t)
10 nF
-
-
Z ins ( s) = R + sL + Z in ( f ) = R + j 2πfL + 1 sC
1 sC
1 1 = 100 + j 0.5πf + j 2πfC j 2π × 10 −8 f
1 1 1 = 1 s LC + sCR + 1 LC s2 + s R + 1 R + sL + L LC sC 1 1 1 9 = 2 × 10 H( f ) = 2 2 1 R −4π f + 2 × 10 9 + j 4000πf LC j 2πf 2 + j 2πf + ( ) L LC H s ( s) =
=
2
|Z ( f )|
|H( f )|
in
100000
-40000
25
40000
f
-40000
Phase of Zin( f )
40000
Phase of H( f ) π
π
-40000
f
40000
f
-40000
40000 -π
-π
Solutions 6-37
f
M. J. Roberts - 7/12/03
28. The signal, x(t), in Exercise 23 is the excitation of an RC lowpass filter with R = 1kΩ and C = 0.3 µF. Sketch the excitation and response voltages versus time on the same scale. From Exercise 23,
x( t) = 500 rect (1000 t) ∗ comb(500 t) 1 ∞ n X( f ) = ∑ sinc δ ( f − 500 n ) 2 2 n =−∞
The transfer function is H( f ) =
1 j 2πfRC + 1
Therefore the output is Y( f ) =
∞ 1 1 n sinc δ ( f − 500 n ) ∑ 2 2 j 2πfRC + 1 n =−∞
1 ∞ 1 n δ ( f − 500 n ) Y( f ) = ∑ sinc 2 n =−∞ 2 j1000πnRC + 1 Converting to the time domain, n sinc ∞ 2 1 e j1000πnt y( t) = ∑ 2 n =−∞ j1000πnRC + 1 or ∞ e j1000πnt e − j1000πnt 1 n y( t) = 1 + ∑ sinc + 2 j1000πnRC + 1 − j1000πnRC + 1 2 n =1
− j1000πnt ∞ − e j1000πnt ) + e j1000πnt + e − j1000πnt 1 n j1000πnRC (e y( t) = 1 + ∑ sinc 2 2 2 n =1 nRC 1000 1 π ( ) + ∞ 1 n 2000πnRC sin(1000πnt) + 2 cos(1000πnt) y( t) = 1 + ∑ sinc 2 2 n =1 (1000πnRC) 2 + 1 1
-2 ms
2 ms
t
29. Draw asymptotic and exact magnitude and phase Bode diagrams for the frequency responses of the following circuits and systems. (a)
Solutions 6-38
M. J. Roberts - 7/12/03
R1 = 1 kΩ
R 2 = 10 kΩ
+
+ C1 = 1 µF
vi (t)
vC2(t)
C2 = 0.1 µF
-
-
From Exercise 26(b) H( jω ) =
1 1 − ω R1R2C1C2 + jω (C1 + C2 ) R1 + R2C2
[
2
H( jω ) =
]
1 1 − 10 ω + j 2.1 × 10 −3 ω −6
2
0 -20
|H(jω)|dB
-40 -60 -80
-100 -120 -140 1 10
2
10
3
10
4
ω
5
10
10
6
10
Phase of H(jω)
0 -0.5 -1 -1.5 -2 -2.5 -3 -3.5 1 10
2
10
3
10
4
ω
5
10
10
6
10
(b) X(jω)
10 jω+10
jω jω+10
Y(jω)
zero at jω = 0
10 jω j10ω H( jω ) = = = jω + 10 jω + 10 ( jω + 10) 2
} jω ω ω j + 1 j + 1 frequency 10 10 independent 1 4 24 3 14 24 3 gain 1 10 {
pole at jω =−10pole at jω =−10
Solutions 6-39
M. J. Roberts - 7/12/03
0
|H(jω)|dB
-10 -20 -30 -40 -50 -1 10
0
1
10
2
10
3
10
ω
10
Phase of H(jω)
2
1
0
-1
-2 -1 10
(c)
0
1
10
2
10
A system whose transfer function is H( jω ) =
H( jω ) =
j 20ω = ( jω + 10 − j 99.5)( jω + 10 + j 99.5)
H( jω ) =
1 500
3
10
ω
10
j 20ω 10, 000 − ω 2 + j 20ω
j 20ω ω ω2 − 10000 1 + j 500 10000
jω
ω ω2 1+ j − 500 10000
0
|H(jω)|dB
-10 -20 -30 -40 -50 -60 0 10
1
10
2
10
ω
3
10
4
10
Phase of H(jω)
2
1
0
-1
-2 0 10
1
10
2
10
ω
3
10
4
10
30. Find the transfer function for the following circuit. What function does it perform?
Solutions 6-40
M. J. Roberts - 7/12/03
if(t)
Rf
i i(t) C i +
vx(t)
vi (t)
+
vo(t)
-
-
Summing currents at the virtual ground, j 2πfCVi ( f ) = − or
Vo ( f ) R
Vo ( f ) = − j 2πfRCVi ( f ) = − RC × j 2πfVi ( f ) 1424 3 derivative of Vi ( f )
This is a differentiator. 31. Design an active highpass filter using an ideal operational amplifier, two resistors and one capacitor and derive its transfer function to verify that it is high pass.
if(t) i i(t)
Ri
+
Rf
Ci vx(t)
vi (t) -
+
vo(t) -
Summing currents at the virtual ground,
Vi ( f ) V (f) =− o 1 Rf Ri + j 2πfCi
or
j 2πfCi R f Vo ( f ) =− Vi ( f ) j 2πfCi Ri + 1 Vo ( f ) , of these active filters and identify them as Vi ( f ) lowpass, highpass, bandpass or bandstop.
32. Find the transfer functions, H( f ) =
(a)
Solutions 6-41
M. J. Roberts - 7/12/03
C4
R5
R1
vx(t)
+
C3 +
vi (t)
vo(t)
R2
-
-
Vx ( f )(G1 + G2 + j 2πfC3 + j 2πfC4 ) − Vi ( f )G1 − Vo ( f ) j 2πfC4 = 0 − Vx ( f ) j 2πfC3 − Vo ( f )G5 = 0
(G1 + G2 + j 2πfC3 + j 2πfC4 ) − j 2πfC4 Vx ( f ) G1 = Vi ( f ) j 2πfC3 G5 Vo ( f ) 0 ∆ = (G1 + G2 + j 2πfC3 + j 2πfC4 )G5 − (2πf ) C3C4 2
∆ = −(2πf ) C3C4 + j 2πfG5 (C4 + C3 ) + (G1 + G2 )G5 2
Vo ( f ) = H( f ) = H( f ) =
1 (G1 + G2 + j 2πfC4 + j 2πfC3 ) G1 VI ( f ) ∆ 0 j 2πfC3 − j 2πfG1C3
−(2πf ) C3C4 + j 2πfG5 (C4 + C3 ) + (G1 + G2 )G5 2
j 2πfR5C3
(2πf ) 2 R1R5C3C4 − j 2πfR1(C4 + C3 ) − 1 +
R1 R2
Bandpass (b)
C1 +
C4 vx(t)
R5 C3 +
vi (t) -
vo(t)
R2
-
Solutions 6-42
M. J. Roberts - 7/12/03
[
]
Vx ( f ) j 2πf (C1 + C4 + C3 ) + G2 − j 2πfC4 Vo ( f ) − j 2πfC1 Vi ( f ) = 0
− j 2πfC3 Vx ( f ) − G5 Vo ( f ) = 0 j 2πf (C1 + C4 + C3 ) + G2 j 2πfC3
− j 2πfC4 Vx ( f ) j 2πfC1 = Vi ( f ) G5 Vo ( f ) 0
∆ = −(2πf ) C3C4 + j 2πfG5 (C1 + C4 + C3 ) + G2G5 2
Vo ( f ) =
1 j 2πf (C1 + C4 + C3 ) + G2 ∆ j 2πfC3
j 2πfC1 Vi ( f ) 0
V (f) (2πf ) C1C3 H( f ) = o = 2 Vi ( f ) −(2πf ) C3C4 + j 2πfG5 (C1 + C4 + C3 ) + G2G5 2
H( f ) =
(2πf ) 2 R2 R5C1C3
−(2πf ) R2 R5C3C4 + j 2πfR2 (C1 + C4 + C3 ) + 1 2
Highpass (c)
R4
C5
R1
vx(t)
+
R3 +
vi (t) -
vo(t)
C2
-
Vx ( f )(G1 + G3 + G4 + j 2πfC2 ) − G1 Vi ( f ) − G4 Vo ( f ) = 0 −G3 Vx ( f ) − j 2πfC5 Vo ( f ) = 0 G1 + G3 + G4 + j 2πfC2 G3
−G4 Vx ( f ) G1 V(f) = j 2πfC5 Vo ( f ) 0 i
∆ = −(2πf ) C2C5 + j 2πf (G1 + G3 + G4 )C5 + G3G4 2
Vo ( f ) =
1 G1 + G3 + G4 + j 2πfC2 G3 ∆
G1 0
Solutions 6-43
Vi ( f )
M. J. Roberts - 7/12/03
H( f ) =
Vo ( f ) G1G3 =− 2 Vi ( f ) −(2πf ) C2C5 + j 2πf (G1 + G3 + G4 )C5 + G3G4
H( f ) = −
R4
−(2πf ) R1R3 R4 C2C5 + j 2πf ( R3 R4 + R1R4 + R1R3 )C5 + R1 2
Lowpass 33. When music is recorded on analog magnetic tape and later played back, a high-frequency noise component, called tape “hiss” is added to the music. For purposes of analysis assume that the spectrum of the music is flat at –30 dB across the audio spectrum from 20 Hz to 20 kHz. Also assume that the spectrum of the signal played back on the tape deck has an added component making the playback signal have a Bode diagram as illustrated in Figure E33. -24 dB -30 dB
f 200
2 kHz
20 kHz
6 kHz 12 kHz
Figure E33 Bode diagram of playback signal The extra high-frequency noise could be attenuated by a lowpass filter but that would also attenuate the high-frequency components of the music, reducing its fidelity. One solution to the problem is to “pre-emphasize” the high-frequency part of the music during the recording process so that when the lowpass filter is applied to the playback the net effect on the music is zero but the “hiss” has been attenuated. Design an active filter which could be used during the recording process to do the pre-emphasis. The pre-emphasis active filter should have a low-frequency gain of one and a highfrequency gain of 6 dB and should transition between those gains between 6 and 12 kHz with a slope (asymptotic slope) of 6 dB/octave. So the pre-emphasis filter should have one real zero to create the first corner and one real pole to create the second corner. If we use the inverting amplifier configuration with an op-amp and make the feedback impedance a resistor then the source impedance can create the zero and pole at the required locations. The design idea is to make the source impedance have the required low-frequency value and then to make its high-frequency value lower, increasing the overall gain at high frequencies, as illustrated in Figure S33.
Solutions 6-44
M. J. Roberts - 7/12/03
Rf
R s1 R s2
Cs
+ vi (t) -
+ vo(t) Figure S33 Pre-emphasis filter
The source impedance is 1 1 Rs1 Rs2 + j 2πf + j 2πfCs j 2πfCsRs1Rs2 + Rs1 R R CsRs2 = s1 s2 Zs ( f ) = = 1 1 j 2πfCs ( Rs1 + Rs2 ) + 1 Rs1 + Rs2 j 2πf + Rs1 + Rs2 + j 2πfCs Cs ( Rs1 + Rs2 ) The numerator provides the zero of the source impedance (the pole of the overall gain) and the denominator provides the pole of the source impedance (the zero of the overall gain). So we want the gain-pole location to be set by 1 = 2π × 12000 CsRs2 and the gain-zero location to be set by 1 = 2π × 6000 . Cs ( Rs1 + Rs2 ) There is no unique solution so let’s arbitrarily set Rs2 = 10 kΩ. Cs = 1.33 nF and Rs1 = 10 kΩ. The overall gain is
Rf
H( f ) = − Rs1Rs2 Rs1 + Rs2
1 j 2πf + CsRs2 1 j 2πf + Cs ( Rs1 + Rs2 )
= −Rf
Rs1 + Rs2 Rs1Rs2
At low frequencies, H( f ) = −
Rf . Rs1
To make the low-frequency gain one, set R f = Rs1 = 10 kΩ .
Solutions 6-45
Then it follows that
1 Cs ( Rs1 + Rs2 ) 1 j 2πf + CsRs2
j 2πf +
M. J. Roberts - 7/12/03
34. One problem with causal CT filters is that the response of the filter always lags the excitation. This problem cannot be eliminated if the filtering is done in real time but if the signal is recorded for later “off-line” filtering one simple way of eliminating the lag effect is to filter the signal, record the response and then filter that recorded response with the same filter but playing the signal back through the system backward. Suppose the filter is a single-pole filter with a transfer function of the form, H( jω ) =
1
,
ω 1+ j ωc
where ω c is the cutoff frequency (half-power frequency) of the filter. (a) What is the effective transfer function of the entire process of filtering the signal forward, then backward? (b)
h(− t) ∗ h( t)
What is the effective impulse response?
The impulse response of the filter is h( t) = F −1[H( jω )] . The response of the filter on the first pass through the filter forward is y1 ( t) = x( t) ∗ h( t) . The response of the filter on the second pass through the filter backward is y 2 ( t) = y1 (− t) ∗ h( t) . According to the definition of convolution, y1 ( t) = Therefore
∞
∞
−∞
−∞
∫ x(τ ) h(t − τ )dτ = ∫ h(τ ) x(t − τ )dτ ∞
∞
−∞
−∞
.
y1 (− t) = ∫ x(τ ) h(− t − τ ) dτ = ∫ h(τ ) x(− t − τ ) dτ or Then
y1 (− t) = x( t) ∗ h(− t) = x(− t) ∗ h( t) . y 2 ( t) = x( t) ∗ h(− t) ∗ h( t) .
F → Taking the Fourier transform of both sides and using the property, x( at) ←
Y2 ( jω ) = X( jω ) H(− jω ) H( jω ) .
1 f X , a a
For all real systems the transfer function has the quality, H* ( jω ) = H(− jω ) . Therefore Y2 ( jω ) = X( jω ) H* ( jω ) H( jω ) = X( jω ) H( jω ) . 2
The effective transfer function is the square of the magnitude of the filter’s transfer function. The effective impulse response is h(− t) ∗ h( t) .
Solutions 6-46
M. J. Roberts - 7/12/03
Just as there is no phase shift in the frequency domain, the effective impulse response, h(− t) ∗ h( t) is symmetrical about t = 0 which also means there is no lag when it is convolved with the excitation. 35. Repeat Exercise 18 but with the second cos(2πf c t) replaced by sin(2πf c t) . X t ( f ) = rect ( f )
x t ( t) = sinc( t)
Modulation |Xt( f )|
x (t) t
1
1
-2
2
f
Phase of Xt( f ) -4
4
π
t -2
2
-0.5
-π
Yt ( f ) = rect ( f ) ∗
y t ( t) = sinc( t) cos(20πt)
f
Yt ( f ) =
1 [δ ( f − 10) + δ ( f + 10)] 2
1 [rect( f − 10) + rect( f + 10)] 2
Modulated Carrier |Yt( f )|
yt(t)
0.5
1
-10
-4
4
10
f
Phase of Yt( f )
t
π
-10 -1
10 -π
Solutions 6-47
f
M. J. Roberts - 7/12/03
y d ( t) = sinc( t) cos(20πt) sin(20πt)
Yd ( f ) =
1 rect ( f − 10) j δ ( f + 10) ∗ 2 + rect ( f + 10) 2 −δ ( f − 10)
Yd ( f ) =
j [rect( f + 20) − rect( f − 20)] 4
Demodulated Carrier |Yd( f )|
y (t) d
0.5
0.5
-20
-4
4
20
f
Phase of Yd( f )
t
π
-20
20
-0.5
f
-π
Yf ( f ) = 0
y f ( t) = 0
36. In the system below, x t ( t) = sinc( t) , f c = 10 and the cutoff frequency of the lowpass filter is 1 Hz. Plot the signals, x t ( t) , y t ( t) , y d ( t) and y f ( t) and the magnitudes and phases of their CTFT’s. yt (t) = x r(t)
|H( f )|
x t(t)
fm -fc
fm fc
f
cos(2πfct)
cos(2πfct) x t ( t) = sinc( t)
yd (t)
X t ( f ) = rect ( f )
Solutions 6-48
LPF
yf (t)
M. J. Roberts - 7/12/03
Modulation |Xt( f )|
xt(t)
1
1
-2
2
f
Phase of Xt( f ) -4
4
π
t -2
-0.5
2
f
-π
f − 10.25 rect 10 0.5 1 δ ( f − ) Yt ( f ) = rect ( f ) ∗ 2 10 25 + . f 10 +δ ( f + ) + rect 0.5 f − 10.25 rect 0.5 1 rect ( f − 10) Yt ( f ) = 2 + rect ( f + 10) f + 10.25 + rect 0.5 Yt ( f ) = y t ( t) =
1 f + 10.25 f − 10.25 rect + rect 0.5 0.5 2
1 1 1 t t sinc e j 2π (10.25) t + sinc e − j 2π (10.25) t 2 2 2 2 2
1 t y t ( t) = sinc cos(20.5πt) 2 2
Solutions 6-49
M. J. Roberts - 7/12/03
Modulated Carrier |Yt( f )|
yt(t)
0.5
0.5
-20
-4
4
20
f
Phase of Yt( f )
t
π
-20 -0.5
20
f
-π
1 t y d ( t) = sinc cos(20.5πt) cos(20πt) 2 2 1 δ ( f − 10.25) 1 δ ( f − 10) 1 Yd ( f ) = 2 rect (2 f ) ∗ ∗ 2 +δ ( f + 10.25) 2 +δ ( f + 10) 2 Yd ( f ) =
δ ( f − 20.25) + δ ( f + 0.25) 1 rect (2 f ) ∗ 4 +δ ( f − 0.25) + δ ( f + 20.25)
Yd ( f ) =
1 rect (2( f − 20.25)) + rect (2( f + 0.25)) 4 + rect (2( f − 0.25)) + rect (2( f + 20.25))
Demodulated Carrier |Yd( f )|
yd(t)
0.5
0.5
-20
20
f
Phase of Yd( f ) -4
4
π
t -20
-0.25
20 -π
Solutions 6-50
f
M. J. Roberts - 7/12/03
y f ( t) =
1 sinc( t) 4
[
]
Yf ( f ) =
1 rect (2( f + 0.25)) + rect (2( f − 0.25)) 4
Yf ( f ) =
rect ( f ) 4
Demodulated and Filtered Carrier |Yf( f )|
y (t) f
0.25
0.25
-2
2
f
Phase of Yf( f ) -4
4
π
t -2
-0.125
2
f
-π
37. A quadrature modulator modulates a sine carrier, sin(20πt) , with a signal, x1 (t ) = sinc( t) , and a cosine carrier, cos(20πt) , with a signal, x 2 (t ) = rect ( t) . The π quadrature demodulator has a phase error making its local oscillators be sin 20πt − 6 π and cos 20πt − . Plot the two demodulated and filtered signals, x1 f (t ) and x 2 f (t ) . 6 y( t) = sinc( t) sin(20πt) + rect ( t) cos(20πt)
π X1d ( f ) = F [sinc( t) sin(20πt) + rect ( t) cos(20πt)] sin 20πt − 6 j πf rect ( f ) ∗ 2 [δ ( f + 10) − δ ( f − 10)] j −j X1d ( f ) = ∗ [δ ( f + 10) − δ ( f − 10)]e 60 1 + sinc( f ) ∗ [δ ( f − 10) + δ ( f + 10)] 2 2 X1d ( f ) =
πf −j j j[rect ( f + 10) − rect ( f − 10)] ∗ [δ ( f + 10) − δ ( f − 10)]e 60 4 +[sinc( f − 10) + sinc( f + 10)]
Solutions 6-51
M. J. Roberts - 7/12/03 πf πf −j −j rect ( f + 10) ∗ δ ( f + 10)e 60 − δ ( f − 10)e 60 j πf πf − rect ( f − 10) ∗ δ ( f + 10)e − j 60 − δ ( f − 10)e − j 60 j X1d ( f ) = πf πf 4 −j −j sinc( f − 10) ∗ δ ( f + 10)e 60 − δ ( f − 10)e 60 + πf πf −j −j + sinc( f + 10) ∗ δ ( f + 10)e 60 − δ ( f − 10)e 60 π π −j j 6 rect ( f + 10) ∗ δ ( f + 10)e − rect ( f + 10) ∗ δ ( f − 10)e 6 j π π − rect ( f − 10) ∗ δ ( f + 10)e j 6 − rect ( f − 10) ∗ δ ( f − 10)e − j 6 j X1d ( f ) = π π 4 j − j sinc( f − 10) ∗ δ ( f + 10)e 6 − sinc( f − 10) ∗ δ ( f − 10)e 6 + π π j −j + sinc( f + 10) ∗ δ ( f + 10)e 6 − sinc( f + 10) ∗ δ ( f − 10)e 6 π π π π j −j j −j 6 6 6 − rect ( f )e + rect ( f − 20)e 6 j rect ( f + 20)e − rect ( f )e j X1d ( f ) = π π π π 4 j −j j −j + sinc( f )e 6 − sinc( f − 20)e 6 + sinc( f + 20)e 6 − sinc( f )e 6 π π j −j π j rect ( f + 20)e 6 − 2 cos rect ( f ) + rect ( f − 20)e 6 6 j X1d ( f ) = π π 4 −j j + j 2 sin π sinc( f ) − sinc( f − 20)e 6 + sinc( f + 20)e 6 6
Now apply the lowpass filter, X1 f ( f ) =
j π π − j 2 cos rect ( f ) + j 2 sin sinc( f ) 4 6 6
X1 f ( f ) =
1 π π cos rect ( f ) − sin sinc( f ) 6 2 6
Solutions 6-52
M. J. Roberts - 7/12/03
x1 f ( t) =
1 π π cos sinc( t) − sin rect ( t) 2 6 6 Similarly,
x 2 f ( t) =
1 π π cos rect ( t) + sin sinc( t) 6 2 6
Quadrature Demodulated and Filtered Signal x1f(t) 1
-4
4
-0.25
t
In-phase Demodulated and Filtered Signal x2f(t) 1
-4
4
-0.25
t
38. A pulse train, p( t) = is modulated by a signal,
1 t rect ∗ 4 comb( 4 t) w w x( t) = sinc( t) .
Plot the response of the modulator, y( t) , and the CTFT’s of the excitation and response for
and
(a)
w = 10 ms
(b)
w = 1 ms .
X( f ) = rect ( f )
Solutions 6-53
M. J. Roberts - 7/12/03
PAM Modulator Excitation |X( f )| 1
-2
2
f
Phase of X( f ) π
-2
2
f
-π
1 t y( t) = rect ∗ 4 comb( 4 t) sinc( t) w w n t − 1 4 y( t) = sinc( t) ∑ rect w n =−∞ w ∞
f Y( f ) = sinc( wf ) comb ∗ rect ( f ) 4 ∞ Y( f ) = 4 ∑ sinc( 4 wk )δ ( f − 4 k ) ∗ rect ( f ) k =−∞ ∞
Y( f ) = 4 ∑ sinc( 4 wk ) rect ( f − 4 k ) k =−∞
Solutions 6-54
M. J. Roberts - 7/12/03
PAM Modulator Response (a) |Y( f )| x(t) and y(t)
4
100
-200
200
f
Phase of Y( f ) -3
3
π
t -200
200
-50
f
-π
PAM Modulator Response (b) |Y( f )| x(t) and y(t)
4
1000
-200
200
f
Phase of Y( f ) -3
3
π
t -200
-500
200
f
-π
1 n and the cutoff DT frequency of the 39. In the system below, x t [ n ] = sinc , Fc = 20 4 1 lowpass filter is . Plot the signals, x t [ n ] , y t [ n ] , y d [ n ] and y f [ n ] and the 20 magnitudes and phases of their DTFT’s. x t[n]
yt [n] = xr[n]
cos(2πFc n)
n x t [ n ] = sinc 20
yd [n]
LPF
yf [n]
cos(2πFc n)
X t ( F ) = 20 rect (20 F ) ∗ comb( F )
Solutions 6-55
M. J. Roberts - 7/12/03
Modulation |X ( F )| t
x [n] t
20
1
-1
1
F
Phase of Xt( F ) -80
80
π
n -1
-0.5
1
F
-π
n πn y t [ n ] = sinc cos 20 2 1 1 1 Yt ( F ) = [20 rect (20 F ) ∗ comb( F )] ⊗ comb F − + comb F + 2 4 4 Yt ( F ) = 10[rect (20 F ) ∗ comb( F )] ∗ δ F −
1 1 + δ F + 4 4
∞ 1 1 Yt ( F ) = 10 ∑ rect (20( F − k )) ∗ δ F − + δ F + 4 4 k =−∞ 1 1 Yt ( F ) = 10 ∑ rect 20 F − − k + rect 20 F + − k 4 4 k =−∞ ∞
Modulated Carrier |Yt( F )|
yt[n]
10
1
-1
-80
80
1
F
Phase of Yt( F )
n
π
-1 -1
1 -π
Solutions 6-56
F
M. J. Roberts - 7/12/03
n πn y d [ n ] = sinc cos2 20 2 ∞ 1 1 1 1 1 Yd ( F ) = 10 ∑ rect 20 F − − k + rect 20 F + − k ⊗ comb F − + comb F + 2 4 4 4 4 k =−∞ ∞ 1 1 1 1 1 Yd ( F ) = 10 ∑ rect 20 F − − k + rect 20 F + − k ∗ δ F − + δ F + 2 4 4 4 4 k =−∞ 1 1 1 1 rect 20 F − − k ∗ δ F − + rect 20 F − − k ∗ δ F + 4 4 4 4 Yd ( F ) = 5 ∑ 1 1 1 1 k =−∞ + rect 20 F + − k ∗ δ F − + rect 20 F + − k ∗ δ F + 4 4 4 4 ∞
∞ 1 1 Yd ( F ) = 5 ∑ rect 20 F − − k + 2 rect (20( F − k )) + rect 20 F + − k 2 2 k =−∞ ∞ 1 Yd ( F ) = 10 ∑ rect 20 F − − k + rect (20( F − k )) 2 k =−∞
Demodulated Carrier |Yd( F )|
yd[n]
10
1
-1
1
F
Phase of Xd( F ) -80
80
π
n -1
-0.5
1 -π
∞
Yf ( F ) = 10 ∑ rect (20( F − k )) = 10 rect (20 F ) ∗ comb( F ) k =−∞
1 n y f [ n ] = sinc 20 2
Solutions 6-57
F
M. J. Roberts - 7/12/03
Demodulated and Filtered Carrier |Yf( F )|
yf[n]
10
0.5
-1
1
F
Phase of Yf( F ) -80
80
π
n -1
1
-0.25
F
-π
40. Repeat Exercise 22 but with an excitation, x( t) = rect (1000 t) ∗ 20 comb(20 t) . x sh ( t) = [rect (1000 t) ∗ 20 comb(20 t)] cos(2πf c t) ∞
n x sh ( t) = cos(2πf c t) ∑ rect 1000 t − 20 n =−∞ 1 f f 1 X sh ( f ) = sinc comb ∗ δ ( f − f c ) + δ ( f + f c ) 1000 20 2 1000
[
X sh ( f ) =
]
∞ 1 f sinc δ ( f − 20 k ) ∗ δ ( f − f c ) + δ ( f + f c ) ∑ 1000 k =−∞ 8
[
]
1 ∞ k X sh ( f ) = ∑ sinc δ ( f − f c − 20 k ) + δ ( f + f c − 20 k ) 4 8 k =−∞
[
]
1 ∞ k f Y( f ) = ∑ sinc δ ( f − f c − 20 k ) + δ ( f + f c − 20 k ) rect 4 2B 8 k =−∞
[
]
∞ f k rect sinc δ ( f − f c − 20 k ) ∑ 2 B k =−∞ 4 1 Y( f ) = ∞ 8 f k + rect ∑ sinc δ ( f + f c − 20 k ) 2 B k =−∞ 4
1 k Y( f ) = ∑ sinc δ ( f − f c − 20 k ) + 4 8 f c + 20 k < B
k sinc δ ( f + f c − 20 k ) 4 f c − 20 k < B
∑
Solutions 6-58
M. J. Roberts - 7/12/03
Py =
1 2 k ∑ sinc + 64 f c + 20 k < B 4
k sinc 2 4 f c − 20 k < B
∑
Signal Power 0.1
2000
f
c
Signal Power 0.1
2000
fc
2000
fc
Signal Power 0.1
41. The diffraction of light can be approximately described through the use of the Fourier transform. Consider an opaque screen with a small slit being illuminated from the left by a normally-incident uniform plane light wave (Figure E41). Diffracting Screen
Viewing Screen
x1
Propagation Direction
x0 Optical Axis Slit
Wavefronts
z
Figure E41 One-dimensional diffraction of light through a slit
πx12 is a good approximation for any x1 in the slit, then the electric field strength of λ the light striking the viewing screen can be accurately described by
If z >>
j
2πz λ
π
2π
j x 02 ∞ −j x 0 x1 e e λz ∫ E1 ( x1 ) e λz dx1 E 0 ( x0 ) = K jλz −∞
where E1 is field strength at the diffracting screen, E 0 is field strength at the viewing screen, K is a constant of proportionality and λ is the wavelength of the light. The integral
Solutions 6-59
M. J. Roberts - 7/12/03
is a Fourier transform with different notation. The field strength at the viewing screen can be written as j
2πz λ
π
j x 02 e λz e E 0 ( x0 ) = K F[E1 ( t)] f → x 0 . jλz λz
The intensity, I 0 ( x 0 ) , of the light at the viewing screen is the square of the magnitude of the field strength, 2 I( x 0 ) = E 0 ( x 0 ) . (a) Plot the intensity of light at the viewing screen if the slit width is 1 mm, the wavelength of light is 500 nm, the distance, z, is 100 m, the constant of proportionality is V 10 −3 and the electric field strength at the diffraction screen is 1 . m The electric field exiting the diffraction slit is x E ( x1 ) = rect 1 0.001 E 0 ( x 0 ) = 10
j
−3
2πz
j
π
2πz
π
j x 02 j x 02 e λ e λ t λz λz e e F rect = 0.001sinc(0.001 f ) f → x 0 x 0 j 5 × 10 −2 j 5 × 10 −5 0 . 001 λz f→
λz
j
2πz
π
e λ j λz x 02 x0 e E 0 ( x0 ) = sinc 5 × 10 −2 j 50 x0 sinc 2 5 × 10 −2 I0 ( x0 ) = = 4 × 10 −4 sinc 2 (20 x 0 ) 2500 I(x ) 0
0.0004
-0.2
0.2
x
0
(b) Now let the slit be replaced by two slits each 0.1 mm in width, separated by 1 mm (center-to-center) and centered on the optical axis. Plot the intensity of light at the viewing screen if the other parameters are the same as in part (a). x + 0.001 x − 0.001 E ( x1 ) = rect 1 + rect 1 0.0001 0.0001
Solutions 6-60
M. J. Roberts - 7/12/03
E 0 ( x 0 ) = 10 j
j
2πz λ
π
j x 02 e t + 0.001 t − 0.001 λz e F rect + rect −5 0.0001 f → x 0 j 5 × 10 0.0001
−3
λz
2πz λ
π
[
j x 02 e λz e 0.0001sinc(0.0001 f ) e − j 2πf ( 0.001) + e + j 2πf ( 0.001) E 0 ( x0 ) = −2 j 5 × 10
j
2πz λ
]
f→
x0 λz
π
j x 02 e e λz sinc(0.0001 f )2 cos(2πf (0.001)) f → x 0 E 0 ( x0 ) = j 500 λz
j
2πz
π
e λ j λz x 02 e E 0 ( x0 ) = sinc(2 x 0 )2 cos( 40πx 0 ) j 500 I0 ( x0 ) =
4 sinc 2 (2 x 0 ) cos2 ( 40πx 0 ) = 1.6 × 10 −5 sinc 2 (2 x 0 ) cos2 ( 40πx 0 ) 5 2.5 × 10 I(x ) 0
1.6e-05
-1
x0
1
42. In Figure 42-1 is a circuit diagram of a half-wave rectifier followed by a capacitor to smooth the response voltage. Model the diode as ideal and let the excitation be a cosine at 60 Hz with an amplitude of 120 2 volts. Let the RC time constant be 0.1 seconds. Then the response voltage will look as illustrated in Figure E42-2 . Find and plot the magnitude of the CTFT of the response voltage.
+ v i (t)
+ R
C
-
vo (t) -
Figure E42-1 A half-wave rectifier with a capacitive smoothing filter
Solutions 6-61
M. J. Roberts - 7/12/03
175
vo(t) vi (t) 0.05
t
-175
Figure E42-2 Excitation and response voltages The response voltage has two parts, the exponential decay time and the cosinusoidal charging time. The dividing time, td , between these two parts is set by the intersection of the cosine and the exponential decay. The peak of the cosine is 120 2 . The decay time constant is 0.1 seconds. Therefore the dividing time is the solution of 120 2 cos(120πtd ) = 120 2e
−
td 0.1
or cos(120πtd ) = e
−
td 0.1
This is a transcendental equation best solved numerically. This equation is simple enough that a trial-and-error method converges very quickly to a solution. That solution is td = 15.23906 ms . Therefore the description of the response voltage over one period is − 0t.1 , 0 < t < 15.23906 ms e v o1 ( t) = 120 2 2 cos(120πt) , 15.23906 < t < 16 ms 3 or − 0t.1 t − 0.00761953 t − 0.01595286 v o1 ( t) = 120 2 e rect + cos(120πt) rect 0.01523906 0.001427607 The CTFT of the response is the CTFT of this voltage convolved with a comb to make it periodically repeat. The CTFT of one period is 0.01523906 − t − j 2πft dt ∫ e 0.1e 0 Vo1 ( f ) = 120 2 1 + [δ ( f − 60) + δ ( f + 60)] ∗ 0.001427607 sinc(0.001427607 f )e − j 2πf ( 0.01595286) 2
Solutions 6-62
M. J. Roberts - 7/12/03 0.01523906 − t j ft 2 − π e 0.1e −10 − j 2πf 0 1 − j 2π ( f − 60)( 0.01595286) Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e
− 0.01523906 − j 2πf 0.01523906 0.1 e 1 e − −10 − j 2πf −10 − j 2πf 1 − j 2π ( f − 60)( 0.01595286) Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e 0.01523906 − 0.1 e − j 2πf 0.01523906 1 − e 10 + j 2πf 1 − j 2π ( f − 60)( 0.01595286) Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e
The CTFT of the actual periodic response is the product of this CTFT with the CTFT of f 60 comb(60 t) which is comb . Therefore 60 0.01523906 − 0.1 1 − e e − j 2πf 0.01523906 10 + j 2πf 1 f − j 2π ( f − 60)( 0.01595286) Vo ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e comb 60 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e
Solutions 6-63
M. J. Roberts - 7/12/03
|V ( f )| o
dB
50
-600
600
f
Phase of V ( f ) o
π
-600
600
f
-π
43. Create a discrete-space image consisting of 96 by 96 pixels. Let the image be a “checkerboard” consisting of 8 by 8 alternating black-and-white squares. (a) Filter the image row-by-row and then column-by-column with a DT filter whose impulse response is n h[ n ] = 0.2(0.8) u[ n ] and display the image on the screen using the imagesc command in MATLAB. After defining the checkerboard we can filter it by convolving it with the impulse response using the MATLAB conv function.
(b) Filter the image row-by-row and then column-by-column with a DT filter whose impulse response is n h[ n ] = δ [ n ] − 0.2(0.8) u[ n ] and display the image on the screen using the imagesc command in MATLAB.
Solutions 6-64
M. J. Roberts - 7/12/03
f 44. In the system of Figure E44 let the CTFT of the the excitation be X( f ) = tri . This fc system is sometimes called a scrambler because it moves the frequency components of a signal to new locations making it unintelligible. (a) Using only an analog multiplier and an ideal filter, design a “descrambler” which would recover the original signal. (b) Sketch the magnitude spectrum of each of the signals in the scrambler-descrambler system.
Multiplier ys (t)
x(t) cos(2πfct)
Figure E44 A “scrambler” The scrambler-descrambler system is illustrated in Figure S44 . x(t)
ys (t)
cos(2πfct)
yss (t)
LPF
y(t)
cos(2πfct)
The cutoff frequency of the lowpass filter is f c . Figure S44 Scrambler-descrambler system
Solutions 6-65
M. J. Roberts - 7/12/03
|X( f )| A
-fc
fc
f
|Ys ( f )| A 2
-fc
fc
f
|Yss( f )| A 2
-fc
fc
f
|Y( f )| A 2
-fc
fc
f
45. Electronic amplifiers that handle very-low-frequency signals are difficult to design because thermal drifts of offset voltages cannot be distinguished from the signals. For this reason a popular technique for designing low-frequency amplifiers is the so called “chopper-stabilized” amplifier (Figure E45).
Typical Amplifier + vi (t) -
+ vo(t) Chopper-Stabilized Amplifier
+ vi (t) -
BPF
LPF
+ vo(t) -
Figure E45 A chopper-stabilized amplifier A chopper-stabilized amplifier “chops” the excitation signal by switching it on and off periodically. This action is equivalent to a pulse amplitude modulation in which the pulse train being modulated by the excitation is a 50% duty-cycle square wave which alternates between zero and one. Then the “chopped” signal is bandpass filtered to remove any slow thermal drift signals from the first amplifier. Then the amplified signal is “chopped” again at Solutions 6-66
M. J. Roberts - 7/12/03
exactly the same rate and in phase with the chopping signal used at the input of the first amplifier. Then this signal may be further amplified. The last step is to lowpass filter the signal out of the last amplifier to recover an amplified version of the original signal. (This is a simplified model but it illustrates the essential features of a chopper-stabilized amplifier.) Let the following be the parameters of the chopper-stabilized amplifier: Chopping frequency Gain of the first amplifier Bandpass filter Gain of the second amplifier Lowpass filter
500 Hz 100 V/V Unity-gain, ideal, zero-phase. Passband 250 < f < 750 10 V/V Unity-gain, ideal, zero-phase. Bandwidth 100 Hz
Let the excitation signal have a 100 Hz bandwidth. What is the effective DC gain of this chopper-stabilized amplifier? Let the excitation be v i ( t) . Then the signal after the first amplifier is v1 ( t) = 100 v i ( t)[rect (1000 t) ∗ 500 comb(500 t)] V1 ( f ) = 100 Vi ( f ) ∗
1 f f sinc comb 500 1000 1000
∞ 1 f V1 ( f ) = 100 Vi ( f ) ∗ sinc δ ( f − 500 m) 1000 m∑ 2 =−∞
V1 ( f ) = 50 Vi ( f ) ∗ V1 ( f ) = 50
∞
∞
m
∑ sinc 2 δ ( f − 500m)
m =−∞
m
∑ sinc 2 V ( f − 500m)
m =−∞
i
The response signal from the bandpass filter is that part of the spectrum lying between 250 and 750 Hz which is 1 Vbpf ( f ) = 50 sinc [Vi ( f − 500) + Vi ( f + 500)] . 2 The response of the second amplifier is 1 f 1 f V2 ( f ) = 50 sinc [Vi ( f − 500) + Vi ( f + 500)] ∗ sinc comb 2 1000 500 1000 ∞ 1 m V2 ( f ) = 25 sinc [Vi ( f − 500) + Vi ( f + 500)] ∗ ∑ sinc δ ( f − 500 m) 2 2 m =−∞ ∞ 1 m V2 ( f ) = 25 sinc ∑ sinc [Vi ( f − 500 m − 500) + Vi ( f − 500 m + 500)] 2 m =−∞ 2
Solutions 6-67
M. J. Roberts - 7/12/03
If the excitation is dc, then Vi ( f ) = Aδ ( f ) and ∞ 1 m V2 ( f ) = 25 A sinc ∑ sinc [δ ( f − 500 m − 500) + δ ( f − 500 m + 500)] . 2 m =−∞ 2
The response of the lowpass filter is that part of the signal below 100 Hz, which is 1 1 1 V2 ( f ) = 25 A sinc sinc [δ ( f ) + δ ( f )] = 50 A sinc 2 δ ( f ) 2 2 2 and the effective dc gain is 1 50 A sinc 2 2 DC Gain = = 20.264 . A 46. A common problem in over-the-air television signal transmission is “multipath” distortion of the received signal due to the transmitted signal bouncing off structures. Typically a strong “main” signal arrives at some time and a weaker “ghost” signal arrives later. So if the transmitted signal is x t ( t) , the recieved signal is
(
x r ( t) = K m x t ( t − tm ) + K g x t t − tg
)
where K m >> K g and tg > tm . (a)
What is the transfer function of this communication channel?
(b) What would be the transfer function of an “equalization” system that would compensate for the effects of multipath? The impulse response of the system is
(
)
h( t) = K mδ ( t − tm ) + K gδ t − tg . Therefore its transfer function is
H( f ) = K m e − j 2πft m + K g e
− j 2πft g
.
An equalization system would then have a transfer function of the form, Ae − j 2πft 0 H eq ( f ) = . − j 2πft g K m e − j 2πft m + K g e
Solutions 6-68
M. J. Roberts - 7/13/03
Chapter 7 - Sampling and the DFT Solutions (In this solution manual, the symbol, ⊗, is used for periodic convolution because the preferred symbol which appears in the text is not in the font selection of the word processor used to create this manual.) 1. Sample the signal,
x( t) = 10 sinc(500 t)
by multiplying it by the pulse train p( t) = rect (10 4 t) ∗ 1000 comb(1000 t) to form the signal, x p ( t) . Sketch the magnitude of the CTFT, X p ( f ) , of x p ( t) .
[
]
x p ( t) = 10 sinc(500 t) rect (10 4 t) ∗ 1000 comb(1000 t) Xp( f ) =
1 f 1 f f rect ∗ 4 sinc 4 comb 500 10 10 1000 50
∞ 1 f f Xp( f ) = rect ∗ sinc 4 ∑ δ ( f − 1000 k ) 500 10 k =−∞ 500
Xp( f ) =
∞ 1 f k rect ∗ ∑ sinc δ ( f − 1000 k ) 500 k =−∞ 10 500
Xp( f ) =
1 ∞ k f sinc rect ∗ δ ( f − 1000 k ) ∑ 10 500 500 k =−∞
Xp( f ) =
1 ∞ k f − 1000 k sinc rect ∑ 500 500 k =−∞ 10 |X( f )| 0.002
-20000
2. Let
20000
x( t) = 10 sinc(500 t)
Solutions 7-1
f
M. J. Roberts - 7/13/03
as in Exercise 1 and form a signal, x p ( t) = [1000 x( t) comb(1000 t)] ∗ rect (10 4 t) . Sketch the magnitude of the CTFT, X p ( f ) , of x p ( t) and compare it to the result of Exercise 1. x p ( t) = [10, 000 sinc(500 t) comb(1000 t)] ∗ rect (10 4 t) 1 f f 1 f X p ( f ) = rect ∗ comb 4 sinc 4 10 500 1000 10 50 Xp( f ) =
∞ 1000 f f rect ∗ δ ( f − 1000 k ) sinc 4 ∑ 5 10 500 k =−∞ 5 × 10
∞ 1 f f − 1000 k Xp( f ) = sinc 4 ∑ rect 500 500 10 k =−∞
|X( f )| 0.002
-20000
f
20000
3. Given a CT signal, x( t) = tri(100 t) , form a DT signal, x[ n ] , by sampling x( t) at a rate, f s = 800, and form an informationequivalent CT impulse signal, xδ ( t) , by multiplying x( t) by a periodic sequence of unit impulses whose fundamental frequency is the same, f 0 = f s = 800 . Sketch the magnitude of the DTFT of x[ n ] and the CTFT of xδ ( t) . Change the sampling rate to f s = 5000 and repeat. x[ n ] = tri(100 nTs ) X( F ) =
∞
∑ tri(100nT )e
n =−∞
X( F ) = 1 +
− j 2πFn
s
=
∑ (1 − 100nT )e
100 nTs <1
∑ (1 − 100nT )(e
0 <100 nTs <1
− j 2πFn
s
j 2πFn
s
fs 100
+ e − j 2πFn )
X( F ) = 1 + 2 ∑ (1 − 100 nTs ) cos(2πFn ) n =1
Solutions 7-2
M. J. Roberts - 7/13/03
where
fs 100
means the greatest integer in
fs 100
(same as the “floor” function in MATLAB).
xδ ( t) = tri(100 t) f s comb( f st) Xδ ( f ) =
f 1 f sinc 2 ∗ comb 100 100 fs
∞ fs 2 f Xδ ( f ) = sinc δ ( f − kf s ) ∗ 100 k∑ 100 =−∞
Xδ ( f ) =
fs ∞ f − kf s sinc 2 ∑ 100 100 k =−∞
Also, using X DTFT ( F ) = Xδ ( f sF ) we get f f F − k 2 F − k X( F ) = s ∑ sinc 2 f s = s ∑ sinc 100 100 100 k =−∞ 100 k =−∞ fs ∞
X( F ) =
∞
fs f F sinc 2 s ∗ comb( F ) 100 100
If both DTFT’s are correct then 100 F f f F n ←→ s sinc 2 s ∗ comb( F ) tri 100 100 fs and, more generally, n F → w sinc 2 ( wF ) ∗ comb( F ) . tri ← w
Solutions 7-3
M. J. Roberts - 7/13/03
fs = 800
fs = 5000
|X ( f )|
|X ( f )|
δ
δ
8
50
-1600
f
1600
-10000
|X( F )|
10000
f
|X( F )|
8
50
-2
F
2
-2
2
F
4. Given a bandlimited CT signal, t x( t) = sinc cos(2πt) , 4 form a DT signal, x[ n ] , by sampling x( t) at a rate, f s = 4, and form an information-equivalent CT impulse signal, xδ ( t) , by multiplying x( t) by a periodic sequence of unit impulses whose fundamental frequency is the same, f 0 = f s = 4 . Sketch the magnitude of the DTFT of x[ n ] and the CTFT of xδ ( t) . Change the sampling rate to f s = 2 and repeat. nT x[ n ] = sinc s cos(2πnTs ) 4 Using 1 n F sinc ← → rect ( wF ) ∗ comb( F ) w w and F cos(2πF0 n ) ← →
[
]
1 comb( F − F0 ) + comb( F + F0 ) 2
4 1 4F X( F ) = rect ∗ comb( F ) ⊗ comb( F − Ts ) + comb( F + Ts ) Ts Ts 2
[
X( F ) = X( F ) =
]
4F 2 rect ∗ comb( F − Ts ) + comb( F + Ts ) Ts Ts
[
]
4F ∞ 2 rect ∗ ∑ δ ( F − Ts − k ) + δ ( F + Ts − k ) Ts Ts k =−∞
Solutions 7-4
M. J. Roberts - 7/13/03
X( F ) = 2 f s
∞
∑ rect (4 f (F − T − k )) + rect (4 f (F + T − k )) s
k =−∞
s
s
s
t xδ ( t) = sinc cos(2πt) f s comb( f st) 4 Xδ ( f ) = 4 rect ( 4 f ) ∗
f 1 δ ( f − 1) + δ ( f + 1) ∗ comb 2 fs
[
]
f − 1 f + 1 Xδ ( f ) = 2 rect ( 4 f ) ∗ comb + comb fs fs ∞ ∞ f −1 f +1 Xδ ( f ) = 2 rect ( 4 f ) ∗ ∑ δ − k + ∑ δ − k k =−∞ f s k =−∞ f s ∞ ∞ Xδ ( f ) = 2 f s rect ( 4 f ) ∗ ∑ δ ( f − 1 − kf s ) + ∑ δ ( f + 1 − kf s ) k =−∞ k =−∞
Xδ ( f ) = 2 f s
∞
∑ rect (4( f − 1 − kf )) + rect (4( f + 1 − kf )) s
k =−∞
s
fs = 4
fs = 2
|Xδ( f )|
|Xδ( f )|
8
-8
8
8
f
-4
|X( F )|
f
|X( F )|
8
-2
4
8
2
F
-2
5. Find the Nyquist rates for these signals. (a)
x( t) = sinc(20 t)
X( f ) =
1 f rect ⇒ f Nyq = 2 f m = 20 20 20
(b)
x( t) = 4 sinc 2 (100 t)
X( f ) =
4 f tri ⇒ f Nyq = 2 f m = 200 100 100
Solutions 7-5
2
F
M. J. Roberts - 7/13/03
X( f ) = j 4[δ ( f + 25) − δ ( f − 25)] ⇒ f Nyq = 2 f m = 50
(c)
x( t) = 8 sin(50πt)
(d)
x( t) = 4 sin( 30πt) + 3 cos( 70πt)
j 2[δ ( f + 15) − δ ( f − 15)] X( f ) = 3 ⇒ f Nyq = 2 f m = 70 35 35 + − + + δ δ f f ) ( )] [ ( 2 (e)
x( t) = rect ( 300 t)
(f)
x( t) = −10 sin( 40πt) cos( 300πt)
Not Bandlimited. Nyquist rate is infinite.
[
] 12 [δ ( f − 150) + δ ( f + 150)]
X( f ) = − j 5 δ ( f + 20) − δ ( f − 20) ∗
5 δ ( f + 20) ∗ δ ( f − 150) − δ ( f − 20) ∗ δ ( f − 150) X( f ) = − j 2 +δ ( f + 20) ∗ δ ( f + 150) − δ ( f − 20) ∗ δ ( f + 150) X( f ) = − j
5 [δ ( f − 130) − δ ( f − 170) + δ ( f + 170) − δ ( f + 130)] 2
f Nyq = 2 f m = 340 6. Sketch these time-limited signals and find and sketch the magnitude of their CTFT’s and confirm that they are not bandlimited. (a)
t x( t) = 5 rect 100
X( f ) = 500 sinc(100 f ) x(t)| 5
-200
200
t
|X( f )| 500
-0.04
(b)
x( t) = 10 tri(5 t)
0.04
f
f X( f ) = 2 sinc 2 5
Solutions 7-6
M. J. Roberts - 7/13/03
x(t)| 10
-0.4
0.4
t
|X( f )| 2
-20
(c)
20
f
x( t) = rect ( t)[1 + cos(2πt)] 1 X( f ) = sinc( f ) ∗ δ ( f ) + [δ ( f − 1) + δ ( f + 1)] 2 X( f ) = sinc( f ) +
1 [sinc( f − 1) + sinc( f + 1)] 2 x(t)| 2
-1
1
t
|X( f )| 1
-5
(d)
5
f
x( t) = rect ( t)[1 + cos(2πt)] cos(16πt) 1 1 X( f ) = sinc( f ) ∗ δ ( f ) + [δ ( f − 1) + δ ( f + 1)] ∗ [δ ( f − 8) + δ ( f + 8)] 2 2
Solutions 7-7
M. J. Roberts - 7/13/03
δ ( f − 1) ∗ δ ( f − 8) 1 8 + − ∗ + δ f δ f ( ) ( ) 1 1 X( f ) = sinc( f ) ∗ δ ( f − 8) + δ ( f + 8) + 2 +δ ( f + 1) ∗ δ ( f − 8) 2 +δ ( f + 1) ∗ δ ( f + 8) 1 δ ( f − 9) + δ ( f + 7) 1 X( f ) = sinc( f ) ∗ δ ( f − 8) + δ ( f + 8) + 2 +δ ( f − 7) + δ ( f + 9) 2 1 sinc( f − 9) + sinc( f + 7) 1 X( f ) = sinc( f − 8) + sinc( f + 8) + 2 + sinc( f − 7) + sinc( f + 9) 2 x(t)| 2
-1
1
t
-2
|X( f )| 1
-12
12
f
7. Sketch the magnitudes of these bandlimited-signal CTFT’s and find and sketch their inverse CTFT’s and confirm that they are not time limited. (a)
X( f ) = rect ( f )e − j 4 πf
x( t) = sinc( t − 2) |X( f )| 1
-1
1
f
x(t)| 1
-6
6
t
-0.5
Solutions 7-8
M. J. Roberts - 7/13/03
(b)
1 t + 1 2 x( t) = sinc 2 100 100
X( f ) = tri(100 f )e jπf |X( f )| 1
-0.02
0.02
f
x(t)| 0.01
-400
400
t
-0.005
(c)
X( f ) = δ ( f − 4 ) + δ ( f + 4 )
x( t) = 2 cos(8πt) |X( f )| 1
-4
4
f
x(t)| 2
-1
1
t
-2
(d)
X( f ) = j[δ ( f + 4 ) − δ ( f − 4 )] ∗ rect (8 f ) 1 t 1 t x( t) = 2 sin(8πt) sinc = sinc sin(8πt) 8 8 8 4
Solutions 7-9
M. J. Roberts - 7/13/03
|X( f )| 1
-6
6
f
x(t)| 0.25
-16
16
t
-0.25
8. Sample the CT signal,
x( t) = sin(2πt) ,
at a sampling rate, f s. Then, using MATLAB, plot the interpolation between samples in the time range, −1 < t < 1, using the approximation, f x( t) ≅ 2 c fs
∑ x(nT ) sinc(2 f (t − nT )) N
s
n =− N
c
with these combinations of f s, f c , and N. (a)
fs = 4 , fc = 2 , N = 1 x(t) 1 -1
1
t
-1
(b)
fs = 4 , fc = 2 , N = 2 x(t) 1 -1
1
t
-1
(c)
fs = 8 , fc = 4 , N = 4 x(t) 1 -1
1
t
-1
(d)
fs = 8 , fc = 2 , N = 4 Solutions 7-10
s
,
M. J. Roberts - 7/13/03
x(t) 1 -1
1
t
-1
(e)
f s = 16 , f c = 8 , N = 8 x(t) 1 -1
1
t
-1
(f)
f s = 16 , f c = 8 , N = 16 x(t) 1 -1
1
t
-1
(a) fs = 4 , fc = 2 , N = 1 x(t)
(b) fs = 4 , fc = 2 , N = 2 x(t)
1 -1
1 1
t
-1
-1
t
-1
(c) fs = 8 , fc = 4 , N = 4 x(t)
(d) fs = 8 , fc = 2 , N = 4 x(t)
1 -1
1
1 1
t
-1
-1
1
t
-1
(e) fs = 16 , fc = 8 , N = 8 (f) fs = 16 , fc = 8 , N = 16 x(t)
x(t)
1
1
-1
1
t
-1
-1
1
t
-1
9. For each signal and specified sampling rate, plot the original signal and an interpolation between samples of the signal using a zero-order hold, over the time range, −1 < t < 1. (The MATLAB function “stairs” could be useful here.) (a)
x( t) = sin(2πt) , f s = 8
Solutions 7-11
M. J. Roberts - 7/13/03
x(t) 1
-1
1
t
-1
(b)
x( t) = sin(2πt) , f s = 32 x(t) 1
-1
1
t
-1
(c)
x( t) = rect ( t) , f s = 8 x(t) 1
-1
1
t
-1
(d)
x( t) = tri( t) , f s = 8 x(t) 1
-1
1
t
-1
10. For each signal in Exercise 9, lowpass filter the zero-order-hold-interpolated signal with a single-pole lowpass filter whose -3 dB frequency is one-fourth of the sampling rate. x(t)
x(t)
1
1
-1
(a)
1 -1
t
-1
(b)
1 -1
Solutions 7-12
t
M. J. Roberts - 7/13/03
x(t)
x(t)
1
1
-1
(c)
1
t
-1
1
(d)
-1
t
-1
11. Repeat Exercise 9 except use a first-order hold instead of a zero-order hold. x(t)
x(t)
1
1
-1
(a)
1
t
-1
(b)
-1
x(t)
1
1
1
t
-1
1
(d)
-1
t
-1
x(t)
-1
(c)
1
t
-1
12. Sample the two signals, x1 ( t) = e − t
2
x 2 ( t) = e − t + sin(8πt) 2
and
in the time interval, −3 < t < 3 , at 8 Hz and demonstrate that the sample values are the same. x1[ n ] = e −( nTs ) x1[ n ] = e
n − 8
2
x 2 [ n ] = e −( nTs ) + sin(8πnTs ) 2
2
x2[n] = e
n − 8
2
+ sin(πn ) = e 123
n − 8
2
= x1[ n ]
=0
13. For each pair of signals below, sample at the specified rate and find the DTFT of the sampled signals. In each case, explain, by examining the DTFT’s of the two signals, why the samples are the same. (a)
x1 ( t) = 4 cos(16πt) and x 2 ( t) = 4 cos( 76πt)
,
x1[ n ] = 4 cos(16πnTs ) and x 2 [ n ] = 4 cos( 76πnTs )
[
]
X1 ( F ) = 2 comb( F − 8Ts ) + comb( F + 8Ts )
8 8 X1 ( F ) = 2 comb F − + comb F + 30 30 Solutions 7-13
f s = 30
M. J. Roberts - 7/13/03
Similarly, 38 38 X 2 ( F ) = 2 comb F − + comb F + 30 30 8 8 X 2 ( F ) = 2 comb F − − 1 + comb F + + 1 = X1 ( F ) 30 3 30 3 144 144 42444 42444 8 8 = comb F − = comb F + 30 30 (b)
x1 ( t) = 6 sinc(8 t) and x 2 ( t) = 6 sinc(8 t) cos( 400πt)
, f s = 100
x1[ n ] = 6 sinc(8 nTs ) and x 2 [ n ] = 6 sinc(8 nTs ) cos( 400πnTs ) X1 ( F ) =
F 3 rect ∗ comb( F ) 4 Ts 8Ts
25 X1 ( F ) = 75 rect F ∗ comb( F ) 2 3 1 F X 2 (F ) = rect ∗ comb( F ) ⊗ comb( F − 200Ts ) + comb( F + 200Ts ) 8Ts 4 Ts 2
[
(c)
]
X 2 (F ) =
F 3 rect ∗ comb( F − 200Ts ) + comb( F + 200Ts ) 8Ts 8Ts
X 2 (F ) =
75 25 rect F ∗ comb( F − 2) + comb( F + 2) 4244 3 14 4244 3 2 14 2 = comb( F ) = comb( F )
X 2 (F ) =
75 25 25 rect F ∗ [2 comb( F )] = 75 rect F ∗ comb( F ) = X1 ( F ) 2 2 2
[
]
x1 ( t) = 9 cos(14πt) and x 2 ( t) = 9 cos(98πt) x1[ n ] = 9 cos(14πnTs )
,
f s = 56
x 2 [ n ] = 9 cos(98πnTs )
[
]
X1 ( F ) =
9 comb( F − 7Ts ) + comb( F + 7Ts ) 2
X1 ( F ) =
9 1 1 comb F − + comb F + 2 8 8
Solutions 7-14
M. J. Roberts - 7/13/03
X 2 (F ) =
[
]
9 comb( F − 49Ts ) + comb( F + 49Ts ) 2
9 7 7 X 2 ( F ) = comb F − + comb F + 2 14 83 144244 83 4244 1 = comb F + 1 = comb F − 8 8 X 2 (F ) =
9 1 1 comb F + + comb F − = X1 ( F ) 2 8 8
14. For each sinusoid, find the two other sinusoids whose frequencies are nearest the frequency of the given sinusoid and which, when sampled at the specified rate, have exactly the same samples. (a)
x( t) = 4 cos(8πt) , f s = 20 4 cos( 48πt) and 4 cos( 32πt)
(b)
x( t) = 4 sin(8πt) , f s = 20 4 sin( 48πt) and − 4 sin( 32πt)
(c)
x( t) = 2 sin(−20πt) , f s = 50 −2 sin(−80πt) and 2 sin(−120πt)
(d)
x( t) = 2 cos(−20πt) , f s = 50 2 cos(−80πt) and 2 cos(−120πt)
(e)
π x( t) = 5 cos 30πt + , f s = 50 4 π π 5 cos130πt + and 5 cos −70πt + 4 4
15. For each DT signal, plot the original signal and the sampled signal for the specified sampling interval. 2πn (a) x[ n ] = sin , Ns = 4 24 2πn x s[ n ] = sin comb 4 [ n ] 24
Solutions 7-15
M. J. Roberts - 7/13/03
x[n] 1
-24
24
n
-1
x [n] s
1
-24
24
n
-1
(b)
x[ n ] = rect 9 [ n ]
, Ns = 2
x s[ n ] = rect 9 [ n ] comb 2 [ n ] x[n] 1
-20
20
n
xs[n] 1
-20
(c)
2πn 2πn x[ n ] = cos cos 48 8
20
n
, Ns = 2
2πn 2πn x s[ n ] = cos cos comb 2 [ n ] 48 8
Solutions 7-16
M. J. Roberts - 7/13/03
x[n] 1
-24
24
n
-1
x [n] s
1
-24
24
n
-1 n
(d)
9 x[ n ] = u[ n ] 10
, Ns = 6
n
9 x s[ n ] = u[ n ] comb 6 [ n ] 10 x[n] 1
-5
40
n
x [n] s
1
-5
40
n
16. For each signal in Exercise 15Error! Reference source not found., sketch the magnitude of the DTFT of the original signal and the sampled signal. (a)
2πn x[ n ] = sin 24
2πn x s[ n ] = sin comb 4 [ n ] 24
X( F ) =
j 1 1 comb F + − comb F − 2 24 24
X( F ) =
∞ j ∞ 1 1 δ F + − k − δ F − − k ∑ ∑ k =−∞ 2 k =−∞ 24 24
Solutions 7-17
M. J. Roberts - 7/13/03
j ∞ 1 1 X( F ) = ∑ δ F + − k − δ F − − k 2 k =−∞ 24 24 X s (F ) =
j 1 1 comb F + − comb F − ⊗ comb( 4 F ) 2 24 24
X s (F ) =
j 1 1 δ F + − δ F − ∗ comb( 4 F ) 2 24 24
X s (F ) =
j 1 1 comb 4 F + − comb 4 F − 24 24 2
X s (F ) =
j ∞ 1 δ F + − ∑ 8 k =−∞ 24
∞ k 1 − − ∑ δ F − 4 k =−∞ 24
k 4
|X(F)| 0.5
-1
1
F
|X (F)| s
0.5
-1
(b)
x[ n ] = rect 9 [ n ]
Using F rect N w [ n ] ← →
X( F ) =
1
F
x s[ n ] = rect 9 [ n ] comb 2 [ n ] sin(πF (2 N w + 1)) sin(πF )
sin(19πF ) sin(πF )
X s (F ) =
sin(19πF ) 1 sin(19πF ) 1 ⊗ comb(2 F ) = ∗ δ ( F ) + δ F − sin(πF ) 2 sin(πF ) 2
Solutions 7-18
M. J. Roberts - 7/13/03
1 sin19π F − 2 1 sin(19πF ) X s (F ) = + 1 2 sin(πF ) sinπ F − 2 |X(F)| 20
-1
1
F
|X (F)| s
20
-1
1
F
(c)
2πn 2πn x[ n ] = cos cos 48 8
2πn 2πn x s[ n ] = cos comb 2 [ n ] cos 48 8
X( F ) =
1 1 1 1 1 1 comb F − + comb F + ⊗ comb F − + comb F + 2 48 48 2 8 8
X( F ) =
1 1 1 1 1 1 comb F − + comb F + ∗ δ F − + δ F + 2 48 48 2 8 8
X( F ) =
1 7 5 5 7 comb F − + comb F + + comb F − + comb F + 4 48 48 48 48
7 5 comb F − + comb F + 48 48 1 ⊗ comb(2 F ) X s (F ) = 4 5 7 + comb F − 48 + comb F + 48 7 5 comb F − + comb F + 48 48 1 ∗ 1 δ ( F ) + δ F − 1 X s (F ) = 4 2 5 7 2 + comb F − 48 + comb F + 48
Solutions 7-19
M. J. Roberts - 7/13/03
7 31 5 19 comb F − + comb F − + comb F + + comb F − 48 48 48 48 1 X s (F ) = 8 5 29 7 17 + comb F − 48 + comb F − 48 + comb F + 48 + comb F − 48 |X(F)| 0.25
-1
1
F
|X (F)| s
0.25
-1 n
(d)
9 x[ n ] = u[ n ] 10
1
F n
9 x s[ n ] = u[ n ] comb 6 [ n ] 10
Using F α n u[ n ] ← →
X( jΩ) =
1 1 − αe − jΩ
10 1 = 9 − jΩ 10 − 9e − jΩ 1− e 10
X s ( jΩ) =
X s ( jΩ) =
1 10 Ω − jΩ ⊗ comb 6 2π 2π 10 − 9e
1 10 2π 10 − 9e − jΩ
π 2π δ (Ω) + δ Ω − + δ Ω − 3 3 π ∗ 3 π π 4 5 +δ (Ω − π ) + δ Ω − 3 + δ Ω − 3
10 10 10 π + 2π 10 − 9e − jΩ + − j Ω− − j Ω− 1 10 − 9e 3 10 − 9e 3 X s ( jΩ) = 10 10 10 6 + + − j (Ω − π ) + 4π 5π − j Ω− − j Ω− 10 − 9e 10 − 9e 3 10 − 9e 3
Solutions 7-20
M. J. Roberts - 7/13/03
1 1 1 π + 2π 10 − 9e − jΩ + − j Ω− − j Ω− 3 3 5 10 − 9e 10 − 9e X s ( jΩ) = 1 1 1 3 + + π + 2π − j Ω− − j Ω− 10 + 9e − jΩ 10 + 9e 3 10 + 9e 3 |X(jΩ)| 10
-2 π
2π
jΩ
|Xs(jΩ)| 10
-2 π
2π
jΩ
17. For each DT signal, plot the original signal and the decimated signal for the specified sampling interval. Also plot the magnitudes of the DTFT’s of both signals. (a)
n x[ n ] = tri 10 Using
, Ns = 2
n x d [ n ] = tri 5
n F → w sinc 2 ( wF ) ∗ comb( F ) tri ← w X( F ) = 10 sinc 2 (10 F ) ∗ comb( F ) or X( F ) =
∞
10 10 n j 2πFn n − j 2πFn n − j 2πFn 1 tri e = tri e = + + e − j 2πFn ) 1 − (e ∑ ∑ ∑ 10 10 10 n =−∞ n =−10 n =1
Similarly, X d ( F ) = 5 sinc 2 (5 F ) ∗ comb( F ) or X d ( F ) = 1 + ∑ 1 − n =1 5
n j 2πFn + e − j 2πFn ) (e 5
Solutions 7-21
M. J. Roberts - 7/13/03
x[n]
|X(F)|
1
10
-20
20
n
-1
x [n]
d
1
(b)
10
20
n
-1
2πn n x[ n ] = (0.95) sin u[ n ] 10
Using
F
|X (F)|
d
-20
1
1
F
, Ns = 2
α sin(Ω0 )e − jΩ α sin(Ω0 n ) u[ n ] ←→ 1 − 2α cos(Ω0 )e − jΩ + α 2e − j 2Ω n
F
π 0.95 sin e − jΩ 5 X( jΩ) = π 1 − 1.9 cos e − jΩ + 0.9025e − j 2Ω 5 2πn 2πn 2n n x d [ n ] = (0.95) sin u[2 n ] = (0.9025) sin u[ n ] 5 5 Similarly
2π 0.9025 sin e − jΩ 5 X d ( jΩ) = 2π 1 − 1.805 cos e − jΩ + 0.8145e − j 2Ω 5
Solutions 7-22
M. J. Roberts - 7/13/03
x[n]
|X(F)|
1
100
-5
40
n
-1
-1
x [n]
d
1
100
-5
40
n
-1
2πn x[ n ] = cos 8
F
|X (F)|
d
(c)
1
-1
1
F
, Ns = 7
14πn 7πn 7πn 8πn πn 2πn x d [ n ] = cos − = cos − = cos = cos = cos 8 4 4 4 8 4 x d [ n ] = x[ n ] X( F ) =
1 1 1 comb F − + comb F + 2 8 8
X d (F ) =
1 7 7 comb F − + comb F + 2 8 8
X d (F ) =
1 1 1 comb F + + comb F − = X( F ) 2 8 8 x[n]
|X(F)|
1
0.5
-20
20
n
-1
-1
x [n]
d
1
0.5
20 -1
F
|X (F)|
d
-20
1
n
-1
Solutions 7-23
1
F
M. J. Roberts - 7/13/03
18. For each signal in Exercise 17, insert the specified number of zeros between samples, lowpass DT filter the signals with the specified cutoff frequency and plot the resulting signal and the magnitude of its DTFT. (a)
Insert 1 zero between points. Cutoff frequency is Fc = 0.1. n n an integer x d , x s[ n ] = N s N s 0 , otherwise n n n n an integer tri , an integer tri , Ns 2 x s[ n ] = 5 N s = 10 0 0 , otherwise , otherwise n x s[ n ] = tri comb 2 [ n ] 10 Using n F → w sinc 2 ( wF ) ∗ comb( F ) tri ← w
[
]
X s ( F ) = 10 sinc 2 (10 F ) ∗ comb( F ) ⊗ comb(2 F ) 1 1 X s ( F ) = 10 sinc 2 (10 F ) ∗ comb( F ) ∗ δ ( F ) + δ F − 2 2
[
]
1 X s ( F ) = 5 sinc 2 (10 F ) ∗ comb( F ) + comb F − 2 ∞ 1 X s ( F ) = 5 ∑ sinc 2 (10( F − k )) + sinc 2 10 F − − k 2 k =−∞
1 X i ( F ) = [rect (5 F ) ∗ comb( F )] × 5 sinc 2 (10 F ) ∗ comb( F ) + comb F − 2 ∞
1 X i ( F ) = [rect (5 F ) ∗ comb( F )] × 5 ∑ sinc 2 (10( F − k )) + sinc 2 10 F − − k 2 k =−∞ n F F → w sinc 2 ( wF ) ∗ comb( F ) Using x[ n ] ∗ y[ n ] ← → X( F ) Y( F ) , tri ← w F and e j 2πF0 n x[ n ] ← → X( F − F0 )
Solutions 7-24
M. J. Roberts - 7/13/03
n 1 n 1 n x i [ n ] = sinc ∗ tri + tri e jπn 5 10 10 10 10 πn πn πn j n 1 n n 1 n j − j x i [ n ] = sinc ∗ tri (1 + e jπn ) = sinc ∗ tri e 2 e 2 + e 2 5 10 10 5 10 10 πn 1 n n j2 πn x i [ n ] = sinc ∗ tri e cos 2 5 5 10 πm 1 ∞ m j 2 πm n − m x i [ n ] = ∑ tri e cos sinc 5 5 m =−∞ 10 2
xi[n] 0.5
-20
20
n
|Xi(F)| 5
-1
(b)
1
F
Insert 4 zeros between points. Cutoff frequency is Fc = 0.2 . 2πn n x d [ n ] = (0.9025) sin u[ n ] 5 n 2π n n Ns n n n an integer (0.9025) N s sin an integer x d , u , N N 5 = x s[ n ] = N s N s s s 0 , otherwise 0 , otherwise 2πn n x s[ n ] = (0.9797) sin u[ n ] comb 5 [ n ] 25
Using F α n sin(Ω0 n ) u[ n ] ← →
α sin(Ω0 )e − jΩ 1 − 2α cos(Ω0 )e − jΩ + α 2e − j 2Ω
Solutions 7-25
M. J. Roberts - 7/13/03
2π 0.9797 sin e − jΩ 25 1 Ω X s ( jΩ) = ⊗ comb 5 2π 2π 2π 2 1 − 2(0.9797) cos e − jΩ + (0.9797) e − j 2Ω 25
2π 0.9797 sin e − jΩ 25 1 X s ( jΩ) = 2π 2π 2 1 − 2(0.9797) cos e − jΩ + (0.9797) e − j 2Ω 25
2π 0.9797 sin e − jΩ 25 X s ( jΩ) = 2π 2 1 − 2(0.9797) cos e − jΩ + (0.9797) e − j 2Ω 25
Ω 2 Ω 1 δ 2π + 5 + δ 2π + 5 Ω Ω 1 ∗ +δ + δ − 2π 2π 5 Ω 2 +δ 2π − 5
4π 2π δ Ω + 5 + δ Ω + 5 2π ∗ +δ (Ω) + δ Ω − 5 4π +δ Ω − 5
4π 2π δ Ω + 5 + δ Ω + 5 2π 0.9797 sin e − jΩ Ω 25 Ω 2π X i ( jΩ) = rect ∗ +δ (Ω) + δ Ω − ∗ comb 2 5 2π π 2 2Ωc 1 − 2(0.9797) cos e − jΩ + (0.9797) e − j 2Ω 25 4π +δ Ω − 5
4π 2π δ Ω + 5 + δ Ω + 5 2π 0.9797 sin e − jΩ ∞ 25 2π Ω − 2πk X i ( jΩ) = 2π ∑ rect ∗ +δ (Ω) + δ Ω − 0.8π 5 2π 2 k =−∞ 1 − 2(0.9797) cos e − jΩ + (0.9797) e − j 2Ω 25 4π +δ Ω − 5
Ω Ω x i [ n ] = x s[ n ] ∗ F −1 rect ∗ comb 2π 2Ωc 2 2πn 2n n x i [ n ] = (0.9797) sin u[ n ] comb 5 [ n ] ∗ sinc 25 5 5 ∞ 2 2πn 2n n x i [ n ] = sinc ∗ (0.9797) sin u[ n ] ∑ δ [ n − 5 m] 25 5 5 m =−∞
Solutions 7-26
M. J. Roberts - 7/13/03 ∞ 2 10πm 2n 5m x i [ n ] = sinc ∗ ∑ (0.9797) sin u[5 m]δ [ n − 5 m] 25 5 5 m =−∞ ∞ 2 2πm 2n m x i [ n ] = sinc ∗ ∑ (0.9025) sin u[ m]δ [ n − 5 m] 5 5 m =−∞ 5
xi[n] =
2πm 2 ∞ 2n (0.9025) m sin u[ m] sinc ∗ δ [ n − 5 m] ∑ 5 5 5 m =−∞
xi[n] =
2πm 2 ∞ 2( n − 5 m) (0.9025) m sin u[ m] sinc ∑ 5 5 5 m =−∞
2 ∞ 2( n − 5 m) 2πm m x i [ n ] = ∑ (0.9025) sin sinc 5 5 5 m =0 xi[n] 0.5 40 -5
n
-0.5
|Xi(F)| 200
-2 π
(c)
2π
F
Insert 4 zeros between points. Cutoff frequency is Fc = 0.02 . 7πn 14πn x d [ n ] = cos = cos 8 4 7πn x s[ n ] = cos comb 5 [ n ] 20 X s (F ) =
1 7 7 comb F − + comb F + ⊗ comb(5 F ) 2 8 8
Solutions 7-27
M. J. Roberts - 7/13/03
X s (F ) =
1 comb F − 10
7 8
1 δ ( F ) + δ F − 5 2 3 7 + comb F + ∗ +δ F − + δ F − 5 5 8 4 +δ F − 5
7 43 51 59 comb F − 8 + comb F − 40 + comb F − 40 + comb F − 40 67 7 27 19 1 X s ( F ) = + comb F − + comb F + + comb F + + comb F + 40 8 40 40 10 3 11 + comb F + 40 + comb F + 40 7 43 51 59 comb F − 8 + comb F − 40 + comb F − 40 + comb F − 40 67 7 27 19 1 X i ( F ) = [rect (25 F ) ∗ comb( F )]+ comb F − + comb F + + comb F + + comb F + 40 8 40 40 10 11 3 + comb F + 40 + comb F + 40 1 1 3 3 comb F − 8 + comb F + 8 + comb F − 40 + comb F + 40 11 11 13 13 1 X i ( F ) = [rect (25 F ) ∗ comb( F )]+ comb F − + comb F + + comb F − + comb F + 40 40 40 40 10 19 19 + comb F − 40 + comb F + 40
comb( F − 0.125) + comb( F + 0.125) + comb( F − 0.075) + comb( F + 0.075) 1 X i ( F ) = [rect (25 F ) ∗ comb( F )]+ comb( F − 0.275) + comb( F + 0.275) + comb( F − 0.325) + comb( F + 0.325) 10 + comb( F − 0.475) + comb( F + 0.475)
None of these impulses passes through the ideal lowpass filter. X i (F ) = 0
and
xi[n] = 0
No graph needed. 19. Sample the following CT signals, x( t) , to form DT signals, x[ n ] . Sample at the Nyquist rate and then at the next higher rate for which the number of samples per cycle is an integer. Plot the CT and DT signals and the magnitudes of the CTFT’s of the CT signals and the DTFT’s of the DT signals. (a)
x( t) = 2 sin( 30πt) + 5 cos(18πt)
Solutions 7-28
M. J. Roberts - 7/13/03
[
] 52 [δ ( f + 9) + δ ( f − 9)]
X( f ) = j δ ( f + 15) − δ ( f − 15) + Nyquist rate is 30 Hz. Period is
1 second. Ten samples are required. 3
At Nyquist rate:
3πn x Nyq [ n ] = 2 sin( 30πnTs ) + 5 cos(18πnTs ) = 2 sin(πn ) + 5 cos 5 1 1 5 3 3 X Nyq ( F ) = j comb F + − comb F − + comb F − + comb F + 2 2 2 10 10 144444 42444444 3 = 0 due to aliasing
X Nyq ( F ) =
5 3 3 comb F − + comb F + 2 10 10
At the next higher rate 11 samples are required and the sampling rate is 33 Hz. 10πn 6πn x11[ n ] = 2 sin( 30πnTs ) + 5 cos(18πnTs ) = 2 sin + 5 cos 11 11 5 5 5 3 3 X11 ( F ) = j comb F + − comb F − + comb F − + comb F + 11 11 2 11 11 x(t)
|X(f )|
8 -0.25
3 0.25
t
-8
-15
xNyq[n] 5 -8
t
-5
-1
x [n]
F
11
5
(b)
1
|X (F )|
11
3 8
-5
f
3 8
-8
15
|XNyq(F )|
t -1
1
F
x( t) = 6 sin(6πt) cos(24πt)
[
] [
]
X( f ) = j
1 3 δ ( f + 3) − δ ( f − 3) ∗ δ ( f − 12) + δ ( f + 12) 2 2
X( f ) = j
3 [δ ( f − 9) + δ ( f + 15) − δ ( f − 15) − δ ( f + 9)] 4
Solutions 7-29
M. J. Roberts - 7/13/03
Nyquist rate is 30 Hz. Period is
1 second. Ten samples are required. 3
At Nyquist rate: 2πn 8πn x Nyq [ n ] = 6 sin(6πnTs ) cos(24πnTs ) = 6 sin cos 10 10 1 1 1 4 4 X Nyq ( F ) = j 3comb F + − comb F − ⊗ comb F − + comb F + 10 10 2 10 10 3 1 1 4 4 X Nyq ( F ) = j comb F + − comb F − ∗ δ F − + δ F + 2 10 10 10 10 3 3 5 5 3 X Nyq ( F ) = j comb F − + comb F + − comb F − − comb F + 2 10 144444 104 10 24444410 4 3 = 0 due to aliasing 3 3 3 X Nyq ( F ) = j comb F − − comb F + 2 10 10 At next higher rate 11 samples are required and the sampling rate is 33 Hz. 2πn 8πn x11[ n ] = 6 sin(6πnTs ) cos(24πnTs ) = 6 sin cos 11 11 1 1 1 4 4 X11 ( F ) = j 3comb F + − comb F − ⊗ comb F − + comb F + 11 11 2 11 11 3 1 1 4 4 X11 ( F ) = j comb F + − comb F − ∗ δ F − + δ F + 2 11 11 11 11 3 3 5 5 3 X11 ( F ) = j comb F − + comb F + − comb F − − comb F + 2 11 11 11 11
Solutions 7-30
M. J. Roberts - 7/13/03
x(t)
|X(f )|
8
3
-0.25
0.25
t
-8
x
-15
[n]
15
|X
Nyq
1.5 8
t
-8
-1
x [n]
1
F
|X (F )|
11
11
8 -8
(F )|
Nyq
8 -8
f
1.5 8
-8
t -1
1
F
N 20. For each of these signals find the DTFS over one period and show that X 0 is real. 2 (a)
x[ n ] = rect 2 [ n ] ∗ comb12 [ n ]
kπ sin (2 N w + 1) N0 1 FS Using rect N w [ n ] ∗ comb N 0 [ n ] ←→ kπ N0 sin N0 kπ sin 5 12 1 X[ k ] = 12 kπ sin 12
6π 5π sin 5 sin 12 1 2 1 1 X[6] = = = 12 6π 12 π 12 sin sin 12 2 (b)
,
Real.
x[ n ] = rect 2 [ n + 1] ∗ comb12 [ n ]
kπ sin 5 πk 12 j 6 1 X[ k ] = e 12 kπ sin 12 6π 5π sin 5 π 6 sin 12 j 6 2 jπ 1 1 1 X[6] = e = e =− , 12 12 12 6π π sin sin 12 2 Solutions 7-31
Real
M. J. Roberts - 7/13/03
(c)
14πn 2πn x[ n ] = cos cos 16 16 Period is 16. X[ k ] =
1 1 comb16 [ k − 7] + comb16 [ k + 7]) ⊗ (comb16 [ k − 1] + comb16 [ k + 1]) ( 2 2
X[ k ] =
1 (comb16[k − 7] + comb16[k + 7]) ∗ (δ[k − 1] + δ[k + 1]) 4
X[ k ] =
1 (comb16[k − 8] + comb16[k − 6] + comb16[k + 6] + comb16[k + 8]) 4
X[8] =
1 1 1 comb16 [0] + comb16 [2] + comb16 [14 ] + comb16 [16]) = (1 + 1) = ( 2 4 4
Real. (d)
2π ( n − 3) 12πn x[ n ] = cos cos 14 14 Period is 14.
3πk 1 j 7 1 1 1 6 6 X[ k ] = (comb14 [ k − ] + comb14 [ k + ]) ⊗ (comb14 [ k − ] + comb14 [ k + ])e 2 2 3πk j 1 6 6 1 1 X[ k ] = {(comb14 [ k − ] + comb14 [ k + ]) ∗ (δ [ k − ] + δ [ k + ])}e 7 4
X[ k ] =
3πk j 1 comb14 [ k − 7] + comb14 [ k − 5] + comb14 [ k + 5] + comb14 [ k + 7])e 7 ( 4
X[ 7] =
1 1 comb14 [0] + comb14 [2] + comb14 [12] + comb14 [14 ])e j 3π = − ( 2 4
Real 21. Start with a signal, x( t) = 8 cos( 30πt) , and sample, window and periodically-repeat it using a sampling rate of f s = 60 and a window width of N F = 32. For each signal in the process, plot the signal and its transform, either CTFT or DTFT. F x( t) = 8 cos( 30πt) ← → X( f ) = 4[δ ( f − 15) + δ ( f + 15)]
Solutions 7-32
M. J. Roberts - 7/13/03
Using x s[ n ] = x( nTs ) and X s ( F ) = f s X( f sF ) ∗ comb( F ) = f s
∞
∑ X( f (F − n)) ,
n =−∞
s
1 1 2πn F x s[ n ] = 8 cos ←→ X s ( F ) = 4 comb F − + comb F + 4 4 4 |X( f )|
x(t)
4
8
-0.3
0.8
t
-15
15
f
15
f
1
F
1
F
Phase of X( f ) π
-8
-15
-π
|Xs(F )|
xs[n]
4
8 -1 -16
48
t
π -1
-8
Phase of Xs(F )
-π
1 , 0 ≤ n < N F , x sw [ n ] = w[ n ] x s[ n ] , X sw ( F ) = W ( F ) ⊗ X s ( F ) Using w[ n ] = 0 , otherwise and W ( F ) = e − jπF ( N F −1)
sin(πFN F ) , sin(πF ) 2πn , 0 ≤ n < 32 8 cos 4 x sw [ n ] = 0 , otherwise
sin( 32πF ) 1 sin( 32πF ) 1 X sw ( F ) = 4 comb F − ⊗ e − j 31πF + comb F + ⊗ e − j 31πF sin(πF ) 4 sin(πF ) 4 X sw ( F ) = 4 δ F −
1 − j 31πF sin( 32πF ) 1 sin( 32πF ) + δ F + ∗ e − j 31πF ∗e 4 sin(πF ) 4 sin(πF )
1 1 1 sin 32π F + − j 31π F − 1 sin 32π F − − j 31π F + 4 4 4 4 X sw ( F ) = 4 e +e 1 1 sinπ F − sinπ F + 4 4 Using
Solutions 7-33
M. J. Roberts - 7/13/03
x sws[ n ] =
∞
∑ x [n − mN ]
m =−∞
sw
F
and X sws[ k ] =
1 4 − j 31π F − 4 X sws[ k ] = e 32
k 1 X sw , k an integer NF NF
2πn x sws[ n ] = 8 cos 4 1 1 sin 32π F − 1 sin 32π F + − j 31π F + 4 4 4 +e 1 1 sinπ F − sinπ F + 4 F → k 4
32
k −8 k +8 − 31 π − 31 π j j sin(πk ) sin(πk ) 1 32 32 +e X sws[ k ] = e k−8 k +8 8 sin sin π π 32 32 The quantity,
sin(πk ) k−8 is zero unless is an integer. If k−8 32 sin π 32
integer, lim k→ 8
If
k−8 is an even 32
sin(πk ) π cos(πk ) = lim = 32 . k → 8 π k−8 k−8 sin cos π π 32 32 32
k−8 is an odd integer, 32 lim
k → 40
sin(πk ) π cos(πk ) = lim = −32 . k → 8 − 8 k π k−8 sin cos π π 32 32 32 k −8 32
− j 31π k−8 is an even integer, e If 32 Therefore
e
k −8 − j 31π 32
= 1.
k −8 32
− j 31π k−8 is an odd integer, e If 32
sin(πk ) = 32 comb 32 [ k − 8] k−8 sin π 32
Similarly,
Solutions 7-34
= −1.
M. J. Roberts - 7/13/03
e
k +8 − j 31π 32
sin(πk ) = 32 comb 32 [ k + 8] k +8 sin π 32
Then 2πn F x sws[ n ] = 8 cos ←→ X sws[ k ] = 4 (comb 32 [ k − 8] + comb 32 [ k + 8]) 4 Using X( F ) =
∞
∑ X[k ]δ (F − kF ) 0
k =−∞
∞ k X( F ) = 4 ∑ (comb 32 [ k − 8] + comb 32 [ k + 8])δ F − 32 k =−∞
(Impulses of strength, 4, at k = L − 40, −24, −8, 8, 24, 40,L or at 5 3 1 1 3 5 F = L − , − , − , , , ,L ) 4 4 4 4 4 4 |Xsw(F )|
xsw[n]
128
8
-16
48
t
-1
Phase of Xsw(F )
1
F
1
F
1
F
1
F
π
-8
-1
-π
|X
(F )|
sws 4
xsws[n] 8 -1 -16
48
t
π -1
-8
Phase of Xsws(F )
-π
22. Sometimes window shapes other than a rectangle are used. Using MATLAB, find and plot the magnitudes of the DFT’s of these window functions, with N = 32. (a)
von Hann or Hanning w[ n ] =
1 2πn 1 − cos , 0≤ n< N N − 1 2
Solutions 7-35
M. J. Roberts - 7/13/03
x[n]
|X[k]|
1
16
31
(b)
n
-32
32
k
Bartlett N −1 2n , 0≤ n≤ N − 1 2 w[ n ] = 2 − 2 n , N − 1 ≤ n < N N −1 2 x[n]
|X[k]|
1
16
31
(c)
Hamming
-32
32
k
2πn w[ n ] = 0.54 − 0.46 cos , 0≤ n< N N − 1
x[n]
|X[k]|
1
16
31
(d)
n
n
-32
32
k
Blackman 2πn 4πn w[ n ] = 0.42 − 0.5 cos + 0.08 cos , 0≤ n< N N − 1 N − 1 x[n]
|X[k]|
1
16
31
n
-32
32
k
23. Sample the following signals at the specified rates for the specified times and plot the N N magnitudes of the DFT’s versus harmonic number in the range, − F < k < F − 1. 2 2 (a) x( t) = cos(2πt) , f s = 2 , N F = 16
Solutions 7-36
M. J. Roberts - 7/13/03
x[n]
|X[k]|
1
16
15
n
-1
(b)
x( t) = cos(2πt)
-8
,
fs = 8
,
|X[k]|
1
16
n
-1
x( t) = cos(2πt)
-8
,
f s = 16
256
255
n
-1
-128
,
fs = 2
,
1
16
x( t) = cos( 3πt)
n
-8
,
fs = 8
,
7
|X[k]|
1
16
-1
x( t) = cos( 3πt)
n
-8
,
f s = 16
k
N F = 16
x[n]
15
k
N F = 16 |X[k]|
-1
(f)
127
x[n]
15
(e)
k
|X[k]|
1
x( t) = cos( 3πt)
7
, N F = 256
x[n]
(d)
k
N F = 16
x[n]
15
(c)
7
, N F = 256
Solutions 7-37
7
k
M. J. Roberts - 7/13/03
x[n]
|X[k]|
1
256
255
n
-1
24. Sample the magnitudes N − F
-128
127
following signals at the specified rates for the specified times and plot the and phases of the DFT’s versus harmonic number in the range, NF − 1. 2 x( t) = tri( t − 1)
,
fs = 2
N F = 16
,
f = 2 , N = 16 s
F
|X[k]| 2
-8
7
k
Phase of X[k] π
-8
7
k
-π
(b)
k
x( t) = tri( t − 1)
,
fs = 8
N F = 16
,
fs = 8 , NF = 16 |X[k]| 8
-8
7
k
Phase of X[k] π
-8
7 -π
Solutions 7-38
k
M. J. Roberts - 7/13/03
(c)
x( t) = tri( t − 1)
,
f s = 16
N F = 256
,
f = 16 , N = 256 s
F
|X[k]| 16
-128
127
k
Phase of X[k] π
-128
127
k
-π
(d)
x( t) = tri( t) + tri( t − 4 )
,
fs = 2
,
NF = 8
fs = 2 , NF = 8 |X[k]| 2
-4
3
k
Phase of X[k] π
-4
(e)
x( t) = tri( t) + tri( t − 4 )
3 -π
,
fs = 8
,
N F = 32
Solutions 7-39
k
M. J. Roberts - 7/13/03
f = 8 , N = 32 s
F
|X[k]| 8
-16
15
k
Phase of X[k] π
-16
15
k
-π
(f)
x( t) = tri( t) + tri( t − 4 )
,
f s = 64
,
N F = 256
fs = 64 , NF = 256 |X[k]| 64
-128
127
k
Phase of X[k] π
-128
127
k
-π
25. Sample each CT signal, x( t) , N F times at the rate, f s, creating the DT signal, x[ n ] . Plot x( t) vs. t and x[ n ] vs. nTs over the time range, 0 < t < N F Ts. Find the DFT, X[ k ], of the N F samples. Then plot the magnitude and phase of X( f ) vs. f and Ts X[ k ] vs. k∆f over f f f the frequency range, − s < f < s , where ∆f = s . Plot Ts X[ k ] as a continuous NF 2 2 function of k∆f using the MATLAB “plot” command. (a)
x( t) = 5 rect (2( t − 2))
,
f s = 16
,
N F = 64
5 f X( f ) = sinc e − j 4 πf 2 2
Solutions 7-40
M. J. Roberts - 7/13/03
|X( f )|
x(t)
2.5
5 -8
8
f
8
f
8
kfs/NF
Phase of X( f ) π 4
t
-8
-π
|TsXs[k]| xs(nTs)
2.5
5 -8
Phase of T X [k] s s
π
(b)
t − 20 x( t) = 3 sinc 5
4
nT
-8
fs = 1 ,
,
8
-π
s
kf /N s
F
N F = 40
X( f ) = 15 rect (5 f )e − j 40πf |X( f )|
x(t)
15
3 -0.5
0.5
f
0.5
f
Phase of X( f ) 40
π
t
-1
-0.5
-π
|TsXs[k]| xs(nTs)
15
3 -0.5
40
nTs
-1
(c)
x( t) = 2 rect ( t − 2) sin(8πt) X( f ) = 2 sinc( f )e − j 4 πf ∗
,
[
Phase of TsXs[k]
0.5
kf /N s
F
π -0.5
f s = 32
0.5
-π
,
kfs/NF
N F = 128
]
j δ ( f + 4) − δ ( f − 4) 2
[
X( f ) = j sinc( f + 4 )e − j 4 π ( f + 4 ) − sinc( f − 4 )e − j 4 π ( f − 4 )
Solutions 7-41
]
M. J. Roberts - 7/13/03
|X( f )|
x(t)
1
2
4
-16
t
16
f
16
f
16
kfs/NF
16
kfs/NF
Phase of X( f ) π
-2
-16
-π
|TsXs[k]| xs(nTs)
1
2 -16 4
nTs
π -16
-2
(d)
Phase of TsXs[k]
t − 2 t − 6 x( t) = 10 tri − tri 2 2
fs = 8
,
-π
N F = 64
,
[
X( f ) = 10 2 sinc 2 (2 f )e − j 4 πf − 2 sinc 2 (2 f )e − j12πf
]
X( f ) = 20 sinc 2 (2 f )(e − j 4 πf − e − j12πf ) X( f ) = 20 sinc 2 (2 f )e − j 8πf (e j 4 πf − e − j 4 πf ) = j 40 sinc 2 (2 f )e − j 8πf sin( 4πf ) |X( f )|
x(t)
40
10
8
-4
t
4
f
4
f
4
kfs/NF
4
kfs/NF
Phase of X( f ) π
-10
-4
-π
|TsXs[k]| xs(nTs)
40
10 -4 8
nT
x( t) = 5 cos(2πt) cos(16πt)
[
π
s -4
-10
(e)
Phase of TsXs[k]
,
f s = 64
,
-π
N F = 128
]
X( f ) =
1 5 δ ( f − 1) + δ ( f + 1) ∗ [δ ( f − 8) + δ ( f + 8)] 2 2
X( f ) =
5 [δ ( f − 9) + δ ( f − 7) + δ ( f + 7) + δ ( f + 9)] 4
Solutions 7-42
M. J. Roberts - 7/13/03
|X( f )|
x(t)
1.25
5 -32
t
2
32
f
32
f
32
kfs/NF
32
kfs/NF
Phase of X( f ) π
-5
-32
-π
|TsXs[k]| xs(nTs)
2.5
5 -32
nTs
2
Phase of TsXs[k] π
-32
-5
-π
26. Sample each CT signal, x( t) , N F times at the rate, f s, creating the DT signal, x[ n ] . Plot x( t) vs. t and x[ n ] vs. nTs over the time range, 0 < t < N F Ts. Find the DFT, X[ k ], of the X[ k ] vs. k∆f over the N F samples. Then plot the magnitude and phase of X( f ) vs. f and NF f f f X[ k ] as an impulse function of frequency range, − s < f < s , where ∆f = s . Plot NF NF 2 2 k∆f using the MATLAB “stem” command to represent the impulses. (a)
x( t) = 4 cos(200πt)
,
f s = 800
[
N F = 32
,
]
X( f ) = 2 δ ( f − 100) + δ ( f + 100)
x( t) = 4cos(200πt) |X( f )|
x(t)
2
4
0.04
-400
t
-400
-π
f = 800 , N = 32 s
s
f
400
f
π
-4
x (nT )
400
Phase of X( f )
|Xs[k]/NF|
F
2
s
4 -400 0.04
nTs
x( t) = 6 rect (2 t ) ∗ comb( t)
,
kfs/NF
π -400
-4
(b)
Phase of Xs[k]/NF 400
f s = 16
400
-π
,
kfs/NF
N F = 128
∞ f k X( f ) = 3 sinc comb( f ) = 3 ∑ sinc δ ( f − k ) 2 2 k =−∞
Solutions 7-43
M. J. Roberts - 7/13/03
x( t) = 6rect(2t)*comb(t)
|X( f )|
x(t)
3
6 -8
8
f
8
f
8
kfs/NF
8
kfs/NF
Phase of X( f )| π 8
t
-8
-π
fs = 16 , NF = 128
|Xs[k]/NF|
xs(nTs)
3
6 -8
Phase of Xs[k]/NF π
8
(c)
-8
nTs
x( t) = 6 sinc( 4 t ) ∗ comb( t)
,
-π
f s = 16
N F = 128
,
3 3 ∞ f k X( f ) = rect comb( f ) = ∑ rect δ ( f − k ) 4 4 2 2 k =−∞ x( t) = 6sinc(4t)*comb(t)
|X( f )|
x(t)
1.5
6 -8
8
f
8
f
8
kfs/NF
8
kfs/NF
Phase of X( f )| 8
π
t
-2
-8
-π
fs = 16 , NF = 128
|Xs[k]/NF|
xs(nTs)
1.5
6 -8
8
nTs
-2
(d)
x( t) = 5 cos(2πt) cos(16πt)
[
,
Phase of Xs[k]/NF π
-8
f s = 64
-π
,
N F = 128
]
X( f ) =
1 5 δ ( f − 1) + δ ( f + 1) ∗ [δ ( f − 8) + δ ( f + 8)] 2 2
X( f ) =
5 [δ ( f − 9) + δ ( f − 7) + δ ( f + 7) + δ ( f + 9)] 4
Solutions 7-44
M. J. Roberts - 7/13/03
x( t) = 5cos(2πt)cos(16πt) |X( f )|
x(t)
1.25
5
2
-32
t
-32
f = 16 , N = 128 s
f
32
f
32
kfs/NF
32
kfs/NF
π
-5
xs(nTs)
32
Phase of X( f )|
-π
|X [k]/N |
F
s 1.25
F
5 -32 2
nTs
Phase of Xs[k]/NF π
-32
-5
-π
27. Using MATLAB (or an equivalent mathematical computer tool) plot the signal, x(t ) = 3 cos(20πt ) − 2 sin(30πt ) over a time range of 0 < t < 400 ms . Also plot samples of this function taken at the following 1 1 1 1 sampling intervals: a) Ts = s , b) Ts = s, c) Ts = s and d) Ts = s. Based on what 120 60 30 15 you observe what can you say about how fast this signal should be sampled so that it could be reconstructed from the samples? Ts = 1/120
Ts = 1/60
x[n]
x[n]
5
5
48
n
-5
24 -5
Ts = 1/30
Ts = 1/15
x[n]
x[n]
5
5
12 -5
n
n
6
n
-5
28. A signal, x(t ) = 20 cos(1000πt ) is impulse sampled at a sampling rate of 2 kHz. Plot two periods of the impulse-sampled signal, xδ ( t) . (Let the one sample be at time, t = 0.) Then plot four periods, centered at zero Hz, of the CTFT, X n ( f ) , of the impulse-sampled signal, xδ ( t) . Change the sampling rate to 500 Hz and repeat.
Solutions 7-45
M. J. Roberts - 7/13/03
[
]
A δ ( f − f0 ) + δ ( f + f0 ) 2 xδ ( t) = A cos(2πf 0 t) × f s comb( f st) ∞ n xδ ( t) = A cos(2πf 0 t) ∑ δ t − fs n =−∞ f A Xδ ( f ) = δ ( f − f 0 ) + δ ( f + f 0 ) ∗ comb 2 fs ∞ A Xδ ( f ) = δ ( f − f 0 ) + δ ( f + f 0 ) ∗ f s ∑ δ ( f − nf s ) 2 n =−∞ ∞ A Xδ ( f ) = f s ∑ δ ( f − nf s − f 0 ) + δ ( f − nf s + f 0 ) 2 n =−∞ X( f ) =
[
]
[
]
[
x[n]
]
fs = 2 kHz
x[n]
20
fs = 500 Hz
20 8
n
2
-20
n
-20
|X( f )|
|X( f )|
20000
-4500
10000
4500
f
-1500
Phase of X( f )
f
Phase of X( f )
π -4500
1500 π
4500
f
-1500
-π
1500
f
-π
t 29. A signal, x(t ) = 10 rect , is impulse sampled at a sampling rate of 2 Hz. Plot the 4 impulse-sampled signal, xδ ( t) on the interval, −4 < t < 4 . Then plot three periods, centered at f = 0 , of the CTFT, Xδ ( f ) , of the impulse-sampled signal, xδ ( t) . Change the sampling rate to 1/2 Hz and repeat. t X( f ) = 40 sinc( 4 f ) , xδ ( t) = 10 rect f s comb( f st) 4 ∞ f Xδ ( f ) = 40 sinc( 4 f ) ∗ comb , Xδ ( f ) = 40 f s sinc( 4 f ) ∗ ∑ δ ( f − kf s ) fs k =−∞ Xδ ( f ) = 40 f s ∞
∞
∑ sinc[4( f − kf )]
k =−∞
s
For a sampling rate of 2 Hz, Xδ ( f ) = 80 ∑ sinc[ 4 ( f − 2 k )] k =−∞ ∞
k For a sampling rate of 1/2 Hz, Xδ ( f ) = 20 ∑ sinc 4 f − 2 k =−∞
Solutions 7-46
M. J. Roberts - 7/13/03
fs = 2 Hz xδ(t)
fs =1/ 2 Hz xδ(t) 10
10
-8
8
t
-2
|Xδ( f )|
20
5
f
-2
Phase of Xδ( f
2
f
Phase of Xδ( f )
π
-5
t
|Xδ( f )|
80
-5
2
π
5
f
-2
-π
2
f
-π
30. A signal, x(t ) = 4 sinc(10t ) , is impulse sampled at a sampling rate of 20 Hz. Plot the impulse-sampled signal, xδ ( t) on the interval, -0.5 < t < 0.5. Then plot three periods, centered at f = 0 , of the CTFT, Xδ ( f ) , of the impulse-sampled signal, xδ ( t) . Change the sampling rate to 4 Hz and repeat. 2 f X( f ) = rect 10 5 x n ( t) = 4 sinc(10 t) f s comb( f st) f 2 f X n ( f ) = rect ∗ comb 10 5 fs ∞ 2 f − nf s X n ( f ) = f s ∑ rect 10 5 n =−∞ fs = 20 Hz x[n]
fs = 4 Hz x[n]
4 -10
4 10
n
-2
-4
|X( f )|
8
8
40
f
-8
Phase of X( f )
f
π 40
-π
8
Phase of X( f )
π -40
n
-4
|X( f )|
-40
2
f
-8
8 -π
f
31. A DT signal, x[ n ] , is formed by sampling a CT signal, x( t) = 20 cos(8πt) , at a sampling rate of 20 Hz. Plot x[ n ] over 10 periods versus discrete time. Then do the same for sampling frequencies of 8 Hz and 6 Hz.
Solutions 7-47
M. J. Roberts - 7/13/03
fs = 20
x[n] 20
n
50 -20
fs = 8
x[n] 20
n
20 -20
fs = 6
x[n] 20
n
15 -20
32. A DT signal, x[ n ] , is formed by sampling a CT signal, x(t ) = −4 sin(200πt ) , at a sampling rate of 400 Hz. Plot x[ n ] over 10 periods versus discrete time. Then do the same for sampling frequencies of 200 Hz and 60 Hz. fs = 400
x[n] 4
40
n
-4
fs = 200
x[n] 4
20
n
-4
fs = 60
x[n] 4
6
n
-4
33. Find the Nyquist rates for these signals. (a)
x( t) = 15 rect ( 300 t) cos(10 4 πt)
(b)
x( t) = 7 sinc( 40 t) cos(150πt)
Not Bandlimited. Nyquist rate is infinite.
X( f ) =
7 f 1 rect ∗ δ ( f − 75) + δ ( f + 75) 40 2 40
X( f ) =
7 f + 75 f − 75 rect ⇒ f Nyq = 2 f m = 190 + rect 40 80 40
[
]
Solutions 7-48
M. J. Roberts - 7/13/03
(c)
x( t) = 15[rect (500 t) ∗ 100 comb(100 t)] cos(10 4 πt) Not Bandlimited. Nyquist rate is infinite.
(d)
x( t) = 4[sinc(500 t) ∗ comb(200 t)] 1 1 f f 1 f f X( f ) = 4 rect comb rect = comb 200 500 200 200 25000 500 500 ∞ 200 f X( f ) = rect δ ( f − 200 k ) 500 k∑ 25000 =−∞
X( f ) = (e)
1 1 ∑ δ ( f − 200k ) ⇒ f Nyq = 2 f m = 400 125 k =−1
x( t) = −2[sinc(500 t) ∗ comb(200 t)] cos(10 4 πt) 1 f 1 f 1 X( f ) = −2 rect comb ∗ δ ( f − 5000) + δ ( f + 5000)] 500 200 200 2 [ 500 f f rect comb ∗ δ ( f − 5000) 500 200 1 X( f ) = − 100000 f f + rect 500 comb 200 ∗ δ ( f + 5000) ∞ 2k ∑ rect 5 δ ( f − 200 k ) ∗ δ ( f − 5000) 1 k =−∞ X( f ) = − 500 ∞ 2 k + ∑ rect δ ( f − 200 k ) ∗ δ ( f + 5000) 5 k =−∞ 1 ∞ 2 k δ ( f − 5000 − 200 k ) X( f ) = − rect ∑ 5 +δ ( f + 5000 − 200 k ) 500 k =−∞ X( f ) = −
1 1 δ ( f − 5000 − 200 k ) ⇒ f Nyq = 2 f m = 10, 400 ∑ 500 k =−1 +δ ( f + 5000 − 200 k )
34. On one graph, plot the DT signal formed by sampling the following three CT functions at a sampling rate of 30 Hz. (a)
x1 (t ) = 4 sin(20πt )
(b)
x 2 (t ) = 4 sin(80πt )
(c)
Solutions 7-49
x 2 (t ) = −4 sin( 40πt )
M. J. Roberts - 7/13/03
f = 30
x[n]
s
4
n
9 -4
35. Plot the DT signal, x[ n ] , formed by sampling the CT signal, x(t ) = 10 sin(8πt ) , at twice the Nyquist rate and x( t) itself. Then on the same graph plot at least two other CT sinusoids which would yield exactly the same samples if sampled at the same times. f = 16
x[n]
s
10
n
8 -10
36. Plot the magnitude of the CTFT of t x(t ) = 25 sinc 2 . 6 What is the minimum sampling rate required to exactly reconstruct x(t) from its samples? Infinitely many samples would be required to exactly reconstruct x(t) from its samples. If one were to make a practical compromise in which he sampled over the minimum possible time which could contain 99% of the energy of this waveform, how many samples would be required? (25sinc2(t/6))2 600
10
X( f ) = 150 tri(6 f )
Solutions 7-50
t
M. J. Roberts - 7/13/03
The maximum frequency present in s(t) occurs where 6 f = ±1 or f = ±
1 . 6
The total energy is most easily found in the frequency domain, ∞
EX =
∫ X( f )
2
∞
df =
−∞ 1 6
∫ 150 tri(6 f )
2
−∞
1 6
df = 150 2 ∫ tri(6 f ) df −
2
1 6
1 6
1 6
E X = 150 2 × 2 ∫ tri(6 f ) df = 150 2 × 2 ∫ (1 − 6 f ) df = 150 2 × 2 ∫ (1 − 12 f + 36 f 2 ) df 2
2
0
0
0
[
E X = 150 2 × 2 f − 6 f 2 + 12 f
]
1 3 6 0
12 1 6 12 = 150 2 × 2 − + = 150 2 × 2 = 2500 6 36 216 216
From MATLAB simulation and trapezoidal-rule integration the minimum possible time that would contain 99% of the energy of the signal would be from -3.9 s to +3.9 s. Sample at times, -3 s, 0 s and 3 s. 37. Plot the magnitude of the CTFT of x(t ) = 8 rect(3t ) . This signal is not bandlimited so it cannot be sampled adequately to exactly reconstruct the signal from the samples. As a practical compromise, assume that a bandwidth which contains 99% of the energy of x(t) is great enough to practically reconstruct x(t) from its samples. What is the minimum required sampling rate in this case? |X( f )|2 7
8
f
8 f X( f ) = sinc 3 3 The total signal energy can be found most simply in the time domain. ∞
Ex =
∫ x(t)
−∞
2
∞
dt =
∫ 8 rect (3t)
−∞
2
1 6
dt = 64 ∫ dt = −
Solutions 7-51
1 6
64 3
M. J. Roberts - 7/13/03
From MATLAB simulation and trapezoidal-rule integration the minimum possible frequency range that would contain 99% of the energy of the signal would be from -30.9 Hz to +30.9 Hz. totalArea=64/3 ; %From analytical solution in time domain. ptsPerLobe=40 ; df=3/ptsPerLobe ; %First zero at 3 Hz, 20 pts per lobe. nLobes=4 ; nPts=ptsPerLobe*nLobes ; f=[0:df:nPts*df] ; X=abs((8/3)*sinc(f/3)).^2 ; p1=plot(f,S,'k') ; grid ; set(p1,'LineWidth',2) ; title('Problem 9.3.11','FontName','Times','FontSize',18) ; xlabel('Frequency, f (Hz)','FontName','Times') ; ylabel('|(8/3)*sinc(f/3)|^2','FontName','Times') ; set(gca,'Position',[0.1,0.6,0.6,0.3],'FontName','Times') ; loop='y' ; area=0 ; f1=0 ; f2=df ; while loop=='y', area=area+(abs(8/3)^2)*(sinc(f1/3)^2+sinc(f2/3)^2)*df/2 ; disp(['f2 = ',num2str(f2),', Area = ',num2str(area)]) ; if area>.99*totalArea/2, loop='n' ; else f1=f1+df ; f2=f2+df ; end end
38. A signal, x( t) , is periodic and one period of the signal is described by 3t , 0 < t < 5.5 x( t ) = . 0 , 5.5 < t < 8 Find the samples of this signal over one period sampled at a rate of 1 Hz (beginning at time, t = 0). Then plot, on the same scale, two periods of the original signal and two periods of a periodic signal which is bandlimited to 0.5 Hz or less that would have these same samples. x(0) = 0 , x(1) = 3 , x(2) = 6 , x(3) = 9 , x( 4) = 12 , x(5) = 15 , x(6) = 0 , x(7) = 0 Using MATLAB, 45.0000 , − 26.8492 + j3.8787 , 6.0000 + j9.0000 , 2.8492 − j8.1213 , X= −9.0000 , 2.8492 + j8.1213 , 6.0000 − j9.0000 , − 26.8492 − 3.8787i
Solutions 7-52
M. J. Roberts - 7/13/03
x(t), x (t) and x[n] bl
20
16
%
t
Solution to Exercise 38 in Sampling and the DFT
close all ; fs = 1 ; Ts = 1/fs ; T = 8 ; N = T/Ts ; %
Set up a vector of sampling times, ts.
ts = [0:Ts:(N-1)*Ts]' ; %
Set up a vector of corresponding sample values, xs.
xs = 3*ramp(ts).*rect((ts-2.75)/5.5) ; % %
Set up vectors of times and signal values much closer for plotting the continuous signal, x.
nPts = 256 ; dt = T/nPts ; t = [0:dt:(nPts-1)*dt]' ; x = 3*ramp(t).*rect((t-2.75)/5.5) ; %
Find the CTFS of the signal, x(t).
[Xs,k] = CTFS(xs,ts,k) ; % % %
Generate the bandlimited signal, xbl, which passes through all the samples, xs[n]. Sum all the complex frequency components from n = -N/2 to n = +N/2.
xbl = zeros(nPts,1) ; f0 = 1/T ; for nn = -N/2:N/2-1, xbl = xbl + Xs(nn+N/2+1)*exp(j*2*nn*pi*f0*t) ; end % %
Clean up any small imaginary parts left over due to round off error.
xbl=real(xbl) ; % %
Form two periods from the one period computed so far for each time-domain function computed; s, x and sbl.
xbl2 = [xbl;xbl] ; t2 = [t;t+T] ; x2 = [x;x] ; xs2 = [xs;xs] ; ts2 = [ts;ts+T] ;
Solutions 7-53
M. J. Roberts - 7/13/03
%
Plot the original signal, samples and bandlimited signal.
p = xyplot({t2,t2,ts2},{x2,xbl2,xs2},[0,16,0,20],'\itt',... 'x({\itt}),x_b_l({\itt}) and x[{\itn}]','Times',18,'Times',14,... '','Times',24,{'n','n','n'},{'c','c','d'},{'k','k','k'}) ; set(p{1},'LineWidth',0.5) ;
39. How many sample values are required to yield enough information to exactly describe these bandlimited periodic signals? (a) x(t ) = 8 + 3 cos(8πt ) + 9 sin( 4πt ) , f m = 4, f Nyq = 8 T0 = least common multiple of
1 1 1 s and s which is s. 4 2 2
At the Nyquist rate we would have 4 samples. We must have an integer number of samples in one period, sampled above the Nyquist rate therefore we need 5 samples and f s = 10. (b) x(t ) = 8 + 3 cos(7πt ) + 9 sin( 4πt ) , f m = 3.5 , f Nyq = 7 T0 = least common multiple of (1/3.5) s and 0.5 s which is 2 s. At the Nyquist rate we would have 14 samples. We must have an integer number of samples in one period, sampled above the Nyquist rate therefore we need 15 samples and f s = 7.5 . 40. Sample the CT signal, 1 t x( t) = 15 sinc(5 t) ∗ comb sin( 32πt) 2 2 to form the DT signal, x[ n ] . Sample at the Nyquist rate and then at the next higher rate for which the number of samples per cycle is an integer. Plot the CT and DT signals and the magnitude of the CTFT of the CT signal and the DTFT of the DT signal. 1 t x( t) = 15 sinc(5 t) ∗ comb sin( 32πt) 2 2 x( t) = 15 sin( 32πt)
∞
∑ sinc(5(t − 2m))
m =−∞
1 j f X( f ) = 15 rect comb(2 f ) ∗ [δ ( f + 16) − δ ( f − 16)] 5 5 2 X( f ) = j
3 ∞ 2k rect δ ( f − 2 k ) ∗ [δ ( f + 16) − δ ( f − 16)] ∑ 5 4 k =−∞
Solutions 7-54
M. J. Roberts - 7/13/03
3 ∞ 2k X( f ) = j ∑ rect [δ ( f + 16 − 2 k ) − δ ( f − 16 − 2 k )] 5 4 k =−∞ 3 1 X( f ) = j ∑ [δ ( f + 16 − 2 k ) − δ ( f − 16 − 2 k )] 4 k =−1 Nyquist rate is 36 Hz. The period is 2. Therefore 72 samples are required. Using X DTFT ( F ) = f s
∞
∑X
k =−∞
CTFT
( f s (F − k ))
we get X( F ) = f s
∞
∑ ∑ [δ ( f (F − q) + 16 − 2k ) − δ ( f (F − q) − 16 − 2k )] 1
q =−∞ k =−1
s
s
or 1 X( F ) = f s ∑ δ ( f sF + 16 − 2 k ) − δ ( f sF − 16 − 2 k ) ∗ comb( F ) k =−1
[
]
or 1 X( F ) = 36 ∑ [δ ( 36 F + 16 − 2 k ) − δ ( 36 F − 16 − 2 k )] ∗ comb( F ) k =−1 or 1 4 k 4 k X( F ) = ∑ δ F + − − δ F − − ∗ comb( F ) 9 18 9 18 k =−1 1 The impulses at F = ± relative to integer values, cancel each other out leaving only the ones 2 above and below. At the next higher sampling rate 73 samples are required in 2 seconds. Therefore the sampling rate is 36.5 Hz and X( F ) = or
73 1 73 73 ∑ δ F + 16 − 2 k − δ F − 16 − 2 k ∗ comb( F ) 2 2 k =−1 2
1 32 2 k 32 2 k X( F ) = ∑ δ F + − − δ F − − ∗ comb( F ) 73 73 73 73 k =−1
Solutions 7-55
M. J. Roberts - 7/13/03
x(t)
|X(f )|
15
1
-2
2
t
-15
x
-20
20
[n]
|X
Nyq
f
(F )|
Nyq
15
1
-72
72
t
-15
-1
1
x [n]
|X
73
F
(F )|
Nyq
15
1
-73
73
t
-15
-1
1
F
41. Without using a computer, find the forward DFT of the following sequence of data and then find the inverse DFT of that sequence and verify that you get back the original sequence.
{x[0],x[1],x[2],x[3]} = {3, 4 ,1,− 2} X[ k ] =
N 0 −1
∑ x[n]e
−j
2πnk N
n =0
3
3
n =0
n =0
X[0] = ∑ x[ n ] = 3 + 4 + 1 − 2 = 6 , X[1] = ∑ x[ n ]e 3
3
n =0
n =0
−j
πn 2
= 3 − j 4 − 1 − j2 = 2 − j6
X[2] = ∑ x[ n ]e − jπn = 3 − 4 + 1 + 2 = 2 , X[ 3] = ∑ x[ n ]e
−j
3πn 2
= 3 + j 4 − 1 + j2 = 2 + j6
x[0] =
πk j 1 3 1 1 3 1 X[ k ] = [6 + 2 − j 6 + 2 + 2 + j 6] = 3 , x[1] = ∑ X[ k ]e 2 = [6 + j (2 − j 6) − 2 − j (2 + j 6)] = 4 ∑ 4 k =0 4 4 k =0 4
x[2] =
3πk j 1 3 1 1 3 1 X[ k ]e jπk = [6 − (2 − j 6) + 2 − (2 + j 6)] = 1 , x[ 3] = ∑ X[ k ]e 2 = [6 − j (2 − j 6) − 2 + j (2 + j 6)] = −2 ∑ 4 k =0 4 4 k =0 4
. 42. Redo Example 7-5 except with x(t ) = 1 + sin(8πt ) + cos( 4πt ) as the signal being sampled. Explain any apparent discrepancies that arise. The bandlimited periodic signal, x(t ) = 1 + sin(8πt ) + cos( 4πt ) is sampled at the Nyquist rate. Find the sample values over one period, find the DFT of the sample values and compare with the CFT of the signal. The highest frequency present in the signal is 4 Hz. Therefore the samples must be taken at 8 Hz. The period of the signal is 0.5 second. Therefore 4 samples are required. Assuming that the first sample is taken at time t = 0, the samples are
Solutions 7-56
M. J. Roberts - 7/13/03
{x[0],x[1],x[2],x[3]} = {2,1,0,1}
.
From the DFT definition, X[ k ] =
N 0 −1
∑ x[n]e
−j
2πnk N
,
n =0
3
3
n =0
n =0
X[0] = ∑ x[ n ] = 2 + 1 + 0 + 1 = 4 , X[1] = ∑ x[ n ]e
−j
πn 2
= 2− j +0+ j = 2
3
3
n =0
n =0
X[2] = ∑ x[ n ]e − jπn {2,1, 0,1} = 2 − 1 + 0 − 1 = 0 , X[ 3] = ∑ x[ n ]e
−j
3πn 2
,
= 2+ j +0− j = 2
The CFT of the original signal is 1 j X( f ) = δ ( f ) + [δ ( f − 2) + δ ( f + 2)] + [δ ( f + 4) − δ ( f − 4)] 2 2 or, ordering the impulses with increasing frequency, j j 1 1 X( f ) = δ ( f + 4 ) + δ ( f + 2 ) + δ ( f ) + δ ( f − 2 ) − δ ( f − 4 ) . 2 2 2 2 The CTFS for the bandlimited, periodic signal from which samples (over one period) were taken can be found from X [k ] XCTFS [ k ] = DFT N0 . Using the periodicity of the DFT, the " XCTFS [ k ] 's" needed are X[−2] = 0 , X[−1] =
1 1 , X[0] = 1 , X[1] = , X[2] = 0 2 2
Expressing the signal as a CTFS, x( t) =
N0 2
1 1 XCTFS [ k ]e j 2π ( kf 0 ) t = 0 + e − j 4 πt + 1 + e + j 4 πt + 0 = 1 + cos( 4πt) 2 2 N0
∑ k =−
2
which is the same as the original description of s(t) except that the sine term is missing. The discrepancy lies in the fact that the sine part of the function is sampled at its Nyquist rate (instead of above the Nyquist rate) and therefore does not show up in the DFT because of aliasing. 43. Sample the bandlimited periodic signal, x(t ) = 15 cos(300πt ) + 40 sin(200πt ) at exactly its Nyquist rate over exactly one period of x( t) . Find the DFT of those samples. From the DFT find the CTFS. Plot the CTFS representation of the signal that results and compare it with x(t). Explain any differences. Repeat for a sampling rate of twice the Nyquist rate. The Nyquist frequency of the signal is 150 Hz. Therefore the Nyquist rate is 300 Hz. The period of the signal is the LCM of the two periods of the sinusoidal components, 1/150 and 1/100. The LCM is 1/50. Therefore the period of the signal is 1/50 second. Sampling for 1/50
Solutions 7-57
M. J. Roberts - 7/13/03
second at 300 Hz requires 6 samples. If the samples begin at t = 0, they are {15,19.641 , − 19.641 , − 15 , 49.6410 , − 49.641} . The X DFT [ k ] ’s are {0 , 0 , − j120 , 90 , + j120 , 0} . The CTFS XCTFS [ k ] ’s are {7.5 , j20 , 0 , 0 , 0 , − j20 , 7.5} . x(t) and x[n] 50
t
0.02 -50
x(t) from CTFS 50
t
0.02 -50
At a sampling rate of twice the Nyquist rate (600 Hz): The X DFT [ k ] ’s are {0 , 0 , − j240 , 90 , 0 , 0 , 0 , 0 , 0 , 90 , + j240 , 0} and the XCTFS [ k ] 's are {0 , 0 , 0 , 7.5 , j20 , 0 , 0 , 0 , − j20 , 7.5 , 0 , 0 , 0} . x(t) and x[n] 50 0.02
t
-50
x(t) from CTFS 50 0.02
t
-50
44. Sample the bandlimited periodic signal, x(t ) = 8 cos(50πt ) − 12 sin(80πt ) at exactly its Nyquist rate over exactly one period of x(t). Find the DFT of those samples. From the DFT find the CTFS. Plot the CTFS representation of the signal that results and compare it with x(t). Explain any differences. Repeat for a sampling rate of twice the Nyquist rate. The first sampling is at the Nyquist rate, 80 Hz, for one period, 1/5 s. The X DFT [ k ] ’s are {0 , 0 , 0 , 0 , 0 , 64 , 0 , 0 , 0 , 0 , 0 , 64 , 0 , 0 , 0 , 0} The XCTFS [ k ] 's are {0 , 0 , 0 , 4 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 4 , 0 , 0 ,0} x(t) and x[n] 20 0.2
t
-20
x(t) from CTFS 20 0.2 -20
Solutions 7-58
t
M. J. Roberts - 7/13/03
The two signals are different because the signal contains a sine wave at the Nyquist frequency. For the second sampling at twice the Nyquist rate, The X DFT [ k ] ’s are 0 , 0 , 0 , 0 , 0 , 128 , 0 , 0 , j192 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , − j192 , 0 , 0 , 128 , 0 , 0 , 0 , 0 The XCTFS [ k ] 's are 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , − j6 , 0 , 0 , 4 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 4 , 0 , 0 , j6 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 x(t) and x[n] 20 0.2
t
-20
x(t) from CTFS 20 0.2
t
-20
45. Using MATLAB, (a)
Generate a pseudo-random sequence of 256 data points in a vector, "x", using the "randn" function which is built in to MATLAB.
(b)
Find the DFT of that sequence of data and put it in a vector, "X".
(c)
Set a vector, "Xlpf", equal to "X".
(d)
Change all the values in "Xlpf" to zero except the first 8 points and the last 8 points.
(e)
Take the real part of the inverse DFT of "Xlpf" and put it in a vector, "xlpf".
(f)
Generate a set of 256 sample times, "t", which begin with "0" and are uniformly separated by "1".
(g)
Plot "x" and "xlpf" versus "t" on the same scale and compare.
What kind of effect does this operation have on a set of data? Why is the output array called "xlpf"?
Solutions 7-59
M. J. Roberts - 7/13/03
x(t) and x (t) lpf
3
t
256
-3
%
Solution to exercise 45 in Sampling and the DFT
close all ; x = randn(256,1) ; X = fft(x) ; mask = [ones(8,1);zeros(240,1);ones(8,1)] ; Xlpf = mask.*X ; xlpf = real(ifft(Xlpf)) ; t = [0:255]' ; p = xyplot({t,t},{x,xlpf},[0,256,-3,3],'\itt',... 'x({\itt}) and x_l_p_f({\itt})','Times',18,... 'Times',14,'','Times',24,{'n','n'},{'c','c'},{'k','k'}) ; set(p{1},'LineWidth',0.5) ;
46. Sample the signal, x( t) = rect ( t) , at three different frequencies, 8 Hz, 16 Hz and 32 Hz for 2 seconds. Plot the magnitude of the DFT in each case. Which of these sampling frequencies yields a magnitude plot that looks most like the magnitude of the CTFT of x(t)? x[n]
|X[k]|
1
-1
1
n
16
x[n]
|X[k]|
1
16
-1
-1
8
1
n
32
x[n]
|X[k]|
1
32
1
n
64
k
k
k
47. Sample the signal, x(t ) = rect(t ) , at 8 Hz for three different total times, 2 seconds, 4 seconds and 8 seconds. Plot the magnitude of the DFT in each case. Which of these total
Solutions 7-60
M. J. Roberts - 7/13/03
sampling times yields a magnitude plot that looks most like the magnitude of the CTFT of x(t)? x[n]
|X[k]|
1
-1
8
1
n
x[n]
8
2
n
x[n]
32
k
|X[k]|
1
-4
k
|X[k]|
1
-2
16
8
4
n
64
k
48. Sample the signal, x(t ) = cos(πt ) at three different frequencies, 2 Hz, 4 Hz and 8 Hz for 5 seconds. Plot the magnitude of the DFT in each case. Which of these sampling frequencies yields a magnitude plot that looks most like the magnitude of the CTFT of x(t)? x[n]
|X[k]|
1
5 5
n
-1
10
x[n]
|X[k]|
1
10 5
n
-1
20
x[n]
|X[k]|
1
20 5
k
k
n
-1
40
k
49. Sample the signal, x(t ) = cos(πt ) , at 8 Hz for three different total times, 5 seconds, 9 seconds and 13 seconds. Plot the magnitude of the DFT in each case. Which of these total sampling times yields a magnitude plot that looks most like the magnitude of the CTFT of x(t)?
Solutions 7-61
M. J. Roberts - 7/13/03
x[n]
|X[k]|
1
20 5
n
-1
40
x[n]
|X[k]|
1
36 9
n
-1
72
x[n]
|X[k]|
1
52 13
k
k
n
-1
104
Solutions 7-62
k
M. J. Roberts - 7/12/03
Chapter 8 - Correlation, Energy Spectral Density and Power Spectral Density Solutions (In this solution manual, the symbol, ⊗, is used for periodic convolution because the preferred symbol which appears in the text is not in the font selection of the word processor used to create this manual.) 1. Plot correlograms of the following pairs of CT and DT signals. (a)
x1 ( t) = cos(2πt)
x 2 ( t) = 2 cos( 4πt)
and x2 2
-2
2
x1
-2
(b)
2πn x1[ n ] = sin 16
2πn x 2 [ n ] = 2 cos 8
and x2 2
-2
2
x1
-2
(c)
x1 ( t) = e − t u( t)
x 2 ( t) = e −2 t u( t)
and x2 1
-1
1
x
1
-1
(d)
x1[ n ] = e
−
n 10
2πn cos u[ n ] 10
and
x1[ n ] = e
Solutions 8-1
−
n 10
2πn sin u[ n ] 10
M. J. Roberts - 7/12/03
x
2
1
-1
1
x1
-1
2.
Plot correlograms of the following pairs of CT and DT signals. (a)
x1 ( t) = cos(2πt)
x 2 ( t) = cos2 (2πt)
and x
2
1
-1
1
x1
-1
(b)
x1[ n ] = n
x2[n] = n 3
and
−10 < n < 10
,
x2 1000
-10
10
x1
-1000
(c)
x1 ( t) = t
x 2 ( t) = 2 − t 2 ,
and
x
−4 < t < 4
2
20
-5
5
x
1
-20
3. In MATLAB generate two vectors, x1 and x2, representing DT signals using the following code fragment, x1 = randn(100,1) ; x2 = randn(100,1) ; x3 = randn(100,1) ;
Then plot correlograms of the following pairs of DT signals. (a)
x1
and
x2
Solutions 8-2
M. J. Roberts - 7/12/03
x
2
3
-3
3
x
1
-3
(b)
x1
and
x1+x2 x1 + x2 3
-3
3
x1
-3
(c)
x1+x2 and
x1+x3 x1 + x3 3
-3
3
x1 + x2
3
x1 +
-3
(d)
x1+x2/10
and
-x1+x3/10 -x1 +
x3 10
3
x -3
2
10
-3
4. Plot the correlation function for each of the following pairs of energy signals. (a)
x1 ( t) = 4 rect ( t) Using
and
x 2 ( t) = −3 rect (2 t)
R xy (τ ) = x(−τ ) ∗ y(τ ) ,
R12 (τ ) = 4 rect (−τ ) ∗ (−3 rect (2τ )) = −12 rect (τ ) ∗ rect (2τ ) Express the wider rectangle as the sum of two narrower rectangles, each of which is the same width as the other rectangle, then do two easy convolutions instead of one hard one.
Solutions 8-3
M. J. Roberts - 7/12/03
1 1 R12 (τ ) = −12 rect 2τ + + rect 2τ − ∗ rect (2τ ) 4 4 1 R12 (τ ) = −12 rect 2τ + ∗ rect (2τ ) + rect 2τ − 4
1 ∗ rect (2τ ) 4
πf πf 1 1 f j 1 f 1 f j F rect 2τ + ∗ rect (2τ ) ← → sinc e 2 sinc = sinc 2 e 2 2 2 4 2 2 2 4
rect 2τ −
πf πf 1 1 f −j 1 f 1 F 2 f − j ∗ rect (2τ ) ←→ sinc e 2 sinc = sinc e 2 2 2 4 2 2 2 4
πf j π2f −j 1 1 F 2 f −12rect 2τ + + rect 2τ − ∗ rect (2τ ) ←→ −3 sinc e + e 2 2 4 4
1 −3 × 2 tri(2τ ) ∗ δ τ + + δ τ − 4
πf j π2f −j 1 F 2 f 3 ←→ − sinc e + e 2 2 4
1 R12 (τ ) = −6 tri 2τ + + tri 2τ − 4
1 4
R (τ) 12
-2
2
τ
-6
(b)
x1[ n ] = 2 rect 3 [ n ] Using
and
x 2 [ n ] = 5 rect 8 [ n ]
R xy [ m] = x[− m] ∗ y[ m]
R12 [ m] = 2 rect 3 [− m] ∗ 5 rect 8 [ m] = 10 rect 3 [− m] ∗ rect 8 [ m] ∞
∞
n =−∞
n =−∞
R12 [ m] = 10 ∑ rect 3 [− n ] rect 8 [ m − n ] = 10 ∑ rect 3 [ n ] rect 8 [ m − n ] 3
R12 [ m] = 10 ∑ rect 8 [ m − n ] n =−3
Solutions 8-4
M. J. Roberts - 7/12/03
R12[m] 70
-15
(c)
x1 ( t) = 4 e − t u( t)
15
m
x 2 ( t) = 4 e − t u( t)
and
Using R xy (τ ) = F −1 ( F ( x(−τ ) ∗ y(τ ))) = F −1 ( X* ( jω ) Y( jω )) 4 * 4 1 4 4 −1 −1 R12 (τ ) = F = 16 F =F 1+ ω2 1 − jω 1 + jω 1 + jω 1 + jω Using 2a F e − a t ← → 2 , Re( a) > 0 a + ω2 −1
2 −τ R12 (τ ) = 8 F −1 = 8e 1 + ω 2 R (τ) 12
8
-4
(d)
x1[ n ] = 2e
−
n 16
2πn sin u[ n ] 8
4
− 2πn R12 [ m] = −6 ∑ e16 sin − u[− n ]e 8 n =−∞ ∞
n
2πn − R12 [ m] = −6 ∑ e sin − e 8 n =−∞ n 16
0
R12 [ m] = −6e
−
m 16
x 2 [ n ] = −3e
and
R12 [ m] = x1[− m] ∗ x 2 [ m] =
−
n 16
2πn π sin − u[ n ] 8 4
∞
∑ x [− n] x [m − n]
n =−∞
( m − n)
( m − n) 16
τ
16
1
2
2π ( m − n ) π sin − u[ m − n ] 8 4
2π ( m − n ) π sin − u[ m − n ] 8 4
n 2πn 2π ( m − n ) π 8 e sin − − u[ m − n ] sin ∑ 8 8 4 n =−∞ 0
Solutions 8-5
M. J. Roberts - 7/12/03
R12 [ m] = −3e
−
m 16
n 2πm π 4πn 2πm π 8 e + − cos − + − u[ m − n ] cos − ∑ 8 8 4 8 4 n =−∞ 0
For m ≥ 0, R12 [ m] = −3e
−
m 16
n 2πm π 4πn 2πm π 8 e + − cos − + − cos − ∑ 8 8 4 8 4 n =−∞
m 16
n 2πm π 4πn 2πm π 8 e + − cos − + − cos − ∑ 8 8 4 8 4 n =−∞
0
For m < 0, R12 [ m] = −3e
−
m
R [m] 12
25
-80
80
m
-25
5. Plot the correlation function for each of the following pairs of power signals. (a)
x1 ( t) = 6 sin(12πt)
and
LCM of the two periods is CTFS representation.
x 2 ( t) = 5 cos(12πt)
1 . Therefore both signals are at the fundamental of the 6
FS R12 (τ ) ← → X1* [ k ] X 2 [ k ]
X1[ k ] = j 3(δ [ k + 1] − δ [ k − 1]) FS R12 (τ ) ← →− j
and
X 2[k ] =
5 (δ[k − 1] + δ[k + 1]) 2
15 (δ[k + 1] − δ[k − 1])(δ[k − 1] + δ[k + 1]) 2
FS R12 (τ ) ← →− j
15 (δ[k + 1] − δ[k − 1]) 2
R12 (τ ) = −15 sin(12πτ ) Alternate Solution:
Solutions 8-6
M. J. Roberts - 7/12/03
1 30 6 sin(12πt)5 cos(12π ( t + τ )) dt = lim ∫ sin(12πt) cos(12π ( t + τ )) dt ∫ T →∞ T T T →∞ T T
R12 (τ ) = lim
15 15 sin(−12πτ ) + sin(24πt + 12πτ )]dt = lim [ ∫ T →∞ T T T →∞ T
R12 (τ ) = lim
R12 (τ ) = lim
T →∞
T 2
∫ [sin(−12πτ ) + sin(24πt + 12πτ )]dt −
T 2
T 2
cos(24πt + 12πτ ) 15 t sin(−12πτ ) − T 24π T −
2
24πT 24πT cos + 12πτ cos − + 12πτ 2 T 15 T 2 R12 (τ ) = lim sin(−12πτ ) − + sin(−12πτ ) + T →∞ T 24π 24π 2 2 R12 (τ ) = 15 sin(−12πτ ) = −15 sin(12πτ ) R (τ) 12
15
-0.5
0.5
τ
-15
(b)
2πn x1[ n ] = 6 sin 12
and
2πn x 2 [ n ] = 5 sin 12
LCM of the two periods is 12. Therefore both signals are at the fundamental of the DTFS representation. FS R12 [ m] ← → X1* [ k ] X 2 [ k ]
X1[ k ] = j 3(comb12 [ k + 1] − comb12 [ k − 1]) and X 2[k ] = j FS R12 [ m] ← →
5 (comb12[k + 1] − comb12[k − 1]) 2
15 (comb12[k + 1] − comb12[k − 1])(comb12[k + 1] − comb12[k − 1]) 2 FS R12 [ m] ← →
15 (comb12[k + 1] + comb12[k − 1]) 2
Solutions 8-7
M. J. Roberts - 7/12/03
2πm R12 [ m] = 15 cos 12 Alternate Solution: 1 N →∞ N
R12 [ m] = lim
2πn
1
2π ( n + m) 12
∑ x [n] x [n + m] = lim N ∑ 6 sin 12 5 sin
n= N
1
2
N →∞
15 ∑ N →∞ N n= N
R12 [ m] = lim
n= N
2πm 4πn + 2πm cos − 12 − cos 12
As N tends to infinity, the second cosine sums to a finite number which is then divided by N and the contribution is zero. Therefore 2πm 2πm R12 [ m] = 15 cos − = 15 cos . 12 12 R12[m] 15
-24
24
m
-15
(c)
x1 ( t) = 6 sin(12πt)
and
LCM of the two periods is CTFS representation.
π x 2 ( t) = 5 sin12πt − 4
1 . Therefore both signals are at the fundamental of the 6
FS R12 (τ ) ← → X1* [ k ] X 2 [ k ]
X1[ k ] = j 3(δ [ k + 1] − δ [ k − 1]) and
1 π − j 2π ( kf 0 ) − jk 5 5 48 1 1 1 1 X 2 [ k ] = j (δ [ k + ] − δ [ k − ])e = j (δ [ k + ] − δ [ k − ])e 4 2 2
[
]
FS R12 (τ ) ← → j 3(δ [ k + 1] − δ [ k − 1]) × j *
π − jk 5 δ [ k + 1] − δ [ k − 1])e 4 ( 2
π − jk 15 1 1 1 1 R12 (τ ) ←→ (δ [ k + ] − δ [ k − ])(δ [ k + ] − δ [ k − ])e 4 2 FS
FS R12 (τ ) ← →
π − jk 15 δ [ k + 1] + δ [ k − 1])e 4 ( 2
Solutions 8-8
M. J. Roberts - 7/12/03
1 π R12 (τ ) = 15 cos12π τ − = 15 cos12πτ − 48 4 Alternate Solution: R12 (τ ) = lim
T →∞
π π 1 30 6 sin(12πt)5 sin12π ( t + τ ) − dt = lim ∫ sin(12πt) sin12π ( t + τ ) − dt ∫ T T T →∞ T T 4 4 π π 15 cos −12πτ + + cos 24πt + 12πτ − dt ∫ T →∞ T T 4 4
R12 (τ ) = lim
T
π 2 sin 24πt + 12πτ − π 15 4 R12 (τ ) = lim t cos −12πτ + + T →∞ T 4 24π T −
π π R12 (τ ) = 15 cos −12πτ + = 15 cos12πτ − 4 4
2
R (τ) 12
15
-0.5
0.5
τ
-15
6. Find the autocorrelations of the following CT and DT energy and power signals and show that, at zero shift, the value of the autocorrelation is the signal energy or power and that all the properties of autocorrelation functions are satisfied. (a)
x( t) = e −3 t u( t) R x (τ ) = x(−τ ) ∗ x(τ ) = e 3τ u(−τ ) ∗ e −3τ u(τ ) F R x (τ ) ← →
1 1 1 = 3 − jω 3 + jω 9 + ω 2
F e − a t ← →
2a a + ω2 2
1 R x (τ ) = e −3 t 6 Even function, maximum at τ = 0. The total signal energy is
Solutions 8-9
M. J. Roberts - 7/12/03 ∞
Ex =
∫e
−3 t
−∞
∞
∞
1 1 1 u( t) dt = ∫ e dt = − e −6 t = − (0 − 1) = = R x (0) . Check. 6 6 6 0 0 2
−6 t
x[ n ] = rect 5 [ n − 5]
(b)
R x [ m] = rect 5 [− m − 5] ∗ rect 5 [ m − 5] sin(−11πF ) j10πF sin(11πF ) − j10πF sin(11πF ) sin(11πF ) sin(11πF ) e e R x [ m] ←→ = = sin(−πF ) sin(πF ) sin(πF ) sin(πF ) sin(πF )
2
F
m R x [ m] = rect 5 [ m] ∗ rect 5 [ m] = 11tri 11 Even function, maximum at m = 0. Energy is Ex = x( t) = rect 2 t −
(c)
∞
∑ rect [n − 5]
n =−∞
5
1 − rect 2 t − 4
2
= 11 = R x [0] . Check.
3 4
3 1 3 1 R x (τ ) = x(−τ ) ∗ x(τ ) = rect 2 −τ − − rect 2 −τ − ∗ rect 2τ − − rect 2τ − 4 4 4 4 F R x (τ ) ← → X( f ) X( f ) *
*
3πf πf 1 f −j 2 1 f −j 2 R x (τ ) ←→ sinc e − sinc e 2 2 2 2 F
3πf πf 1 f −j 2 1 f −j 2 − sinc e sinc e 2 2 2 2
3πf πf πf πf 1 f j2 1 f −j 2 1 f j2 1 f −j 2 − sinc e sinc e sinc e sinc e 2 2 2 2 2 2 2 2 F R x (τ ) ←→ 3πf 3πf 3πf πf 1 j − j j − j 1 1 1 f f f f − sinc e 2 sinc e 2 + sinc e 2 sinc e 2 2 2 2 2 2 2 2 2
f sinc 2 2 F 1 − e − jπf − e jπf + 1) R x (τ ) ← → ( 4 1 1 1 R x (τ ) = tri(2 t) − tri 2 t − + tri 2 t + 2 2 2
Solutions 8-10
M. J. Roberts - 7/12/03
∞
∞
E x = ∫ x( t) dt = ∫ rect 2 t − −∞ −∞ 2
1 − rect 2 t − 4
2
3 dt = ∫ dt = 1 = R x (0) . Check. 4 0 1
7. Find the autocorrelation functions of the following power signals. (a)
x( t) = 5 sin(24πt) − 2 cos(18πt) The period of this signal is X[ k ] = j
1 second. Therefore f 0 = 3 and its CTFS is 3
5 (δ[k + 4] − δ[k − 4]) − (δ[k − 3] + δ[k + 3]) 2
The autocorrelation can be found from 5 R x (τ ) ←→ X [ k ] X[ k ] = j (δ [ k + 4 ] − δ [ k − 4 ]) − (δ [ k − 3] + δ [ k + 3]) 2 FS
FS R x (τ ) ← →
25 (δ[k + 4] + δ[k − 4]) + (δ[k − 3] + δ[k + 3]) 4
R x (τ ) = (b)
2
*
25 cos(24πt) + 2 cos(18πt) 2
2πn 2πn x[ n ] = −4 sin − 2 cos 36 40 The period of this signal is 360. Therefore F0 =
1 and its DTFS is 360
X[ k ] = − j 2(comb 360 [ k + 10] − comb 360 [ k − 10]) − (comb 360 [ k − 9] + comb 360 [ k + 9]) R xy [ m] ←→ X [ k ] Y[ k ] = FS
*
− j 2(comb 360 [ k + 10] − comb 360 [ k − 10])
2
−(comb 360 [ k − 9] + comb 360 [ k + 9])
FS R xy [ m] ← → 4 (comb 360 [ k + 10] − comb 360 [ k − 10]) + (comb 360 [ k − 9] + comb 360 [ k + 9])
2πn 2πn R xy [ m] = 8 cos + 2 cos 36 40
8. A signal is sent from a transmitter to a receiver and is corrupted by noise along the way. The signal shape is of the functional form, f 1 x( t) = A sin(2πf 0 t) rect 0 t − 4 2 f0
Solutions 8-11
M. J. Roberts - 7/12/03
What is the transfer function of a matched filter for this signal? f 1 h( t) = K sin(2πf 0 ( t0 − t)) rect 0 t0 − t − 2 f 0 4 The sine is odd and the rect is even. Therefore f 1 h( t) = −K sin(2πf 0 ( t − t0 )) rect 0 t − t0 + 2 f 0 4
1
4 f − j 2πf t 0 − 2 f 0 j 4 H( f ) = −K (δ ( f + f 0 ) − δ ( f − f 0 ))e − j 2πft 0 ∗ sinc e 2 f0 f0
1
4 f − j 2πf t 0 − 2 f 0 K 4 H( f ) = j (δ ( f − f 0 )e − j 2πf 0 t 0 − δ ( f + f 0 )e j 2πf 0 t 0 ) ∗ sinc e 2 f0 f0 1 1 4 ( f − f 0 ) − j 2π ( f − f 0 ) t 0 − 2 f 0 4 ( f + f 0 ) − j 2π ( f + f 0 ) t 0 − 2 f 0 2K − j 2πf 0 t 0 j 2πf 0 t 0 −e H( f ) = j sinc sinc e e e f 0 f0 f0 1 1 1 1 4 ( f − f 0 ) − j 2πf t 0 − 2 f 0 j 2πf 0 − 2 f 0 4 ( f + f 0 ) − j 2πf t 0 − 2 f 0 − j 2πf 0 − 2 f 0 2K e14243 − sinc e14243 H( f ) = j sinc e e f 0 f0 f0 =−1 =−1 1 4( f + f 0 ) 4( f − f 0 ) 2K − j 2πf t 0 − 2 f 0 e − sinc H( f ) = j sinc f0 f0 f0
9. Find the signal power of the following sums or differences of signals and compare it to the power in the individual signals. How does the comparison relate to the correlation between the two signals that are summed or differenced? (a)
x( t) = sin(2πt) + cos(2πt)
1 1 2 2 2 Px = ∫ sin(2πt) + cos(2πt) dt = ∫ sin (2πt) + cos (2πt) + 2 sin(2πt) cos(2πt) dt T0 T0 T0 T0 144424443 =1
Px = ∫ dt + 2 ∫ sin(2πt) cos(2πt) dt = 1 1 1 4442444 1 3 =0
Solutions 8-12
M. J. Roberts - 7/12/03
1 1 and the power of the cosine is . The power of the sum 2 2 equals the sum of the powers because these two signals are uncorrelated at zero shift. The power of the sin is
(b)
π x( t) = sin(2πt) + cos 2πt − 4
π 2 sin (2πt) + cos2 2πt − 4 π 1 1 Px = ∫ sin(2πt) + cos 2πt − dt = ∫ dt T0 T0 T0 T0 4 π +2 sin(2πt) cos 2πt − 4 2
π 1 − cos( 4πt) + 1 + cos 4πt − 2 1 2 π dt = ∫ 1 + sin dt = 1 + Px = ∫ ≅ 1.707 1 4 2 1 π 2 π +2 sin + sin 4πt − 4 4 1 1 and the power of the cosine is . The power of the sum 2 2 exceeds the sum of the powers because these two signals are partially correlated at zero shift. n (c) x[ n ] = rect 2 [ n ] ∗ comb10 [ n ] − tri ∗ comb10 [ n ] 2 The power of the sin is
1 1 4 2 n Px = x[ n ] = rect 2 [ n ] − tri ∑ ∑ 2 10 n =−5 N 0 n = N0
2
2 2 2.5 1 4 2 1 1 2 2 2 2 2 2 2 1 0 0 0 0 0 0 1 Px = + + + + + + + + + ∑ = 10 = 0.25 2 2 10 n =−5
1 and the triangle-wave signal has 2 a signal power of 0.15. The signal power of the sum is less than sum of the signal powers because the two signals are negatively correlated at zero shift.
The rectangle-wave signal has a signal power of
(d)
n − 5 x[ n ] = rect 2 [ n ] ∗ comb10 [ n ] + tri ∗ comb10 [ n ] 2 1 1 7 2 n − 5 Px = x[ n ] = rect 2 [ n ] + tri ∑ ∑ 2 10 n =−2 N 0 n = N0
2
2 2 6.5 1 2 2 2 2 2 1 1 2 2 Px = 1 + 1 + 1 + 1 + 1 + 0 + + 1 + + 0 2 = = 0.65 2 2 10 10
Solutions 8-13
M. J. Roberts - 7/12/03
1 and the triangle-wave signal has a 2 signal power of 0.15. The power of the sum equals the sum of the powers because the two signals are uncorrelated at zero shift. The rectangle-wave signal has a signal power of
10. Find the crosscorrelation functions of the following pairs of periodic signals. (a)
t − 3 t x 2 ( t) = rect ∗ comb 6 24
t t x1 ( t) = rect ∗ comb and 6 24
FS R12 (τ ) ← → X1* [ k ] X 2 [ k ] ∞
k k X1 ( f ) = 6 sinc(6 f ) × 24 comb(24 f ) = 6 ∑ sinc δ f − 4 24 k =−∞ X 2 ( f ) = 6 sinc(6 f ) × 24 comb(24 f )e
−6πf
πk k k − 4 = 6 ∑ sinc e δ f − 4 24 k =−∞ ∞
Therefore
πk
k X1[ k ] = 6 sinc 4
k − X 2 [ k ] = 6 sinc e 4 . 4
and
πk
k − FS R12 (τ ) ← → X1* [ k ] X 2 [ k ] = 36 sinc 2 e 4 4 Using
t FS w w t 1 → sinc 2 k tri ∗ comb ← w T0 T0 T0 T0 and
FS x( t − t0 ) ← → X[ k ]e − j 2π ( kf 0 ) t 0 ∞ τ − 3 τ − 3 τ R12 (τ ) = 6 tri ∗ ∑ δ (τ − 24 n ) ∗ comb = 144 tri 6 n =−∞ 6 24
(b)
2πn x1[ n ] = sin 2 8
and
1 4πn x1[ n ] = 1 − cos 8 2
2πn x 2 [ n ] = sin 2 10 and
1 4πn x 2 [ n ] = 1 − cos 10 2
The LCM of the two periods is 40. 1 1 X1[ k ] = comb 40 [ k ] − (comb 40 [ k − 5] + comb 40 [ k + 5]) 2 2
Solutions 8-14
M. J. Roberts - 7/12/03
1 1 X 2 [ k ] = comb 40 [ k ] − (comb 40 [ k − 4 ] + comb 40 [ k + 4 ]) 2 2 FS R12 [ m] ← → X1* [ k ] X 2 [ k ] =
R12 [ m] = (c)
1 comb 40 [ k ] 4
1 4
x1 ( t) = e − j10πt and
x 2 ( t) = cos(10πt)
1 The two signals are both periodic with the same period, T0 = . 5 FS Using e j 2π ( k0 f 0 ) t ← → δ [ k − k0 ]
X1[ k ] = δ [ k + 1]
X 2[k ] =
and
1 (δ[k − 1] + δ[k + 1]) 2
1 FS R12 (τ ) ← → X1* [ k ] X 2 [ k ] = δ [ k + 1] 2 1 R12 (τ ) = e − j10πτ 2 11. Find the ESD’s of the following energy signals. (a)
x[ n ] = Aδ [ n − n 0 ] X( F ) = A ⇒ Ψx ( F ) = A 2
(b)
x( t) = e −100 t u( t) X( jω ) =
1 1 ⇒ Ψx ( jω ) = 4 100 + jω 10 + ω 2
2πn 7 x[ n ] = 10 sin u[ n ] 8 12 n
(c)
Using F α n sin(Ω0 n ) u[ n ] ← →
α sin(Ω0 )e − jΩ 1 − 2α cos(Ω0 )e − jΩ + α 2e − j 2Ω
Solutions 8-15
M. J. Roberts - 7/12/03
7 π − jΩ sin e 0.458e − jΩ 8 6 10 X( jΩ) = 10 = 2 1 − 1.515e − jΩ + 0.7656e − j 2Ω 7 π − jΩ 7 − j 2Ω 1 − cos e + e 6 8 4 0.458e − jΩ X( jΩ) = 10 1 − 1.515 cos(Ω) + 0.7656 cos(2Ω) + j1.515 sin(Ω) − j 0.7656 sin(2Ω) Ψx ( jΩ) = 100
(0.458) 2 [1 − 1.515 cos(Ω) + 0.7656 cos(2Ω)]2 + [1.515 sin(Ω) − 0.7656 sin(2Ω)]2
t − t0 x( t) = A tri w
(d)
X( f ) = Aw sinc 2 ( wf )e − j 2πft 0 ⇒ Ψx ( f ) = ( Aw ) sinc 4 ( wf ) 2
12. Find the ESD of the response, y( t) or y[ n ] , of each system with impulse response, h( t) or h[ n ], to the excitation, x( t) or x[ n ] . n
9 x[ n ] = δ [ n ] , h[ n ] = − u[ n ] 10
(a)
X( jΩ) = 1 ⇒ Ψx ( jΩ) = 1
Ψy ( jΩ) = Ψx ( jΩ) H( jΩ) =
and
2
(b)
H( jΩ) =
100
(10 + 9 cos(Ω))
+ 81sin (Ω)
2
2
=
100 181 + 180 cos(Ω)
x( t) = e −100 t u( t) , h( t) = e −100 t u( t) X( jω ) =
1 1 ⇒ Ψx ( jω ) = 4 100 + jω 10 + ω 2 Ψy ( jΩ) = Ψx ( jΩ) H( jΩ)
(c)
10 1 = − jΩ 9 1 + e − jΩ 10 + 9e 10
2
1 = 4 2 10 + ω
x[ n ] = rect 3 [ n ] , h[ n ] = rect 2 [ n − 2]
Using F rect N w [ n ] ← →
H( jω ) =
and
sin(πF (2 N w + 1)) sin(πF )
Solutions 8-16
2
1 100 + jω
M. J. Roberts - 7/12/03
and
X( F ) =
sin( 7πF ) sin 2 ( 7πF ) F ⇒ Ψx ( ) = sin(πF ) sin 2 (πF )
H( F ) =
sin(5πF ) − j 4 πF e sin(πF )
sin 2 ( 7πF ) sin 2 (5πF ) Ψy ( F ) = sin 2 (πF ) sin 2 (πF ) (d)
1 x( t) = 4 e − t cos(2πt) u( t) , h( t) = rect t − 2
Using F e − at cos(ω 0 t) u( t) ← →
X( jω ) =
jω + 1
( jω + 1) 2 + (2π ) 2
=
jω + a
( jω + a) 2 + ω 02
jω + 1 1+ ω2 ⇒ Ψ j ω = ( ) 2 x 2 1 − ω 2 + j 2ω + (2π ) (2π ) 2 + 1 − ω 2 + 4ω 2
[
]
ω ω −j H( jω ) = sinc e 2 2π
Ψy ( jω ) =
[(2π )
1+ ω2 2
+ 1− ω2
]
2
ω sinc 2 2π + 4ω 2
13. Find the PSD’s of these signals. (a)
x( t) = A cos(2πf 0 t + θ ) R x (τ ) =
(b)
[
]
A2 A2 cos(2πf 0τ ) ⇒ Gx ( f ) = δ ( f − f0 ) + δ ( f + f0 ) 2 4
x( t) = 3 rect (100 t) ∗ comb(25 t) Using R xy (τ ) = R x (τ ) =
x(−τ ) ⊗ y(τ ) T
[3rect (−100τ ) ∗ comb(25τ )] ⊗ [3rect (100τ ) ∗ comb(25τ )] 1 25
R x (τ ) = 225 rect (100τ ) ∗ rect (100τ ) ∗ comb(25τ ) = 22500 tri(100τ ) ∗ comb(25τ )
Solutions 8-17
M. J. Roberts - 7/12/03
G x ( f ) = 22500
1 f 1 f f 2 f sinc 2 comb = 9 sinc comb 100 25 25 100 25 100
2πn x[ n ] = 8 sin 12
(c)
1 1 2πm R x [ m] = 32 cos ⇒ G x ( F ) = 16 comb F − + comb F + 12 12 12 (d)
x[ n ] = 3 rect 4 [ n ] ∗ comb 20 [ n ]
x[− m] ⊗ x[ m] ( 3 rect 4 [− m] ∗ comb 20 [ m]) ⊗ ( 3 rect 4 [ m] ∗ comb 20 [ m]) = N0 20 3 rect 4 [ m] ∗ 3 rect 4 [ m] ∗ comb 20 [ m] 9 R x [ m] = = rect 4 [ m] ∗ rect 4 [ m] ∗ comb 20 [ m] 20 20 R x [ m] =
9 sin 2 (9πF ) comb(20 F ) G x (F ) = 20 sin 2 (πF ) 14. Find the PSD of the response, y( t) or y[ n ] , of each system with impulse response, h( t) or h[ n ], to the excitation, x( t) or x[ n ] . − π x( t) = 4 cos 32πt − , h( t) = e 10 u( t) 4 t
(a)
[
]
G x ( f ) = 4 δ ( f − 16) + δ ( f + 16) and
H( f ) =
1 10 = 1 + j 2πf 1 + j 20πf 10
G y ( f ) = 4[δ ( f − 16) + δ ( f + 16)] G y ( f ) = 400 (b)
100
1 + (20πf )
2
δ ( f − 16) δ ( f + 16) = 400 + 2 2 1 + (20πf ) 1 + (20πf )
δ ( f − 16) + δ ( f + 16) 2 1 + ( 320π )
x( t) = 4 comb(2 t) , h( t) = rect ( t − 1) R x (τ ) =
4 comb(2 t) ⊗ 4 comb(2 t) = 32[comb(2 t) ⊗ comb(2 t)] 1 2
Solutions 8-18
M. J. Roberts - 7/12/03
1 R x (τ ) = 32 δ ( t) ∗ comb(2 t) = 16 comb(2 t) 2 f G x ( f ) = 8 comb 2 H( f ) = sinc( f )e − j 2πf f G y ( f ) = 8 comb sinc 2 ( f ) 2 n
(c)
11 x[ n ] = 2 comb 8 [ n ] , h[ n ] = u[ n − 1] 12 R x [ m] =
2 comb 8 [− m] ⊗ 2 comb 8 [ m] 1 1 = (δ [ m] ⊗ comb 8 [ m]) = comb 8 [ m] 8 2 2
1 G x ( F ) = comb(8 F ) 2 (d)
x[ n ] = (−0.9) u[ n ] , h[ n ] = (0.5) u[ n ] n
n
G x ( F ) = 0 because x[ n ] is an energy signal, not a power signal. Therefore G y ( F ) = 0. 15. Plot correlograms of the following pairs of CT and DT signals. (a)
x1 ( t) = tri 4 t −
1 − tri 4 t − 4 x
3 ∗ comb( t) 4
and
2
1
-1
1
x
1
-1
(b)
n − 8 n − 24 x1[ n ] = tri ∗ comb 32 [ n ] − tri 8 8 2πn x 2 [ n ] = cos 32
Solutions 8-19
and
x 2 ( t) = sin(2πt)
M. J. Roberts - 7/12/03
x
2
1
-1
1
x1
-1
(c)
1 3 x1 ( t) = tri 4 t − − tri 4 t − ∗ comb( t) 4 4 1 x 2 ( t) = tri( 4 t) − tri 4 t − ∗ comb( t) 2 x
and
2
1
-1
1
x
1
-1
(d)
n − 24 n − 8 x1[ n ] = rect ∗ comb 32 [ n ] − rect 16 16 2πn x 2 [ n ] = sin 32 x
and
2
1
-1
1
x1
-1
16. Plot a correlogram for the following sets of samples from two signals, x and y. In each case, from the nature of the correlogram what relationship, if any, exists between the two sets of data? (a)
x = {6,5,8,-2,3,-10,9,-2,-4,3,-2,6,0,-5,-7,1,9,9,4,-6} y = {-1,-10,-4,4,5,-2,-3,-5,-9,2,6,-5,-1,-10,-9,0,4,-10,9,-1}
(b) x = {4,6,0,0,5,-6,8,-9,0,8,7,2,-5,-3,-4,-4,8,0,4,7} y = {-11,-13,3,-1,-8,10,-16,16,1,-17,-14,-3,9,7,12,9,-17,1,-8,-17} (c) x = {0,6,11,16,19,20,19,16,11,6,-0,-7,-12,-17,-20,-20,-20,-17,-12,-7} y = {19,15,10,8,3,-9,-12,-19,-19,-25,-19,-17,-12,-5,-1,5,8,12,17,20}
Solutions 8-20
M. J. Roberts - 7/12/03
(a)
(b)
y
y
20
20
-20
20
x
-20
x
20
-20
-20
(c) y 20
-20
20
x
-20
17. Plot the correlation function for each of the following pairs of energy signals. x1 ( t) = rect ( t) sin(10πt)
(a)
and
x 2 ( t) = rect ( t) cos(10πt)
X1 ( f ) = sinc( f ) ∗
j j δ ( f + 5) − δ ( f − 5)] = [sinc( f + 5) − sinc( f − 5)] [ 2 2
X 2 ( f ) = sinc( f ) ∗
1 1 δ ( f − 5) + δ ( f + 5)] = [sinc( f − 5) + sinc( f + 5)] [ 2 2
X1* ( f ) X 2 ( f ) =
j [sinc( f + 5) − sinc( f − 5)][sinc( f − 5) + sinc( f + 5)] 4
X1* ( f ) X 2 ( f ) = R12 (τ ) =
[
]
j sinc 2 ( f + 5) − sinc 2 ( f − 5) 4
tri(τ ) sin(10πτ ) j j tri(τ )e − j10πτ − tri(τ )e j10πτ = tri(τ )(− j 2 sin(10πτ )) = 2 4 4
[
]
R12(τ) 0.5
-2
2
τ
-0.5
(b)
x1[ n ] = δ [ n − 1] − δ [ n + 1]
and
x 2 [ n ] = −δ [ n − 1] + δ [ n + 1]
R12 [ m] = x1[− m] ∗ x 2 [ m] R12 [ m] = (δ [− m − 1] − δ [− m + 1]) ∗ (−δ [ m − 1] + δ [ m + 1])
Solutions 8-21
M. J. Roberts - 7/12/03
−δ [− m − 1] ∗ δ [ m − 1] + δ [− m + 1] ∗ δ [ m − 1] R12 [ m] = +δ [− m − 1] ∗ δ [ m + 1] − δ [− m + 1] ∗ δ [ m + 1] R12 [ m] = −δ [−( m − 1) − 1] + δ [−( m − 1) + 1] + δ [−( m + 1) − 1] − δ [−( m + 1) + 1] R12 [ m] = −δ [− m] + δ [− m + 2] + δ [− m − 2] − δ [− m] R12 [ m] = −2δ [ m] + δ [− m + 2] + δ [− m − 2] R12 [ m] = δ [ m + 2] − 2δ [ m] + δ [ m − 2] R [m] 12 2
-5
5
m
-2
(c)
x1 ( t) = e − t x1 ( t) = e
X1 ( f ) = π e
−π
(
2
t −π π
π f)
2
x 2 ( t) = e −2 t
and
2
2
= πe
x 2 ( t) = e
and −π f 2
2
2
π R12 (τ ) = 2
f2
2
π X2( f ) = e 2
and
2 2 π − π2 R12 (τ ) = F −1 π e −π f e 2
2 −π t π
π −π f 2
2
π − π2 = e 2
3 −π −π 2 f 2 −1 π −1 π 2 e =F = F 2 e 2
2 −π e 3π
2 τ 3π
2
=
2
3π 2
f
π − 23τ 2 e 3
R (τ) 12
1
-4
4
τ
18. Plot the correlation function for each of the following pairs of power signals. (a)
2πn x1[ n ] = −3 sin 20
and
2πn x 2 [ n ] = 8 sin 10
Solutions 8-22
2
f2
M. J. Roberts - 7/12/03
LCM of the two periods is 20. Therefore x1[ n ] is at the fundamental and x 2 [ n ] is at the second harmonic of the DTFS representation. FS R12 [ m] ← → X1* [ k ] X 2 [ k ]
X1[ k ] = − j and
3 (comb20[k + 1] − comb20[k − 1]) 2
X 2 [ k ] = j 4 (comb 20 [ k + 2] − comb 20 [ k − 2]) FS R12 [ m] ← →−
12 (comb20[k + 1] − comb20[k − 1])(comb20[k + 2] − comb20[k − 2]) 2 FS R12 [ m] ← → 0 ⇒ R12 [ m] = 0
Old Solution: 1 N →∞ N
R12 [ m] = lim
2πn
1
2π ( n + m) 10
∑ x [n] x [n + m] = lim N ∑ −3sin 20 8 sin
n= N
1
2
N →∞
R12 [ m] = − lim
N →∞
12 ∑ N →∞ N n= N
R12 [ m] = − lim
n= N
24 2πn 2π ( n + m) sin sin ∑ 20 N n= N 10
2πn 2πm 6πn 2πm cos − 20 − 10 − cos 20 + 10 = 0 R12[m] 1
-20
20
m
-1
(b)
x1 ( t) = rect ( 4 t) ∗ comb( t)
and
x 2 ( t) = rect ( 4 t) ∗ comb( t)
LCM of the two periods is 1. Therefore both signals are at the fundamental of the CTFS representation. FS R12 (τ ) ← → X1* [ k ] X 2 [ k ] Using t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 X1[ k ] =
1 k sinc 4 4
and
X 2[k ] =
1 k sinc 4 4
Solutions 8-23
M. J. Roberts - 7/12/03
FS R12 (τ ) ← →
1 k sinc 2 4 16
Then, using t FS w w t 1 → sinc 2 k tri ∗ comb ← w T0 T0 T0 T0 R12 (τ ) =
tri( 4 t) ∗ comb( t) 4 R12(τ) 0.25
-2
2
τ
(c) 1 t x1 ( t) = 4 rect ( t) ∗ comb − 2 2 2
and
1 t x 2 ( t) = 4 rect ( t − 1) ∗ comb − 2 . 2 2
FS R12 (τ ) ← → X1* [ k ] X 2 [ k ]
LCM of the two periods is 2. The fundamental frequencies of both signals are at the fundamental frequency in the CTFS. Using t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 k X1[ k ] = 2 sinc − 2δ [ k ] 2
and
k X 2 [ k ] = 2 sinc e − jπk − 2δ [ k ] 2
k k FS R12 (τ ) ← → 2 sinc − 2δ [ k ] 2 sinc e − jπk − 2δ [ k ] 2 2 k k k k FS R12 (τ ) ← → 4 sinc sinc e − jπk − sinc δ [ k ] − δ [ k ] sinc e − jπk + δ [ k ]δ [ k ] 2 2 2 2 k FS R12 (τ ) ← → 4 sinc 2 e − jπk − δ [ k ] 2 Using t FS w w t 1 → sinc 2 k tri ∗ comb ← w T0 T0 T0 T0
Solutions 8-24
M. J. Roberts - 7/12/03
1 t − 1 R12 (τ ) = 8tri( t) ∗ comb −4 2 2 ∞
R12 (τ ) = −4 + 8 ∑ tri( t − 1 − 2 n ) n =−∞
Graphical Solution: 1 t x1 ( t) = 4 rect ( t) ∗ comb − 2 2 2
1 t x 2 ( t) = 4 rect ( t − 1) ∗ comb − 2 . 2 2
and
The easiest way to find the correlation function for these two signals is to first graph them, then find the correlation graphically.
x1(t)
x2(t) 2
2 -2
2
t
2
t
-2
For periodic functions, the correlation function is the average value of the product of the two functions over one period of the product of the two functions. In this case the periods of the two functions are the same, 2, and the period of the product is also 2. With zero shift of x 2 ( t) the product is -4 everywhere and therefore the average of the product is also -4. If x 2 ( t) is shifted by exactly half a period, 1, the product is +4 everywhere and the average value is +4. For any shift between these two shifts, the average value of the product varies linearly with the shift amount between these two extreme values of the correlation. Also the correlation function is periodic with the same period as these two functions. Therefore the correlation function is all illustrated below.
R12(τ) 4
2 -4
τ
∞
R12 (τ ) = −4 + 8 ∑ tri( t − 1 − 2 n ) . Check n =−∞
19. Find the autocorrelations of the following CT and DT energy and power signals and show that, at zero shift, the value of the autocorrelation is the signal energy or power and that all the properties of autocorrelation functions are satisfied. (a)
x[ n ] = δ [ n ] + δ [ n − 1] + δ [ n − 2] + δ [ n − 3]
δ [− m] + δ [− m − 1] δ [ m] + δ [ m − 1] R x [ m] = x[− m] ∗ x[ m] = ∗ +δ [− m − 2] + δ [− m − 3] +δ [ m − 2] + δ [ m − 3] Solutions 8-25
M. J. Roberts - 7/12/03
δ [− m] + δ [− m + 1] + δ [− m + 2] + δ [− m + 3] +δ [− m − 1] + δ [− m] + δ [− m + 1] + δ [− m + 2] R x [ m] = +δ [− m − 2] + δ [− m − 1] + δ [− m] + δ [− m + 1] +δ [− m − 3] + δ [− m − 2] + δ [− m − 1] + δ [− m] 4δ [ m] + 3δ [− m + 1] + 2δ [− m + 2] + δ [− m + 3] R x [ m] = +3δ [− m − 1] + 2δ [− m − 2] + δ [− m − 3] R x [ m] = δ [ m + 3] + 2δ [ m + 2] + 3δ [ m + 1] + 4δ [ m] + 3δ [ m − 1] + 2δ [ m − 2] + δ [ m − 3] Ex =
∞
∑
x[ n ] = 2
n =−∞
∞
∑ δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3]
2
n =−∞
= 4 = R x [0] . Check.
x( t) = A cos(2πf 0 t + θ )
(b)
FS R x (τ ) ← → X* [ k ] X[ k ]
θ
− j 2π ( kf 0 ) A A 2πf 0 X[ k ] = (δ [ k − 1] + δ [ k + 1])e = (δ [ k − 1] + δ [ k + 1])e − jkθ 2 2 FS R x (τ ) ← →
A A δ [ k − 1] + δ [ k + 1])e + jkθ (δ [ k − 1] + δ [ k + 1])e − jkθ ( 2 2 FS R x (τ ) ← →
A2 (δ[k − 1] + δ[k + 1]) 4
A2 R x (τ ) = cos(2πf 0τ ) 2
Old Solution: R x (τ ) = lim
T →∞
1 T
T 2
∫ x(t) x(t + τ )dt = lim
T →∞
T − 2
R x (τ ) = lim
T →∞
2
A 2T
T 2 2
A T
∫ cos(2πf t + θ ) cos(2πf (t + τ ) + θ )dt 0
0
T − 2
T 2
∫ [cos(−2πf τ ) + cos(4πf t + 2πf τ + 2θ )]dt 0
−
0
o
T 2
T
sin( 4πf 0 t + 2πf oτ + 2θ ) 2 A t cos(−2πf 0τ ) + R x (τ ) = lim T →∞ 2T 4πf 0 − T 2
2
Solutions 8-26
M. J. Roberts - 7/12/03
A2 R x (τ ) = cos(2πf 0τ ) 2 R x (0) =
A2 which is the average signal power of any sinusoid of amplitude A 2
(c)
x[ n ] = comb12 [ n ] FS R x [ m] ← → X* [ k ] X[ k ]
FS Using comb N 0 [ n ] ← →
1 N0 *
1 1 1 R x [ m] ←→ = 2 N0 N0 N0 FS
R x [ m] = R x [0] = Old Solution: 1 N →∞ N
R x [ m] = lim
comb12 [ m] 12
1 . Check. 12
1 ∑ x[n] x[n + m] = lim N ∑ comb [n]comb [n + m] N →∞
n= N
n= N
12
12
The product, comb12 [ n ] comb12 [ n + m], is non-zero only if m is a integer multiple of 12. The summation is over a period of length N. There are two cases. Case I. N even Let the summation be from −
N N to − 1. Then 2 2
R x [ m] = lim
N →∞
1 N
N −1 2
∑ comb [n]comb [n + m] . 12
n =−
12
N 2
In that range, for m an integer multiple of 12, the number of impulses in the product N is the greatest integer in . Therefore 12 N 1 Greatest Integer in , m an integer multiple of 12 12 R x [ m] = lim N →∞ N 0 , otherwise
Solutions 8-27
M. J. Roberts - 7/12/03
1 , m an integer multiple of 12 comb12 [ m] R x [ m] = 12 = 12 0 , otherwise Case II.
N odd
Let the summation be from −
N −1 N −1 to . Then 2 2
1 N →∞ N
R x [ m] = lim
N −1 2
∑
N −1 n =− 2
comb12 [ n ] comb12 [ n + m] .
The rest of the analysis is the same and so is the answer (in the limit). 1 , m an integer multiple of 12 comb12 [ m] R x [ m] = 12 . = 12 0 , otherwise
20. Find and sketch the autocorrelation function of 1 t x( t) = 10 rect (2 t) ∗ comb . 4 4 FS R12 (τ ) ← → X1* [ k ] X 2 [ k ]
The period is 4. Using t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 X[ k ] =
5 k sinc 8 4 2
k 5 R12 (τ ) ←→ sinc 2 4 8 FS
Using t FS 1 w 1 t 1 tri ∗ comb ← → sinc 2 k w w T0 T0 T0 T0 R12 (τ ) =
1 25 t tri(2 t) ∗ comb 4 4 2
Solutions 8-28
M. J. Roberts - 7/12/03
R12 (0) =
25 . Check. 2
Old Solution: Check to be sure that its value at a shift of zero is the same as its average signal power. The signal is graphed below.
x(t) 10 1 4
4
t
This is a periodic signal. Therefore its autocorrelation function is also periodic with the same period. The maximum autocorrelation value occurs at zero shift and that value is the average of the square of the signal (its average power), 12.5. The autocorrelation varies linearly with shift up to a shift of one-half at which time it is zero. This is true for positive 1 or negative shifts of up to one-half. The autocorrelation for shifts, < τ < 2 , is zero. This 2 pattern repeats periodically. Therefore the autocorrelation function is as illustrated below.
Rx(τ) 12.5 1 2 R12 (τ ) =
4
τ
1 25 t tri(2 t) ∗ comb 4 4 2
21. Find all cross correlation and autocorrelation functions for these three signals: x1 ( t) = cos(2πt) , x 2 ( t) = sin(2πt) , x 3 ( t) = cos( 4πt) Check your autocorrelation answers by finding the average power of each signal. FS R12 (τ ) ← → X1* [ k ] X 2 [ k ]
The LCM of the periods is 1 second. Therefore x1 ( t) and x 2 ( t) are at the fundamental of the CTFS and x 3 ( t) is at the second harmonic.
Solutions 8-29
M. J. Roberts - 7/12/03
X1[ k ] =
j 1 1 δ [ k − 1] + δ [ k + 1]) , X 2 [ k ] = (δ [ k + 1] − δ [ k − 1]) , X 3 [ k ] = (δ [ k − 2] + δ [ k + 2]) ( 2 2 2 FS R12 (τ ) ← →
j 1 δ [ k − 1] + δ [ k + 1]) (δ [ k + 1] − δ [ k − 1]) ( 2 2
FS R12 (τ ) ← →
j (δ[k + 1] − δ[k − 1]) 4
1 R12 (τ ) = sin(2πτ ) 2 FS R13 (τ ) ← → X1* [ k ] X 3 [ k ] FS R13 (τ ) ← →
1 1 δ [ k − 1] + δ [ k + 1]) (δ [ k − 2] + δ [ k + 2]) = 0 ( 2 2 R13 (τ ) = 0 FS R 23 (τ ) ← → X*2 [ k ] X 3 [ k ]
FS R 23 (τ ) ← →−
1 j δ [ k + 1] − δ [ k − 1]) (δ [ k − 2] + δ [ k + 2]) = 0 ( 2 2
R 23 (τ ) = 0 FS R1 (τ ) ← → X1* [ k ] X1[ k ] FS R1 (τ ) ← →
1 (δ[k − 1] + δ[k + 1]) 4
R1 (τ ) =
cos(2πτ ) 2
FS R 2 (τ ) ← → X*2 [ k ] X 2 [ k ] FS R 2 (τ ) ← →−
1 j j δ [ k + 1] − δ [ k − 1]) (δ [ k + 1] − δ [ k − 1]) = (δ [ k + 1] + δ [ k − 1]) ( 4 2 2
R 2 (τ ) =
cos(2πτ ) 2
FS R 3 (τ ) ← → X*3 [ k ] X 3 [ k ] FS R 3 (τ ) ← →
1 1 1 δ [ k − 2] + δ [ k + 2]) (δ [ k − 2] + δ [ k + 2]) = (δ [ k − 2] + δ [ k + 2]) ( 4 2 2
Solutions 8-30
M. J. Roberts - 7/12/03
R 3 (τ ) = Old Solutions:
cos( 4πτ ) 2
Cross correlation of x1 with x 2 : 1 T →∞ T
T 2
∫ x (t) x (t + τ )dt = lim T ∫ cos(2πt) sin[2π (t + τ )]dt
R12 (τ ) = lim
1 T →∞ 2T
R12 (τ ) = lim
T 2
1
1
−
2
T 2
T →∞
−
T 2
T 2
T 2
1 ∫ [sin(2πτ ) + sin(4πt + 2πτ )]dt = lim 2T ∫ [sin(2πτ ) + sin(2πτ + 4πt)]dt
−
T →∞
T 2
−
R12 (τ ) = lim
T →∞
1 2T
T 2
T 2
cos(2πτ + 4πt) t sin(2πτ ) − T 4π −
2
cos(2πτ + 2πT ) T cos(2πτ − 2πT ) T + sin(2πτ ) + 2 sin(2πτ ) − 4π 4π 2 sin(2πτ ) R12 (τ ) = 2 Cross correlation of x1 with x 3 : 1 R12 (τ ) = lim T →∞ 2T
1 T →∞ T
R13 (τ ) = lim
T 2
T 2
∫ x (t) x (t + τ )dt = lim T ∫ cos(2πt) cos[4π (t + τ )]dt 1
1
−
3
T →∞
T 2
−
1 T →∞ 2T
R13 (τ ) = lim
R13 (τ ) = lim
T →∞
T 2
T 2
∫ [cos(−2πt − 4πτ ) + cos(6πt + 4πτ )]dt −
T 2
T 2
1 sin(−2πt − 4πτ ) sin(6πt + 4πτ ) + T 2T 6π −2π −
2
1 sin(−πT − 4πτ ) sin( 3πT + 4πτ ) sin(πT + 4πτ ) sin(−3πT + 4πτ ) + − R13 (τ ) = lim − T →∞ 2T 6π 6π −2π −2π R13 (τ ) = 0 Cross correlation of x 2 with x 3 : 1 T →∞ T
R23 (τ ) = lim
T 2
T 2
∫ x (t) x (t + τ )dt = lim T ∫ sin(2πt) cos[4π (t + τ )]dt 1
2
−
3
T 2
R23 (τ ) = lim
T →∞
T →∞
−
1 2T
T 2
T 2
∫ [sin(−2πt − 4πτ ) + sin(6πt + 4πτ )]dt −
T 2
Solutions 8-31
M. J. Roberts - 7/12/03
T
1 R23 (τ ) = lim T →∞ 2T
cos(−2πt − 4πτ ) cos(6πt + 4πτ ) 2 − − T 6π −2π −
2
1 R23 (τ ) = lim T →∞ 2T
cos(−πT − 4πτ ) cos( 3πT + 4πτ ) cos(πT − 4πτ ) cos(−3πT + 4πτ ) − + + − 6π 6π −2π −2π R23 (τ ) = 0
Autocorrelation of x1: 1 T →∞ T
R1 (τ ) = lim
T 2
T 2
∫ x (t) x (t + τ )dt = lim T ∫ cos(2πt) cos[2π (t + τ )]dt 1
1
−
1
T →∞
T 2
−
1 T →∞ 2T
T 2
T 2
∫ [cos(−2πτ ) + cos(4πt + 2πτ )]dt
R1 (τ ) = lim
T 2
−
T
sin( 4πt + 2πτ ) 2 2 t cos ( πτ ) + T 4π −
1 R1 (τ ) = lim T →∞ 2T
2
R1 (τ ) = lim
T →∞
sin(2πT + 2πτ ) sin(−2πT + 2πτ ) 1 T cos(2πτ ) + − 4π 4π 2T cos(2πτ ) R1 (τ ) = 2
Autocorrelation of x 2 : 1 T →∞ T
T 2
T 2
∫ x (t) x (t + τ )dt = lim T ∫ sin(2πt) sin[2π (t + τ )]dt
R2 (τ ) = lim
1
2
−
2
T →∞
T 2
−
1 T →∞ 2T
T 2
T 2
∫ [cos(−2πτ ) − cos(4πt + 2πτ )]dt
R2 (τ ) = lim
−
T 2
T
1 R2 (τ ) = lim T →∞ 2T
sin( 4πt + 2πτ ) 2 2 t cos ( πτ ) − T 4π −
2
R2 (τ ) = lim
T →∞
sin(2πT + 2πτ ) sin(−2πT + 2πτ ) 1 T cos(2πτ ) − + 4π 4π 2T cos(2πτ ) R2 (τ ) = 2
Autocorrelation of x 3 : 1 T →∞ T
R3 (τ ) = lim
T 2
T 2
∫ x (t) x (t + τ )dt = lim T ∫ cos(4πt) cos[4π (t + τ )]dt 1
3
−
T 2
3
T →∞
−
T 2
Solutions 8-32
M. J. Roberts - 7/12/03
1 T →∞ 2T
T 2
∫ [cos(−4πτ ) + cos(8πt + 4πτ )]dt
R3 (τ ) = lim
R3 (τ ) = lim
T →∞
−
1 2T
T 2
T 2
sin(8πt + 4πτ ) t cos( 4πτ ) + T 8π −
2
1 R3 (τ ) = lim T →∞ 2T
sin( 4πT + 4πτ ) sin(−4πT + 4πτ ) − T cos( 4πτ ) + 8π 8π cos( 4πτ ) R3 (τ ) = 2
22. Find and sketch the crosscorrelation between a unit-amplitude, one Hz cosine and a 50% duty-cycle square wave which has a peak-to-peak amplitude of two, a period of one, an average value of zero and is an even function. FS R12 (τ ) ← → X1* [ k ] X 2 [ k ]
x1 ( t) = cos(2πt)
and
x 2 ( t) = 2 rect (2 t) ∗ comb( t) − 1
Using t FS w w t 1 → sinc k rect ∗ comb ← w T0 T0 T0 T0 X1[ k ] =
1 ( δ[k − 1] + δ[k + 1]) 2 FS R12 (τ ) ← →
and
k X 2 [ k ] = sinc − δ [ k ] 2
1 k δ [ k − 1] + δ [ k + 1]) sinc − δ [ k ] ( 2 2
1 1 1 FS R12 (τ ) ← → sinc δ [ k − 1] + sinc − δ [ k + 1] 2 2 2 1 1 FS R12 (τ ) ← → sinc ( δ [ k − 1] + δ [ k + 1]) 2 2 1 R12 (τ ) = sinc cos(2πτ ) 2 Old Solution: T0 4
T0
4 4 sin(2π ( kf 0 ) t) 4 2 kπ k X c [ k ] = ∫ cos(2π ( kf 0 ) t) dt = sin = sinc = 2 T0 0 T0 2π ( kf 0 ) kπ 2 0
Solutions 8-33
M. J. Roberts - 7/12/03
X s[ k ] = 0 , because the function is even All the cosines in the trigonometric expression for the square wave, except the one at the fundamental frequency of one Hz, are orthogonal to the one-Hz cosine over one period. Therefore the crosscorrelation is simply R(τ ) =
1 2
1 2
1 1 1 cos(2πt) sinc cos(2π ( t + τ )) dt = 2 sinc ∫ [cos(−2πτ ) + cos( 4πt + 2πτ )]dt ∫ 2 2 0 1 1 −
2 1
sin( 4πt + 2πτ ) 2 1 1 = sinc cos(2πτ ) R(τ ) = 2 sinc t cos(2πτ ) + 2 2 4π 0
R(τ) 2 π
1
τ
23. Find and sketch the ESD of each of these signals: (a)
t x( t) = A rect w
Ψx ( f ) = X( f ) = Aw sinc( wf ) = A 2 w 2 sinc 2 ( wf ) 2
2
Ψx ( f ) A2w 2 f
1 w (b)
t + 1 x( t) = A rect w Ψx ( f ) = X( f ) = Aw sinc( wf )e j 2πf = A 2 w 2 sinc 2 ( wf ) 2
2
same as (a)
Solutions 8-34
M. J. Roberts - 7/12/03
(c)
t x( t) = A sinc w Ψx ( f ) = X( f ) = Aw rect ( wf ) = A 2 w 2 rect ( wf ) 2
2
Ψx ( f ) A2w 2 f
1 2w 2
(d)
x( t) =
1 − t2 e 2π
From a table of Fourier pairs, F e −πt ← → e −πf 2
Using the linearity property,
2
1 −πt 2 F 1 −πf 2 e ←→ e 2π 2π
Using the time-scaling property, 2
1 − t2 e = 2π
2π t
1 −π e 2π
2 −π Ψx ( f ) = X( f ) = e (
2π f
)2
2
2
−π F ← →e (
=e
−2π
(
2π f
)2
2π f
)2
= e −( 2πf )
Ψx ( f ) 1
1 2 2π 24. Find the PSD's of (a)
x( t ) = A
Solutions 8-35
f
2
M. J. Roberts - 7/12/03
1 T →∞ T
T 2
Rx (τ ) = lim
T 2 2
A T →∞ T
∫ x(t) x(t + τ )dt = lim −
T 2
∫ dt = A
−
Gx ( f ) = F [ Rx (τ )] = A2δ ( f ) (b)
2
T 2
x(t ) = A cos(2πf0 t ) FS R x (τ ) ← → X* [ k ] X[ k ]
X[ k ] =
A (δ[k − 1] + δ[k + 1]) 2
FS R x (τ ) ← →
A2 (δ[k − 1] + δ[k + 1]) 4
A2 R x (τ ) = cos(2πf 0τ ) 2 Gx ( f ) =
[
]
A2 δ ( f − f0 ) + δ ( f + f0 ) 4
Old Solution: 1 T →∞ T
Rx (τ ) = lim
T 2
T 2 2
A T →∞ T
∫ x(t) x(t + τ )dt = lim
T − 2
0
−
0
T 2
T 2
∫ [cos(−2πf τ ) + cos(4πf t + 2πf τ )]dt
2
A T →∞ 2T
Rx (τ ) = lim
∫ cos(2πf t) cos[2πf (t + τ )]dt
o
−
o
o
T 2
T
A2 Rx (τ ) = lim T →∞ 2T
sin( 4πf o t + 2πf oτ ) 2 t f 2 cos − π τ + ( ) o 4πf o − T
2
sin(2πf oT + 2πf oτ ) − sin(−2πf oT + 2πf oτ ) A T cos(2πf oτ ) + Rx (τ ) = lim T →∞ 2T 4πf o A2 A2 Rx (τ ) = δ ( f − f0 ) + δ ( f + f0 ) cos(2πfoτ ) , Gx ( f ) = 2 4 2
[
(c)
x(t ) = A sin(2πf0 t ) FS R x (τ ) ← → X* [ k ] X[ k ]
X[ k ] = j
A (δ[k + 1] − δ[k − 1]) 2
Solutions 8-36
]
M. J. Roberts - 7/12/03
A2 R x (τ ) ←→ (δ [ k − 1] + δ [ k + 1]) 4 FS
R x (τ ) = Gx ( f ) =
A2 cos(2πf 0τ ) 2
[
]
A2 δ ( f − f0 ) + δ ( f + f0 ) 4
Old Solution: Rx (τ ) = lim
T →∞
1 T
T 2
∫ x(t) x(t + τ )dt = lim
T →∞
T − 2
Rx (τ ) = lim
T →∞
2
A 2T
T 2 2
A T
∫ sin(2πf t) sin[2πf (t + τ )]dt 0
0
T − 2
T 2
∫ [cos(−2πf τ ) − cos(4πf t + 2πf τ )]dt o
−
o
o
T 2
and by the same logic as in b Rx (τ ) =
A2 cos(2πfoτ ) 2
and Gx ( f ) =
[
]
A2 δ ( f − f0 ) + δ ( f + f0 ) 4
25. Which of the following functions could not be the autocorrelation function of a real signal and why? (a) (b) (c) (d)
R(τ ) = tri(τ ), Even function, Maximum at zero, Transform is non-negative, ok. R(τ ) = A sin(2πf0 t ) , Odd function, cannot be an autocorrelation. R(τ ) = rect (τ ) , Even function, Maximum at zero Transform is not non-negative, cannot be an autocorrelation. R(τ ) = A sinc( Bτ ) , Even function, Maximum at zero, Transform is nonnegative, ok.
Solutions 8-37
Chapter 9 - The Laplace Transform Solutions 1. Sketch the pole-zero plot and region of convergence (if it exists) for these signals.
ω
ROC
(a)
x( t) = e
(b)
x( t) = e 3 t cos(20πt) u(− t)
−8 t
[s]
u( t)
σ
s = −8
e j 20πt − e − j 20πt e 3 t e j 20πt − e 3 t e − j 20πt x( t) = e u(− t) = u(− t) 2 2 3t
ω
[s] s = 3+j20π
e( 3 + j 20π ) t − e( 3 − j 20π ) t x( t) = u(− t) 2
ROC σ
s = 3-j20π
Solutions 9-1
ω
(c)
x( t) = e u(− t) − e 2t
−5 t
[s]
ROC
u( t)
σ
s = −5 s = 2
2. Starting with the definition of the Laplace transform, L (g( t)) = G( s) =
∞
∫ g(t)e
− st
dt ,
0−
find the Laplace transforms of these signals. (a)
x( t) = e t u( t) X( s) =
∞
∫ x(t)e
∞
− st
dt =
0−
∫ e u(t)e t
∞
− st
dt =
0−
∫e
−( s −1) t
dt
0−
∞
e −( s−1) t 1 1 = X( s) = , Re( s) = σ > 1 = 0 − −( s − 1) 0 − −( s − 1) s − 1 (b) X( s) =
x( t) = e 2 t cos(200πt) u( t) ∞
− st ∫ x(t)e dt =
0−
∞
2t − st ∫ e cos(200πt) u(t)e dt =
0−
X( s) =
∞
2t ∫e
0−
e j 200πt + e − j 200πt − st e dt 2
∞
1 −( s− 2 − j 200π ) t e + e −( s− 2 + j 200π ) t dt 2 0∫− ∞
e −( s− 2 + j 200π ) t 1 e −( s− 2 − j 200π ) t X( s) = + 2 −( s − 2 − j 200π ) −( s − 2 + j 200π ) 0 −
Solutions 9-2
X( s) =
1 1 1 + , Re( s) = σ > 2 2 s − 2 − j 200π s − 2 + j 200π X( s) =
(c)
x( t) = ramp( t) X( s) =
s− 2 , Re( s) = σ > 2 (s − 2) + (200π ) 2 2
∞
− st ∫ x(t)e dt =
0
−
∞
− st ∫ ramp(t)e dt =
0
−
∞
∫ te
0
− st
dt
−
Using e ax (ax − 1) a2 ∞ e − st 1 X( s) = = 2 , Re( s) = σ > 0 2 (− st − 1) (− s) 0− s ax ∫ xe dx =
(d)
x( t) = te t u( t) X( s) =
∞
∫ x(t)e
∞
− st
dt =
0−
∫ te u(t)e t
∞
− st
dt =
0−
∫ te
−( s −1) t
dt
0−
∞
e −( s−1) t 1 X( s) = − ( s − 1 ) t − 1 = ( ) 2 2 , Re( s) = σ > 1 0 − ( s − 1) (−( s − 1)) 3. Using the time-shifting property, find the Laplace transform of these signals. (a)
(b)
(c)
x( t) = u( t) − u( t − 1)
1 1 1 − e−s X( s) = − e − s = s s s −3( t − 2) x( t) = 3e u( t − 2) 3e −2 s X( s) = s+3 x( t) = 3e −3 t u( t − 2) x( t) = 3e −6e −3( t − 2) u( t − 2) X( s) =
(d)
3e −6e −2 s 3e −2 s− 6 = s+3 s+3
x( t) = 5 sin(π ( t − 1)) u( t − 1)
Using
Solutions 9-3
sin(βt) u( t) ← L→
β , Re( s) > 0 s + β2 2
5 sin(πt) u( t) ← L→
5π s +π2 2
5 sin(π ( t − 1)) u( t − 1) ← L→
5πe − s s2 + π 2
4. Using the complex-frequency-shifting property, find and sketch the inverse Laplace transform of 1 1 + . X( s) = (s + j 4) + 3 (s − j 4) + 3 e s0 t g( t) ← L→ G( s − s0 ) e −3 t u( t) ← L→ e −( 3 + j 4 ) t u( t) ← L→
[
1 s + j4 + 3
and
1 s+3 e −( 3 − j 4 ) t u( t) ← L→
1 s − j4 + 3
]
x( t) = e −( 3 + j 4 ) t + e −( 3 − j 4 ) t u( t) = e −3 t (e − j 4 t + e j 4 t ) u( t) = 2e −3 t cos( 4 t) u( t) x(t) 2
-0.1
2
t
-2
5. Using the time-scaling property, find the Laplace transforms of these signals. (a)
x( t) = δ ( 4 t)
1 s g( at) ← L→ G , a > 0 a a
δ ( t) ← L→1 , All s δ ( 4 t) ← L→ (b)
1 , All s 4
x( t) = u( 4 t) u( t) ← L→
1 , Re( s) > 0 s
Solutions 9-4
u( 4 t) ← L→
1 1 1 = , Re( s) > 0 4 s s 4
6. Using the time-differentiation property, find the Laplace transforms of these signals. (a)
x( t) =
d (u(t)) dt
d g( t)) ← L→ s G( s) − g(0 − ) ( dt u( t) ← L→
1 , Re( s) > 0 s
d 1 u( t)) ← L→ s − u(0 − ) = 1 , All s ( 3 dt s 12 =0
(b)
x( t) =
d −10 t (e u(t)) dt e −αt u( t) ← L→
1 , Re( s) > −α s+α
s d −10 t 1 −α 0 − e u( t)) ← , Re( s) > −10 L→ s − e ( ) u(0 − ) = ( 12 3 s + 10 dt s + 10 =0
Alternate Solution: 10 d −10 t s e u( t)) = e −10 tδ ( t) − 10e −10 t u( t) ← L→1 − = Check. ( dt s + 10 s + 10 (c)
x( t) =
d (4 sin(10πt) u(t)) dt sin(βt) u( t) ← L→
β , Re( s) > 0 s + β2
4 sin(10πt) u( t) ← L→ 4
2
10π 2 , Re( s) > 0 s + (10π ) 2
10π 40πs d − − 4 sin(10πt) u( t)) ← L→ 4 s 2 ( 2 − 4 sin 10π (0 ) u(0 ) = 2 2 , Re( s) > 0 1 2 3 dt s + (10π ) s + (10π )
(
)
=0
Alternate Solution:
Solutions 9-5
d (4 sin(10πt) u(t)) = 4 sin(10πt)δ (t) + 40π cos(10πt) u(t) = 40π cos(10πt) u(t) dt 40π cos(10πt) u( t) ← L→
(d)
x( t) =
40πs 2 . Check. s + (10π )
d (10 cos(15πt) u(t)) dt cos(βt) u( t) ← L→ 10 cos(15πt) u( t) ← L→
s , Re( s) > 0 s + β2 2
10 s 2 , Re( s) > 0 s + (15π ) 2
(
)
10 s 10 s2 d − − 10 cos(15πt) u( t)) ← 10 15 0 0 L→ s 2 − cos π u = ( ( ) 1(23) s2 + (15π )2 , Re(s) > 0 2 dt s + (15π ) =0
Alternate Solution: d (10 cos(15πt) u(t)) = 10 cos(15πt)δ (t) − 150π sin(15πt) u(t) = 10δ (t) − 150π sin(15πt) u(t) dt
[
]
10 s2 + (15π ) − 10(15π ) 15π 10 s2 d L 10 cos(15πt) u( t)) ← →10 − 150π 2 = = ( 2 2 2 dt s + (15π ) s2 + (15π ) s2 + (15π ) 2
2
Check. 7. Using multiplication-convolution duality, find the Laplace transforms of these signals and sketch the signals versus time. (a)
x( t) = e − t u( t) ∗ u( t) e − t u( t) ∗ u( t) ← L→
1 1 s s +1
1 1 e − t u( t) ∗ u( t) ← L→ − s s +1
Therefore
1 1 u( t) − e − t u( t) ← L→ − s s +1 e − t u( t) ∗ u( t) = (1 − e − t ) u( t)
Solutions 9-6
x(t) 1
-1
(b)
5
t
x( t) = e − t sin(20πt) u( t) ∗ u( t) e − t sin(20πt) u( t) ∗ u( t) ← L→
20π 1 2 (s + 1) + (20π ) s 2
20π 1 20π 1 As + B = + 2 2 2 (s + 1) + (20π ) s 1 + (20π ) s (s + 1) 2 + (20π ) 2 Multiplying through by s and letting s approach infinity, 0=
20π 20π 2 + A⇒ A= − 2 1 + (20π ) 1 + (20π )
20π 2 s+ B 20π 1 20π 1 1 + (20π ) = + (s + 1) 2 + (20π ) 2 s 1 + (20π ) 2 s (s + 1) 2 + (20π ) 2 −
Now, letting s = 1 and solving for B, B=− X( s) = X( s) =
40π 2 1 + (20π )
1 20π s+2 2 − 2 2 1 + (20π ) s ( s + 1) + (20π )
1 20π 1 20π s +1 − − 2 2 2 20π ( s + 1) 2 + (20π ) 2 1 + (20π ) s ( s + 1) + (20π )
x( t) =
20π 1 −t −t e sin(20πt) u( t) 2 1 − e cos(20πt) − 20 π 1 + (20π )
x( t) =
sin(20πt) 20π −t u( t) 2 1 − e cos(20πt) + 20π 1 + (20π )
Solutions 9-7
x(t) 0.025
-1
(c)
5
t
πt x( t) = 8 cos u( t) ∗ [u( t) − u( t − 1)] 2
π 1 e s 1− e 2 2 X( s) = 8 1 − e−s) ( =8 2 − 2 = 8 2 π π π s s π s+ s+ s+ 2 2 2 −s
x( t) =
−s
16 π π sin t u( t) − sin ( t − 1) u( t − 1) 2 π 2 x(t) 10
-1
8
t
-10
(d)
x( t) = 8 cos(2πt) u( t) ∗ [u( t) − u( t − 1)] 1 e−s 2π s 1 − e−s 1 1 − e−s) − 8 = ( 2 2 = 8 2 2π s + (2π ) s s s + (2π ) s + (2π )
X( s) = 8 x( t) =
[
]
4 sin(2πt) u( t) − sin(2π ( t − 1)) u( t − 1) π
After time, t = 1, the solution is zero. x(t) 2
-1
4
t
-2
8. Using the initial and final value theorems, find the initial and final values (if possible) of the signals whose Laplace transforms are these functions.
Initial Value Theorem
g(0 + ) = lim s G( s) s→∞
Solutions 9-8
Final Value Theorem
limg( t) = lim s G( s) t →∞
(a)
X( s) =
10 s+8
, if limg( t) exists
s→ 0
t →∞
, One pole in open LHP x(0 + ) = lim s s→∞
10 = 10 s+8
lim x( t) = lim s t →∞
s→ 0
10 =0 s+8
and the limit exists because the only pole of X( s) = (b)
X( s) =
s+3 , Poles at −3 ± j 2 (s + 3) 2 + 4 x(0 + ) = lim s s→∞
lim x( t) = lim s t →∞
(c)
10 is in the open LHP s+8
s→ 0
s+3 =1 (s + 3) 2 + 4
s+3 =0 (s + 3) 2 + 4
and the limit exists because both poles are in the open LHP. s , Poles at ± j2 X( s) = 2 s +4 x(0 + ) = lim s s→∞
s =1 s +4 2
Final-value theorem does not apply because there are two poles on the ω axis (d)
X( s) =
10 s , Poles at −5 ± j16.583 s + 10 s + 300 2
x(0 + ) = lim s s→∞
10 s = 10 s + 10 s + 300
lim x( t) = lim s t →∞
(e)
X( s) =
s→ 0
2
10 s =0 s + 10 s + 300 2
8 , Poles at 0 and -20 s( s + 20) 8 x(0 + ) = lim s =0 s→∞ s( s + 20)
Solutions 9-9
lim x( t) = lim s t →∞
(f)
X( s) =
s→ 0
8 2 = s( s + 20) 5
8 , Double pole at zero. s ( s + 20) 2
x(0 + ) = lim s s→∞
8 =0 s ( s + 20) 2
Final-value theorem does not apply because of the double pole at zero 9. Find the inverse Laplace transforms of these functions. (a)
X( s) =
24 s( s + 8) X( s) =
3 −3 + s s+8
x( t) = 3(1 − e −8 t ) u( t) (b)
X( s) =
20 s + 4s + 3 2
X( s) =
20 10 10 = − (s + 1)(s + 3) s + 1 s + 3
x( t) = 10(e − t − e −3 t ) u( t) (c)
X( s) =
5 s + 6 s + 73 2
X( s) =
5 5 8 = 2 (s + 3) + 64 8 (s + 3) 2 + 64 5 x( t) = e −3 t sin(8 t) u( t) 8
(d)
X( s) =
10 s( s + 6 s + 73) 2
10 Bs + C X( s) = = 73 + 2 s ( s + 3) 2 + 64 s ( s + 3) + 64
[
10
]
In Solutions 9-10
10 10 Bs + C = 73 + 2 s ( s + 3) 2 + 64 s ( s + 3) + 64
[
]
multiply through by s and then let s approach infinity. Then 0= To find C let s = 1 in
10 10 +B⇒B=− . 73 73
10 10 − s+C 73 = 73 + 2 s (s + 3) 2 + 64 s ( s + 3) + 64 10
[
yielding
]
10 +C 10 10 60 73 = + ⇒C=− . 80 73 80 73 −
Then X( s) = and
10 1 10 1 3 8 s+6 s+3 = − − − 73 s ( s + 3) 2 + 64 73 s ( s + 3) 2 + 64 8 ( s + 3) 2 + 64 x( t) =
or, using
3 10 1 − e −3 t cos(8 t) + sin(8 t) u( t) 8 73
B A 2 + B 2 cos x − tan −1 A 73 −3 t 10 x( t) = e cos(8 t − 0.3588) u( t) 1 − 64 73 4 X( s) = 2 2 s ( s + 6 s + 73) A cos( x ) + B sin( x ) =
(e)
X( s) =
4 A B Cs + D = 2+ + 2 s s + 6 s + 73 s ( s + 6 s + 73) s 2
2
Using the “cover up” method, A =
K D ,k
B=
4 ≅ 0.0548 . Using 73
[
]
1 d m −k m = m − k ( s − pD ) H ( s) (m − k )! ds
1 d 4 2 1! ds ( s + 6 s + 73)
[
= −4 ( s2 + 6 s + 73) s→ 0
−2
Solutions 9-11
s→ p D
(2s
2
, k = 1, 2,L, m
]
+ 6)
s→ 0
=−
24 ≅ −0.0045 (73) 2
24 4 4 Cs + D (73) 2 73 X( s) = 2 2 = 2 − + 2 s s + 6 s + 73 s ( s + 6 s + 73) s Multiply through by s and let s approach infinity, 0=−
24 24 ≅ 0.0045 2 +C ⇒C = (73) (73) 2
24 24 4 s+D 2 4 (73) (73) 2 73 X( s) = 2 2 = − + s s ( s + 6 s + 73) s2 (s + 3) 2 + 64 Then let s = 1. 24 +D 4 4 24 21464 148 (73) 2 ⇒ D= 4− = − = −0.0278 2 + 2 = − 80 73 ( 73) 80 (73) (73) 2 24 24 148 4 37 2 2 s− 2 s − 1 292 24 (73) (73) (73) 6 = X( s) = 73 + 2 + 24 2 2 − 2 2 − s s s + 6 s + 73 s (73) s (s + 3) + 64 X( s) =
s+3 1 292 24 55 8 − + − 24 2 2 s (73) 2 s2 ( s + 3) + 64 48 ( s + 3) + 64
x( t) =
(f)
X( s) =
1 55 −3 t cos(8 t) − sin(8 t) u( t) 2 292 t − 24 + 24 e 48 (73) 2s s + 2 s + 13 2
X( s) = 2
s +1 1 12 = 2 − 2 2 (s + 1) + 12 (s + 1) + 12 12 (s + 1) + 12 s
2
1 x( t) = 2 e − t cos 12 t − sin 12 t u( t) 12
(
(g)
X( s) =
)
s s+3
Solutions 9-12
(
)
3 ⇒ x( t) = δ ( t) − 3e −3 t u( t) s+3
X( s) = 1 − (h)
X( s) =
s s + 4s + 4 2
X( s) = Using te −αt u( t) ← L→
1 s s −2 = 2 = 2 + s + 4 s + 4 ( s + 2) (s + 2) s + 2 2
1 , Re( s) > −α , (s + α )2
x( t) = (−2 te −2 t + e −2 t ) u( t) = e −2 t (1 − 2 t) u( t) Alternate Solution: Using te −αt u( t) ← L→ X( s) =
1 , Re( s) > −α , and the differentiation property (s + α )2
s d −2 t te u( t)) = ( te −2 t )δ ( t) + (−2 te −2 t + e −2 t ) u( t) ( 2 ⇒ x( t) = dt (s + 2) x( t) = e −2 t (1 − 2 t) u( t) . Check.
(i)
X( s) =
s2 s2 − 4 s + 4
4s − 4 s2 s−1 =1+ 4 X( s) = 2 =1+ 2 s − 4s + 4 s − 4s + 4 (s − 2) 2 1 1 2t 2t 2t X( s) = 1 + 4 + ⇒ x( t) = δ ( t) + 4 ( te + e ) u( t) = δ ( t) + 4 e ( t + 1) u( t) 2 s− 2 ( s − 2) (j)
X( s) =
10 s s + 4 s2 + 4 4
X( s) =
(s
10 s 2
+ 2)
2
=
10 s
(s − j 2 ) (s + j 2 ) 2
j10 2 j10 2 0 8 8 X( s) = + 2 + s− j 2 s− j 2 s+ j 2 −
(
)
(
Using
Solutions 9-13
)
2
+
2
0 s+ j 2
te −αt u( t) ← L→ j5 2 4 X( s) = s+ j 2
(
x( t) = −
j5 2 4 − s− j 2
) ( 2
(
j5 2 t ej 4
2t
1 , Re( s) > −α (s + α )2
)
2
⇒ x( t) =
− e− j
2t
(
j5 2 − j te 4
2t
− te j
2t
) u(t)
) u(t) = 5 22 t sin( 2t) u(t)
10. Using a table of Laplace transforms, find the CTFT’s of these signals. (a)
x( t) = 10e −100 t u( t) X( s) =
(b)
10 10 ⇒ X( jω ) = s + 100 jω + 100
x( t) = 3e −50 t cos(100πt) u( t) X( s) = 3
s + 50 jω + 50 2 2 ⇒ X ( jω ) = 3 (s + 50) + (100π ) ( jω + 50) 2 + (100π ) 2
11. Using the Laplace transform, solve these differential equations for t ≥ 0. (a)
x′ ( t) + 10 x( t) = u( t)
,
x(0 − ) = 1
s X( s) − x(0 − ) + 10 X( s) =
1 s
1 +1 s +1 s X( s) = = s + 10 s( s + 10) 1 9 1 + 9e −10 t 10 10 X( s) = + ⇒ x( t) = u( t) 10 s s + 10 Checking initial conditions,
x(0 + ) = 1
which agrees with the initial condition, x(0 − ) = 1. For this system and this excitation the response cannot change instantaneously.
Solutions 9-14
x′′ ( t) − 2 x′ ( t) + 4 x( t) = u( t) ,
(b)
d x(0 − ) = 0 , x( t) =4 dt t = 0−
1 d s2 X( s) − s x(0 − ) − ( x( t)) − 2 s X( s) − 2 x(0 − ) + 4 X( s) = s dt t = 0− s2 X( s) − 4 − 2 s X( s) + 4 X( s) =
1 s
1 +4 1 4s + 1 1 4s + 1 s = X( s) = 2 = 2 s − 2 s + 4 s s − 2 s + 4 s ( s − 1) 2 + 3 1 Bs + C X( s) = 4 + s ( s − 1) 2 + 3 In the equality, 1 1 4s + 1 Bs + C = 4 + 2 s ( s − 1) + 3 s ( s − 1) 2 + 3 multiply through by s and let s approach infinity yielding 0= Then
1 1 +B⇒B=− . 4 4
1 5 1 −4 +C 27 = + ⇒C= 3 4 3 6 1 1 27 − s+ 1 1 17 3 s − 18 1 1 s−1 X( s) = 4 + 4 2 6 = − = − + 2 2 2 s ( s − 1) + 3 4 s ( s − 1) + 3 4 s ( s − 1) + 3 3 ( s − 1) + 3 x( t) =
1 t 1 − e cos 4
( 3t) + 173 e sin( 3t) u(t) t
( 3t) − e cos( 3t) + 173 [
d 1 x( t)) = 3e t sin ( dt 4
t
( 3t) + e sin( 3t)] u(t)
3e t cos
e t 20 d (x(t)) = 4 3 sin dt
( 3t) + 16 cos( 3t) u(t)
d dt ( x( t)) + = 4 . Check. t =0 Solutions 9-15
t
(c)
x′ ( t) + 2 x( t) = sin(2πt) u( t)
,
s X( s) − x(0 − ) + 2 X( s) = X( s) =
x(0 − ) = −4 2π 2 s + (2π ) 2
2π
(s + 2)(s + (2π ) 2
2
)
−
4 s+2
2π 2 4 Bs + C 4 + (2π ) + 2 X( s) = 2 − s+2 s+2 s + (2π ) In 2π
(s + 2)(s2 + (2π ) 2 )
2π 2 Bs + C 4 + (2π ) + 2 = 2 s+2 s + (2π )
multiply through by s and let s approach infinity 0=
2π 2π 2 + B⇒ B= − 2 4 + (2π ) 4 + (2π )
In 2π
(s + 2)(s2 + (2π ) 2 )
2π 2π − 2 2 s+C 4 + (2π ) 4 + (2π ) + = 2 s+2 s2 + (2π )
let s = 1 and solve for C,
(
2π
3 1 + (2π )
2
)
2π 2π − 2 2 +C 4 + (2π ) 4 + (2π ) = + 2 3 1 + (2π ) C=
4π 2 4 + (2π )
2π 2π 4π − 2 2 s+ 2 4 4 + (2π ) 4 + (2π ) 4 + (2π ) + X( s) = − 2 2 s+2 s+2 s + (2π )
Solutions 9-16
X( s) = X( s) =
2π 1 2πs − 4π 4 − 2 2 2 − 4 + (2π ) s + 2 s + (2π ) s + 2
2π 2π 4 1 s − 2π 2 2 2 +2 2 2 − 4 + (2π ) s + 2 s + (2π ) s + (2π ) s + 2
2πe −2 t − 2π cos(2πt) + 2 sin(2πt) x( t) = − 4 e −2 t u( t) 2 4 + (2π ) x(0 + ) = −4 . Check. 12. Using the Laplace transform, find and sketch the time-domain response, y( t) , of the systems with these transfer functions to the sinusoidal excitation, x( t) = A cos(10πt) u( t) . (Both graphs at the end) (a)
H( s) =
1 s +1
Using cos(ω 0 t) u( t) ← L→ X( s) =
Y( s) = A
s , σ >0 s + ω 02 2
As 2 , σ > 0 s + (10π ) 2
1 s 2 s + 1 s + (10π ) 2
To find C let s be zero. Then
1 − 1 + (10π ) 2 Bs + C + 2 = A 2 s +1 s + (10π )
1 C (10π ) 0=− 2 + 2 ⇒C = 2 1 + (10π ) 1 + (10π ) (10π ) 2
and
1 (10π ) Bs + 2 2 1 + (10π ) 1 + (10π ) + = . 2 s +1 s2 + (10π ) 2
1 s 2 s + 1 s + (10π ) 2
−
Now multiply through by s and let s approach infinity. 0=−
1 1 2 + B⇒ B= 2 1 + (10π ) 1 + (10π ) Solutions 9-17
Then
2 1 1 10π A s + (10π ) A s Y( s) = + 2 + 2 2 − 2= 2 − 2 + 10π 2 2 s + (10π ) 1 + (10π ) s + 1 s + (10π ) 1 + (10π ) s + 1 s + (10π )
y( t) =
[
]
A −t u( t) 2 cos(10πt) + 10π sin(10πt) − e 1 + (10π )
or, using A cos( x ) + B sin( x ) =
y( t) = A Alternate solution: Y( s) =
1 As 2 s + 1 s + (10π ) 2
B A 2 + B 2 cos x − tan −1 A
1 + (10π ) cos(10πt − 1.539) − e − t 2
1 + (10π )
2
u( t)
A − Aj10π Aj10π 2 j 20π (1 + j10π ) − j 20π (1 − j10π ) 1 + (10π ) + + = s +1 s − j10π s + j10π −
j10πA Ae − t j10πA e − j10πt u( t) e j10πt + y( t) = − 2 + j 20π (1 − j10π ) j 20π (1 + j10π ) 1 + (10π ) e−t e j10πt e − j10πt y( t) = A − + u( t) 2 + 2(1 + j10π ) 2(1 − j10π ) 1 + (10π ) e−t (1 − j10π )e j10πt + (1 + j10π )e − j10πt u(t) y( t) = A − + 2 2(1 + j10π )(1 − j10π ) 1 + (10π ) j10πt + e − j10πt − j10π (e j10πt − e − j10πt ) e−t 1e y( t) = A − u( t) 2 + 2 2 1 + (10π ) 1 + (10π )
cos(10πt) + 10π sin(10πt) − e − t y( t) = A u( t) 2 1 + (10π ) Using A cos( x ) + B sin( x ) =
y( t) = A
B A 2 + B 2 cos x − tan −1 A
1 + (10π ) cos(10πt − 1.539) − e − t 2
1 + (10π )
2
As a check on the steady state solution, use the CTFT.
Solutions 9-18
u( t)
X( f ) =
A [δ ( f − 5) + δ ( f + 5)] 2
Y( f ) =
H( f ) =
and
1 j 2πf + 1
1 A A δ ( f − 5) δ ( f + 5) = + δ ( f − 5) + δ ( f + 5)] [ 2 j 2πf + 1 2 j 2πf + 1 j 2πf + 1
Y( f ) =
A δ ( f − 5) δ ( f + 5) A δ ( f − 5) δ ( f + 5) + = − 2 j10π + 1 − j10π + 1 2 j10π + 1 j10π − 1
Y( f ) = − Y( f ) =
A j10πδ ( f − 5) − j10πδ ( f + 5) − δ ( f − 5) − δ ( f + 5) 2 2 1 + (10π )
A [ j10πδ ( f + 5) − j10πδ ( f − 5)] + [δ ( f − 5) + δ ( f + 5)] 2 2 1 + (10π ) y( t) = A
10π sin(10πt) + cos(10πt) . Check. 2 1 + (10π )
y(t) 0.033333 1
t
-0.033333
(b)
H( s) =
s− 2 (s − 2) 2 + 16
s− 2 As s2 − 2 s Y( s) = =A 2 2 2 2 2 (s − 2) 2 + 16 s2 + (10π ) 2 s ( s − 2) + 16 s2 + (10π ) ( s − 2) + 16(10π ) Y( s) =
(
s2 − 2 s
)
s4 − 4 s3 + s2 20 + (10π ) − 4 (10π ) s + 20(10π ) 2
2
s2 − 2 s Y( s) = 4 s − 4 s3 + 1006.97 s2 − 3947.84 s + 19739.2 Using MATLAB, »X Transfer function: s ---------
Solutions 9-19
2
s^2 + 987 »H Transfer function: s - 2 -------------s^2 - 4 s + 20 »Y Transfer function: s^2 - 2 s -----------------------------------------s^4 - 4 s^3 + 1007 s^2 - 3948 s + 1.974e04 »[z,p,k] = zpkdata(Y,'v') ; »z z = 0 2 »p p = -0.00000000000000 -0.00000000000000 2.00000000000000 2.00000000000000
+31.41592653589795i -31.41592653589795i + 4.00000000000000i - 4.00000000000000i
r = -0.00105906221326 -0.00105906221326 0.00105906221326 0.00105906221326
+ + -
0.01610704734047i 0.01610704734047i 0.00203398509622i 0.00203398509622i
−0.00106 − j 0.01611 −0.00106 + j 0.01611 + s − j10π s + j10π Y( s) = A 0.00106 + j 0.00203 0.00106 − j 0.00203 + + s − 2 − j4 s − 2 + j4 −0.00212 s + 1.0122 0.00212 s − 0.02048 Y( s) = A + 2 s2 + (10π ) (s − 2) 2 + 16 s − 9.66 s + 477.45 Y( s) = 0.00212 A − 2 2 2 ( s − 2) + 16 s + (10π )
Solutions 9-20
s− 2 s 7.66 4 477.45 10π Y( s) = 0.00212 A − − 2 2 2 2 − 2 2 4 ( s − 2) + 16 s + (10π ) 10π s + (10π ) ( s − 2) + 16
{
}
y( t) = 0.00212 A e 2 t [cos( 4 t) − 1.915 sin( 4 t)] − cos(10πt) − 15.2 sin(10πt) u( t)
y(t) 5 -1
5
t
-10
13. Write the differential equations describing these systems and find and sketch the indicated responses. (a)
x( t) = u( t)
,
y( t) is the response, y(0 − ) = 0
x(t)
∫
y(t)
4 y′ ( t) = x( t) − 4 y( t) or
y′ ( t) + 4 y( t) = x( t) y′ ( t) + 4 y( t) = u( t) s Y( s) − y(0 − ) + 4 Y( s) =
1 s
Y( s) =
1 1 1 1 = − s( s + 4 ) 4 s s + 4
y( t) =
1 1 − e −4 t ) u( t) , t > 0 ( 4
Solutions 9-21
y(0 + ) = 0 = y(0 − ) . Check. The solution is continuous at t = 0 because, if it were not the discontinuity would cause an impulse on the LHS of the equation which could not be equated to the step excitation on the RHS. x(t) 0.25
t
1
(b)
v(0 − ) = 10 , v( t) is the response + C = 1 µF v(t)
R = 1 kΩ
C v′ ( t) +
[
v( t) =0 R
]
C s V( s) − v(0 − ) + V( s) =
v( t) = 10e
−
t RC
10C sC +
1 R
V( s) =0 R
= 10
1 s+
1 RC
u( t) = 10e −1000 t u( t) , t > 0
v(0 + ) = 10 = v(0 − ) . Check. The solution is continuous at t = 0 because the capacitor voltage cannot change instantaneously. x(t) 10
0.004
14.
t
Find the three parts, x ac ( t), x 0 ( t) and x c ( t) , of the following signals. (a)
x( t) = e −10 t u( t) − e 2 t u(− t)
Solutions 9-22
x ac ( t) = −e 2 t u(− t) , x 0 ( t) = 0 , x c ( t) = e −10 t u( t) (b)
x( t) = K x ac ( t) = K u(− t) , x 0 ( t) = 0 , x c ( t) = K u( t)
(c)
x( t) = u( t) x ac ( t) = 0 , x 0 ( t) = 0 , x c ( t) = u( t)
(d)
x( t) =
d (u(t)) dt
x( t) = δ ( t) x ac ( t) = 0 , x 0 ( t) = δ ( t) , x c ( t) = 0
15. Find the bilateral Laplace transforms of these signals. (a)
x( t) = 3e −7 t u( t) − 12e 4 t u(− t) x c ( t) = 3e −7 t u( t) ⇒ X c ( s) =
3 , Re( s) > −7 s+7
x 0 ( t) = 0 ⇒ X 0 ( s) = 0 x ac ( t) = −12e 4 t u(− t) ⇒ x ac (− t) = −12e −4 t u( t) ⇒ X ac (− s) = − X ac ( s) = X( s) = (b)
12 , Re( s) > −4 s+4
12 , Re( s) < 4 s− 4
5 s + 24 3 12 + =3 2 , − 7 < Re( s) < 4 s+ 7 s− 4 s + 3s − 28
x( t) = 50e −10 t x c ( t) = 50e −10 t u( t) ⇒ X c ( s) =
50 , Re( s) > −10 s + 10
x 0 ( t) = 0 ⇒ X 0 ( s) = 0 x ac ( t) = 50e10 t u(− t) ⇒ x ac (− t) = 50e −10 t u( t) ⇒ X ac (− s) = X ac ( s) = −
50 , Re( s) < 10 s − 10
Solutions 9-23
50 , Re( s) > −10 s + 10
X( s) =
50 50 1000 − =− 2 , − 10 < Re( s) < 10 s + 10 s − 10 s − 100
16. Find the responses, y( t) , of these systems to these excitations. (a)
h( t) = e −5 t u( t) X( s) =
3 12 + , − 7 < Re( s) < 4 s+ 7 s− 4
H( s) = Y( s) = Y( s) =
x( t) = 3e −7 t u( t) − 12e 4 t u(− t)
,
1 , Re( s) > −5 s+5
1 3 12 + , − 5 < Re( s) < 4 s + 5 s + 7 s − 4 3
(s + 5)(s + 7)
+
12 , − 5 < Re( s) < 4 (s + 5)(s − 4)
3 1 1 4 1 1 Y( s) = − − + , − 5 < Re( s) < 4 2 s+5 s+7 3 s − 4 s + 5 y( t) =
3 −5 t 4 4 e − e −7 t ) u( t) − e −5 t u( t) − e 4 t u(− t) ( 2 3 3 1 3 4 y( t) = e −5 t − e −7 t u( t) − e 4 t u(− t) 6 2 3
(b)
h( t) = tri( t)
x( t) = e − t u( t)
,
Using s − 2s 2 e − e tri( t) ← L→ s s − 2s 2 e −e H( s) = s Therefore
2
, All s
2
, All s
and
X( s) =
2
1 , Re( s) > −1 s +1
s − 2s 2 1 e − e es − 2 + e−s = 2 Y( s) = , Re( s) > −1 s + 1 s s ( s + 1)
Solutions 9-24
1 1 1 Y( s) = (e s − 2 + e − s ) 2 − + , Re( s) > −1 s s s + 1 ramp( t + 1) − 2 ramp( t) + ramp( t − 1) y( t) = − u( t + 1) + 2 u( t) − u( t − 1) + e −( t +1) u( t + 1) − 2e − t u( t) + e −( t −1) u( t − 1)
[
]
ramp( t + 1) − 1 + e −( t +1) u( t + 1) −t y( t) = −2 ramp( t) − 1 + e u( t) −( t −1) u( t − 1) + ramp( t − 1) − 1 + e
[
(c)
h( t) = e −10 t u( t)
[
]
,
]
x( t) = 50e −10 t
Using e −α t ← L→
1 1 − , − α < Re( s) < α s+α s−α
1 1 1 Y( s) = 50 − , − 10 < Re( s) < 10 s + 10 s − 10 s + 10 1 1 Y( s) = 50 , − 10 < Re( s) < 10 2 − (s − 10)(s + 10) ( s + 10)
Using
1 1 1 20 Y( s) = 50 − 20 , − 10 < Re( s) < 10 2 − ( s + 10) s − 10 s + 10 te −αt u( t) ← L→
1 , Re( s) > −α (s + α )2
1 1 y( t) = 50te −10 t u( t) − − e t u(− t) − e − t u( t) 20 20 1 t y( t) = 50te −10 t u( t) + e u(− t) + e − t u( t) 20
[
]
17. Sketch the pole-zero plot and region of convergence (if it exists) for these signals.
Solutions 9-25
ω
(a)
x( t) = e u(− t) − e
(b)
x( t) = e −2 t u(− t) − e t u( t)
−t
−4 t
[s]
ROC
u( t)
σ
s = −4 s = −1
No ROC 18. Using the integral definition find the the unilateral Laplace transform of these time functions. (a)
g( t) = e − at u( t) G( s) =
∞
∫e
0
− at
u( t)e dt = − st
−
∫e
∞
−( s + a ) t
0−
G( s) = (b)
∞
e −( s + a ) t dt = − s + a 0−
1 , Re( s) > − a s+a
g( t) = e − a ( t −τ ) u( t − τ ) , τ > 0 G( s) =
∞
∫e
− a ( t −τ )
u( t − τ )e − st dt , τ > 0
0−
∞
G( s) = e aτ
∫e
τ
−( s + a ) t
dt , τ > 0
−
∞
−( s + a )τ e −( s + a ) t e − sτ aτ e G( s) = e − , τ>0 = =e s+a s+a s + a τ − aτ
(c)
g( t) = e − a ( t +τ ) u( t + τ ) , τ > 0
Solutions 9-26
G( s) =
∞
∫e
− a ( t +τ )
u( t + τ )e − st dt , τ > 0
0−
G( s) = e
− aτ
∞
∫e
−( s + a ) t
dt , τ > 0
0−
G( s) = e
(d)
∞
− aτ
e − aτ 1 −( s + a ) t − s + a e − = s + a , τ > 0 0
g( t) = sin(ω 0 t) u( t) ∞
∞
e jω 0 t − e − jω 0 t − st e dt G( s) = ∫ sin(ω 0 t) u( t)e dt = ∫ j 2 − − 0 0 − st
∞
∞ 1 1 e −( s− jω 0 ) t e −( s+ jω 0 ) t −( s − jω 0 ) t −( s + jω 0 ) t G( s) = e − e dt = + − j 2 0∫− j 2 s − jω 0 s + jω 0 0 −
[
G( s) = (e)
]
1 1 1 ω − = 2 0 2 , Re( s) > 0 j 2 s − jω 0 s + jω 0 s + ω 0
g( t) = rect ( t) 1 2
∞
1
e − st 2 1 − e G( s) = ∫ rect ( t)e − st dt = ∫ e − st dt = − = s s 0− 0− 0− (f)
−
s 2
, Re( s) > 0
1 g( t) = rect t − 2 ∞
1 G( s) = ∫ rect t − e − st dt = 2 − 0
1
e − st 1 − e−s e dt = − , Re( s) > 0 = s ∫− s − 0 0 1
− st
19. Using MATLAB (or any other appropriate computer mathematics tool) do the inversion integral of 1 G( s ) = s + 10 numerically. That is, approximate the inversion integral with a summation of the form g(t ) = L-1 (G(s)) =
1 j 2π
1 e sn t ∆sn = ∑ j 2π n = − N sn + 10 N
e(σ + jn∆ω )t j∆ω , σ > 0 . ∑ n = − N σ + jn∆ω + 10 N
Solutions 9-27
Choose the combination of large N and small ∆ω so that the summation will range over a contour from well below to well above the real axis. Plot g(t) versus t by computing the value of g(t) at every value of t from the above summation approximation to the inversion integral. Compare to the analytical result. Try at least three different values of σ to see the effect on the result. (Ideally there is no effect of changing σ as long as it is greater than -10, but actually, in this numerical approximation, there will be some small effects.) %
Program to demonstrate the inverse Laplace transform numerically.
close all ; tau = 0.1 ; dw = 1 ; p = -1/tau ; w = dw*[-2000:2000]' ; ds = j*dw*ones(length(w),1) ; allint = [] ; for sigma = 0:5, s = sigma + j*w ; int = [] ; tv = [] ; for t = -tau*2:tau/20:tau*4, f = exp(s.*t)./(s - p) ; tv = [tv;t] ; int = [int;sum(f.*ds)/(j*2*pi)] ; end int = real(int) ; allint = [allint,int] ; end subplot(3,2,1) ; h = plot(tv, allint(:,1), 'k') ; set(h, 'LineWidth',2) ; xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 0') ; grid ; axis([-0.2, 0.4, -0.1, 1]) ; subplot(3,2,2) ; h = plot(tv, allint(:,2), 'k') ; set(h, 'LineWidth',2) ; xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 1') ; grid ; axis([-0.2, 0.4, -0.1, 1]) ; subplot(3,2,3) ; h = plot(tv, allint(:,3), 'k') ; set(h, 'LineWidth',2) ; xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 2') ; grid ; axis([-0.2, 0.4, -0.1, 1]) ; subplot(3,2,4) ; h = plot(tv, allint(:,4), 'k') ; set(h, 'LineWidth',2) ; xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 3') ; grid ; axis([-0.2, 0.4, -0.1, 1]) ; subplot(3,2,5) ; h = plot(tv, allint(:,5), 'k') ; set(h, 'LineWidth',2) ; xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 4') ; grid ; axis([-0.2, 0.4, -0.1, 1]) ; subplot(3,2,6) ; h = plot(tv, allint(:,6), 'k') ; set(h, 'LineWidth',2) ; xlabel('Time, t (s)') ; ylabel('h(t)') ; title('sigma = 5') ; grid ; axis([-0.2, 0.4, -0.1, 1]) ;
20. Using a table of unilateral Laplace transforms and the properties find the unilateral Laplace transforms of the following functions. (a)
g( t) = 5 sin(2π ( t − 1)) u( t − 1)
β s + β2 2π sin(2πt) u( t) ← L→ 2 2 s + (2π ) sin(βt) u( t) ← L→
2
Solutions 9-28
(b)
10π 2 s + (2π )
Linearity
5 sin(2πt) u( t) ← L→
Time Shifting
10πe − s 5 sin(2π ( t − 1)) u( t − 1) ← → 2 2 s + (2π )
2
L
g( t) = 5 sin(2πt) u( t − 1)
sin(2πt) = sin(2π ( t − 1)) Therefore 5 sin(2πt) u( t − 1) ← L→ (c)
10πe − s 2 s2 + (2π )
g( t) = 2 cos(10πt) cos(100πt) u( t) 1 [cos(10πt − 100πt) + cos(10πt + 100πt)] 2 1 cos(10πt) cos(100πt) = [cos(−90πt) + cos(110πt)] 2 2 cos(10πt) cos(100πt) u( t) = [cos(90πt) + cos(110πt)] u( t) s cos(βt) u( t) ← L→ 2 s + β2 s [cos(90πt) + cos(110πt)]u(t) ←L→ s2 + (−s90π )2 + s2 + (110 2 π) s s 2 cos(10πt) cos(100πt) u( t) ← L→ 2 2 + 2 2 s + (90π ) s + (110π ) cos(10πt) cos(100πt) =
(d)
g( t) =
d (u(t − 2)) dt u( t) ← L→ Time Shifting
Time Differentiation Once
(e)
g( t) =
1 s
u( t − 2) ← L→
e −2 s s
e −2 s d L u( t − 2)) ← → s − u(−2 − ) = e −2 s ( s dt
t
∫ u(τ )dτ
0−
t
L
0−
(f)
1
∫ u(τ )dτ ←→ s
Integration d − t −2τ g( t) = 5e u( t − τ ) , τ > 0 dt
Solutions 9-29
2
e −αt u( t) ← L→ Frequency Shifting
−
t 2
1 s+α
e u( t) ← L→ t −τ 2
1 s+
1 2
e −τs Time Shifting e u( t − τ ) ← → 1 s+ 2 t −τ −τs − 5 e Linearity 5e 2 u( t − τ ) ← L→ 1 s+ 2 t −τ −τ − − L 5e −τs d − 2 Time Differentiation Once u( t − τ ) ← → s − 5e 2 u(−τ − ) 5e 1 123 dt s+ =0 2 L 5 se −τs d − t −2τ e t 5 u ( − τ ) → ← 1 dt s+ 2 −
(g)
L
g( t) = 2e −5 t cos(10πt) u( t) s+α (s + α )2 + β 2 s+5 e −5 t cos(10πt) u( t) ← L→ (s + 5) 2 + (10π ) 2 e −αt cos(βt) u( t) ← L→
2e −5 t cos(10πt) u( t) ← L→ 2
Linearity
(h)
s+5 (s + 5) 2 + (10π ) 2
π x( t) = 5 sinπt − u( t) 8 1 1 − jπ t − 5 jπ t − 8 1 x( t) = 5 sinπ t − u( t) = − e 8 u( t) e 8 j 2
x( t) =
jπ 5 jπt − j8π − jπt 8 e e − e e u( t) j2
jπ jπ jπ − − jπ 5 e 8 e 8 5 ( s + jπ )e 8 − ( s − jπ )e 8 X( s) = − = j 2 s − jπ s + jπ j 2 s2 + π 2
Solutions 9-30
jπ jπ − j8π − j8π π π 8 s e − e + jπ e +e 8 − j 2 sin s + j 2π cos 8 8 5 5 = X( s) = 2 2 2 2 j2 s +π j2 s +π
π π π cos − sin s 8 8 X( s) = 5 2 2 s +π 21. Given g( t) ← L→
s +1 s( s + 4 )
find the Laplace transforms of (a)
g(2t) s +1 1 2 Time Scaling g(2 t) ← L→ 2 ss + 4 22 s+2 g(2 t) ← L→ s( s + 8)
(b)
d (g(t)) dt
s +1 d g( t)) ← L→ s − g(0 − ) ( s( s + 4 ) dt
Time Differentiation Once
s +1 d − g(0 − ) g( t)) ← L→ ( s+4 dt
g(0 + ) = lim s G( s) = 1
Initial Value Theorem
s→∞
s +1 d −1 g( t)) ← L→ ( s+4 dt d 3 g( t)) ← L→ − ( dt s+4 (This is correct if g(0 − ) = g(0 + ) . That is, if g is continuous at time, t = 0.) (c)
g( t − 4 ) Time Shifting
(d)
g( t − 4 ) ← L→
g( t) ∗ g( t)
Solutions 9-31
s + 1 −4 s e s( s + 4 )
g( t) ∗ g( t) ← L→
Time-domain Convolution
(s + 1) 2 g( t) ∗ g( t) ← → 2 2 s (s + 4)
s +1 s +1 s( s + 4 ) s( s + 4 )
L
22. Find the time-domain functions which are the inverse Laplace transforms of these functions. Then, using the initial and final value theorems verify that they agree with the time-domain functions. (a)
G( s) =
4s (s + 3)(s + 8) 12 32 G( s) = 5 + 5 s+3 s+8 −
32 12 g( t) = − e −3 t + e −8 t u( t) 5 5 12 32 limg( t) = lim − e −3 t + e −8 t u( t) = 0 t →∞ t →∞ 5 5 2 4s = 0 , Check. lim+ s G( s) = lim+ s→ 0 s→ 0 ( s + 3)( s + 8) 12 32 lim+ g( t) = lim+ − e −3 t + e −8 t u( t) = 4 t→0 t → 0 5 5 2 4s = 4 , Check. lim s G( s) = lim s→∞ s→∞ ( s + 3)( s + 8) (b)
G( s) =
4
(s + 3)(s + 8) 4 4 G( s) = 5 − 5 s+3 s+8 4 4 g( t) = e −3 t − e −8 t u( t) 5 5 4 4 limg( t) = lim e −3 t − e −8 t u( t) = 0 t →∞ t →∞ 5 5 lim+ s G( s) = lim+
s→ 0
s→ 0
4s = 0 , Check . (s + 3)(s + 8)
Solutions 9-32
4 4 lim+ g( t) = lim+ e −3 t − e −8 t u( t) = 0 t→0 t → 0 5 5 4s lim s G( s) = lim = 0 , Check . s→∞ s→∞ ( s + 3)( s + 8) (c)
G( s) =
s s + 2s + 2 2
G( s) =
s s +1 1 = − 2 2 (s + 1) + 1 (s + 1) + 1 (s + 1) 2 + 1 g( t) = e − t [cos( t) − sin( t)] u( t)
Alternate Inverse:
1− j 1+ j 2 2 G( s) = + s + (1 + j ) s + (1 − j ) 1 − j −(1+ j ) t 1 + j −(1− j ) t g( t) = e + e u( t) 2 2 g( t) = e − t (cos( t) − sin( t)) u( t)
[
]
limg( t) = lim e − t (cos( t) − sin( t)) u( t) = 0 t →∞
t →∞
s2 lim s G( s) = lim+ 2 = 0 , Check . s→ 0 + s→ 0 s + 2 s + 2
[
]
lim g( t) = lim+ e − t (cos( t) − sin( t)) u( t) = 1
t→0+
t→0
s2 lim s G( s) = lim 2 = 1 , Check . s→∞ s→∞ s + 2 s + 2
(d)
G( s) =
e −2 s s2 + 2 s + 2
G( s) =
[
]
g( t) = e − t sin( t) u( t)
e −2 s (s + 1) 2 + 1
t→t −2
= e −( t − 2) sin( t − 2) u( t − 2)
Alternate Inverse: 1 1 − j2 j2 G( s) = e −2 s + s + (1 + j ) s + (1 − j ) −( t − 2) g( t) = e sin( t − 2) u( t − 2) Solutions 9-33
[
]
limg( t) = lim e −( t − 2) sin( t − 2) u( t − 2) = 0 t →∞
t →∞
e −2 ss lim s G( s) = lim+ 2 = 0 , Check . s→ 0 + s→ 0 s + 2 s + 2
[
]
lim g( t) = lim+ e −( t − 2) sin( t − 2) u( t − 2) = 0
t→0+
t→0
e −2 ss lim s G( s) = lim 2 = 0 , Check . s→∞ s→∞ s + 2 s + 2 23. Given
e −4 t u( t) ← L→ G( s)
find the inverse Laplace transforms of (a)
s G 3 s 3e −12 t u( 3t) ← L→ G 3
Frequency Scaling
3e −12 t u( 3t) = 3e −12 t u( t) (b)
G( s − 2) + G( s + 2) e 2 t e −4 t u( t) ← L→ G( s − 2) Frequency Shifting e −2 t u( t) ← L→ G( s − 2) e −6 t u( t) ← L→ G( s + 2) Frequency Shifting Linearity (e −2t + e −6t ) u(t) ←L→ G(s − 2) + G(s + 2)
(c)
G( s) s t
∫e
Integration
−4 λ
u(λ ) ← L→
0−
e −4 λ G( s) L→ − 4 u( t) ← s 0− t
G( s) 1 − e −4 t u( t) ← L→ 4 s 24. The CTFT of
x( t) = e − t
exists but the (unilateral) Laplace transform does not. Why? Solutions 9-34
G( s) s
This is an energy signal so the CTFT exists in the strict sense. But the signal is not causal so the unilateral transform does not exist. 25. Compare the CTFT and the Laplace transform of a unit step. Why can the CTFT not be found from the Laplace transform? F u( t) ← →
1 + πδ (ω ) jω
u( t) ← L→
and
1 s
The pole of the Laplace transform lies on the ω axis (at the origin). Therefore the CTFT does not exist in the strict sense and cannot be found from the Laplace transform. 26. Show that the common Laplace transform pairs u( t) ← L→
1 1 1 , e −αt u( t) ← L→ , te −αt u( t) ← L→ s s+α (s + α )2
sin(ω 0 t) u( t) ← L→ e −αt sin(ω 0 t) u( t) ← L→
ω0 s L→ 2 2 , cos(ω 0 t) u( t) ← s + ω 02 s + ω0 2
ω0 s+α , e −αt cos(ω 0 t) u( t) ← L→ 2 2 (s + α ) + ω 0 (s + α ) 2 + ω 02
can be derived from only the impulse transformation, δ ( t) ← L→1, and the properties of the Laplace transform. u( t) ← L→
1 : s t
∫ g(τ )dτ ←→
Integration
L
0− t
G( s) s
1
∫ δ (λ )dλ ←→ s L
0−
u( t) ← L→ e −αt u( t) ← L→
1 s
1 : s+α e −αt u( t) ← L→
Frequency Shifting te −αt u( t) ← L→
1 : (s + α )2
Complex Frequency Differentiation
− t g( t) ← L→
Solutions 9-35
d (G(s)) ds
1 s+α
d 1 ds s
− t u( t) ← L→
− t u( t) ← L→ −
1 s2 t u( t) ← L→
Linearity
te −αt u( t) ← L→
Frequency Shifting sin(ω 0 t) u( t) ← L→
1 (s + α )2
ω0 : s + ω 02 2
e −αt u( t) ← L→ e − jω 0 t u( t) ← L→
1 s+α 1 s + jω 0
Linearity
(e
− jω 0 t
Linearity
(e
jω 0 t
Linearity
e
jω 0 t
sin(ω 0 t) u( t) ← L→ cos(ω 0 t) u( t) ← L→
1 s2
− e jω 0 t ) u( t) ← L→
1 1 − s + jω 0 s − jω 0
− e − jω 0 t ) u( t) ← L→
1 1 − s − jω 0 s + jω 0
−e j2
− jω 0 t
1 1 − s − jω 0 s + jω 0 L→ u( t) ← j2
ω0 s + ω 02 2
s : s + ω 02 2
e −αt u( t) ← L→ e − jω 0 t u( t) ← L→
Linearity
(e
1 s+α 1 s + jω 0
jω 0 t
+ e − jω 0 t ) u( t) ← L→
Solutions 9-36
1 1 + s − jω 0 s + jω 0
e
Linearity
jω 0 t
cos(ω 0 t) u( t) ← L→ e −αt sin(ω 0 t) u( t) ← L→
+e 2
− jω 0 t
1 1 + s − jω 0 s + jω 0 L→ u( t) ← 2
s s + ω 02 2
ω0 : (s + α ) 2 + ω 02
ω0 s + ω 02 ω0 e −αt sin(ω 0 t) u( t) ← L→ (s + α ) 2 + ω 02 sin(ω 0 t) u( t) ← L→
Frequency Shifting e −αt cos(ω 0 t) u( t) ← L→
2
s+α : (s + α ) 2 + ω 02 cos(ω 0 t) u( t) ← L→
s s + ω 02 s+α e −αt cos(ω 0 t) u( t) ← L→ (s + α ) 2 + ω 02
Frequency Shifting
2
27. Given an LTI system transfer function, H( s) , find the time-domain response, y( t) to the excitation, x( t) . (a)
x( t) = sin(2πt) u( t) , H( s) =
1 s +1 X( s) =
Y( s) = X( s) H( s) = Y( s) = Y( s) =
2π 2 s + (2π ) 2
2π 1 2π 1 = 2 s + (2π ) s + 1 ( s + j 2π )( s − j 2π ) s + 1 2
1− s 2π 1 2 2 2 + s + 1 1 + (2π ) s + (2π )
1 2π 2π s 1 2 2 − 2 2 + 2 s + 1 1 + (2π ) 2π s + (2π ) s + (2π )
y( t) =
sin(2πt) 2π − cos(2πt) + e − t u( t) 2 1 + (2π ) 2π
Solutions 9-37
(b)
x( t) = u( t) , H( s) =
3 s+2 X( s) =
Y( s) = X( s) H( s) = y( t) = (c)
x( t) = u( t) , H( s) =
1 s
3 = s( s + 2)
31 1 − 2 s s + 2
3 (1 − e −2t ) u(t) 2
3s s+2 X( s) =
1 s
Y( s) = X( s) H( s) =
3 s+2
y( t) = 3e −2 t u( t) (d)
x( t) = u( t) , H( s) =
5s s + 2s + 2 2
1 s 1 1 5s 5 Y( s) = 2 = 2 =5 s s + 2s + 2 s + 2s + 2 (s + 1) 2 + 1 X( s) =
y( t) = 5e − t sin( t) u( t) (e)
5s s + 2s + 2
x( t) = sin(2πt) u( t) , H( s) =
2
X( s) = Y( s) =
2π 2 s + (2π ) 2
s 2π 5s = 10π 2 2 2 2 s + (2π ) s + 2 s + 2 s + (2π ) ( s2 + 2 s + 2)
[
2
Y( s) = 10π
[s
s
2
+ (2π )
2
](s
2
+ 2 s + 2)
=
Letting s be zero, 0=
B D 2 + (2π ) 2
Solutions 9-38
]
As + B Cs + D 2 + 2 s + 2s + 2 s + (2π ) 2
Multiplying through by s and letting s approach infinity, 0= A+C Letting s = 1,
A+B C+D 2π . 2 = 2 + 5 1 + (2π ) 1 + (2π )
Letting s = −1,
−A + B −10π 2 = 2 −C + D . 1 + (2π ) 1 + (2π )
Arranging the equations in matrix form, 0 1 1 1 + (2π ) 2 1 − 2 1 + (2π ) Solving,
1 (2π ) 2 0 1 2 1 + (2π ) 1 2 1 + (2π )
1 0 2 A 0 1 0 B 2π 1 1 = 2 + ( ) π 1 2 5 5 C D − 10π 2 −1 1 1 + (2π ) 0
A −0.7535 B 1.5875 = C 0.7535 D −0.0804
where Y( s) =
−0.7535 s + 1.5875 0.7535 s − 0.0804 + 2 s2 + 2 s + 2 s2 + (2π )
s +1 s 2.107 2π 1 + Y( s) = −0.7535 2 0 . 7535 − 1 . 1067 2 − 2 2 2 2π s2 + (2π ) (s + 1) + 1 s + (2π ) ( s + 1) + 1
{
}
y( t) = −0.7535[cos(2πt) − 0.3353 sin(2πt)] + 0.7535e − t [cos( t) − 1.1067 sin( t)]
28. Write the differential equations describing these systems and find and sketch the indicated responses. (a)
x( t) = u( t) , y( t) is the response ,
y(0 − ) = −5
Solutions 9-39
d , ( y( t)) = 10 dt t = 0−
∫
x(t)
∫
y(t)
2
10 y′′ ( t) = x( t) − (2 y′ ( t) + 10 y( t)) y′′ ( t) + 2 y′ ( t) + 10 y( t) = x( t)
[
]
1 d s2 Y( s) − s y(0 − ) − ( y( t)) + 2 s Y( s) − y(0 − ) + 10 Y( s) = s dt t = 0− s2 Y( s) + 5 s − 10 + 2 s Y( s) + 10 + 10 Y( s) =
1 s
1 s2 Y( s) + 2 s Y( s) + 10 Y( s) = − 5 s s 1 − 5s 1 1 − 5 s2 s Y( s) = 2 = s + 2 s + 10 s ( s + 1) 2 + 9 1 Bs + C Y( s) = 10 + s ( s + 1) 2 + 9 In 1 1 1 − 5s Bs + C = 10 + 2 s ( s + 1) + 9 s ( s + 1) 2 + 9 2
multiply through by s and let s approach infinity yielding −5 =
1 51 . +B⇒B=− 10 10
Then in 1 51 − s+C 1 1 − 5 s2 = 10 + 10 2 s ( s + 1) 2 + 9 s ( s + 1) + 9 let s be one and solve for C,
Solutions 9-40
C=−
1 . 5
1 51 1 − s− 1 1 51s + 2 Y( s) = 10 + 10 2 5 = − s ( s + 1) + 9 10 s ( s + 1) 2 + 9 Y( s) = y( t) =
s +1 1 1 49 3 − + 51 10 s (s + 1) 2 + 9 3 (s + 1) 2 + 9
49 1 −t 1 − e 51cos( 3t) − sin( 3t) u( t) , t > 0 3 10 y(0 + ) = −5 = y(0 − ) . Check.
49 d 1 y( t)) = −51(−3e − t sin( 3t) − e − t cos( 3t)) + ( 3e − t cos( 3t) − e − t sin( 3t)) , t > 0 ( 3 dt 10 d d dt ( y( t)) + = 10 = dt ( y( t)) − , t > 0 . Check. t =0 t =0 Both the response and its first derivative must be continuous in response to a step excitation because of the double integration between excitation and response. x(t) 5
4
t
-5
(b)
is ( t) = u( t) , v( t) is the response , No initial energy storage i(t) i s(t)
C 1 = 3 µF
R 1 = 2 kΩ + R 2 = 1 kΩ
C 2 = 1 µF v(t) -
Solutions 9-41
v( t) = C1 v{ is ( t) − C2 v′ ( t) + ′s ( t) R2 144244 voltage 3 across i( t )
current source
v( t) R v s ( t) = v( t) + C2 v′ ( t) + R2 1 144 42444 3 voltage across R1
Combining equations,
or
v( t) v′ ( t) R = C1 v′ ( t) + C2 v′′ ( t) + is ( t) − C2 v′ ( t) + R2 1 R2 R1R2C1C2 v′′ ( t) + ( R2C2 + ( R2 + R1 )C1 ) v′ ( t) + v( t) = R2 is ( t) R1R2C1C2 s2 V( s) + ( R2C2 + ( R2 + R1 )C1 ) s V( s) + V( s) = V( s) = V( s) =
R2
[
R2 s
]
s R1R2C1C2 s + ( R2C2 + ( R2 + R1 )C1 ) s + 1 2
1 1 R1C1C2 2 1 1 R + R2 ss + + 1 s + R1C1 R1R2C2 R1R2C1C2 V( s) =
1.667 × 10 8 s( s2 + 1667 s + 1.667 × 10 5 )
1.667 × 10 8 1000 73.52 1073.52 V( s) = = + − s( s + 1560)( s + 106.8) s s + 1560 s + 106.8 v( t) = (1000 + 73.52e −1560 t − 1073.52e −106.8 t ) u( t) v(0 + ) = 0 . Check (−1560 × 73.52e −1560 t + 106.8 × 1073.52e −106.8 t ) u( t) d (v(t)) = −1560 t dt − 1073.52e −106.8 t )δ ( t) +(1000 + 73.52e d dt ( v( t)) + = 0 . Check. t =0
Solutions 9-42
(d) x(t) 1000
0.04
t
is ( t) = cos(2000πt) u( t) , v( t) is the response , No initial energy storage
(c)
R 1 = 2 kΩ
i(t) C 1 = 3 µF
i s(t)
+ C 2 = 1 µF v(t)
R 2 = 1 kΩ
-
From part (b) R1R2C1C2 v′′ ( t) + ( R2C2 + ( R2 + R1 )C1 ) v′ ( t) + v( t) = R2 is ( t) R1R2C1C2 s2 V( s) + ( R2C2 + ( R2 + R1 )C1 ) s V( s) + V( s) = R2 V( s) = V( s) =
[s
2
+ (2000π )
1 R1C1C2
R2 s
2
][R R C C s + (R C 2
1
2
1
2
2
2
s 2 s + (2000π ) 2
]
+ ( R2 + R1 )C1 ) s + 1
s
[s
2
1 1 R + R2 2 + 1 + (2000π ) s2 + s + R1C1 R1R2C2 R1R2C1C2
]
1.667 × 10 8 s V( s) = 2 (s + 3.948 × 107 )(s2 + 1667s + 1.667 × 105 ) Expanding in partial fractions, V( s) = − V( s) = −3.96
3.96 s − 6626 4.269 0.3104 − 2 + 2 s + 1560 s + 106.84 s + (2000π )
s 6626 2000π 4.269 0.3104 − 2 + 2 + 2 2000π s + (2000π ) s + 1560 s + 106.84 s + (2000π ) 2
[
]
v( t) = −3.96 cos(2000πt) + 1.054 sin(2000πt) + 4.269e −1560 t − 0.3104 e −106.84 t u( t)
Solutions 9-43
v(0 + ) = 0 . Check. 2000π × 3.96 sin(2000πt) + 2000π × 1.054 cos(2000πt) t ( ) u −1560 t + 106.84 × 0.3104 e −106.84 t −1560 × 4.269e d v( t)) = ( dt −3.96 cos(2000πt) + 1.054 sin(2000πt) δ ( t) + +4.269e −1560 t − 0.3104 e −106.84 t d dt ( v( t)) + = 0 . Check. t =0 (e) x(t) 5
0.01
t
-5
29.
Find the three parts, x ac ( t), x 0 ( t) and x c ( t) , of the following signals. (a)
x( t) = t x ac ( t) = t u(− t) , x 0 ( t) = 0 , x c ( t) = t u( t)
(b)
x( t) = sin(ωt) x ac ( t) = sin(ωt) u(− t) , x 0 ( t) = 0 , x c ( t) = sin(ωt) u( t)
(c)
x( t) =
d (sgn(t)) dt x( t) = 2δ ( t) x ac ( t) = 0 , x 0 ( t) = 2δ ( t) , x c ( t) = 0
30. Find the bilateral Laplace transforms of these signals. (a)
x( t) = rect ( t) 1 2
x c ( t) = rect ( t) u( t) 1
−
s
e − st 2 e 2 − 1 1 − e − st X c ( s) = ∫ e dt = = −s = s − s 0+ 0+ x 0 ( t) = 0 ⇒ X 0 ( s) = 0
Solutions 9-44
−
s 2
, Any s
x ac ( t) = rect ( t) u(− t) ⇒ x ac (− t) = rect (− t) u( t) X ac (− s) =
1 2
∫e
− st
0+
X( s) =
1− e s
s 2
+
s 2
−
s 2
e e − 1 1− e dt = = = s −s − s 0−
X ac ( s) = −
1 2
− st
s 2
e −1 e − e = s s
−
s 2
s 2
−
s 2
, Any s
s 2
1− e e −1 = , Any s −s s
=
e
−j
js 2
−e s
j
js 2
js sin − 2 js = = sinc − , Any s s 2π −j 2
ω Notice that if we make the change of variable, s = jω , we get X( jω ) = sinc 2π ( ) = ( ) which is the CTFT of x t rect t and this is allowed because the region of convergence is the entire s plane. x( t) = rect ( t) sin(20πt)
(b)
x c ( t) = rect ( t) sin(20πt) u( t) X c ( s) =
1 2
∫ sin(20πt)e
0+
− st
1 2
dt =
∫
0+
e
j 20πt
−e j2
− j 20πt
e − st dt =
1 2
(
)
1 e −( s− j 20π ) t − e −( s+ j 20π ) t dt j 2 0∫+
1 −( s + j 20π ) −( s− j 20π ) 12 2 e e − −( s − j 20π ) t −( s + j 20π ) t + 1 e 1 s − j 20π e 20 π s + j X c ( s) = + − = j 2 s − j 20π s + j 20π 0 + j 2 1 1 + − s − j 20π s + j 20π 1 2
s − j 20π s + j 20π − − 1 1 − e 2 1− e 2 − X c ( s) = j 2 s − j 20π s + j 20π
s − j 20π s + j 20π − − 1 − e 2 ( s + j 20π ) − 1 − e 2 ( s − j 20π ) 1 X c ( s) = j2 (s − j 20π )(s + j 20π )
Solutions 9-45
1 j 40π − j 20πe X c ( s) = j2
X c ( s) =
20π − 2e
−
s 2
−
s 2
(e j10π + e − j10π ) + se s2 + (20π )
−
s 2
(e
− e j10π )
2
[sin(10π )s + 10π cos(10π )] s2 + (20π )
− j10π
2
, Any s
x 0 ( t) = 0 ⇒ X 0 ( s) = 0 x ac ( t) = rect ( t) sin(20πt) u(− t) ⇒ x ac (− t) = rect (− t) sin(−20πt) u( t) x ac (− t) = − rect ( t) sin(20πt) u( t) Then, using the previous result, X ac (− s) = −
20π − 2e
−
s 2
[sin(10π )s + 10π cos(10π )] s2 + (20π )
2
and s
X ac ( s) = −
X( s) = 2
20π − 2e 2 [− sin(10π ) s + 10π cos(10π )] 2
, Any s
s s −s −s − e 2 + e 2 sin(10π ) s − 10π cos(10π ) e 2 − e 2
X( s) = 4
(c)
s2 + (20π )
s2 + (20π )
2
s s 10π cos(10π ) sinh − sin(10π ) cosh s 2 2
[
s2 + (20π )
2
]
x( t) = e −2 t u( t) − e 2 t u(− t) sin(2πt) x c ( t) = e −2 t sin(2πt) u( t) ⇒ X c ( s) =
2π , Re( s) > −2 (s + 2) 2 + (2π ) 2
x 0 ( t) = 0 ⇒ X 0 ( s) = 0 x ac ( t) = −e 2 t sin(2πt) u(− t) ⇒ x ac (− t) = −e −2 t sin(−2πt) u( t) = e −2 t sin(2πt) u( t) Using the previous result, X ac (− s) =
2π 2π , Re( s) < 2 2 2 ⇒ X ac ( s) = (s + 2) + (2π ) (− s + 2) 2 + (2π ) 2 Solutions 9-46
X( s) =
2π 2π , − 2 < Re( s) < 2 2 2 + (s + 2) + (2π ) (− s + 2) 2 + (2π ) 2
Solutions 9-47
M. J. Roberts - 7/12/03
Chapter 10 -
Laplace Transform Analysis of Signals and Systems Solutions
1. For each circuit write the transfer function between the indicated excitation and indicated response. Express each transfer function in the standard form, sN + bN −1sN −1 + L + b2 s2 + b1s + b0 . H( s) = A D s + aD −1sD −1 + L + a2 s2 + a1s + a0 Excitation: v s ( t)
(a)
Response: v o ( t)
R1
L +
vs(t)
C
R2
vo(t) -
H( s) =
R2 Z RLC ( s) Z RLC ( s) + R1 sL + R2
where
(R2 + sL) sC
R s+ 2 sL + R2 1 L = = 2 Z RLC ( s) = R 1 1 + + s LC sR C C 1 2 2 2 s +s + R2 + sL + L LC sC 1
sL + R2 R2 R2 s LC + sR2C + 1 = H( s) = sL + R2 sL + R2 R2 + sL + R1 ( s2 LC + sR2C + 1) + R 1 s2 LC + sR2C + 1 2
H( s) =
R2 R = 2 s R1LC + s( L + R1R2C ) + R2 + R1 R1LC
(b)
Excitation: is ( t)
2
Response: v o ( t)
10-1
1 1 R R + R1 + 2 + 2 s2 + s R1LC R1C L
M. J. Roberts - 7/12/03
C2 R2 R1 + i s(t)
C1
vo(t) -
The current flowing toward the op-amp through resistor, R1 , is
I1 ( s) =
1 sC1 1 + R1 sC1
I s ( s) =
I s ( s) 1 + sR1C1
That current must all flow through the feedback network and the inverting input must be at virtual ground. Therefore 1 R 1 R2 sC2 2 =− Vo ( s) = − I1 ( s) I ( s) 1 sR1C1 + 1 sR2C2 + 1 s R2 + sC2 1 Vo ( s) =− I s ( s) R1C1C2
(c)
Excitation: v s ( t)
1 1 1 1 + s2 + s + R2C2 R1C1 R1R2C1C2 Response: i1 ( t) C2 R1 i1(t) C1
vs(t)
I s ( s) =
Vs ( s) 1 R2 + sC2 R1 + 1 R2 + + sC2
10-2
R2
1 sC1 1 sC1
M. J. Roberts - 7/12/03
I1 ( s) = I s ( s)
R2 + R2 +
1 sC2
1 1 + sC2 sC1
1 I ( s) 1 sC2 H( s) = 1 = = Vs ( s) 1 1 R + 1 + 1 1 1 1 1 R R + + + R + 2 R2 + 1 2 2 sC2 sC1 sC2 sC1 sC2 sC1 sC2 sC1 R1 + 1 1 R2 + + sC2 sC1 R2 +
1 sC2
R2 +
1 s2 R2C1C2 + sC1 sC2 H( s) = = 2 R R R 1 s R1R2C1C2 + sR1C1 + sR1C2 + sR2C2 + 1 R1R2 + 1 + 1 + 2 + 2 sC2 sC1 sC1 s C1C2 R2 +
1 1 R2C2 H( s) = R1 2 1 1 1 1 + + s + s + R2C2 R2C1 R1C1 R1R2C1C2 s2 + s
2. For each block diagram write the transfer function between the excitation , x( t) , and the response, y( t) . (a) x(t)
∫
∫
∫
8
2 s2 Y( s) = H( s) =
X( s) − [8 s Y( s) + 2 Y( s)] s
Y( s) 1 1 = = 3 2 X( s) s( s + 8 s + 2) s + 8 s2 + 2 s
(b)
10-3
y(t)
M. J. Roberts - 7/12/03
x(t)
-1
∫
∫
-4
∫
y(t)
-10 s2 Y( s) =
X( s) − X( s) − 4 s Y( s) − 10 Y( s) s
H( s) =
s−1 Y( s) =− 3 s + 4 s2 + 10 s X( s)
3. Evaluate the stability of each of these system transfer functions. 100 s + 200
(a)
H( s) = −
(b)
H( s) =
80 s− 4
Pole at s = 4. Unstable.
(c)
H( s) =
6 s( s + 1)
Poles at s = 0 and s = −1. Marginally stable
Pole at s = −200. Stable.
(therefore unstable) (d)
H( s) = −
15 s s + 4s + 4
(e)
H( s) = 3
s − 10 Poles at s = −2 ± j 5. Stable. s + 4 s + 29
(f)
H( s) = 3
s2 + 4 Poles at s = 2 ± j 5. Unstable. s2 − 4 s + 29
(g)
H( s) =
(h)
H( s) =
2
Double pole at s = −2. Stable.
2
1 s + 64 2
Poles at s = ± j 8. Marginally stable (therefore unstable)
10 Poles at s = 0 and s = −2 ± j 5. Unstable. s + 4 s2 + 29 s 3
4. Find the overall transfer functions of these systems in the form of a single ratio of polynomials in s. 10-4
M. J. Roberts - 7/12/03
(a)
s2
X(s)
10 s2 + 3s + 2
s2 + 3s + 2 H( s) =
Y(s)
s2 s2 10 10 = s2 + 3s + 2 s2 + 3s + 2 s4 + 6 s3 + 13s2 + 12 s + 4
(b) s+1 s 2 + 2s + 13 X(s)
Y(s) 1 s + 10
H( s) =
s +1 1 (s + 1)(s + 10) + s2 + 2s + 13 + = s2 + 2 s + 13 s + 10 s3 + 12 s2 + 33s + 130
13 23 s + 2 s + 13s + 23 2 2 =2 3 H( s) = 3 s + 12 s2 + 33s + 130 s + 12 s2 + 33s + 130 s2 +
2
(c) s
X(s)
2
s +s+5
Y(s)
s s H( s) = s + s + 5 = 2 s s + 2s + 5 1+ 2 s +s+5 2
(d) X(s)
20s s + 200s + 290000 2
1 s + 400
10-5
Y(s)
M. J. Roberts - 7/12/03
20 s 20 s( s + 400) s + 200 s + 290000 H( s) = = 2 20 s 1 (s + 200s + 290000)(s + 400) + 20s 1+ 2 s + 200 s + 290000 s + 400 2
s2 + 400 s H( s) = 20 3 s + 600 s2 + 370020 s + 1.16 × 10 8 5. In the feedback system below, find the overall system transfer function for these values of forward-path gain, K. H( s) =
K 1 + 0.1K
10 6 ≅ 10 1 + 10 5
(a)
K = 10 6
H( s) =
(b)
K = 10
10 5 H( s) = ≅ 10 1 + 10 4
(c)
K = 10
H( s) =
10 =5 1+1
(d)
K =1
H( s) =
1 = 0.909 1 + 0.1
(e)
K = −1
H( s) =
−1 = −1.11K 1 − 0.1
(f)
K = −10
H( s) → −∞
5
X(s)
K
Y(s)
0.1
6. In the feedback system below, plot the response of the system to a unit step, for the time period, 0 < t < 10 , then write the expression for the overall system transfer function and draw a pole-zero diagram, for these values of K.
10-6
M. J. Roberts - 7/12/03
X(s)
Y(s)
K e -s
0.1
One-second time delay
(All plots at the end). H( s) = (a)
K = 20
H( s) = 20
K 1 + 0.1e − sK
1 1 + 2e − s
H( s) = 20(1 − 2e − s + 4 e −2 s − 8e −3 s + L) h( t) = 20[δ ( t) − 2δ ( t − 1) + 4δ ( t − 2) − 8δ ( t − 3) + L] h −1 ( t) = 20[u( t) − 2 u( t − 1) + 4 u( t − 2) − 8 u( t − 3) + L] Poles at 1 + 2e − s = 0 . Solving for s, s = log(−2) = ln(2) + jπ (2 n + 1)
-
Unstable ω
h- 1(t)
8π
8000 -1
10
t
-3
K = 10
σ
3
σ
-8 π
-8000
(b)
3
H( s) = 10
1 1 + e−s
H( s) = 10(1 − e − s + e −2 s − e −3 s + L) h( t) = 10[δ ( t) − δ ( t − 1) + δ ( t − 2) − δ ( t − 3) + L] h −1 ( t) = 10[u( t) − u( t − 1) + u( t − 2) − u( t − 3) + L] Poles at 1 + e − s = 0 . Solving for s, s = log(−1) = jπ (2 n + 1)
-
Marginally stable
h- 1(t)
ω
10
8π -3
-1
10
t
-8 π
10-7
M. J. Roberts - 7/12/03
(c)
H( s) =
K =1
1 1 + 0.1e − s
H( s) = 1 − 0.1e − s + 0.01e −2 s − 0.001e −3 s + L h( t) = δ ( t) − 0.1δ ( t − 1) + 0.01δ ( t − 2) − 0.001δ ( t − 3) + L h −1 ( t) = u( t) − 0.1u( t − 1) + 0.01u( t − 2) − 0.001u( t − 3) + L Poles at 1 + 0.1e − s = 0 . Solving for s, s = log(−0.1) = ln(0.1) + jπ (2 n + 1)
-
Stable
h- 1(t)
ω
1
8π -3
-1
(d)
K = −1
10
H( s) = −
t
3
σ
-8 π
1 1 − 0.1e − s
H( s) = −1 − 0.1e − s − 0.01e −2 s − 0.001e −3 s + L h( t) = −δ ( t) − 0.1δ ( t − 1) − 0.01δ ( t − 2) − 0.001δ ( t − 3) + L h −1 ( t) = − u( t) − 0.1u( t − 1) − 0.01u( t − 2) − 0.001u( t − 3) + L Poles at 1 − 0.1e − s = 0 . Solving for s, s = log(0.1) = ln(0.1) + j 2 nπ
-
Stable
h (t)
ω
-1
-1
10
t
8π -3 -8 π
-2
(e)
K = −10
3
H( s) = −10
1 1 − e−s
H( s) = −10(1 + e − s + e −2 s + e −3 s + L) h( t) = −10[δ ( t) + δ ( t − 1) + δ ( t − 2) + δ ( t − 3) + L] h −1 ( t) = −10[u( t) + u( t − 1) + u( t − 2) + u( t − 3) + L]
10-8
σ
M. J. Roberts - 7/12/03
Poles at 1 − e − s = 0 . Solving for s, s = log(1) = j 2 nπ
-
Marginally stable ω
h- 1(t) -1
10
8π
t -3
3 -8 π
-100
(f)
H( s) = −20
K = −20
σ
1 1 − 2e − s
H( s) = −20(1 + 2e − s + 4 e −2 s + 8e −3 s + L) h( t) = −20[δ ( t) + 2δ ( t − 1) + 4δ ( t − 2) + 8δ ( t − 3) + L] h −1 ( t) = −20[u( t) + 2 u( t − 1) + 4 u( t − 2) + 8 u( t − 3) + L] Poles at 1 − 2e − s = 0 . Solving for s, s = log(2) = ln(2) + j 2 nπ
-
Unstable
h- 1(t) -1
10
ω
t
8π -3
-32000
3
σ
-8 π
7. For what range of values of K is the system below stable? Plot the step responses for K = 0, K = 4 and K = 8. 1 s2 - 4s + 4
X(s)
Y(s)
Ks
1 1 H( s) = s − 4 s + 4 = 2 Ks s + s(K − 4 ) + 4 1+ 2 s − 4s + 4 2
4 − K ± K 2 − 8K . The system is unstable if either pole is in the 2 RHP. If K 2 − 8K < 0 the poles are complex. If K (K − 8) < 0 then 0 < K < 8 . In this range of K, if K < 4, the real part of the poles is in the RHP and the system is unstable. Poles at s =
10-9
M. J. Roberts - 7/12/03
For 4 < K < 8 , the poles are both complex with real parts in the LHP and the system is 4 − K − K 2 − 8K stable. For K > 8, the poles are both real, the pole, s = in the LHP is 2 4 − K + K 2 − 8K , is in the LHP approaching negative infinity and the pole, s = 2 approaching the origin. For any K < 0, K 2 − 8K > −K > 0 , therefore 2 4 − K ± K − 8K > 0 and both poles are in the RHP. So the system is unstable if K < 4. For K = 0, H( s) = H −1 ( s) = For K = 4,
1 1 = ⇒ h( t) = te 2 t u( t) s − 4 s + 4 ( s − 2) 2 2
1 1 1 1 2 1 1 = + ⇒ h −1 ( t) = (1 + 2 te 2 t − e 2 t ) u( t) 2 − 2 s s − 4 s + 4 4 s ( s − 2) s − 2 4 H( s) = H −1 ( s) =
For K = 8,
1 1 1 1 s 1 − cos(2 t) = − 2 u( t) ⇒ h −1 ( t) = 2 s s + 4 4 s s + 4 4
H( s) = H −1 ( s) =
sin(2 t) 1 ⇒ h( t) = u( t) 2 s +4 2
1 1 −2 t = u( t) 2 ⇒ h( t) = te s + 4 s + 4 ( s + 2) 2
1 1 1 1 2 1 1 −2 t −2 t = − − ⇒ h −1 ( t) = (1 − 2 te − e ) u( t) 2 2 s s + 4 s + 4 4 s ( s + 2) s + 2 4 h- 1(t)
K=0
3000
-1
h- 1(t)
K=4
4
t
0.5 -1
4
t
-0.5
h- 1(t)
K=8
0.25
-1
4
t
8. Plot the impulse response and the pole-zero diagram for the forward-path and the overall system below. 10-10
M. J. Roberts - 7/12/03
100 s + 2s + 26
X(s)
2
Y(s)
10 s + 20 H1 ( s) = 100
(
)
1 100 − t 1 e sin 24 t u( t) = 100 ⇒ h1 ( t) = 2 s + 2 s + 26 24 (s + 1) + 24 2
Poles at s = −1 ± j 5. 100 100( s + 20) s + 2 s + 26 H( s) = = 2 100 10 (s + 2s + 26)(s + 20) + 1000 1+ 2 s + 2 s + 26 s + 20 2
H( s) = 100
s + 20 s + 22 s + 66 s + 1520 3
2
Poles at s = −22.12, 0.0612 ± j 8.29 . Using MATLAB to do the partial fraction expansion, H( s) = − H( s) = −
0.3785 0.3785 s + 91.58 + 2 s + 22.12 s − 0.1224 s + 68.71
242.02 8.29 0.3785 s − 0.0612 + 0.3785 + 2 2 8.29 ( s − 0.0612) + 68.71 s + 22.12 ( s − 0.0612) + 68.71
{
}
h( t) = −0.3785e −22.12 t + 0.3785e 0.0612 t [cos(8.29 t) + 29.19 sin(8.29 t)] u( t)
10-11
M. J. Roberts - 7/12/03
ω [s]
5 h1(t)
Forward Path
H1(s)
20
σ
-1
-0.5
4
t
-5
-20
ω Overall System
[s]
8.29
h(t)
H(s)
30
-22.12 -0.5
8
t
σ
-20
0.0612 -8.29
-30
9. Using the Routh-Hurwitz method, evaluate the stability of the system whose transfer function is s3 + 3s + 10 H( s) = 5 s + 2 s4 + 10 s3 + 4 s2 + 8 s + 20 5 4 3
1 10 8 2 4 20 −2 0 8 9 20 0 2 2 338 0 0 1 − 9 0 20 0 0 This result indicates that there are two poles in the RHP. That is confirmed by finding the poles which lie at s = −0.9432 + j 2.9107 s = −0.9432 − j 2.9107 s = 0.5502 + j1.207 s = 0.5502 − j1.207 s = −1.214 10. Using the Routh-Hurwitz stability test, evaluate the stability of the system whose transfer function is of the general form, H( s) =
N( s) . s + a2 s2 + a1s + a0 3
What are the relations among a2 , a1 and a0 that ensure stability? 10-12
M. J. Roberts - 7/12/03
3 2
1 a2 a0 − a1a2 1 − a2 0 a0 −
a0 − a1a2 a2
For stability, a2 > 0 , −
a1 a0 0 0
a0 − a1a2 > 0 ⇒ a1a2 − a0 > 0 ⇒ a1a2 > a0 , a0 > 0 . a2
11. Plot the root locus for each of the systems which have these loop transfer functions and identify the transfer functions that are stable for all positive real values of K. (a)
T( s) =
K (s + 3)(s + 8) ω
σ
-3
(b)
T( s) =
Ks (s + 3)(s + 8) ω
-8 -3
(c)
T( s) =
σ
Ks2 (s + 3)(s + 8) ω
-8
(d)
T( s) =
-3
K (s + 1)(s + 4 s + 8) 2
10-13
σ
M. J. Roberts - 7/12/03
ω
-1
σ
12. Use the block diagram, of an inverting amplifier using an operational amplifier,
Z f(s) Z i(s) + Z f(s)
Vi (s)
Ve(s)
-
A0
s 1- p
Vo(s)
Z i(s) Z i(s) + Z f(s)
,
with A0 = 10 4 , p = −2000π , Z f = 10kΩ and Zi = 1kΩ, to find the gain and phase margins of the amplifier. A0 A0 Z i ( s) Z i ( s) T( s) = − − = s Z i ( s) + Z f ( s) s 1− 1 − Z i ( s) + Z f ( s) p p
[
T( s) =
]
10 4 × 10 3 10 7 1 = 4 s 3 1.1 × 10 ( s + 2000π ) 4 1 + (10 + 10 ) 2000π 2000π T( s) =
0.5712 × 10 7 s + 6283
Finding the zero-dB frequency, T( jω 0 dB ) = 0.5712 × 10 7
ω
2 0 dB
+ (6283)
2
0.5712 × 10 7 =1 jω 0 dB + 6283
= 1 ⇒ ω 0 dB = 5.712 × 10 6 ⇒ f 0 dB = 909.1 kHz
At that frequency the phase of the loop transfer function is π 0.5712 × 10 7 ∠ T( s) = ∠ = − or − 90° 6 2 j 5.712 × 10 + 6283
10-14
M. J. Roberts - 7/12/03
So the phase margin is 90°. The phase approaches 90° as the frequency approaches infinity. The loop transfer function magnitude approaches zero as the frequency approaches infinity. The loop transfer function magnitude in dB therefore approaches minus infinity. That means the gain margin is infinite. 13. Plot the unit step and ramp responses of unity-gain feedback systems with these forward-path transfer functions. The following relationships are very handy in solving these problems: B − Aα As + B e −αt A cos(βt) + L→ sin(βt) u( t) ← β (s + α )2 + β 2 e
e
e
−
C t 2
−
C t 2
−αt
2 As + B B − A α 2 −1 B − Aα A + β cos t − tan L→ u( t) ← β Aβ (s + α )2 + β 2
2 2 As + B A cos D − C t + 2 B − AC sin D − C t u( t) ← L→ 2 2 s + Cs + D 2 2 4D − C
2 2 As + B A 2 + 2 B − AC cos D − C t − tan −1 2 B − AC u( t) ← L→ 2 2 2 2 s + Cs + D 4D − C A 4D − C
(a)
H1 ( s) =
100 s + 10
H( s) =
100 s + 110
Unit step response: 100 10 1 1 10 −110 t H −1 ( s) = = − ⇒ h −1 ( t) = (1 − e ) u(t) s( s + 110) 11 s s + 110 11 Unit ramp response: 1 1 100 10 1 110 10 1 − e −110 t H −2 ( s) = 2 = 2− + 110 ⇒ h −2 ( t) = t − u( t) s ( s + 110) 11 s s s + 110 11 110 Unit-Ramp Response Unit-Step Response h (t) -1
h (t)
1
0.1
-2
0.1
t
10-15
0.1
t
M. J. Roberts - 7/12/03
(b)
H1 ( s) =
100 s( s + 10)
H( s) =
100 s + 10 s + 100 2
Unit step response: As + B 100 1 H −1 ( s) = = + 2 2 s( s + 10 s + 100) s s + 10 s + 100 Multiplying through by s and letting s approach infinity we get A = −1. Then letting s = 1 we get B = −10. Then 1 1 5 75 s + 10 s+5 = − H −1 ( s) = − 2 − 2 2 s s + 10 s + 100 s ( s + 5) + 75 75 ( s + 5) + 75 h −1 ( t) = 1 − e −5 t cos
(
)
75 t +
(
5 sin 75
75 t u( t)
)
Unit ramp response: 1 As + B 100 1 H −2 ( s) = 2 2 = 2 − 10 + 2 s s + 10 s + 100 s ( s + 10 s + 100) s 1 1 s 1 10 1 1 5 75 s+5 H −2 ( s) = 2 − + 2 10 = 2 − 10 + − 2 2 s s s + 10 s + 100 s s 10 ( s + 5) + 75 75 ( s + 5) + 75 1 e −5 t h −2 ( t) = t − + 10 10
cos
(
)
(
5 sin 75
75 t −
Unit-Step Response
75 t u( t)
)
Unit-Ramp Response h- 2(t)
h- 1(t)
2
1
2
t
2
100 s ( s + 10)
(c)
H1 ( s) =
H( s) =
100 100 = 2 2 s + 10 s + 100 ( s + 10.85)( s − 0.8496 s + 9.216)
2
3
10-16
t
M. J. Roberts - 7/12/03
Unit step response: H −1 ( s) =
As + B 100 1 0.0677 = − + 2 s( s + 10.85)( s − 0.8496 s + 9.216) s s + 10.85 s − 0.8496 s + 9.216 2
Multiplying through by s and letting s approach infinity we get A = −0.9323 . Then, solving for B we get B = 0.0584 . −0.9323s + 0.0575 1 0.0677 H −1 ( s) = − + 2 s s + 10.85 s − 0.8495 s + 9.216 s − 0.0617 1 0.0677 H −1 ( s) = − − 0.9323 2 s s + 10.85 s − 0.8495 s + 9.216 3.006 1 0.0677 s − 0.4248 . 0 1208 H −1 ( s) = − − 0.9323 + 2 s s + 10.85 (s − 0.4248) 2 + 9.0355 ( s − 0.4248) + 9.0355
{
}
h −1 ( s) = 1 − 0.0677e −10.85 t − 0.9323e 0.4248 t [cos( 3.006 t) + 0.1208 sin( 3.006 t)] u( t) Unit ramp response: Using MATLAB, H −2 ( s) = H −2 ( s) = H −2 ( s) =
1 0.00624 0.00624 s + 0.927 − 2 2 + s s + 10.85 s − 0.8495 s + 9.217
s − 148.56 1 0.00624 − 0.00624 2 2 + s s + 10.85 s − 0.8495 s + 9.217
3.006 1 0.00624 s − 0.8496 . . 0 00624 49 70 + − − 2 s2 s + 10.85 (s − 0.4248) 2 + 9.036 ( s − 0.4248) + 9.036
{
}
h −2 ( t) = t + 0.00624 e −10.85 t − 0.00624 e 0.4248 t [cos( 3.006 t) − 49.7 sin( 3.006 t)] u( t) Unit-Ramp Response
Unit-Step Response
h- 2(t)
h (t) -1
20 40
10
t -1 -5
-40
(d)
H1 ( s) =
20 (s + 2)(s + 6)
10-17
10
t
M. J. Roberts - 7/12/03
20 20 20 (s + 2)(s + 6) = 2 H( s) = = 20 (s + 2)(s + 6) + 20 s + 8s + 32 1+ (s + 2)(s + 6) Unit step response: H −1 ( s) =
20 s( s + 8 s + 32) 2
Using MATLAB to do the partial fraction expansion H −1 ( s) =
0.625 0.625 s + 5 − 2 s s + 8 s + 32
H −1 ( s) =
0.625 4 s+4 − 0.625 + 2 2 s ( s + 4 ) + 16 ( s + 4 ) + 16
{
}
h −1 ( t) = 0.625 − 0.625e −4 t [cos( 4 t) + sin( 4 t)] u( t) Unit ramp response: H −2 ( s) =
20 s ( s + 8 s + 32) 2
2
Using MATLAB to do the partial fraction expansion H −2 ( s) = H −2 ( s) =
0.625 0.1562 0.1562 s + 0.625 − + s2 s s2 + 8 s + 32
0.625 0.1562 s+4 − + 0.1562 2 s s (s + 4) 2 + 16
{
]}
[
h −2 ( t) = 0.625 t − 0.1562 1 − e −4 t cos( 4 t) u( t) Unit-Ramp Response
Unit-Step Response h (t)
h- 2(t)
-1 1
1
1
t
1
t
14. Reduce these block diagrams to a single block by block-diagram reduction. Check the answer using Mason’s theorem.
10-18
M. J. Roberts - 7/12/03
(a)
-2 s+3
X(s)
10 s + 20
Y(s)
1 s -2 s+3
X(s)
-20s (s + 3) (s 2 + 20s + 10)
X(s) Mason’s Theorem: P1 ( s) =
10s s + 20s + 10
−20 (s + 3)(s + 20)
Y(s)
2
and T1 ( s) =
Y(s)
10 s( s + 20)
10 and ∆1 ( s) = 1 s( s + 20) Np −20 ∑=1 Pi (s)∆ i (s) (s + 3)(s + 20) −20 s . Check. H( s) = i = = 2 10 ∆( s) 3 20 10 + ( ) + + s s s ( ) 1+ s( s + 20) and
∆( s) = 1 +
(b)
s
X(s)
s s+1
5 s + 20 1 s
10-19
Y(s)
M. J. Roberts - 7/12/03
Y(s)
s
X(s)
s s+1
5 s + 20 1 s
X(s)
5 s + 20
s s+1
s 2 +20s - 5 s + 20
Y(s)
s 2 +20s - 5 s + 20
Y(s)
s 2 +20s - 5 s + 20
Y(s)
5 s (s + 20)
X(s)
s s+1 5 s (s + 20)
X(s)
s (s + 20) s 2 +20s + 5
s s+1
s 2 (s 2 +20s - 5) (s + 1) (s 2 +20s + 5)
X(s)
Y(s)
Mason’s Theorem: P1 ( s) =
−5 s (s + 1)(s + 20)
T1 ( s) =
5 s( s + 20)
and
and
and
∆1 ( s) = 1
P2 ( s) =
∆( s) = 1 + and
s2 s +1
5 s( s + 20)
∆ 2 ( s) = 1
(Forward path #2 and the only feedback loop do share a signal.)
10-20
M. J. Roberts - 7/12/03
Np
H( s) =
∑ Pi (s)∆ i (s) i =1
∆( s)
H( s) =
−5 s −5 s + s2 ( s + 20) s2 + (s + 1)(s + 20) s + 1 (s + 1)(s + 20) = = 5 s2 + 20 s + 5 1+ s( s + 20) s( s + 20) s2 ( s2 + 20 s − 5)
(s + 1)(s2 + 20s + 5)
. Check.
15. Find the responses of the systems with these transfer functions to a unit-step and a suddenly-applied, unit-amplitude, 1 Hz cosine. Also find the responses to a true unitamplitude, 1 Hz cosine (not suddenly applied) using the CTFT and compare to the steady-state part of the total solution found using the Laplace transform. Use the results of the chapter in which if a cosine is suddenly applied to a system N( s) , the response is whose transfer function is H( s) = D( s) N ( s) y( t) = L−1 1 + H ( jω 0 ) cos(ω 0 t + ∠H ( jω 0 )) u( t) D( s) where
N1 ( s) is the partial fraction involving the transfer function poles. D( s) (a)
H( s) =
1 s
Unit-step response: H −1 ( s) =
1 ⇒ h −1 ( t) = t u( t) = ramp( t) s2
Unit-amplitude 1 Hz cosine response: Y( s) =
sin(2πt) u( t) 1 1 2π s 2 = 2 ⇒ y( t) = 2 2 2π s + (2π ) 2π s s + (2π )
Using the CTFT: H( jω ) =
1 jω
Y( jω ) =
δ (ω − 2π ) δ (ω + 2π ) 1 π [δ (ω − 2π ) + δ (ω + 2π )] = π + jω jω jω δ (ω − 2π ) δ (ω + 2π ) Y( jω ) = π + − j 2π j 2π 10-21
M. J. Roberts - 7/12/03
Y( jω ) = (b)
H( s) =
jπ sin(2πt) . Check. δ (ω + 2π ) − δ (ω − 2π )] ⇒ y( t) = [ 2π 2π s s +1
Unit step response: H −1 ( s) =
1 ⇒ h −1 ( t) = e − t u( t) s +1
Unit-amplitude 1 Hz cosine response:
Y( s) =
s s 2 s + 1 s + (2π ) 2
1 2 As + B 1 + (2π ) + 2 = 2 s +1 s + (2π )
N ( s) y( t) = L−1 1 + H( jω 0 ) cos(ω 0 t + ∠ H( jω 0 )) u( t) D( s) 1 2 1 + (2π ) y( t) = L−1 + H( j 2π ) cos(2πt + ∠ H( j 2π )) u( t) s + 1 1 2 1 j 2π j 2π 1 + (2π ) y( t) = L−1 cos t + ∠ + u( t) j 2π + 1 2π s + 1 j 2π + 1 e−t 2π u( t) y( t) = + cos ( 2 π t + 0 . 157 ) 2 2 1 + (2π ) 1 + 2 π ( ) Alternate solution: In Y( s) =
s s 2 s + 1 s + (2π ) 2
1 2 As + B 1 + (2π ) + 2 = 2 s +1 s + (2π )
let s be zero.
10-22
M. J. Roberts - 7/12/03
1 B (2π ) 0= 2 + 2 ⇒ B= − 2 1 + (2π ) 1 + (2π ) (2π ) 2
(2π ) 1 As − 2 2 2 2 1 As 1 + (2π ) − (2π ) 1 1 + (2π ) 1 + (2π ) + + Y( s) = = 2 2 2 s +1 s2 + (2π ) s2 + (2π ) 1 + (2π ) s + 1 2
[
]
Solving for A,
(2π ) 2 A= 2 1 + (2π ) and Y( s) =
1 1 s−1 2 + (2π ) 2 2 2 1 + (2π ) s + 1 s + (2π )
{
}
y( t) =
1 2 −t + (2π ) cos(2πt) − 2π sin(2πt) u( t) 2 e 1 + (2π )
y( t) =
1 −t + 2π 2 e 1 + (2π )
{
}
(2π ) 2 + 1 cos(2πt + 0.157) u(t)
Which agrees with the previous answer. Using the CTFT: H( jω ) = Y( jω ) =
jω jω + 1
j 2π jω j 2π δ (ω + 2π ) δ (ω − 2π ) − π [δ (ω − 2π ) + δ (ω + 2π )] = π 1 − j 2π jω + 1 1 + j 2π
δ (ω − 2π )(1 − j 2π ) − δ (ω + 2π )(1 + j 2π ) δ (ω − 2π ) δ (ω + 2π ) − = j 2π 2 Y( jω ) = j 2π 2 1 − j 2π (1 + j 2π )(1 − j 2π ) 1 + j 2π Y( jω ) = j 2π 2
δ (ω − 2π ) − δ (ω + 2π ) − j 2π [δ (ω − 2π ) + δ (ω + 2π )] 1 + (2π )
2
2π (2π ) cos(2πt) − 2π sin(2πt) y( t) = . Check. 2 [− sin(2πt) + 2π cos(2πt)] = 2 1 + (2π ) 1 + (2π ) 2
10-23
M. J. Roberts - 7/12/03
(c)
H( s) =
s s + 2 s + 40 2
Unit step response: H −1 ( s) =
(
)
e − t sin 39 t 1 1 t ⇒ = h −1 ( ) = u( t) s2 + 2 s + 40 ( s + 1) 2 + 39 39
Unit-amplitude cosine response: Using MATLAB to do the partial fraction expansion, Y( s) =
s s 0.4991s − 0.1302 0.4991s − 0.1319 − 2 = 2 2 s + 2 s + 40 s + (2π ) s2 + 2 s + 40 s2 + (2π ) 2
Y( s) = 0.4991
s − 0.2609 s − 0.2643 2 − 0.4991 2 2 s + 2 s + 40 s + (2π )
s s 0.2609 2π 0.2643 6 Y( s) = 0.4991 2 − + 2 − 2 2 2 2π s2 + (2π ) 6 ( s + 1) + 36 (s + 1) + 36 s + (2π )
{
}
y( t) = 0.4991 cos(2πt) − 0.0415 sin(2πt) − e − t [cos(6 t) + 0.044 sin(6 t)] u( t) H( jω ) = Y( jω ) =
( jω )
( jω )
jω 2
+ j 2ω + 40
jω 2
+ j 2ω + 40
π [δ (ω − 2π ) + δ (ω + 2π )]
j 2πδ (ω − 2π ) j 2πδ (ω + 2π ) − Y( jω ) = π 2 2 ( j 2π ) + j 4π + 40 (− j 2π ) − j 4π + 40 j 2πδ (ω − 2π ) j 2πδ (ω + 2π ) − Y( jω ) = π 0.5216 + j 4π 0.5216 − j 4π Y( jω ) = π
j 3.277[δ (ω − 2π ) − δ (ω + 2π )] + 8π 2 [δ (ω − 2π ) + δ (ω + 2π )] 0.5216 2 + ( 4π )
2
8π 2 cos(2πt) − 3.277 sin(2πt) y( t) = . Check. 158.19 (d)
H( s) =
s2 + 2 s + 40 s2
10-24
M. J. Roberts - 7/12/03
Unit-step response: Using t n u( t) ← L→
n! , Re( s) > 0 sn +1
s2 + 2 s + 40 1 2 40 H −1 ( s) = = + 2 + 3 ⇒ h −1 ( t) = 1 + 2 t + 20 t 2 u( t) 3 s s s s
[
]
Unit-amplitude 1 Hz cosine response: Using MATLAB to find the partial fraction expansion, Y( s) =
s2 + 2 s + 40 s s2 + 2 s + 40 1.013 0.013s − 2 = = − 2 2 2 2 s2 s s2 + (2π ) s + (2π ) s s2 + (2π )
[
Y( s) =
]
1.013 1 2π s − 0.013 2 2 + 2 s π s + (2π ) 2 s + (2π )
y( t) = {1.013 − 0.013 cos(2πt) + 0.318 sin(2πt)} u( t) 2 jω ) + 2 jω + 40 ( H( jω ) = ( jω ) 2 2 jω ) + 2 jω + 40 ( Y( jω ) = π [δ (ω − 2π ) + δ (ω + 2π )] ( jω ) 2 2 2 j 2π ) + j 4π + 40 − j 2π ) − j 4π + 40 ( ( Y( jω ) = π δ (ω − 2π ) + δ (ω + 2π ) ( j 2π ) 2 (− j 2π ) 2
( j 2π ) 2 δ (ω − 2π ) + j 4πδ (ω − 2π ) + 40δ (ω − 2π ) +(− j 2π ) 2 δ (ω + 2π ) − j 4πδ (ω + 2π ) + 40δ (ω + 2π ) Y( jω ) = π −4π 2 Y( jω ) = −π
0.5216[δ (ω − 2π ) + δ (ω + 2π )] + j12.57[δ (ω − 2π ) − δ (ω + 2π )] 39.48 y( t) = [0.3184 sin(2πt) − 0.0132 cos(2πt)]
16. For each pole-zero diagram sketch the approximate frequency response magnitude.
10-25
M. J. Roberts - 7/12/03
ω
[s]
|H( f )| 0.2
(a)
σ
-5
-10
ω
10
f
[s]
|H( f )|
(b)
1
σ
-2
-2
ω
2
f
[s] 4
|H( f )| 0.05
(c)
σ
-3
-2
-4
ω
2
f
[s] 2
|H( f )|
(d)
10
σ
-4
-2
ω
-2
2
f
[s] 10
|H( f )| 0.5
(e)
σ
-1
-20
-10
10-26
20
f
M. J. Roberts - 7/12/03
17. Using only a calculator, find the transfer function of a third-order ( n = 3) lowpass Butterworth filter with cutoff frequency, ω c = 1, and unity gain at zero frequency. H( jω ) =
K = (s − p1)(s − p2 )(s − p3 )
K
(s + 1) s − e
H( jω ) =
K
(s + 1)s2 − s e
2π j 3
+e
2π −j 3
+ 1
=
j
2π 3
2π −j 3 s e −
K K = 3 2 2 (s + 1)(s + s + 1) s + 2s + 2s + 1
For unity gain: H(0) = K ⇒ K = 1. 18. Using MATLAB, find the transfer function of an eighth-order lowpass Butterworth filter with cutoff frequency, ω c = 1, and unity gain at zero frequency. »[z,p,k] = buttap(8) ; »z z = [] »p p = -0.1951 -0.5556 -0.8315 -0.9808 -0.9808 -0.8315 -0.5556 -0.1951
+ + + + -
0.9808i 0.8315i 0.5556i 0.1951i 0.1951i 0.5556i 0.8315i 0.9808i
»k k = 1.00000000000000
H( s) =
1
( s + 0.1951 + j 0.9808)( s + 0.1951 − j 0.9808) ( s + 0.5556 + j 0.8315)( s + 0.5556 + j 0.8315) ( s + 0.8315 + j 0.5556)( s + 0.8315 − j 0.5556) ( s + 0.9808 + j 0.1951)( s + 0.9808 + j 0.1951)
10-27
M. J. Roberts - 7/12/03
H( s) =
1 (s + 0.3902s + 1)(s + 1.1112s + 1)(s2 + 1.663s + 1)(s2 + 1.9616s + 1) 2
2
Multiplying the first factor by the last and then the two middle factors, H( s) = H( s) =
1 (s + 2.3518s + 2.7654 s + 2.3518s + 1)(s4 + 2.7741s3 + 3.8478s2 + 2.7741s + 1) 4
3
2
1 s + 5.126 s + 13.1371s + 21.8462 s + 25.6884 s4 + 21.8462 s3 + 13.1371s2 + 5.126 s + 1 8
7
6
5
19. Find the transfer functions of these Butterworth filters. (a)
Second-order highpass with a cutoff frequency of 20 kHz and a passband gain of 5. The transfer function of the normalized filter is H norm ( s) =
K K = 2 (s + 0.707 + j 0.707)(s + 0.707 − j 0.707) s + 1.414 s + 1 f c = 20kHz ⇒ ω c = 1.257 × 10 5
Making the transformation, s → H( s) =
H( s) =
ωc , s
K
ω ωc + 1.414 c + 1 s s 2
=
s2K ω c2 + 1.414ω c s + s2
s2K
s2 + 1.414 (1.257 × 10 5 ) s + (1.257 × 10 5 )
2
=
s2K s2 + 1.777 × 10 5 s + 1.579 × 1010
The passband is high frequencies (approaching infinity). Therefore H( j∞) = K ⇒ K = 5 H( s) =
(b)
5 s2 s2 + 1.777 × 10 5 s + 1.579 × 1010
Third-order bandpass with a center frequency of 5 kHz, a -3 dB bandwidth of 500 Hz and a passband gain of 1.
10-28
M. J. Roberts - 7/12/03
K
H norm ( s) =
(s + 1) s − e
2π j 3
s − e
2π −j 3
=
K K = 3 2 2 (s + 1)(s + s + 1) s + 2s + 2s + 1
We convert to bandpass with the transformation, s →
s 2 + ω Lω H . s(ω H − ω L )
f 0 = 5000 ⇒ ω 0 = 10000π = 31416 f H = f 0 + 250 = 5250 ⇒ ω H = 32987 f L = f 0 − 250 = 4750 ⇒ ω L = 29845 H( s) =
H( s) =
H( s) =
(s
2
K 3
2
s + ω Lω H s + ω Lω H s 2 + ω Lω H s ω − ω + 2 s ω − ω + 2 s ω − ω + 1 ( H L) ( H ( H L ) L ) 2
+ ω Lω H ) + 2 s(ω H − ω L 3
2
(s(ω )(s + ω ω ) 2
L
− ω L )) K 3
H 2
H
+ 2( s(ω H − ω L )) ( s2 + ω Lω H ) + ( s(ω H − ω L )) 2
3
(1000π ) 3 Ks3 3 2 (s2 + 9.845 × 108 ) + 2000πs(s2 + 9.845 × 108 ) + 2(1000π )2 s2 (s2 + 9.845 × 108 ) + (1000π )3 s3 H( s) =
H( s) =
(1000π ) 3 Ks3 s6 + 3 × 9.845 × 10 8 s4 + 3 × 9.692 × 1017 s2 + 9.542 × 10 26 4 9 2 17 s s s 2000 1 969 10 9 692 10 . . + + × + × π ( ) 2 4 3 3 8 2 +2(1000π ) ( s + 9.845 × 10 s ) + (1000π ) s
(1000π ) 3 Ks3 s6 + 3 × 9.845 × 10 8 s4 + 3 × 9.692 × 1017 s2 + 9.542 × 10 26 5 9 3 17 +2000πs + 2000π × 1.969 × 10 s + 2000π × 9.692 × 10 s 2 4 2 3 3 8 2 +2(1000π ) s + 2(1000π ) × 9.845 × 10 s + (1000π ) s
H( s) =
3.1 × 1010 Ks3 s6 + 2.95 × 10 9 s4 + 2.908 × 1018 s2 + 9.542 × 10 26 5 13 3 21 +6283s + 1.237 × 10 s + 6.09 × 10 s 7 4 16 2 10 3 +1.974 × 10 s + 1.943 × 10 s + 3.1 × 10 s
10-29
M. J. Roberts - 7/12/03
H( s) =
3.1 × 1010 Ks3 s6 + 6283s5 + 2.97 × 10 9 s4 + 1.24 × 1013 s3 + 2.93 × 1018 s2 + 6.09 × 10 21 s + 9.542 × 10 26
From a MATLAB calculation using the symbolic math toolbox, K = 1, (indicating that unity passband gain is preserved in the transformation process). (c)
Fourth-order bandstop with a center frequency of 10 MHz, a -3 dB bandwidth of 50 kHz and a passband gain of 1. H norm ( s) =
1 s + 2.6131s + 3.4142 s2 + 2.6131s + 1 4
3
Making the transformation, s → H( s) =
s(ω H − ω L ) , s 2 + ω Lω H 1
3 2 s(ω − ω ) s(ω H − ω L ) s(ω H − ω L ) H L 2 + 2.6131 s2 + ω ω + 3.4142 s2 + ω ω s + ω Lω H L H L H +2.6131 s(ω H − ω L ) + 1 s2 + ω ω L H 4
f 0 = 10 7 ⇒ ω 0 = 2π × 10 7 = 6.283 × 10 7 f H = f 0 + 25000 = 10, 025, 000 ⇒ ω H = 6.2989 × 10 7 f L = f 0 − 25000 = 9, 975, 000 ⇒ ω L = 6.2675 × 10 7 H( s) =
(s
2
+ ω Lω H )
4
( s(ω − ω )) 4 + 2.6131( s2 + ω ω )( s(ω − ω )) 3 H L L H H L 2 2 2 +3.4142( s + ω ω ) ( s(ω − ω )) L H H L +2.6131 s2 + ω ω 3 s ω − ω + s2 + ω ω 4 ( ( L H) ( H L) L H)
(s + 3.9478 × 10 ) H( s) = (10 πs) + 2.6131( s + 3.9478 × 10 )(10 πs) +3.4142( s + 3.9478 × 10 ) (10 πs) +2.6131 s + 3.9478 × 10 10 πs + s + 3.9478 × 10 ( ) ( ) 15 4
2
5
4
2
15
2
15 2
2
15 3
10-30
5
5
5
3
2
2
15
4
M. J. Roberts - 7/12/03
H( s) =
H( s) =
s8 + 1.579 × 1016 s6 + 9.351 × 10 31 s4 + 2.461 × 10 47 s2 + 2.429 × 10 62
9.741 × 10 21 s4 + ( s2 + 3.9478 × 1015 )8.102 × 1016 s3 +( s4 + 7.8948 × 1015 s2 + 1.5582 × 10 31 ) 3.3697 × 1011 s2 6 +( s + 1.1843 × 1016 s4 + 4.6749 × 10 31 s2 + 6.1514 × 10 46 )8.2093 × 10 5 s 8 16 6 31 4 47 2 62 +( s + 1.579 × 10 s + 9.35 × 10 s + 2.4607 × 10 s + 2.4285 × 10 ) s8 + 1.579 × 1016 s6 + 9.351 × 10 31 s4 + 2.461 × 10 47 s2 + 2.429 × 10 62 9.741 × 10 21 s4 + 8.102 × 1016 s5 + 3.1985 × 10 32 s3 11 6 27 4 42 2 +3.3697 × 10 s + 2.6603 × 10 s + 5.251 × 10 s +8.2093 × 10 5 s7 + 9.7223 × 10 21 s5 + 3..8377 × 10 37 s3 + 5.0499 × 10 52 s + s8 + 1.579 × 1016 s6 + 9.35 × 10 31 s4 + 2.4607 × 10 47 s2 + 2.4285 × 10 62
s8 + 1.579 × 1016 s6 + 9.351 × 10 31 s4 + 2.461 × 10 47 s2 + 2.429 × 10 62 H( s) = 8 s + 8.2093 × 10 5 s7 + 1.579 × 1016 s6 + 9.7223 × 10 21 s5 + 9.35 × 10 31 s4 37 3 47 2 52 62 +3.8377 × 10 s + 2.4607 × 10 s + 5.0499 × 10 s + 2.4285 × 10
20. Draw canonical system diagrams of the systems with these transfer functions. (a)
H( s) =
1 s +1
Y( s) 1 = ⇒ X( s) = s Y( s) + Y( s) ⇒ s Y( s) = X( s) − Y( s) X( s) s + 1
X(s)
1 s
+ -
(b)
H( s) = 4 H( s) =
Y(s)
s+3 s + 10
Y( s) Y1 ( s) Y( s) 1 4 ( s + 3) = = X( s) X( s) Y1 ( s) s + 10
Y1 ( s) 1 = X( s) s + 10
Y( s) = 4 ( s + 3) Y1 ( s)
s Y1 ( s) = X( s) − 10 Y1 ( s)
Y( s) = 4 s Y1 ( s) + 12 Y1 ( s)
10-31
M. J. Roberts - 7/12/03
4
+
12 1 s
+
X(s)
-
Y(s)
+
Y(s) 1
10 21. Draw cascade system diagrams of the systems with these transfer functions. (a)
H( s) =
s s +1
This is a single block in a cascade system realized as a canonical system. s Y1 ( s) = X( s) − Y1 ( s)
Y( s) = s Y1 ( s)
Y(s) 1 s
+
X(s)
-
(b)
H( s) =
X(s)
s+4 (s + 2)(s + 12) 1 s
+ -
1 s
+ -
2 (c)
4
+
+
Y(s)
12
H( s) =
20 s( s + 5 s + 10)
H( s) =
1 20 2 s s + 5 s + 10
X(s)
Y(s) 1
2
1 s
1 s
+ +
1 s
20
Y(s)
5
+
10 22. Draw parallel system diagrams of the systems with these transfer functions.
10-32
M. J. Roberts - 7/12/03
(a)
−12 s + 3s + 10
H( s) =
2
Poles are complex. Therefore one second-order canonical system is the same as the parallel system. X(s)
1 s
+ +
1 s
Y(s)
-12
3
+
10 (b)
H( s) =
2 s2 s2 + 12 s + 32
H( s) =
3s + 8 24 s + 64 2 s2 = 2−8 = 2− 2 2 s + 12 s + 32 s + 12 s + 32 (s + 8)(s + 4) H( s) = 2 −
32 8 + s+8 s+4
2 +
X(s)
-
+
1 s
-32
+
Y(s) +
8 + -
1 s
8
4 23. Write state equations and output equations for the circuit of Figure E23 with the inductor current, iL ( t) , and capacitor voltage, vC ( t) , as the state variables and the voltage at the input, v i ( t) , as the excitation and the voltage at the output, v L ( t) , as the response. R = 10 Ω C = 1 µF +
vi (t)
+
vC (t)
L = 1 mH
+ i L (t) vL(t) -
-
10-33
M. J. Roberts - 7/12/03
Figure E23 An RLC circuit v i ( t) = R iL ( t) + vC ( t) + L i′L ( t) 1 v′C ( t) = iL ( t) C v L ( t ) = v i ( t ) − R i L ( t ) − vC ( t ) State equations, v′C ( t) 0 = i′L ( t) − 1 L
1 0 C vC ( t) + 1 v ( t) i R − iL ( t) L L vC ( t ) v L ( t) = [−1 − R] + v i ( t) i L ( t) 24. Write state equations and output equations for the circuit of Figure E24 with the inductor current, iL ( t) , and capacitor voltage, vC ( t) , as the state variables and the current at the input, ii ( t) , as the excitation and the voltage at the output, v R ( t) , as the response.
+
ii (t)
i L (t)
R = 100 Ω
vR(t)
L = 1 mH + vC (t) -
C = 1 µF
Figure E24 An RLC circuit ii ( t) = iL ( t) + C v′C ( t)
L i′L ( t) = R[ii ( t) − iL ( t)] + vC ( t)
State equations, 1 1 v′C ( t) 0 − C vC ( t) C = + i i ( t) i′L ( t) 1 − R iL ( t) R L L vC ( t) v R ( t) = [0 − R] + R i i ( t) i L ( t) 25. From the system transfer function,
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M. J. Roberts - 7/12/03
H( s) =
s( s + 3) , s + 2s + 9 2
write a set of state equations and output equations using a minimum number of states. This is a second-order system so the minimum number of states is two. We can express the transfer function as Y( s) s( s + 3) = 2 . X( s) s + 2 s + 9 Therefore
s2 Y( s) = s2 X( s) + 3s X( s) − 2 s Y( s) − 9 Y( s)
Let Y( s) be a state, Q1 ( s) and let s Y( s) be a state, Q 2 ( s) . Then s Q1 ( s) = Q 2 ( s) s Q 2 ( s) = −9 Q1 ( s) − 2 Q 2 ( s) + s2 X( s) + 3s X( s) Y( s) = Q1 ( s) and we can write state equations as s Q1 ( s) 0 1 Q1 ( s) 0 = + 2 X( s) s Q 2 ( s) −9 −2 Q 2 ( s) s + 3s
Q1 ( s) Y( s) = [1 0] Q 2 ( s)
26. Write state equations and output equations for the system whose block diagram is in Figure E26 using the responses of the integrators as the state variables. x(t)
∫
∫
∫
8
2 Figure E26 A system Let q1 ( t) = y( t) , q 2 ( t) = y′ ( t) and q′3 ( t) = x( t) . Then
10-35
y(t)
M. J. Roberts - 7/12/03
q1′ ( t) 0 1 0 q1 ( t) 0 q′2 ( t) = −2 −8 1 q 2 ( t) + 0 x( t) q′3 ( t) 0 0 0 q 3 ( t) 1 q1 ( t) y( t) = [1 0 0]q 2 ( t) q 3 ( t) 27. A system is excited by the signal, x( t) = 3 u( t) , and the response is y( t) = 0.961e −1.5 t sin( 3.122 t) u( t) . Write a set of state equations and output equations using a minimum number of states. X( s) =
3 s
3.122 3 Y( s) = 0.961 = 2 2 2 (s + 1.5) + 3.122 s + 3s + 12
.
Then the transfer function is 3 s Y( s) s2 + 3s + 12 H( s) = = = 2 . 3 s + 3s + 12 X( s) s Let Y( s) be a state, Q1 ( s) and let s Y( s) be a state, Q 2 ( s) . Then s Q1 ( s) = Q 2 ( s)
s Q 2 ( s) = −12 Q1 ( s) − 3 Q 2 ( s) + s X( s) Y( s) = Q1 ( s)
and we can write state equations as 1 Q1 ( s) 0 s Q1 ( s) 0 = − + X( s) s Q 2 ( s) 12 −3Q 2 ( s) s
Q1 ( s) Y( s) = [1 0] Q 2 ( s)
28. A system is described by the differential equation, y′′ ( t) + 4 y′ ( t) + 7 y( t) = 10 cos(200πt) u( t) . Write a set of state equations and output equations for this system with two states. Let q1 ( t) = y( t) and let q 2 ( t) = y′ ( t) . Then
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M. J. Roberts - 7/12/03
q1′ ( t) 0 1 q1 ( t) 0 = + 10 cos(200πt) u( t) q′2 ( t) −7 −4 q 2 ( t) 1
q1 ( t) y( t) = [1 0] q 2 ( t)
29. A system is described by the state equations and output equations, q1′ ( t) −2 1 q1 ( t) 1 2 x1 ( t) = + q′2 ( t) 3 0 q1 ( t) −2 0 x 2 ( t)
and
y1 ( t) 3 5 q1 ( t) = y 2 ( t) −2 4 q 2 ( t)
q1 (0 + ) 0 x1 ( t) −δ ( t) With excitation, = . Find the system = and initial conditions, + q 2 (0 ) 3 x 2 ( t) u( t) y1 ( t) response vector, . y 2 ( t) s −1 2 s + − s 2 1 1 1 −1 Φ( s) = ( sI − A ) = = 2 = s + 2s − 3 3 s s + 2 s − 3 3 s + 2 −3 2 s + 2s − 3
[
]
Q( s) = Φ( s) BX( s) + q(0 + )
s s2 + 2 s − 3 = 3 2 s + 2s − 3
s s2 + 2 s − 3 Q( s) = 3 2 s + 2s − 3
1 −1 s + 2 s − 3 1 2 1 + 0 s + 2 −2 0 3 s s2 + 2 s − 3 2
1 2 s + 2 s − 3 −1 + s s + 2 5 2 s + 2s − 3 2
2 5 s s2 + 2 s − 3 −1 + s + s2 + 2 s − 3 Q( s) = 2 3 s+2 −1 + + 5 2 s2 + 2 s − 3 s s + 2 s − 3 5 −s + 2 s2 + 2 s − 3 + s2 + 2 s − 3 Q( s) = −3s + 6 5 s2 + 10 s 2 + 2 s( s + 2 s − 3) s( s + 2 s − 3) 10-37
1 s + 2s − 3 s+2 s2 + 2 s − 3 2
M. J. Roberts - 7/12/03
s− 7 − 2 s + 2s − 3 Q( s) = 5 s2 + 7 s + 6 2 s( s + 2 s − 3) 5 3 −2 + 2 Q( s) = s + 3 s − 1 9 5 2 + 2 − 2 s + 3 s − 1 s 5 −3 t 3 t − e + 2e q( t) = 2 u( t) 5 −3 t 9 t e + e − 2 2 2 s+2 s + 2s − 3 2
5 −3 t 3 t 5e −3 t + 27e t − 10 y1 ( t) 3 5 − 2 e + 2 e u( t) = u( t) = −3 t t y 2 ( t) −2 4 5 e −3 t + 9 e t − 2 15e + 15e − 8 2 2
30. A system is described by the vector state equation and output equation, q′ ( t) = Aq( t) + Bx ( t) and
y ( t) = Cq( t) + Dx ( t) ,
−1 −3 1 0 2 −3 1 0 where A = , B= , C= and D = . Define two new 2 −7 0 1 0 4 0 0 states, in terms of the old states, for which the A matrix is diagonal and re-write the state equations. The eigenvalues can be found from the A matrix. They are the roots of s2 + 8 s + 13 which are s = −2.2679 and s = −5.7321. So the matrix of eigenvalues is 0 −2.2679 Λ= . −5.7321 0 The equation to solve for the transformation matrix that diagonalizes the system is ΛT = TA . Therefore T is the matrix of eigenvectors for the matrix, A. A normalized T is
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M. J. Roberts - 7/12/03
0.8446 −0.5354 T= . −0.3893 0.9211 To verify that this solution is correct, 0 0.8446 −0.5354 0.8446 −0.5354 −1 −3 −2.2679 = 0 −5.7321−0.3893 0.9211 −0.3893 0.9211 2 −7
Using
−1.9155 1.2143 −1.9155 1.2143 2.2315 −5.2798 = 2.2315 −5.2798 Check. q′2 ( t) = Tq1′ ( t)
we get
Using
0.8446 −0.5354 q′2 ( t) = q1′ ( t) . −0.3893 0.9211 q′2 ( t) = TA1T−1q2 ( t) + TB1x ( t) = A 2q2 ( t) + B 2 x ( t)
we get −1
0.8446 −0.5354 −1 −3 0.8446 −0.5354 0.8446 −0.5354 1 0 q′2 ( t) = q 2 ( t) + x ( t) −0.3893 0.9211 2 −7 −0.3893 0.9211 −0.3893 0.9211 0 1 or
0 −2.2679 0.8446 −0.5354 q′2 ( t) = q 2 ( t) + x ( t) −5.7321 0 −0.3893 0.9211
31. For the original state equations and output equations of Exercise 30 write a differentialequation description of the system. The original state equations are
and
q1′ ( t) −1 −3 q1 ( t) 1 0 x1 ( t) = + q′2 ( t) 2 −7 q 2 ( t) 0 1 x 2 ( t) y1 ( t) 2 −3 q1 ( t) 1 0 x1 ( t) = + . y 2 ( t) 0 4 q 2 ( t) 0 0 x 2 ( t)
From the output equation,
or
y1′ ( t) 2 −3 q1′ ( t) 1 0 x1′ ( t) = + y′2 ( t) 0 4 q′2 ( t) 0 0 x′2 ( t)
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y1′ ( t) 2 −3−1 −3 q1 ( t) 1 0 x1 ( t) 1 0 x1′ ( t) = + + y′2 ( t) 0 4 2 −7 q 2 ( t) 0 1 x 2 ( t) 0 0 x′2 ( t) y1′ ( t) −8 15 q1 ( t) 2 −3 x1 ( t) 1 0 x1′ ( t) = + + . y′2 ( t) 8 −28 q 2 ( t) 0 4 x 2 ( t) 0 0 x′2 ( t) Solving the output equations for the states, q1 ( t) 2 −3 y1 ( t) 1 0 x1 ( t) = − q 2 ( t) 0 4 y 2 ( t) 0 0 x 2 ( t) −1
or
1 q1 ( t) 2 = q 2 ( t) 0
3 1 8 y1 ( t) − 1 y 2 ( t) 2 0 4
0 x1 ( t) . t x ( ) 2 0
Then 1 y1′ ( t) −8 15 2 = y′2 ( t) 8 −28 0
3 1 8 y1 ( t) − 1 y 2 ( t) 2 0 4
x ( t) 2 −3 x ( t) 1 0 x′ ( t) 0 1 1 1 + 0 4 + 0 0 t x 2 ( t) x′2 ( t) 0 x 2 ( )
y1′ ( t) −4 3 y1 ( t) 6 −3 x1 ( t) 1 0 x1′ ( t) 4 = + + y ( ) t ′ 2 4 −4 y 2 ( t) −4 4 x 2 ( t) 0 0 x′2 ( t) 3 y ( t) + 6 x1 ( t) − 3 x 2 ( t) + x1′ ( t) 4 2 y′2 ( t) = 4 y1 ( t) − 4 y 2 ( t) − 4 x1 ( t) + 4 x 2 ( t) y1′ ( t) = −4 y1 ( t) +
32. Find the s-domain transfer functions for the circuits below and then draw block diagrams for them as systems with Vi ( s) as the excitation and Vo ( s) as the response. R = 10 kΩ + + vi (t)
L = 5 mH
C = 1 µF
vo(t) −
− (a)
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sL sC H( s) =
sL +
V0 ( s) = Vi ( s)
1 sC
sL sC sL +
1 sC
H( s) = 100
=
+R
s 1 1 1 RC s2 + s + RC LC
s s + 100 s + 2 × 10 8 2
V(s) o
100
V(s) i
1 s
+
1 s
100
+ +
8
2 10
L = 5 mH
C = 1 µF
R = 10 kΩ +
+
vi (t)
vo(t)
− (b)
−
sL sC 1 sL + s Vo ( s) sC = − 1 H( s) = =− 1 R RC 2 Vi ( s) s + LC H( s) = −100
s (s + 2 × 108 ) 2
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V(s) o
-100
1 s
+
V(s) i
1 s
8
2 10
R = 10 kΩ
L = 5 mH +
+ vi (t)
vo(t)
C = 1 µF
−
− (c)
1 Vo ( s) 1 1 sC H( s) = = = R 1 1 Vi ( s) + sL + R LC s2 + s + L LC sC H( s) = 2 × 10 8 V(s) i
1 s + 2 × 10 6 s + 2 × 10 8 2
1 s
+
1 s
8
2 10
V(s) o
6
+
2 10
+
8
2 10
R = 10 kΩ
R = 10 kΩ
+
+
vi (t)
C = 1 µF
C = 1 µF
−
vo(t) −
(d) sRC + 1 2 2 1 V ( s) H( s) = o = 2Cs + s RC sRC 1 + sRC + 1 Vi ( s) R + 2 2 2Cs + s RC
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1 H( s) = RC
2
s2 + s
H( s) = 10 4 V(s) i
1
+
2
1 s + 300 s + 10 4 2
1 s
+
3 1 + RC RC
1 s
4
10
V(s) o
300
+
4
10
33. Determine whether the systems with these transfer functions are stable, marginally stable or unstable. (a)
H( s) =
s( s + 2) s2 + 8
Poles at s = ± j 8
Marginally stable
(b)
H( s) =
s( s − 2) s2 + 8
Poles at s = ± j 8
Marginally stable (Zero in RHP
2
notwithstanding)
(c)
H( s) =
s s + 4s + 8
Poles at s = −2 ± j 2
Stable
(d)
H( s) =
s2 s2 − 4 s + 8
Poles at s = 2 ± j 2
Unstable
(e)
2
s Poles at s = −2 ± j 2 s + 4 s2 + 8 s (Pole zero cancellation at s = 0.) H( s) =
3
Stable
34. Find the expression for the overall system transfer function of the system below. K K Y( s) H( s) = = s + 10 = s + 10 + βK X( s) 1 + βK s + 10 Pole at s = −10 − βK (a)
Let β = 1. For what values of K is the system stable? Pole at s = −10 − K System is stable for K > −10
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M. J. Roberts - 7/12/03
(b)
Let β = −1. For what values of K is the system stable? Pole at s = −10 + K System is stable for K < 10
(c)
Let β = 10. For what values of K is the system stable? Pole at s = −10 − 10K = −10(1 + K ) System is stable for K > −1
X(s)
K s+10
Y(s)
β 35. Find the expression for the overall system transfer function of the system below. For what positive values of K is the system stable? X(s)
H( s) =
K (s+1)(s+2)
Y(s)
K K Y( s) = 2 = X( s) ( s + 1)( s + 2) + K s + 3s + 2 + K
s=
−3 ± 9 − 4 (2 + K ) −3 ± 1 − 4 K = 2 2
System is stable for any positive real value of K. 36. Find the expression for the overall system transfer function of the system below. Using MATLAB plot the paths of the poles of the overall system transfer function as a function of K. For what positive values of K is the system stable? X(s)
H( s) =
K (s+1)(s+2)(s+3)
Y(s)
K K K Y( s) = 2 = = 3 2 X( s) ( s + 1)( s + 2)( s + 3) + K ( s + 3s + 2)( s + 3) + K s + 6 s + 11s + 6 + K
Although this denominator can be factored it is probably easier just to numerically explore the pole locations versus the value of K. A MATLAB program was written to
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M. J. Roberts - 7/12/03
graph the pole locations for a range of K’s and it was found that when K = 60 the poles just touch the ω axis, indicating marginal stability. When K = 60, the poles are at s = −6 , ± j 3.3166 . Therefore for 0 < K < 60 the system is stable.
ω
[s]
-3
-2
-1
σ
37. Thermocouples are used to measure temperature in many industrial processes. A thermocouple is usually mechanically mounted inside a “thermowell”, a metal sheath which protects it from damage by vibration, bending stress, or other forces. One effect of the thermowell is that its thermal mass slows the effective time response of the thermocouple/thermowell combination compared witht the inherent time response of the thermocouple alone. Let the actual temperature on the outer surface of the thermowell in Kelvins be Τs ( t) and let the voltage developed by the thermocouple in response to temperature be v t ( t) . The response of the thermocouple to a one-Kelvin step change in the thermowell outer-surface temperature from T1 to T1 + 1 is t − v t ( t) = K T1 + 1 − e 0.2 u( t)
where K is the thermocouple temperature-to-voltage conversion constant.
µV . Design an active filter which processes K the thermocouple voltage and compensates for its time lag making the overall system have a response to a one-Kelvin step thermowell-surface temperature change that is itself a step of voltage of 1mV. (a)
Let the conversion constant be K = 40
The unit step response of the thermocouple-thermowell combination is v t ( t) = K (1 − e −5 t ) u( t) . The impulse response is the derivative of the step response,
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h t ( t) = 5Ke −5 t u( t) . The transfer function is the transform of the impulse response, H t ( s) =
5K . s+5
The desired overall frequency response is H( s) =
1mV . K
(Here K is the kelvin not the thermocouple gain, K.) Therefore the transfer function of the compensating active filter is s+5 H( s) 10 −3 H f ( s) = = = 2 × 10 −4 . K 5 K H t ( s) s+5 This is a system with a real zero creating a corner frequency, ω c = 5. synthesized by the circuit of Figure S37 . Rf Ri +
C
This can be
+
vi (t)
vo (t)
-
-
Figure S37 Thermocouple-thermowell compensator Rf 10 −3 which is (except for sign) the ratio, . Let The desired low-frequency gain is K Ri R f = 10 kΩ. Then Rs = 400 Ω. To put the corner at ω c = 5 requires a capacitance, C = 500 µF . Although this should work, the capacitance is a little large. Let’s change the design to R f = 1 MΩ, Rs = 40 kΩ and C = 5 µF . These are more reasonable component values. (b) Suppose that the thermocouple also is subject to electromagnetic interference (EMI) from nearby high-power electrical equipment. Let the EMI be modeled as a sinusoid with an amplitude of 20 µV at the thermocouple terminals. Calculate the response of the thermocouple-filter combination to EMI frequencies of 1 Hz, 10 Hz and 60 Hz. How big is the apparent temperature fluctuation caused by the EMI in each case? At 1 Hz:
H f ( j 2π ) = 5( j 2π + 5) = 40.14 ∠51°
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M. J. Roberts - 7/12/03
So the response to the 20 µV excitation is about 800 µV which is equivalent to about 0.8 K. At 10 Hz:
H f ( j 20π ) = 5( j 20π + 5) = 315∠85°
So the response to the 20 µV excitation is about 6.3 mV which is equivalent to about 6.3 K. At 60 Hz:
H f ( j120π ) = 5( j120π + 5) = 1885∠89.2°
So the response to the 20 µV excitation is about 37.7 mV which is equivalent to about 37.7 K. 38. A laser operates on the fundamental principle that a pumped medium amplifies a travelling light beam as it propagates through the medium. Without mirrors a laser becomes a single-pass travelling wave amplifier (Figure E38-1). Pumped Laser Medium Light In
Light Out
Figure E38 -1 A one-pass travelling-wave light amplifier This is a system without feedback. If we now place mirrors at each end of the pumped medium, we introduce feedback into the system. Mirror
Pumped Laser Medium
Mirror
Figure E38-2 A regenerative travelling-wave amplifier When the gain of the medium becomes large enough the system oscillates creating a coherent output light beam. That is laser operation. If the gain of the medium is less that that required to sustain oscillation, the system is known as a regenerative travelling-wave amplifier (RTWA). Let the electric field of a light beam incident on the RTWA from the left be the excitation of the system, E inc ( s) , and let the electric fields of the reflected light, E refl ( s) , and the transmitted light, E trans ( s) , be the responses of the system (Figure E38-3) .
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M. J. Roberts - 7/12/03
E circ (s)
jt i
E inc (s)
ri
E refl (s)
gfp
jt o
E trans(s)
ro
ri grp
jt i
Figure E38-3 Block diagram of an RTWA Let the system parameters be as follows: Electric field reflectivity of the input mirror, ri = 0.99 Electric field transmissivity of the input mirror, ti = 1 − ri2 Electric field reflectivity of the output mirror, ro = 0.98 Electric field transmissivity of the output mirror, to = 1 − ro2
Forward and reverse path electric field gains, g fp ( s) = grp ( s) = 1.01e −10 Find an expression for the frequency response, frequency range, 3 × 1014 ± 5 × 10 8 Hz.
jti E ( s) 1 − g fp ( s) ro grp ( s) ri inc
E trans ( s) = jto g fp ( s) E circ ( s) E trans ( s) = − The transfer function is H( s) =
ti to g fp ( s) E ( s) 1 − ro ri g 2fp ( s) inc
ti to g fp ( s) E trans ( s) =− E inc ( s) 1 − ro ri g 2fp ( s)
and the frequency response is H( jω ) = −
ti to g fp ( jω ) . 1 − ro ri g 2fp ( jω )
10-48
s
E trans ( f ) , and plot its magnitude over the E inc ( f )
E circ ( s) = jti E inc ( s) + g fp ( s) ro grp ( s) ri E circ ( s) E circ ( s) =
−9
M. J. Roberts - 7/12/03
|H(j2πf )| 3
f 2.999995e+14
3.000005e+14
39. A classical example of the use of feedback is the phase-locked loop used to demodulate frequency-modulated signals (Figure E39) . x(t)
+ -
y VCO(t)
Phase Detector
xLF (t)
Loop Filter, H LF (s)
y(t)
VoltageControlled Oscillator Figure E39 A phase-locked loop
The excitation, x( t) , is a frequency-modulated sinusoid. The phase detector detects the phase difference between the excitation and the signal produced by the voltage-controlled oscillator. The response of the phase detector is a voltage proportional to phase difference. The loop filter filters that voltage. Then the loop filter response controls the frequency of the voltage-controlled oscillator. When the excitation is at a constant frequency and the loop is “locked” the phase difference between the two phase-detector excitation signals is zero. (In an actual phase detector the phase difference is 90° at lock. But that is not significant in this analysis since that only causes is a 90° phase shift and has no impact on system performance or stability.) As the frequency of the excitation, x( t) , varies, the loop detects the accompanying phase variation and tracks it. The overall response signal, y( t) , is a signal proportional to the frequency of the excitation. The actual excitation, in a system sense, of this system is not x( t) , but rather the phase of x( t) , φ x ( t) , because the phase detector detects differences in phase, not voltage. Let the frequency of x( t) be fx ( t) . The relation between phase and frequency can be seen by examining a sinusoid. Let x( t) = A cos(2πf 0 t) . The phase of this cosine is 2πf 0 t and, for a simple sinusoid ( f 0 constant), it increases linearly with time. The frequency is f 0 , the derivative of the phase. Therefore the relation between phase and frequency for a frequency-modulated signal is 1 d fx ( t ) = (φ (t)) . 2π dt x Let the frequency of the excitation be 100 MHz. Let the transfer function of the Hz . Let the transfer function of the loop filter be voltage-controlled oscillator be 10 8 V
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M. J. Roberts - 7/12/03
H LF ( s) =
1 . s + 1.2 × 10 5
V . If the frequency of the excitation radian signal suddenly changes to 100.001MHz, plot the change in the output signal, ∆ y( t) .
Let the transfer function of the phase detector be 1
Let φ diff ( t) be phase difference between x( t) and yVCO ( t) . Then the following are the relations among the signals, X LF ( s) = Φ diff ( s) Y( s) =
X LF ( s) s + 1.2 × 10 5
FVCO ( s) = 10 4 Y( s) ΦVCO ( s) = 2π
FVCO ( s) s
Φ diff ( s) = Φ x ( s) − ΦVCO ( s) Combining equations,
X LF ( s) 5 Φ diff ( s) XVCO ( s) = s + 1.2 × 10 = s s( s + 1.2 × 10 5 ) FVCO ( s) = 10 4
ΦVCO ( s) = 2π
10 8
Φ diff ( s) s + 1.2 × 10 5
Φ diff ( s) Φ diff ( s) s + 1.2 × 10 5 = 2π × 10 4 s s( s + 1.2 × 10 5 )
Φ diff ( s) = Φ x ( s) − 2π × 10 8
Φ diff ( s)
s( s + 1.2 × 10 5 )
2π × 10 8 Φ diff ( s) 1 + = Φ x ( s) 5 s( s + 1.2 × 10 ) Φ diff ( s) = Φ x ( s)
s( s + 1.2 × 10 5 ) 1 = 2π × 10 8 s( s + 1.2 × 10 5 ) + 2π × 10 8 1+ s( s + 1.2 × 10 5 )
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Y( s) s = Φ x ( s) s( s + 1.2 × 10 5 ) + 2π × 10 8 Fx ( s) =
sΦ x ( s) 2π
2π Y( s) = . Fx ( s) s( s + 1.2 × 10 5 ) + 2π × 10 8 In steady state with no frequency modulation and a frequency of 100 MHz, y( t) = 1. The response to a step frequency change of 1 kHz, ∆fx ( t) = 1000 u( t) , is ∆ Y( s) =
2π 1000 2000π = 2 5 8 s( s + 1.2 × 10 ) + 2π × 10 s s ( s + 1.2 × 10 5 ) + 2π × 10 8 s ∆ Y( s) =
10 −5 5.033 × 10 −7 1.05 × 10 −5 − + s s + 1.145 × 10 5 s + 5487
(
)
∆ y( t) = 10 −5 + 5.033 × 10 −7 e −1.145 ×10 t − 1.05 × 10 −5 e −5487 t u( t) 5
∆y( t) 1e-05
0.001
t
40. Plot the root locus for each of the systems which have these loop transfer functions and identify the transfer functions that are stable for all positive real values of K. (a)
T( s) =
K ( s + 10) (s + 1)(s2 + 4 s + 8)
ω
-1
(b)
T( s) =
K ( s2 + 10)
(s + 1)(s2 + 4 s + 8)
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M. J. Roberts - 7/12/03 ω
σ
-1
(c)
T( s) =
K s + 37 s + 332 s + 800 3
2
ω
-25
(d)
T( s) =
σ
-8 -4
K (s − 4) s+4 ω 1
-4
4
σ
-1
(e)
T( s) =
K (s − 4) (s + 4)2 ω 1
-4 4
(f)
K ( s + 6) T( s) = (s + 5)(s + 9)(s2 + 4 s + 12)
σ
-1
ω
-9
-6 -5
σ
41. The circuit below is a simple approximate model of an operational amplifier with the inverting input grounded.
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(a) Define the excitation of the circuit as the current of a current source applied to the non-inverting input and define the response as the voltage developed between the noninverting input and ground. Find the transfer function and graph its frequency response. This transfer function is the input impedance. Z i ( s) = Ri = 1MΩ (b) Define the excitation of the circuit as the current of a current source applied to the output and define the response as the voltage developed between the output and ground with the non-inverting input grounded. Find the transfer function and graph its frequency response. This transfer function is the output impedance. Z i ( s) = Ro = 10Ω (c) Define the excitation of the circuit as the voltage of a voltage source applied to the non-inverting input and define the response as the voltage developed between the output and ground. Find the transfer function and graph its frequency response. This transfer function is the voltage gain. 1 1 sCx = A0 Vi ( s) Vo ( s) = VX ( s) = A0 Vi ( s) 1 sRx Cx + 1 + Rx sCx H( s) =
A Vo ( s) 1 1 = 0 = 1.25 × 10 7 1 s + 125 Vi ( s) Rx Cx s + Rx Cx
The corner frequency is approximately 20 Hz.
Rx
Ro
+ + Ri -
vi (t) -
Output
+ A0v i (t)
Cx
vx(t)
v x(t)
-
Ri = 1MΩ , Rx = 1kΩ , Cx = 8 µF , Ro = 10 Ω , A0 = 10 6 42. Change the circuit of Exercise 41 to the circuit below. This is a feedback circuit which establishes a positive closed-loop voltage gain of the overall amplifier. Repeat steps (a), (b) and (c) of Problem #6 for the feedback circuit and compare the results. What are the important effects of feedback for this circuit?
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M. J. Roberts - 7/12/03
Rx
Ro
+ + Ri
+
vi (t)
A0v i (t)
-
-
Output
Cx
Rs
vx(t)
v x(t)
-
Rf
Ri = 1MΩ , Rx = 1kΩ , Cx = 8 µF , Ro = 10 Ω , A0 = 10 6 , R f = 10 kΩ , Rs = 5 kΩ (a)
Z i ( s) =
Vi ( s) + Vs ( s) Ri I i ( s) + Vs ( s) V ( s) = = Ri + s I i ( s) I i ( s) I i ( s) Vs ( s)Gs + [Vs ( s) − Vo ( s)]G f = I i ( s) Vx ( s) = A0
Ri 1 Vi ( s) = A0 I ( s) sRx Cx + 1 i sRx Cx + 1
(
)
Vo ( s) Go + G f − Vs ( s)G f − Vx ( s)Go = 0 Combining the last two equations,
(
)
Vo ( s) Go + G f − Vs ( s)G f − A0
Ri I ( s)Go = 0 sRx Cx + 1 i
Then Gs + G f −G f
(
1 −G f Vs ( s) R I ( s) i = Go + G f Vo ( s) A0 sR C + 1 Go i x x
)(
)
∆ = Gs + G f Go + G f − G 2f = GsGo + GsG f + G f Go RiG f Go −G f 1 Go + G f + A0 Vs ( s) 1 sRx Cx + 1 Ri = = G G G A + GsGo + GsG f + G f Go I i ( s) ∆ 0 o f sRx Cx + 1 o
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M. J. Roberts - 7/12/03
1 1 R f Ro 1 1 Ri Rs + + A0 RsR f + RsR0 + A0 sRx Cx + 1 Vs ( s) Ro R f sRx Cx + 1 = = 1 1 1 1 1 1 I i ( s) R f + Ro + Rs + + Rs Ro Rs R f R f Ro Ri
Zi ( s) = Ri +
Ri Rs sRx Cx + 1 R f + Ro + Rs
RsR f + RsR0 + A0
Substituting in numbers, Zi ( s) = 10 6 +
50 × 10 6 + 50, 000 + 10 6
5 × 10 9 8 × 10 −3 s + 1
15, 010
At low frequencies, 1012 Z i ( s) ≅ 3 This input impedance is much higher than in the open-loop case. (b)
Ground the input terminal for this calculation. Z o ( s) =
Vo ( s) Vs ( s) + V f ( s) = I o ( s) I o ( s)
[V (s) − V (s)]G + [V (s) − V (s)]G o
s
f
o
[V (s) − V (s)]G s
o
f
x
o
= I o ( s)
+ Vs ( s)(Gs + Gi ) = 0
Since the non-inverting input is grounded we can write Vx ( s) = − Vs ( s) A0
1 sRx Cx + 1
Combining equations and solving A0Go Vo ( s) G f + Go + − G f Vs ( s) = I o ( s) sRx Cx + 1
(
)
(
)
− Vo ( s)G f + G f + Gs + Gi Vs ( s) = 0
10-55
M. J. Roberts - 7/12/03
A0Go − G f Vo ( s) I o ( s) sRx Cx + 1 = G f + Gs + Gi Vs ( s) 0
G f + Go −G f
A0Go ∆ = G f + Go G f + Gs + Gi + G f − Gf sRx Cx + 1
(
)(
)
(
)
∆ = G f (Gs + Gi ) + Go G f + Gs + Gi + 1 I o ( s) Vo ( s) = ∆ 0
Z o ( s) = 10
A0Go − G f G f + Gs + Gi I o ( s) = sRx Cx + 1 ∆ G f + Gs + Gi G f + Gs + Gi
Z o ( s) =
Z o ( s) = Ro
A0GoG f sRx Cx + 1
(
)
G f (Gs + Gi ) + Go G f + Gs + Gi +
A0GoG f sRx Cx + 1
RsRi + R f Ri + R f Rs Ro ( Ri + Rs ) + RsRi + R f Ri + R f Rs + RsRi
(
A0 sRx Cx + 1
)
10 6 Rs + R f + R f Rs
(
)
10(10 6 + Rs ) + 10 6 Rs + R f + R f Rs + 10 6 Rs
10 6 8 × 10 −3 s + 1
Substituting in numbers, Z o ( s) ≅ 10
1 15.05 × 10 9 ≅ 15 5 × 10 6 5 × 10 0 1 15.06 × 10 9 + . + 8 × 10 −3 s + 1 8 × 10 −3 s + 1
At low frequencies
Z o ( s) ≅ 0.2 × 10 −6
This output impedance is much lower than in the open-loop case. (c)
(
)
Vo ( s) Go + G f − Vx ( s)Go − Vs ( s)G f = 0
(
)
Vs ( s) Gs + G f + Gi − Vo ( s)G f − Vex ( s)Gi = 0 where Vex ( s) is the overall excitation voltage with respect to ground
10-56
M. J. Roberts - 7/12/03
Vx ( s) = (Vex ( s) − Vs ( s)) A0 Go + G f −G f 0
−Go 0 1
∆ = − A0
1 sRx Cx + 1
Vo ( s) 0 −G f Gs + G f + Gi Vx ( s) = Vex ( s)Gi 1 V ( s) 1 A0 s Vex ( s) A0 sR C + 1 sRx Cx + 1 x x
(
)(
)
G f Go − Go + G f Gs + G f + Gi + G 2f sRx Cx + 1
1 Vo ( s) = ∆
−Go −G f 0 Vex ( s)Gi 0 Gs + G f + Gi 1 1 Vex ( s) A0 A0 1 sRx Cx + 1 sRx Cx + 1
−Go 1 G + G f + Gi Go + G f − Vex ( s) A0 − Vex ( s)Gi A0 sRx Cx + 1 s sRx Cx + 1 Vo ( s) = ∆
(
)
−Go 1 G + G f + Gi Go Gi A0 + G f + A0 sRx Cx + 1 s sRx Cx + 1 Vo ( s) H( s) = = G f Go Vex ( s) A0 + Go + G f Gs + G f + Gi − G f2 sRx Cx + 1
(
(
(
)(
)
)
)
A0 G + G f Go + GiG f sRx Cx + 1 s H( s) = G f Go A0 + Go Gs + G f + Gi + G f (Gs + Gi ) sRx Cx + 1
(
)
(
)
A0 R R + Rs + Ro Rs sRx Cx + 1 i f
H( s) = A0
(
)
RsRi + R f Ri + RsRi + RsR f + Ro ( Rs + Ri ) sRx Cx + 1
Substituting in numbers, 15 × 1013 15 × 1013 4 + 5 × 10 4 + × 5 10 −3 −3 × + 8 10 1 × + 8 10 1 s s ≅ H( s) = 5 × 1013 5 × 1013 7 4 + 20 × 10 7 + 20 × 10 + 15 × 10 −3 −3 8 × 10 s + 1 8 × 10 s + 1
10-57
M. J. Roberts - 7/12/03
H( s) ≅
11 400 s + 15 × 1013 −4 s + 3.75 × 10 ≅ × 2 5 10 . s + 3.125 × 10 7 1.6 × 10 6 s + 5 × 1013
The low-frequency gain is H(0) ≅ 2.5 × 10 −4
R f + Rs 3.75 × 1011 7 = 3= 3.125 × 10 Rs
as it should be. The closed-loop gain has a pole at s = −3.125 × 10 7 which sets a corner frequency of approximately 5 MHz. The open-loop corner frequency was approximately 20 Hz. So the bandwidth has been increased by a factor of approximately 250,000. 43. Plot the unit step and ramp responses of unity-gain feedback systems with these forward-path transfer functions. The following relationships are very handy in solving these problems: B − Aα As + B e −αt A cos(βt) + L→ sin(βt) u( t) ← β (s + α )2 + β 2 e
e
e
−
C t 2
−
C t 2
−αt
2 As + B A 2 + B − Aα cos βt − tan −1 B − Aα u( t) ← L→ 2 β Aβ s + α) + β 2 (
2 2 C As + B C B − AC 2 A cos D − t + sin D − t u( t) ← L→ 2 2 2 2 s + Cs + D 4D − C
2 2 As + B B − AC 2 C −1 2 B − AC 2 A + D t cos tan − − L→ 2 u( t) ← 2 2 s + Cs + D 2 4D − C A 4D − C
(a)
H1 ( s) =
20 s( s + 2)( s + 6)
H( s) =
20 20 = 3 2 s( s + 2)( s + 6) + 20 s + 8 s + 12 s + 20
Unit step response: H( s) =
20 s( s + 8 s + 12 s + 20) 3
2
Using MATLAB to find the partial fraction expansion,
10-58
M. J. Roberts - 7/12/03
1 0.07875 0.9212 s + 1.77 H −1 ( s) = − − 2 s s + 6.647 s + 1.353s + 3.009 1.597 1 0.07875 s + 0.6765 H −1 ( s) = − − 0.9212 + 0.7792 2 2 s s + 6.647 (s + 0.6765) + 2.551 ( s + 0.6765) + 2.551
{
}
h −1 ( t) = 1 − 0.07875e −6.647 t − 0.9212e −0.6765 t [cos(1.597 t) + 0.7792 sin(1.597 t)] u( t) Unit ramp response: 20 s ( s + 8 s + 12 s + 20)
H −2 ( s) =
2
3
2
Using MATLAB to find the partial fraction expansion, H −2 ( s) = H −2 ( s) =
1 0.6 0.01185 0.5882 s − 0.1257 + + 2 2 − s s s + 6.647 s + 1.353s + 3.009
1.597 1 0.6 0.01185 s + 0.6765 + + 0.5882 − 0.5574 2 2 2 − s s s + 6.647 (s + 0.6765) + 2.551 ( s + 0.6765) + 2.551
{
}
h −2 ( t) = t − 0.6 + 0.01185e −6.647 t + 0.5882e −0.6765 t [cos(1.597 t) − 0.5574 sin(1.597 t)] u( t)
Unit-Step Response
Unit-Ramp Response
h (t)
h- 2(t)
-1 1
10
10
(b)
10
t
-5
H1 ( s) =
20 s ( s + 2)( s + 6)
H( s) =
20 20 = 4 3 s ( s + 2)( s + 6) + 20 s + 8 s + 12 s2 + 20
2
2
Unit step response: H −1 ( s) =
20 s( s + 8 s + 12 s2 + 20) 4
3
Using MATLAB to find the partial fraction expansion,
10-59
t
M. J. Roberts - 7/12/03
H −1 ( s) = H −1 ( s) = H −1 ( s) =
1 0.0286 0.2158 0.8128 s − 0.08832 + − − 2 s s + 5.848 s + 2.798 s − 0.6459 s + 1.222
s − 0.1087 1 0.0286 0.2158 + − − 0.8128 2 s s + 5.848 s + 2.798 s − 0.6459 s + 1.222
1.0572 1 0.0286 0.2158 s − 0.323 + − − 0.8128 + 0.2027 2 2 s s + 5.848 s + 2.798 (s − 0.323) + 1.1177 ( s − 0.323) + 1.1177
{
}
h −1 ( t) = 1 + 0.0286e −5.848 t − 0.2158e −2.798 t − 0.8128e 0.323 t [cos(1.0572 t) + 0.2027 sin(1.0572 t)] u( t) Unit ramp response: 20 s ( s + 8 s + 12 s2 + 20)
H −2 ( s) =
2
4
3
Using MATLAB to find the partial fraction expansion, 1 0.004891 0.07715 0.07226 s + 0.7661 + − 2 2 − s s + 5.848 s + 2.798 s − 0.6459 s + 1.222
H −2 ( s) = H −2 ( s) = H −2 ( s) =
s + 10.6 1 0.004891 0.07715 + − 0.07226 2 2 − s s + 5.848 s + 2.798 s − 0.6459 s + 1.222
1.057 1 0.004891 0.07715 s − 0.323 + − 0.07226 + 10.334 2 2 2 − s s + 5.848 s + 2.798 (s − 0.323) + 1.1177 ( s − 0.323) + 1.1177
{
}
h −2 ( t) = t − 0.004891e −5.848 t + 0.07715e −2.798 t − 0.07226e 0.323 t [cos(1.057 t) + 10.334 sin(1.057 t)] u( t) Unit-Ramp Response
Unit-Step Response
h (t)
h (t)
-2
-1
10
20
10
t
-20
(c)
H1 ( s) =
10
100 s + 10 s + 34 2
H( s) =
Unit step response: H −1 ( s) =
100 s( s + 10 s + 134 ) 2
10-60
100 s + 10 s + 134 2
t
M. J. Roberts - 7/12/03
Using MATLAB to find the partial fraction expansion and the inverse transform H −1 ( s) =
0.7463 0.7463s + 7.463 − 2 s s + 10 s + 134
{
}
h −1 ( t) = 0.74627 − 0.74627e −5 t [cos(10.4403t) + 0.4812 sin(10.4403t)] u( t) Unit ramp response: H −2 ( s) =
100 s ( s + 10 s + 134 ) 2
2
Using MATLAB to find the partial fraction expansion and the inverse transform H −2 ( s) =
0.7463 0.05569 0.05569 s − 0.1894 − + s2 s s2 + 10 s + 134
{
}
h −2 ( t) = 0.74267 t − 0.055692 + 0.055692e −5 t [cos(10.4403t) − 0.80457 sin(10.4403t)] u( t)
Unit-Ramp Response
Unit-Step Response
h- 2(t)
h- 1(t)
1
1
1
H1 ( s) =
(d)
t
100 s( s + 10 s + 34 ) 2
1
H( s) =
t
100 s + 10 s + 34 s + 100 3
2
Unit step response: H −1 ( s) =
100 s( s + 10 s + 34 s + 100) 3
2
Using MATLAB to find the partial fraction expansion and the inverse transform 1 0.3036 0.6964 s + 4.133 H −1 ( s) = − − 2 s s + 7.208 s + 2.792 s + 13.87
{
}
h −1 ( t) = 1 − 0.30359e −7.2077 t − 0.69641e −1.3961t [cos( 3.4532 t) + 1.3142 sin( 3.4532 t)] u( t) Unit ramp response:
10-61
M. J. Roberts - 7/12/03
H −2 ( s) =
100 s ( s + 10 s2 + 34 s + 100) 2
3
Using MATLAB to find the partial fraction expansion and the inverse transform 1 0.34 0.04212 0.2979 s + 0.1354 H −2 ( s) = 2 − + + 2 s s s + 7.208 s + 2.792 s + 13.87
{
}
h −2 ( t) = t − 0.34 + 0.042121e −7.2077 t + 0.29788e −1.3961t [cos( 3.4532 t) − 0.27271sin( 3.4532 t)] u( t)
Unit-Ramp Response
Unit-Step Response
h (t)
h (t)
-2
-1
3
1
3
(e)
H1 ( s) =
t
100 s ( s + 10 s + 34 ) 2
2
3
H( s) =
t
100 s + 10 s + 34 s2 + 100 4
3
Unit step response: H −1 ( s) =
100 s( s + 10 s + 34 s2 + 100) 4
3
Using MATLAB to find the partial fraction expansion and the inverse transform 1 0.1126 s + 0.6214 0.8874 s − 0.0409 − 2 H −1 ( s) = − 2 s s + 10.7 s + 38.98 s − 0.7046 s + 2.566 1 − 0.1126e −5.3253 t [cos( 3.214 t) + 0.0512 sin( 3.214 t)] h −1 ( t) = u( t) −0.8874 e 0.35231t [cos(1.5625 t) + 0.196 sin(1.5625 t)] Unit ramp response: H −2 ( s) =
100 s ( s + 10 s + 34 s2 + 100) 2
4
3
Using MATLAB to find the partial fraction expansion and the inverse transform H −2 ( s) =
1 0.01594 s + 0.05802 0.01594 s + 0.8761 − 2 + 2 s2 s + 10.7 s + 38.98 s − 0.7046 s + 2.566
10-62
M. J. Roberts - 7/12/03
t + e −5.3523 t 0.01594[cos( 3.214 t) − 0.533 sin( 3.214 t)] h −2 ( t) = 0.3523 t u( t) 0.01594[cos(1.5625 t) + 35.4 sin(1.5625 t)] −e Unit-Step Response
Unit-Ramp Response
h- 1(t)
h (t)
20
10
-2
10
t 10
t
-5
-20
44. Draw pole-zero diagrams of these transfer functions. ω
1
(a)
H( s) =
(s + 3)(s − 1) s( s + 2)( s + 6)
-6
1
σ
-1
ω 1
(b)
s H( s) = 2 s + s +1
σ
-1 -1
ω
1
(c)
H( s) =
s( s + 10) s + 11s + 10 2
σ
-10
-1
(Pole-zero cancellation.) ω 1
(d)
H( s) =
1 (s + 1)(s + 1.618 + 1)(s2 + 0.618 + 1) 2
σ
-2 -1
45. A second-order system is excited by a unit step and the response is as illustrated in Figure E45 . Write an expression for the transfer function of the system.
10-63
M. J. Roberts - 7/12/03
Step Response 0.2 0.18 0.16
Amplitude
0.14 0.12 0.1 0.08 0.06 0.04 0.02 0
0
10
20
30
40
50
60
Time (sec.) Figure E45 Step response of a second-order system From the graph it is apparent that system is highly underdamped and that the final value of the step response is 0.1. So this second-order system has no zeros at zero. Therefore the general form of the transfer function of this second-order system is H( s) =
Aω 02 s2 + 2ζω o s + ω 02
and A = 0.1. From the graph there are 10 ringing cycles of response between 0 and 10 seconds. Therefore the resonant frequency is approximately 1Hz or 2π radians per second. The impulse response is of the form,
(
)
h( t) = Ke −ζω 0 t cos ω 0 1 − ζ 2 t + θ . 1 . From the graph, the ζω 0 time constant is approximately the time at which the ringing is at 36.8% of its maximum value. That is at about 10 seconds. Therefore So the characteristic exponential decay has a time constant of τ =
ζ=
1 1 = = 0.0159 . ω 0τ 20π
So the transfer function is H( s) =
3.948 . s + 0.2 s + 39.48 2
46. For each of the pole-zero plots below determine whether the frequency response is that of a practical lowpass, bandpass, highpass or bandstop filter.
10-64
M. J. Roberts - 7/12/03
ω
ω [s]
[s]
σ
(a)
σ
(b)
ω
ω [s]
[s]
σ
(c)
(a)
Highpass
(d)
Bandpass
,
(b)
σ
(d)
Lowpass
,
(c)
Bandstop
47. A system has a transfer function, H( s) = (a)
A . s + 2ζω 0 s + ω 02 2
Let ω 0 = 1. Then let ζ vary continuously from 0.1 to 10 and plot in the s-plane the paths that the two poles take while ζ is varying between those limits. ω ζ = 0.1
ζ = 10 -19.9
ζ = 10 -1
-0.1
σ
ζ = 0.1
(b)
Find the real-valued functional form of the impulse response for the case, ω 0 = 1 and ζ = 0.5 . H( s) =
A A A 2 2 = 2 2 = 2 s + 2ζω 0 s + ω 0 ( s + ζω 0 ) + ω 0 − (ζω 0 ) (s + ζω 0 ) + ω 02 (1 − ζ 2 ) 2
10-65
M. J. Roberts - 7/12/03
H( s) =
ω 02 (1 − ζ 2 )
A
ω (1 − ζ 2 0
2
) (s + ζω )
+ ω (1 − ζ
2
2 0
0
h( t) = (c)
2
)
⇒ h( t) =
A
ω (1 − ζ 2 0
2
)
(
e −ζω 0 t sin ω 02 (1 − ζ 2 ) t
)
4 −0.5 t 3 Ae sin t 3 4
Sketch the phase frequency response for the case, ω 0 = 1 and ζ = 0.1. Phase of H( jω ) π
-4
4
ω
-π
(d)
Find the -3 dB bandwidth for for the case, ω 0 = 1 and ζ = 0.1. H( jω ) =
( jω )
A 2
+ 0.2 jω + 1
=
A 1 − ω + 0.2 jω 2
The maximum value of the transfer function occurs at resonance, ω = ω 0 = 1. There H( jω 0 ) =
A = 5A 0.2 jω 02
At the -3 dB points the magnitude of the square of the transfer function is one-half of the square of this value. H( jω ) =
( jω )
H( jω −3 dB ) =
A 2
+ 0.2 jω + 1
=
A 1 − ω + 0.2 jω 2
2
2
1 − ω −23 dB
A 25 A 2 = 2 + 0.2 jω −3 dB
2
1 − ω −23 dB
2 2 1 25 2 = ⇒ (1 − ω −23 dB ) + (0.2ω −3 dB ) = 25 2 + 0.2 jω −3 dB
ω −43 dB − 1.96ω −23 dB + 0.92 = 0 ω
2 −3 dB
=
1.96 ±
(1.96) 2 − 3.68 2
= 0.98 ± 0.2
ω −3 dB = ±1.086, ±0.883
10-66
M. J. Roberts - 7/12/03
So the bandwidth is 0.0323 Hz. (e)
The Q of a system is a measure of how “sharp” its frequency response is near a resonance. It is defined as 1 Q= . 2ζ For very high-Q systems what is the relationship between Q, ω 0 and -3 dB bandwidth?
If the Q is very high that means that the damping factor is very low. The peak of the frequency response occurs at resonance and at that frequency, H( jω 0 ) =
A A 2 = Q 2ζω 0 ω 02
The -3 dB bandwidth is found by solving 2
H( jω −3 dB ) =
A
2
ω 02 − ω −23 dB + j
ω0 ω Q −3 dB
A Q 2 ω0 = 2
2
2
1
ω 02 − ω −23 dB + j
(ω ω +ω 4 0
ω
4 −3 dB
2 0
−ω
4 −3 dB
)
2 2 −3 dB
Q2 2ω 04
2ω 04 ω0 + ω −3 dB = 2 Q Q
− 2ω ω 2 0
ω0 ω Q −3 dB
=
2
2 −3 dB
ω 02 2 2ω 04 + 2 ω −3 dB − 2 = 0 Q Q
2ω 04 1 2 2 4 + 2 − 2 ω 0ω −3 dB + ω 0 − 2 = 0 Q Q
Solving, 4 2ω 04 1 2 1 4 2 2 4 − ω ± − ω − ω 0 − 2 Q2 0 Q2 0 Q 2
ω −23 dB =
2
10-67
M. J. Roberts - 7/12/03
ω −23 dB
1 1 1 1 1 4 +4 2 − 2 ± 2 − 2 ± 4 + 2 Q Q Q Q Q Q2 2 2 = ω0 = ω0 2 2
For very large Q,
ω −23 dB ≅ ω 02
2 Q
2± 2
1 = ω 02 1 ± Q
or
ω −3 dB ≅ ω 0 1 ±
1 1 ≅ ω 0 1 ± 2Q Q
which means that the bandwidth is 1 ω0 1 ∆ω −3 dB ≅ ω 0 1 + . − ω 0 1 − = 2Q 2Q Q That is, for very-high-Q systems the -3 dB bandwidth is approximately the center frequency divided by the Q. 48. Draw canonical system diagrams of the systems with these transfer functions. (a)
s2 + 8 H( s) = 10 3 s + 3s2 + 7 s + 22 s3 Y1 ( s) = X( s) − 3s2 Y1 ( s) − 7 s Y1 ( s) − 22 Y1 ( s) Y( s) = 10 s2 Y1 ( s) + 80 Y1 ( s) 10
+
80 X(s)
1 s
+ +
1 s
1 s
3
+ +
7
+
22
(b)
H( s) = 10
s + 20 (s + 4)(s + 8)(s + 14)
H( s) = 10
s + 20 s + 26 s + 200 s + 448 3
2
10-68
Y(s) 1
+
Y(s)
M. J. Roberts - 7/12/03
s3 Y1 ( s) = X( s) − 26 s2 Y1 ( s) − 200 s Y1 ( s) − 448 Y1 ( s) Y( s) = 10 s Y1 ( s) + 200 Y1 ( s) 10
+
200 X(s)
1 s
+ +
1 s
1 s
Y(s)
+
Y(s) 1
26
+ +
200
+
448
49. Draw cascade system diagrams of the systems with these transfer functions. (a)
H( s) = −50
s2 s3 + 8 s2 + 13s + 40
Factoring the numerator and denominator, H( s) = −50
s s 2 s + 6.958 s + 1.042 s + 5.749
1 s
X(s)
0
1 s
+
6.958
1 s
+
0
-50
+
Y(s)
1.042
+ +
5.749
(b)
X(s)
H( s) =
1 s 10
s3 s s s = 3 2 s + 18 s + 92 s + 120 s + 10 s + 6 s + 2
0
1 s 6
0
1 s 2
50. Draw parallel system diagrams of the systems with these transfer functions.
10-69
0
Y(s)
M. J. Roberts - 7/12/03
(a)
H( s) = 10
s3 39.9 s + 73.84 0.09869 − = 10 − 2 3 2 s + 4 s + 9s + 3 s + 3.604 s + 7.572 s + 0.3962 39.9
73.84
1 s
1 s
Y1 (s)
3.604 7.572
X(s)
Y(s)
10
1 s
0.09869
0.3962
(b)
H( s) =
5 0.01667 0.15 0.1333 = − + s+9 s + 2.333 s + 1.5 6 s + 77 s + 228 s + 189 3
2
1 s
0.01667
9
X(s)
1 s
0.15
Y(s)
2.333
1 s
0.1333
1.5
51. Write state equations and output equations for the circuit of Figure E51 with the two capacitor voltages, vC1 ( t) and vC 2 ( t) , as the state variables and the voltage at the input, v i ( t) , as the excitation and the voltage, v R1 ( t) , as the response. Then, assuming the capacitors are initially uncharged, find the unit step response of the circuit.
10-70
M. J. Roberts - 7/12/03
C1 = 1 µF +
+
vi (t)
R 2 = 10 kΩ
vC1(t)
+ vR1(t)
R 1 = 10 kΩ
+ vC2(t)
-
C2 = 1 µF
Figure E51 A second-order RC circuit v R 1 ( t) + C2 vC′ 2 ( t) R1
C1 v′C1 ( t) =
v i ( t) = vC1 ( t) + v R1 ( t)
v R1 ( t) = vC 2 ( t) + R2C2 vC′ 2 ( t) Therefore C1 v′C1 ( t) =
vC 2 ( t) + R2C2 v′C 2 ( t) + C2 vC′ 2 ( t) R1
v i ( t) = vC1 ( t) + vC 2 ( t) + R2C2 vC′ 2 ( t) which can be rearranged into state equations,
1 1 1 1 1 v′C1 ( t) = − − + vC1 ( t) + − vC 2 ( t) + v i ( t) R1C1 R2C1 R2C1 R1C1 R2C1 v′C 2 ( t) = −
vC1 ( t) vC 2 ( t) v i ( t) − + R2C2 R2C2 R2C2
and an output equation, v R1 ( t) = − vC1 ( t) + v i ( t) . Writing them in standard matrix form, 1 1 − − v′C1 ( t) R1C1 R2C1 = 1 v ( t ) ′ C 2 − R2C2 and
1 1 1 + v ( t) R2C1 C1 R1C1 R2C1 + v i ( t) 1 vC 2 ( t) 1 − R2C2 R2C2 −
v ( t) v R1 ( t) = [−1 0] C1 + v i ( t) . vC 2 ( t )
Substituting numbers,
10-71
M. J. Roberts - 7/12/03
v′C1 ( t) −200 −100 vC1 ( t) 200 = + v′C 2 ( t) −100 −100 vC 2 ( t) 100
and
v ( t) v R1 ( t) = [−1 0] C1 + v i ( t) . vC 2 ( t )
The transfer function is H( s) = C[ sI − A ] B + D −1
which, in this case, is s + 100 −100 −100 s + 200 200 100 200 s + 200 H( s) = [−1 0] 100 + 1 = [−1 0] s2 + 300 s + 10, 000 100 + 1 100 s 100 + −1
or
200 s + 10, 000 100 s + 1 = 1 − 200 s + 10, 000 H( s) = [−1 0] 2 s + 300 s + 10, 000 s2 + 300 s + 10, 000 H( s) =
s( s + 100) s + 300 s + 10, 000 2
The Laplace transform of the step response is H −1 ( s) =
s + 100 0.7236 0.2764 . = + s + 300 s + 10, 000 s + 261.8 s + 38.2 2
The step response is then h −1 ( t) = (0.7326e −261.8 t + 0.2764 e −38.2 t ) u( t) 52. Write state equations and output equations for the circuit of Figure E52 with the two capacitor voltages, vC1 ( t) and vC 2 ( t) , as the state variables and the voltage at the input, v i ( t) , as the excitation and the voltage at the output, v o ( t) , as the response. Then, find and plot the response voltage for a unit step excitation assuming that the initial conditions are vC1 (0) 2 = . vC 2 (0) −1
10-72
M. J. Roberts - 7/12/03
C1 + R1
vC1(t)
R2
K
+
vx(t)
vi (t) -
C2
+ vC2(t) -
+
vo(t) -
R1 = 6.8 kΩ , R2 = 12 kΩ C1 = 6.8 nF , C2 = 6.8 nF K = 3 Figure E52 A constant-K lowpass filter v i ( t) − R1 (C1 vC′ 1 ( t) + C2 vC′ 2 ( t)) − R2C2 v′C 2 ( t) − vC 2 ( t) = 0 v i ( t) − R1 (C1 vC′ 1 ( t) + C2 vC′ 2 ( t)) − vC1 ( t) = K vC 2 ( t) v o ( t ) = K vC 2 ( t )
v i ( t) − vC 2 ( t) R1C1 R1C2 + R2C2 v′C1 ( t) = R C R1C2 1 1 v′C 2 ( t) v i ( t) − K vC 2 ( t) − vC1 ( t) By Cramer’s rule, v′C1 ( t) = − v′C 2 ( t) = −
R1C2 ( v i ( t) − vC 2 ( t)) − ( v i ( t) − K vC 2 ( t) − vC1 ( t))( R1C2 + R2C2 ) R1R2C1C2 R1C1 ( v i ( t) − K vC 2 ( t) − vC1 ( t)) − R1C1 ( v i ( t) − vC 2 ( t)) R1R2C1C2
R C − K ( R1C2 + R2C2 ) R1C2 + R2C2 1 vC1 ( t) + 1 2 vC 2 ( t) + v ( t) R1R2C1C2 R1R2C1C2 R1C1 i K −1 1 v′C 2 ( t) = vC1 ( t) + v ( t) R2C2 C 2 R2C2 vC′ 1 ( t) = −
R1C2 + R2C2 − v′C1 ( t) R1R2C1C2 = 1 v′C 2 ( t) R2C2
R1C2 − K ( R1C2 + R2C2 ) v ( t) 1 R1R2C1C2 C1 + R1C1 v i ( t) K −1 vC 2 ( t ) 0 R2C2
10-73
M. J. Roberts - 7/12/03
v ( t) v o ( t) = [0 K ] C1 vC 2 ( t ) Substituting numbers, vC′ 1 ( t) −33881 −89389 vC1 ( t) 21626 = + v i ( t) v′C 2 ( t) 12255 24510 vC 2 ( t) 0 v ( t) v o ( t) = [0 3] C1 vC 2 ( t ) s − 24510 −89389 12254 89389 s + 33881 s + 33881 −1 Φ( s) = [ sI − A ] = = s2 + 9371s + 2.65 × 10 8 −12254 s − 24510 −1
[
]
Q( s) = Φ( s) BX( s) + q(0 + )
s − 24510 −89389 12254 s + 33881 21626 1 2 = 2 + s + 9371s + 2.65 × 10 8 0 s −1
s − 24510 −89389 12254 s + 33881 21626 + 2 s Q( s) = 2 s + 9371s + 2.65 × 10 8 −1 21626 ( s − 24510) s + 2 + 89389 12254 21626 + 2 − ( s + 33881) s Q( s) = 2 8 s + 9371s + 2.65 × 10 5.3 × 10 8 2 s − + 61995 8 s 2 s − 5.3 × 10 + 61995 8 2 65 10 × . s − s + − 9373 2 s 9371 + s s + 2.65 × 10 8 = Q( s) = 2 8 s + 9371s + 2.65 × 10 8 − s + 2.65 × 10 − 9373 s s2 + 9371s + 2.65 × 10 8 2 s2 + 61995 s − 5.3 × 10 8 s( s2 + 9371s + 2.65 × 10 8 ) Q( s) = s2 + 9373s − 2.65 × 10 8 − 2 9371 2 65 10 8 s+ . × ) s( s + 10-74
M. J. Roberts - 7/12/03
2 s2 + 61995 s − 5.3 × 10 8 s( s2 + 9371s + 2.65 × 10 8 ) s2 + 9373s − 2.65 × 10 8 = − 3 Y( s) = CQ( s) + DX( s) = [0 3] s2 + 9373s − 2.65 × 10 8 s( s2 + 9371s + 2.65 × 10 8 ) − 2 8 s( s + 9371s + 2.65 × 10 ) Y( s) = −3
Y( s) =
s2 + 9373s − 2.65 × 10 8 3 6 s + 56232 = − 2 2 8 s( s + 9371s + 2.65 × 10 ) s s + 9371s + 2.65 × 10 8
3 4686 1.559 × 10 4 s + 4686 − 6 + 2 2 4 8 8 s ( s + 4686) + 2.43 × 10 1.559 × 10 ( s + 4686) + 2.43 × 10
[
)]
(
y( t) = 3 − 6e −4686 t cos(1.559 × 10 4 t) + 0.3 sin(1.559 × 10 4 t) u( t) y(t) 6
0.001 -3
10-75
t
M. J. Roberts - 7/12/03
Chapter 11 -
The z Transform Solutions
1. Using the definition of the z transform and/or the transform pairs,
α n u[ n ] ←Z → and Z sin(Ω0 n ) u[ n ] ← →
z 1 , z>α = z − α 1 − αz −1
z sin(Ω0 ) sin(Ω0 ) z −1 = , z >1 , z 2 − 2 z cos(Ω0 ) + 1 1 − 2 cos(Ω0 ) z −1 + z −2
find the z transforms of these DT signals. (a)
x[ n ] = u[ n ] Z Using α n u[ n ] ← →
Z u[ n ] ← →
(b)
z 1 = , z >1 z − 1 1 − z −1
x[ n ] = e −10 n u[ n ] Z Using α n u[ n ] ← →
Z e −10 n u[ n ] ← →
(c)
z 1 , z > α , let α = 1. Then = z − α 1 − αz −1
z 1 −10 = . Then −1 , z > α , let α = e z − α 1 − αz
z 1 −10 −10 = −10 −1 , z > e z−e 1− e z
x[ n ] = e n sin( n ) u[ n ] e(1+ j ) n − e(1− j ) n e jn − e − jn x[ n ] = e u[ n ] = u[ n ] j2 j2 n
Z Using α n u[ n ] ← →
z 1 , z>α = z − α 1 − αz −1
Z e(1+ j ) n u[ n ] ← →
z 1 , z>e 1+ j = z−e 1 − e1+ j z −1
Z e(1− j ) n u[ n ] ← →
z 1 , z>e 1− j = z−e 1 − e1− j z −1
11-1
M. J. Roberts - 7/12/03
Z e(1+ j ) n u[ n ] − e(1− j ) n u[ n ] ← →
z z , z>e 1+ j − z−e z − e1− j
e(1+ j ) n − e(1− j ) n z 1 z Z → u[ n ] ← , z>e 1+ j − j2 j2 z − e z − e1− j 1− j 1+ j 1 z( z − e ) − z( z − e ) e sin( n ) u[ n ] ←→ , z>e j 2 ( z − e1+ j )( z − e1− j ) Z
n
z(e1+ j − e1− j ) 1 e sin( n ) u[ n ] ←→ , z>e j 2 z 2 − z(e1+ j + e1− j ) + e 2 Z
n
Z e n sin( n ) u[ n ] ← →
(d)
ze sin(1) , z>e z − 2ez cos(1) + e 2 2
x[ n ] = δ [ n ] ∞
X( z) = ∑ x[ n ]z
−n
n =0
∞
= ∑ δ [ n ]z − n = 1
,
Any z
n =0
2. Sketch the region of convergence (if it exists) in the z plane, of the bilateral z transform of these DT signals. (a)
x[ n ] = u[ n ] + u[− n ]
u[ n ] ←→ Z
∞
∑ (u[n] + u[− n])z
−n
∞
∞
n =−∞
u[ n ] ←→ ∑ z Z
n =0
∞
∑ z −n = n =0
=
−n
+
0
∑z
n =−∞
−n
∑ u[n]z
−n
n =−∞
= ∑z n =0
1 , z >1 1 − z −1
∞
−n
∑ u[− n]z
−n
n =−∞
∞
+ ∑ zn n =0 ∞
and
+
∞
∑z n =0
n
=
1 , z <1 1− z
The two regions of convergence do not overlap. Therefore the bilateral transform does not exist. (b)
x[ n ] = u[ n ] − u[ n − 10] ∞
9
−∞
0
Z u[ n ] − u[ n − 10] ← → ∑ (u[ n ] − u[ n − 10]) z − n = ∑ z − n =
11-2
1 − z −10 z − z −9 = , All z 1 − z −1 z −1
M. J. Roberts - 7/12/03
Im(z) [z]
ROC
Re(z)
3. Using the time-shifting property, find the z transforms of these signals. (a)
x[ n ] = u[ n − 5] Z u[ n ] ← →
z , z >1 z −1
Z u[ n − 5] ← →
(b)
z −4 , z >1 z −1
x[ n ] = u[ n + 2] n 0 −1 Z → z n 0 G( z) − ∑ g[ m]z − m , n 0 > 0 Using g[ n + n 0 ] ← m =0
2 1 z z z Z u[ n + 2] ← → z2 − ∑ u[ m]z − m = z 2 − (1 + z −1 ) = z − z − 1 z − 1 m =0 z −1 z −1
z 2 − z( z − 1) − ( z − 1) z = u[ n + 2] ←→ z , z >1 z −1 z −1 Z
Which is the same as the z transform of the unit step, as it should be. n
(c)
2 x[ n ] = u[ n + 2] 3 −2
2 2 x[ n ] = 3 3
n +2
9 2 u[ n + 2] = 4 3
n +2
u[ n + 2]
n 0 −1 Using g[ n + n 0 ] ←→ z G( z) − ∑ g[ m]z − m , n 0 > 0 m =0 Z
n0
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M. J. Roberts - 7/12/03
9 2 4 3
n +2
9 2 4 3
n +2
9 2 4 3
n +2
m 2 −1 9 2 9 z −m u[ n + 2] ←→ z − u[ m]z 2 m∑ =0 4 3 4 z− 3 Z
2
2 9 z 2 9 z Z u[ n + 2] ← → z2 − 1 + z −1 = z 2 − 1 − z −1 2 2 3 3 4 z− 4 z− 3 3 Z u[ n + 2] ← →
z z−
2 3
, z>
2 3 n
2 Which is the same as the z transform of x[ n ] = u[ n ] as it should be. 3 4. Draw system diagrams for these transfer functions. (a)
H( z) =
z2 z+
1 2 Y( z) = z X( z) −
z −1 Y( z) 2
This system is non-causal and cannot be built as a real-time system. But it can be diagrammed for use as an “off-line” system in which all the past and future excitation values are available.
z
X(z)
Y(z) 1 2
z-1
This is not a unique solution. Many other correct diagrams could be drawn. (b)
H( z) =
z z + z +1 2
Y( z) = z −1 X( z) − z −1 Y( z) − z −2 Y( z)
11-4
M. J. Roberts - 7/12/03
z-1
X(z)
Y(z)
z-1 z-1 This is not a unique solution. Many other correct diagrams could be drawn. 5. Using the change-of-scale property, find the z transform of 2πn 2πn x[ n ] = sin cos u[ n ] . 32 8 x[ n ] =
e
j
2πn 32
−e j2
−j
2πn 32
2πn cos u[ n ] 8
[
]
z z − cos(Ω0 ) z Z Z Using α n g[ n ] ← → G and cos(Ω0 n ) u[ n ] ← → 2 , z >1 α z − 2 z cos(Ω0 ) + 1 z z π − cos 2π 2π 4 j 2πn j 32 32 j n 2 π Z e e 32 e cos u[ n ] ←→ 2 8 z z π 2π − 2 2π cos + 1 j 32 4 j e e 32 2π j π z z − e 32 cos 2πn 4 j 2πn Z e 32 cos u[ n ] ←→ 2π 2π 8 j j π z 2 − 2e 32 z cos + e 16 4 Similarly 2π −j π z z − e 32 cos 2πn 4 −j 2πn Z e 32 cos u[ n ] ←→ 2π 2π 8 −j −j π z 2 − 2e 32 z cos + e 16 4
11-5
M. J. Roberts - 7/12/03
e
j
2πn 32
−e j2
−j
2πn 32
2π 2π j −j π π 32 32 − cos cos z z e z z e − 4 4 1 2πn Z cos u[ n ] ←→ 2π 2π − 2π 2π 8 j j −j −j j2 2 π π 2 32 16 32 cos z + e 16 z − 2e cos z + e z − 2e 4 4
2π 2π 2π 2 j −j −j π 2 π z − e 32 cos z z − 2e 32 cos z + e 16 4 4 2π 2π 2π j j −j π π 2 2 − z − e 32 cos z z − 2e 32 cos z + e 16 4 4 1 2πn 2πn Z sin cos u[ n ] ←→ π π π 2 2 2 2π 32 8 j j −j −j j2 2 2 π π 32 16 32 16 z e cos z e z e cos z e 2 2 − + − + 4 4
2π 2π 2π 3 π j 2π −j −j j 32 32 32 32 2 2 − + − + cos z e e e e 4 2π 2π 2π 2π j j −j π 2 − j 16 +z e − e 16 + z cos e 32 − e 32 4 1 2πn 2πn Z sin cos u[ n ] ←→ 2 π 2 π 2 π 2 π 32 8 j2 4 −j −j π j 3 2 j 2 π z − 2 z cos e 32 + e 32 + z e 16 + e 16 + 4 cos 4 4 2π 2π −j π j 32 32 −2 cos z e +e +1 4
2π 2π j π − j 32 3 32 − + cos z e e 4 2π 2π 2π 2π j j −j π 2 − j 16 +z e − e 16 + z cos e 32 − e 32 4 1 2πn 2πn Z sin cos u[ n ] ←→ π π π π 2 2 2 2 32 8 j2 4 −j −j π j 3 2 j 2 π z − 2 z cos e 32 + e 32 + z e 16 + e 16 + 4 cos 4 4 2π 2π −j π j 32 32 −2 cos z e +e +1 4
π 2π 2π π 2π z 2 cos sin − z sin + cos sin 4 32 2 2 π n π n 4 32 16 Z sin cos u[ n ] ←→ z 32 8 4 2π π 2π 2 π 3 2 z − 4 z cos cos + 2 z cos + 2 cos 4 32 16 4 π 2π −4 z cos cos + 1 4 32
11-6
M. J. Roberts - 7/12/03
0.1379 z 2 − 0.3827 z + 0.1379 2πn 2πn Z sin cos u ← → n z [ ] 32 8 z 4 − 2.7741z 3 + 3.8478 z 2 − 2.7741z + 1
6. Using the z-domain-differentiation property find the z transform of n
5 x[ n ] = n u[ n ] . 8 n
5 Z u[ n ] ←→ 8 Z Using − n g[ n ] ← →z
z z−
5 8
d G( z) dz
n
d 5 Z − n u[ n ] ← →z 8 dz
z 5 z− 8
=z
5 z − − z 8 5 z − 8
2
5 z =− 8 2 5 z − 8
5 z 5 Z n u[ n ] ← → 8 2 8 5 z − 8 n
7. Using the convolution property, find the z transforms of these signals. (a)
x[ n ] = (0.9) u[ n ] ∗ u[ n ] n
z z z2 (0.9) u[n] ∗ u[n] ←→ = z − 0.9 z − 1 z 2 − 1.9 z + 0.9 n
(b)
Z
x[ n ] = (0.9) u[ n ] ∗ (0.6) u[ n ] n
n
z z z2 (0.9) u[n] ∗ (0.6) u[n] ←→ = z − 0.9 z − 0.6 z 2 − 1.5 z + 0.54 n
n
Z
8. Using the differencing property and the z transform of the unit sequence, find the z transform of the DT unit impulse and verify your result by checking the z-transform table.
11-7
M. J. Roberts - 7/12/03
Z u[ n ] ← →
Using
z z −1
Z g[ n ] − g[ n − 1] ← → G( z) − z −1 G( z) = (1 − z −1 ) G( z) Z u[ n ] − u[ n − 1] ← →
9. Find the z transform of
z z −1 z = 1. Check − z −1 = z −1 z −1 z −1
x[ n ] = u[ n ] − u[ n − 10]
and, using that result and the differencing property, find the z transform of x[ n ] = δ [ n ] − δ [ n − 10] . Compare this result with the z transform found directly by applying the time-shifting property to a DT impulse. Z u[ n ] − u[ n − 10] ← →
δ [ n ] − δ [ n − 10] ←Z →
z z − z −10 z −1 z −1
z z z z − z −1 − z −10 − z −11 z − 1 z −1 z −1 z −1
δ [ n ] − δ [ n − 10] ←Z →
z 1 1 z − − z −10 − z −1 z −1 z − 1 z − 1
δ [ n ] − δ [ n − 10] ←Z →1 − z −10 10. Using the accumulation property, find the z transforms of these signals. (a)
x[ n ] = ramp[ n ] Using x[ n ] = n
n
Z
m =0
z
1
∑ g[m]←→ z − 1 G(z) = 1 − z Z
m =0
Z ramp[ n ] ← →
(b)
x[ n ] =
1
∑ u[m − 1] and u[n − 1]←→ z − 1 and −1
G( z)
z z 1 = z − 1 z − 1 ( z − 1) 2
n
∑ (u[m] − u[m − 5])
m =0
11-8
M. J. Roberts - 7/12/03
Z u[ n ] ← →
z z −1
Z u[ n − 5] ← → z −5
,
z z −1
z z z2 ∑ u[m]←→ z − 1 z − 1 = (z − 1)2 m =0 n
Z
n
∑ u[m − 5]←Z → z −5
m =0
z2 z z = z −5 z −1 z −1 (z − 1) 2
n
∑ u[m] − u[m − 5]←Z →
m =0
z 2 (1 − z −5 ) z2 z2 −5 − z = (z − 1) 2 (z − 1) 2 (z − 1) 2
11. Using the final-value theorem, find the final value of functions that are the inverse z transforms of these functions (if the theorem applies). z (a) X( z) = z −1 The inverse transform is the unit sequence and its limit exists as t approaches infinity. Applying the final value theorem, lim g[ n ] = lim( z − 1) G( z) ,
n →∞
z →1
we get lim x[ n ] = lim( z − 1)
n →∞
z →1
z =1 z −1
and this checks with our understanding of the unit sequence function.
(b)
X( z) = z
X( z) = z
2z − z2 −
7 3 z+ 4 4
2z −
7 4
3 (z − 1) z −
applies.
7 4
4
3 This function has poles at z = 1, . 4 lim x[ n ] = lim( z − 1) z
n →∞
z →1
11-9
2z −
Therefore the final-value theorem 7 4
(z − 1) z −
3 4
=1
M. J. Roberts - 7/12/03
12. Find the inverse z transforms of these functions in series form by synthetic division. (a)
X( z) =
z z−
1 2
1+ z−
(b)
1 1 1 + 2 +L + +L 2z 4 z (2z) k
1 )z 2 1 z− 2 1 2 1 1 − 2 4z 1 L 4z
X( z) =
z −1 z − 2z + 1 2
1 1 1 1 1 + 2 + 3 +L + k +L = z z z z z −1 2 − + − z 2 z 1) z 1 1 z−2+ z 1 1− z 2 1 1− + 2 z z 1 1 − z z2 Notice that X( z) =
z −1 z 1 = z −1 ⇒ x[ n ] = u[ n − 1] 2 = z −1 (z − 1) z − 1
13. Find the inverse z transforms of these functions in closed form using partial fraction expansions, a z transform table and the properties of the z transform. (a)
X( z) =
1 1 z z − 2 11-10
M. J. Roberts - 7/12/03
1 X( z) = z = 1 2 z− 2 −2
(b)
X( z) =
z
n −2
u[ n − 2]
z2 1 3 z − z − 2 4 X( z) z 2 3 = =− + 1 3 1 3 z z− z− z − z − 2 4 2 4
X( z) = −
(c)
n 3 n 2z 3z 1 + ⇒ x[ n ] = 3 − 2 u[ n ] 1 3 2 4 z− z− 2 4
z2 X( z) = 2 z + 1.8 z + 0.82 Recognizing this as being similar to the forms,
α n sin(Ω0 n ) u[ n ] ←Z →
zα sin(Ω0 ) , z>α z − 2αz cos(Ω0 ) + α 2
α n cos(Ω0 n ) u[ n ] ←Z →
z z − α cos(Ω0 ) , z>α z − 2αz cos(Ω0 ) + α 2
and
2
[
]
2
where α = 0.9055 and Ω0 = 3.031, we can express X( z) as X( z) =
z 2 − zα cos(Ω0 ) cos(Ω0 ) zα sin(Ω0 ) 2 2 + 2 z − 2αz cos(Ω0 ) + α sin(Ω0 ) z − 2αz cos(Ω0 ) + α 2
Then the inverse transform is x[ n ] = α n cos(Ω0 n ) u[ n ] +
cos(Ω0 ) n α sin(Ω0 n ) u[ n ] sin(Ω0 )
cos( 3.031) n sin( 3.031n ) u[ n ] x[ n ] = (0.9055) cos( 3.031n ) u[ n ] + sin( 3.031) x[ n ] = (0.9055) [cos( 3.031n ) − 9.03 sin( 3.031n )] u[ n ] n
14. Using the z transform, find the total solutions to these difference equations with initial conditions, for discrete time, n ≥ 0. 11-11
M. J. Roberts - 7/12/03
(a)
2πn 2 y[ n + 1] − y[ n ] = sin u[ n ] 16
y[0] = 1
,
π z sin 8 2 z( Y( z) − y[0]) − Y( z) = π z 2 − 2 z cos + 1 8
Y( z) =
π z sin 8 π (2z − 1) z 2 − 2z cos + 1 8
Y( z) =
+
2z 2z − 1
+
z
1 π z sin 8 2 1 2 π z − z − 2 z cos + 1 8 2
z−
1 2
1 π sin 4 8 5 π − cos 8 Az + B z 4 + Y( z) = + 1 1 π z− z 2 − 2 z cos + 1 z − 8 2 2 Y( z) =
0.2934 Az + B z + + 1 1 π z− z 2 − 2 z cos + 1 z − 8 2 2
Letting z = 0 in π z sin 8 π (2z − 1) z 2 − 2z cos + 1 8
=
Az + B 0.2934 + 1 π z− z 2 − 2 z cos + 1 8 2
yields B = 0.5868 . Then, letting z = 1 in π z sin 8 π (2z − 1) z 2 − 2z cos + 1 8
=
Az + 0.5868 0.2934 + 1 π z− z 2 − 2 z cos + 1 8 2
11-12
M. J. Roberts - 7/12/03
yields A = −0.2934 . Then Y( z) =
0.2934 0.2934 z − 0.5868 z + − 2 z − 0.5 z − 1.8478 z + 1 z − 0.5
Z Then, using sin(Ω0 n ) u[ n ] ← →
z sin(Ω0 ) , z >1 z − 2 z cos(Ω0 ) + 1 2
and Z cos(Ω0 n ) u[ n ] ← →
[
]
z z − cos(Ω0 ) , z >1 z − 2 z cos(Ω0 ) + 1 2
identify Ω0 = 0.3926 . Y( z) =
2 z −2 z + 0.9239 z 0.2934 −1 z − 0.9239 z z . + 2 0 2934 + − 2 z − 0.5 z − 0.5 z − 1.8478 z + 1 z − 1.8478 z + 1
Y( z) =
z 2 − 0.9239 z z 1.076 0.3827 z 0.2934 − + − 0.2934 z −1 2 2 z − 0.5 z − 0.5 z − 1.8478 z + 1 0.3827 z − 1.8478 z + 1
Y( z) =
z 2 − 0.9239 z z 0.3827 z 0.2934 − 2.812 2 + − 0.2934 z −1 2 z − 1.8478 z + 1 z − 0.5 z − 0.5 z − 1.8478 z + 1
1 y[ n ] = 0.2934 2 (b)
n −1
π π 1 u[ n − 1] + u[ n ] − 0.2934 cos ( n − 1) − 2.812 sin ( n − 1) u[ n − 1] 8 2 8 n
5 y[ n + 2] − 3 y[ n + 1] + y[ n ] = (0.8) u[ n ] n
,
5 z 2 ( Y( z) − y[0] − z −1 y[1]) − 3z( Y( z) − y[0]) + Y( z) = 5 z 2 ( Y( z) + 1 − 10 z −1 ) − 3z( Y( z) + 1) + Y( z) = 5 z 2 Y( z) − 3z Y( z) + Y( z) = Y( z) =
z z − 0.8
z − 5 z 2 + 50 z + 3z z − 0.8
z 5 z 2 − 53z − (z − 0.8)(5z 2 − 3z + 1) 5z 2 − 3z + 1
11-13
y[0] = −1 , y[1] = 10 z z − 0.8
M. J. Roberts - 7/12/03
Y( z) =
1 5
z 3 1 (z − 0.8) z 2 − z + 5 5 z
3 1 (z − 0.8) z 2 − z +
5
5
=
−z
z−
53 5
3 1 z2 − z + 5 5
2.222 Az + B + z − 0.8 z 2 − 3 z + 1 5 5
Letting z = 0 and solving for B, B = −0.5556 . Multiplying through by z and letting z approach infinity and solving for A, A = −2.222 . 53 − z 1 2.222 2.222 z + 0.5556 5 −z Y( z) = − 3 1 3 1 5 z − 0.8 2 2 z − z+ z − z+ 5 5 5 5 53 2 0.4444 0.4444 z + 0.1111 + z − 5 z Y( z) = − 3 1 z − 0.8 z2 − z + 5 5 0.4444 z 2 − 10.1556 z + 0.1111 Y( z) = − 3 1 z − 0.8 z2 − z + 5 5 Let α = 0.4472 and let Ω0 = 0.8355 . Then Y( z) =
0.4444 z 2 − 10.1556 z + 0.1111 − z − 0.8 z 2 − 2α cos(Ω0 ) z + α 2
0.4444 z 2 − 10.1556 z + 0.1111 0.4444 9.5556 z + 0.0889 = − 1 − 2 Y( z) = − z − 0.8 z 2 − 2α cos(Ω0 ) z + α 2 z − 0.8 z − 0.6 z + 0.2 Y( z) =
2 z + 0.009303 0.4444 0.4444 −1 z + 0.009303z z . 1 9 5556 . = − − 1 9 5556 − − z 2 − 0.6 z + 0.2 z − 0.8 z 2 − 0.6 z + 0.2 z − 0.8
Y( z) =
z 2 − 0.3z 0.309303 0.3317 z 0.4444 + − 1 − 9.5556 z −1 2 z − 0.8 0.3317 z 2 − 0.6 z + 0.2 z − 0.6 z + 0.2
11-14
M. J. Roberts - 7/12/03
z 2 − 0.3z 0.3317 z 0.4444 + 0.9325 2 − 1 − 9.5556 z −1 2 z − 0.6 z + 0.2 z − 0.8 z − 0.6 z + 0.2 Then, using zα sin(Ω0 ) α n sin(Ω0 n ) u[ n ] ←Z → 2 , z>α z − 2αz cos(Ω0 ) + α 2 Y( z) =
α n cos(Ω0 n ) u[ n ] ←Z →
[
]
z z − α cos(Ω0 ) , z>α , z − 2αz cos(Ω0 ) + α 2 2
n n −1 cos(0.8355( n − 1)) y[ n ] = 0.4444 (0.8) u[ n ] − δ [ n ] − 9.5556(0.4472) u[ n − 1] +0.9325 sin(0.8355( n − 1)) 15. From each block diagram, write the difference equation and find and sketch the response, y[ n ] , of the system for discrete time, n ≥ 0, assuming no initial energy storage in the system and impulse excitation, x[ n ] = δ [ n ] . (a) x[n]
y[n]
D y[ n ] + y[ n − 1] = δ [ n ] Y( z) =
1 z n ⇒ y[ n ] = (−1) u[ n ] −1 = 1+ z z +1
y[n] 1 -5
20
n
-1
(b) x[n]
y[n] 0.8
D
y[ n ] + 0.8 y[ n − 1] = δ [ n ] Y( z) =
1 z n ⇒ y[ n ] = (−0.8) u[ n ] −1 = 1 + 0.8 z z + 0.8
11-15
M. J. Roberts - 7/12/03
y[n] 1 -5
20
n
-1
(c)
y[n]
-0.5 x[n]
D D
0.9
Designate the lower input to the last summer as w[ n ] . Then the excitation of the delay feeding that point is w[ n + 1] and x[ n ] − 0.9 w[ n − 1] = w[ n + 1]
or
w[ n ] + 0.9 w[ n − 2] = x[ n − 1] y[ n ] = −0.5( x[ n ] − 0.9 w[ n − 1]) + w[ n ] W ( z) + 0.9 z −2 W ( z) = z −1 X( z) W ( z) =
z −1 X( z) 1 + 0.9 z −2
Y( z) = −0.5 X( z) + 0.45 z −1 W ( z) + W ( z) = −0.5 X( z) + (1 + 0.45 z −1 ) W ( z) Y( z) = −0.5 X( z) + (1 + 0.45 z −1 ) Since X( z) = 1, Y( z) = −0.5 +
z −1 + 0.45 z −2 z + 0.45 = −0.5 + 2 −2 z + 0.9 1 + 0.9 z
Y( z) = −0.5 + 1.0541 Using
z −1 X( z) 1 + 0.9 z −2
0.9486 z 0.45 0.9486 z + z −1 2 z + 0.9 0.9486 z 2 + 0.9
α n sin(Ω0 n ) u[ n ] ←Z →
zα sin(Ω0 ) z − 2αz cos(Ω0 ) + α 2 2
( n − 1)π nπ n n −1 y[ n ] = −0.5δ [ n ] + 1.0541(0.9486) sin u[ n ] + 0.4744 (0.9486) sin u[ n − 1] 2 2
11-16
M. J. Roberts - 7/12/03
( n − 1)π nπ n y[ n ] = −0.5δ [ n ] + (0.9486) 1.0541sin u[ n ] + 0.5 sin u[ n − 1] 2 2 y[n] 1 -5
20
n
-1
16. Sketch regions in the z plane corresponding to these regions in the s plane. (a)
0<σ <
1 π , 0<ω < Ts Ts
z = e sTs = e(σ + jω )Ts
, Im(z) [z] 2.718
1
(b)
−
Re(z)
1 π <σ <0 , − <ω <0 Ts Ts Im(z) [z] Re(z) 0.368 1
(c)
−∞ < σ < ∞ , 0 < ω <
2π Ts
The entire z plane. 17. Find the bilateral z transforms and ROC’s of these signals. (a)
x[ n ] = u[− n ] Z u[ n ] u[− n ] ← →1 = X c ( z) , any z Z u[ n ] u[ n ] ← →
z 1 = X ac , z > 1 z z −1
11-17
M. J. Roberts - 7/12/03
1 1 Z u[− n ] ← → z = = X ac ( z) , z < 1 1 − 1 1− z z 1 1 = , z <1 1− z 1− z
X( z) = X c ( z) − x[0] + X ac ( z) = 1 − 1 + Alternate Solution: X( z) = (b)
∞
∑ x[n]z
−n
=
n =−∞
∞
∑ u[− n]z
−n
0
∑z
=
n =−∞
−n
n =−∞
∞
= ∑ zn = n =0
1 , z <1 1− z
x[ n ] = α n u[− n ]
α n u[− n ] u[ n ] ←Z →1 = X c ( z) , any z n
1 α − n u[ n ] u[ n ] = u[ n ] ←Z → α
α n u[− n ] ←Z →
1 z 1 1 − z α
=
1 z 1− α
1 = X ac , z > α 1 z z− α z
= X ac ( z) , z <
X( z) = X c ( z) − x[0] + X ac ( z) = 1 − 1 +
(c)
1 z 1− α
=
1 α 1
z 1− α
, z<
1 α
x[ n ] = (0.5) u[− n ] + (0.3) u[ n ] n
n
Z x[ n ] u[ n ] = δ [ n ] + (0.3) u[ n ] ← →1 + n
z = X c ( z) , z > 0.3 z − 0.3
Z x[− n ] u[ n ] = (0.5) u[ n ] u[ n ] + δ [ n ] = (2) u[ n ] + δ [ n ] ← → −n
n
z 1 + 1 = X ac , z > 2 z z−2
1 1 1 (0.5) n u[− n] + δ [− n] ←Z → 1 z + 1 = + 1 = X ac ( z) , z < 2 1 − 2z −2 z X( z) = X c ( z) − x[0] + X ac ( z) =
1 z 1 + 1− 2 + + 1 , 0.3 < z < 2 z − 0.3 1 − 2z
11-18
M. J. Roberts - 7/12/03
(The two halves overlap at n = 0 causing the value there to be 2.) X( z) = X c ( z) − x[0] + X ac ( z) = X( z) =
z 1 1 + , 0.3 < z < z − 0.3 1 − 2 z 2
1 z(1 − 2 z) + z − 0.3 2 z 2 − 2 z + 0.3 , 0.3 < z < = 2 2 (z − 0.3)(1 − 2z) 2z − 1.6z + 0.3
Alternate Solution: X( z) =
n n ∑ ((0.5) u[− n] + (0.3) u[n])z − n = ∞
n =−∞
∞
∞
0
∞
n =−∞
n =0
n n ∑ (0.5) z − n + ∑ (0.3) z − n
X( z) = ∑ (0.5) z + ∑ (0.3) z − n −n
n
n =0
∞
∑ (0.5)
−n
n =0
zn =
n =0
∞
∑ (0.3)
n
z −n =
n =0
1 1−
z 0.5
, z < 0.5
1 , z > 0.3 0.3 1− z
1 1 z == + , 0.3 < z < 0.5 Check 0.3 z 1 2 0 3 z z − − . 1− 1− 0.5 z 2πn n x[ n ] = (−1.5) cos u[− n ] 8
X( z) =
(d)
n
1
+
2πn n x ac [ n ] = (−1.5) cos u[− n ] 8 xc [n] = δ[n]
x0[n] = δ[n]
2πn −n x ac [− n ] = (−1.5) cos u[ n ] 8 π z z − cos 4 1 X ac = , z > 1.5 z π 2 2 z − 2(−1.5) cos z + (−1.5) 4
11-19
M. J. Roberts - 7/12/03
π z z − cos 4 1 X ac = , z > 1.5 z π 2 z + 3 cos z + 2.25 4 11 π − cos zz 4 2 X ac ( z) = , z< 2 3 1 π 1 + 3 cos + 2.25 z 4 z π 1 − z cos 4 2 X ac ( z) = , z< 3 π 1 + 3 cos z + 2.25 z 2 4 π z cos − 1 4 2 X( z) = X ac ( z) + X c ( z) − X 0 ( z) = X ac ( z) = − , z< 3 π 2.25 z 2 + 3 cos z + 1 4 X( z) = −0.3143
z − 1.414 2 , z< z + 0.9427 z + 0.4444 3 2
18. Using the definition of the z transform verify the z transforms of the following functions: (a) z-Transform Table Entry:
Z u[ n ] ← →
z 1 = z − 1 1 − z −1
∞
∞
n =0
n =0
X( z) = ∑ u[ n ]z − n = ∑ z − n = 1 +
11-20
1 1 + +L z z2
M. J. Roberts - 7/12/03
1+
1 1 + +L z z2
z − 1) z z −1 1
Check.
1−
(b)
1 z 1 z
n2 x[ n ] = u[ n ] 2! n2 z( z + 1) 1 + z −1 Z u[ n ] ← → = 3 2! 2 z(1 − z −1 ) 2( z − 1)
z-Transform Table Entry: ∞
1 4 9 n 2 −n 1 ∞ 2 −n z = ∑n z = 0 + + 2 + 3 +L 2z 2z 2z 2 n =0 n = 0 2!
X( z) = ∑
z( z + 1) z2 + z = 3 2z 3 − 6z 2 + 6z − 2 2( z − 1) 1 4 9 + 2 + 3 +L 2z 2z 2z
)
2z 3 − 6z 2 + 6z − 2 z 2 + z 1 z 1 4z − 3 + z 12 4 4 z − 12 + − 2 z z 11 4 9− + 2 z z 27 27 18 9− + 2− 3 z z z
z 2 − 3z + 3 −
(c)
Check.
x[ n ] = nα n u[ n ]
z-Transform Table Entry:
zα αz −1 nα u[ n ] ←→ = (z − α ) 2 (1 − αz −1) 2 n
Z
11-21
M. J. Roberts - 7/12/03 ∞
X( z) = ∑ nα z
n −n
n =0
α α α α = ∑ n = 0 + + 2 + 3 + L z z z z n =0 ∞
n
2
3
α α α + 2 + 3 + L z z z 2
3
z 2 − 2αz + α 2 ) zα
zα − 2α 2 +
α3 z
α3 z α4 4α 3 2α 2 − +2 2 z z 2α 2 −
Check.
α4 3α 3 −2 2 z z (d)
x[ n ] = α n sin(2πF0 n ) u[ n ]
z-Transform Table Entry:
α n sin(Ω0 n ) u[ n ] ←Z → This can be rewritten as
α n sin(2πF0 n ) u[ n ] ←Z →
zα sin(Ω0 ) α sin(Ω0 ) z −1 = z 2 − 2αz cos(Ω0 ) + α 2 1 − 2α cos(Ω0 ) z −1 + α 2 z −2 zα sin(2πF0 ) α sin(2πF0 ) z −1 = z 2 − 2αz cos(2πF0 ) + α 2 1 − 2α cos(2πF0 ) z −1 + α 2 z −2
Start with the sine function transform, z sin(Ω0 ) sin(Ω0 ) z −1 = sin(Ω0 n ) u[ n ] ←→ 2 z − 2 z cos(Ω0 ) + 1 1 − 2 cos(Ω0 ) z −1 + z −2 Z
or
z sin(2πF0 n ) sin(2πF0 n ) z −1 = sin(2πF0 nn ) u[ n ] ←→ 2 z − 2 z cos(2πF0 n ) + 1 1 − 2 cos(2πF0 n ) z −1 + z −2 Z
Then use the time scaling property of the z transform, z α n g[ n ] ←Z → G . α z sin(2πF0 n ) Z n α α sin(2πF0 nn ) u[ n ] ←→ 2 z z − 2 cos(2πF0 n ) + 1 α α 11-22
M. J. Roberts - 7/12/03
α n sin(2πF0 nn ) u[ n ] ←Z →
αz sin(2πF0 n ) . z − 2αz cos(2πF0 n ) + α 2 2
Check.
19. Sketch the region of convergence (if it exists) in the z plane, of the bilateral z transform of these DT signals. n
(a)
1 x[ n ] = u[ n ] 2 Z Using α n u[ n ] ← →
n
1 Z u[ n ] ←→ 2
z 1 , z>α = z − α 1 − αz −1
z z−
1 2
=
1 1 , z> 1 2 1 − z −1 2 Im(z) [z]
ROC
ROC 1 2
ROC
n
(b)
Re(z) ROC
n
5 10 x[ n ] = u[ n ] + u[− n ] 4 7 n
5 Z u[ n ] ←→ 4 n
z z−
5 4 ∞
, z>
5 4
n
n
10 10 10 10 Z −n u[− n ] ←→ ∑ u[− n ]z = ∑ = ∑ 7 7 n =−∞ n =−∞ 7 z n =∞ 7 z 0
10 10 10 10 Z 7z = 7 , >1 u[− n ] ←→ 10 10 7 7 z −1 −z 7 7z n
11-23
0
−n
∞
10 = ∑ n = 0 7z
−n
M. J. Roberts - 7/12/03
10 10 10 Z , z< u[− n ] ←→ 7 10 7 7 −z 7 n
10 5 10 z 5 10 Z , + 7 < z< u[ n ] + u[− n ] ←→ 10 5 4 7 4 7 z− −z 7 4 n
n
Im(z) [z] ROC 10 7
5 4
Re(z)
20. Using the time-shifting property, find the z transforms of these signals. (a)
2 x[ n ] = 3
n −1
u[ n − 1]
n
2 Z u[ n ] ←→ 3 2 3
n −1
z z−
2 3
Z u[ n − 1] ← → z −1
z z−
2 3
=
1 z−
2 3
n
(b)
2 x[ n ] = u[ n − 1] 3 2 2 x[ n ] = 3 3
n −1
u[ n − 1]
Using the result of part (d) n
2 2 2 x[ n ] = u[ n − 1] = 3 3 3
n −1
2 Z u[ n − 1] ← → 3 2 z− 3
11-24
M. J. Roberts - 7/12/03
(c)
2π ( n − 1) x[ n ] = sin u[ n − 1] 4 Using
Z sin(Ω0 n ) u[ n ] ← →
z sin(Ω0 ) sin(Ω0 ) z −1 = , z >1 z 2 − 2 z cos(Ω0 ) + 1 1 − 2 cos(Ω0 ) z −1 + z −2
π z sin 2 z π Z sin n u[ n ] ← → = 2 , z >1 2 z +1 π 2 z − 2 z cos + 1 2 z 1 2π Z sin ( n − 1) u[ n − 1] ← → z −1 2 = 2 , z >1 4 z +1 z +1 21. Draw system diagrams for these transfer functions.
(a)
2 z z + 3 H( z) = 2 3 z2 + z + 3 4 2 z z + 2 3 2 Y( z) 3 = ⇒ Y( z) z 2 + z + = z z + X( z) 2 3 3 4 3 X( z) z 2 + z + 3 4
Multiplying through by z −2 and rearranging, 2 2 3 Y( z) = X( z) + z −1 X( z) − z −1 Y( z) − z −2 Y( z) 3 3 4 X(z)
Y(z)
z-1
2 3
2 3
z-1
3 4
z-1
This is not a unique solution. Many other correct diagrams could be drawn.
11-25
M. J. Roberts - 7/12/03
(b)
z2 H( z) = (z − 0.75)(z + 0.1)(z − 0.3) z z 1 Y( z) = X( z) z − 0.75 z + 0.1 z − 0.3
X(z)
1 z
1 z
1 z
0.75
0.1
0.3
Y(z)
This is not a unique solution. Many other correct diagrams could be drawn. 22. If the z transform of x[ n ] is X( z) =
what is y[ n ] ?
1 3 4
z−
, and
jπ − jπ Y( z) = j X e 6 z − X e 6 z
3 Doing the inverse z transform, x[ n ] = 4 z Z scale property, α n g[ n ] ← → G , α
or
n −1
u[ n − 1] . Then using the change-of-
n −1 π 3 n −1 − j π n 3 n −1 j π n j n − j π6 n 3 6 6 y[ n ] = j e − e u[ n − 1] = j e − e 6 u[ n − 1] 4 4 4
3 y[ n ] = 2 4
n −1
π sin n u[ n − 1] . 6
23. Using the convolution property, find the z transforms of these signals. (a)
2πn x[ n ] = sin u[ n ] ∗ u[ n ] 8 Z Using sin(Ω0 n ) u[ n ] ← →
z sin(Ω0 ) , z >1 z − 2 z cos(Ω0 ) + 1 2
11-26
M. J. Roberts - 7/12/03
π z sin 4 z 2πn Z sin u[ n ] ∗ u[ n ] ←→ 8 π z −1 z 2 − 2 z cos + 1 4 π z 2 sin 4 2πn Z sin u[ n ] ∗ u[ n ] ←→ 8 π π z 3 − z 2 2 cos − 1 + z 1 + 2 cos − 1 4 4 z 2πn Z sin u[ n ] ∗ u[ n ] ←→ 0.7071 3 8 z − 0.4142 z 2 + 2.4142 z − 1 2
(b)
2πn x[ n ] = sin u[ n ] ∗ (u[ n ] − u[ n − 8]) 8
From part (a), 2πn Z sin u[ n ] ∗ u[ n ] ←→ 8
π z 2 sin 4 π π z 3 − z 2 2 cos − 1 + z 1 + 2 cos − 1 4 4
Then, using the time-shifting property of the z transform, π z 2 sin 4 2πn Z −8 sin u[ n ] ∗ (− u[ n − 8]) ←→ − z 8 π π z 3 − z 2 2 cos − 1 + z 1 + 2 cos − 1 4 4
Therefore π sin 4 2πn Z sin u[ n ] ∗ (u[ n ] − u[ n − 8]) ←→ 8 π π z 3 − z 2 2 cos − 1 + z 1 + 2 cos − 1 4 4
(1 − z )z −8
or
2
z 2 (1 − z −8 ) 2πn Z sin u ∗ u ← → 0 . 7071 n n [ ] [ ] 8 z 3 − 0.4142 z 2 + 2.4142 z − 1
24. Find the inverse z transforms of these functions in closed form using partial fraction expansions, a z transform table and the properties of the z transform. (a)
X( z) =
z −1 z + 1.8 z + 0.82 2
11-27
M. J. Roberts - 7/12/03
X( z) =
z 1 − 2 z + 1.8 z + 0.82 z + 1.8 z + 0.82 2
The denominators are of the form, z 2 − 2α cos(Ω0 ) z + α 2 where α = 0.9055 and Ω0 = 3.031. Therefore zα sin(Ω0 ) zα sin(Ω0 ) 1 z −1 − X( z) = 2 2 α sin(Ω0 ) z + 1.8 z + 0.82 α sin(Ω0 ) z + 1.8 z + 0.82 or 0.1z 0.1z X( z) = 10 2 − 10 z −1 2 z + 1.8 z + 0.82 z + 1.8 z + 0.82 Then, using zα sin(Ω0 ) α n sin(Ω0 n ) u[ n ] ←Z → 2 , z>α z − 2αz cos(Ω0 ) + α 2
{
X( z) = 10 (0.9055) sin( 3.031n ) u[ n ] − (0.9055) X( z) =
(b)
n
n −1
}
sin( 3.031( n − 1)) u[ n − 1]
z −1 z( z + 1.8 z + 0.82) 2
The answer to this part is the same as the answer to the previous part except delayed by 1 in discrete time.
{
X( z) = 10 (0.9055) (c)
n −1
X( z) =
sin( 3.031( n − 1)) u[ n − 1] − (0.9055)
n −2
z2 z2 − z +
1 4
X( z) =
z
2
1 z −z+ 4
= 2z
2
Z Using nα n u[ n ] ← →
1 z 2 1 z − 2
2
zα , z>α (z − α )2
n 0 −1 Z and g[ n + n 0 ] ← → z n 0 G( z) − ∑ g[ m]z − m , n 0 > 0 m =0 Z g[ n + 1] ← → z(G( z) − g[0])
11-28
}
sin( 3.031( n − 2)) u[ n − 2]
M. J. Roberts - 7/12/03
1 (n + 1) 2
n +1
1 z 0 2 1 Z 0 0 u[ n + 1] ← → z − ( ) u [ ] 2 2 1 14243 z − =0 2
1 2( n + 1) 2
n +1
or 1 2( n + 1) 2
Z u[ n + 1] ← → 2z
n +1
Z u[ n + 1] ← →
11-29
1 z 2 1 z − 2
2
z2 z2 − z +
1 4
M. J. Roberts - 7/12/03
Chapter 12 -
z Transform Analysis of Signals and Systems Solutions
1. Find the transfer functions for these systems by block diagram reduction. (a)
1 z X(z)
Y(z)
0.2
1 z 0.5 1 1+0.2z -1
X(z)
Y(z) 1 1+0.5z -1
1 1+0.2z -1
X(z)
X(z)
1 1+0.5z -1
Y(z)
0.3z z2 + 0.7z + 0.1
Y(z)
(b) X(z)
Y(z) 1 z
1 z 0.7
0.9
12-1
M. J. Roberts - 7/12/03
X(z)
1 1+0.9z -1
1 1+0.7z -1
z2 z + 1.6z + 0.63
X(z)
Y(z)
Y(z)
2
(c)
X(z)
Y(z) 1 z -0.75
2 1 z
0.3
3
X(z)
Y(z) 1 z
1 z
-0.75
2 1 z
1 z
0.3
3
X(z)
Y(z) 1 z
1 z 3z-1+ 2
0.3z-1-0.75
12-2
M. J. Roberts - 7/12/03
1 1 - 0.3z -2 + 0.75z -1
X(z)
X(z)
1 + 3z -2 + 2z -1
z2 + 2z + 3 z + 0.75z - 0.3
Y(z)
Y(z)
2
2. Evaluate the stability of the systems with each of these transfer functions. (a)
H( z) =
z z−2
(b)
H( z) =
z
(c)
(d)
H( z) =
H( z) =
z2 −
Pole at z = 2. Outside the unit circle. Unstable.
7 8
Poles at z = ±
7 . Both inside the unit circle. 8
Stable.
z 3 9 z2 − z + 2 8
Poles at z =
3 3 ± j . Both outside the 4 4
unit circle. Unstable.
z2 − 1 z 3 − 2 z 2 + 3.75 z − 0.5625
Poles at z = 0.9185 ± j1.6146 , 0.163 Two poles outside the unit circle. Unstable.
3. A feedback DT system has a transfer function, H( z) =
K z 1+ K z − 0.9
.
For what range of K’s is this system stable? H( z) =
Pole at z =
K ( z − 0.9) K ( z − 0.9) K z − 0.9 = = z − 0.9 + Kz (1 + K ) z − 0.9 1 + K z − 0.9 1+ K
0.9 0.9 < 1 or 0.9 < 1 + K . . For stability, 1+ K 1+ K
If K > −0.1 or K < −1.9 the pole lies inside the unit circle.
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M. J. Roberts - 7/12/03
4. Find the overall transfer functions of these systems in the form of a single ratio of polynomials in z. (a)
X(z)
Y(z) 1 z 0.3 H( z) =
1 z −1 = 1 + 0.3z z + 0.3
(b)
X(z)
Y(z) 1 z 0.3 H( z) =
1 z 0.9
z2 1 = (1 + 0.3z −1)(1 + 0.9z −1) z 2 + 1.2z + 0.27
5. Find the DT-domain responses, y[ n ] , of the systems with these transfer functions to the unit sequence excitation, x[ n ] = u[ n ] . (a)
H( z) =
z z −1
Y( z) =
z z z =z z −1 z −1 (z − 1) 2
Z ramp[ n ] ← →
z (z − 1) 2
z Z ramp[ n + 1] ← → z − ramp 0 [ ] 2 4 4 3 ( z − 1) 12 =0 Z y[ n ] = ramp[ n + 1] ← →z
z (z − 1) 2
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M. J. Roberts - 7/12/03
(b)
H( z) =
z −1 1 z− 2
z −1 z z 1 = ⇒ y[ n ] = u[ n ] Y( z) = 1 1 z −1 2 z− z− 2 2 n
6. Find the DT-domain responses, y[ n ] , of the systems with these transfer functions to the 2πn excitation, x[ n ] = cos u[ n ] . Then show that the steady-state response is the same 8 as would have been obtained by using DTFT analysis with an excitation, 2πn x[ n ] = cos . 8 z (a) H( z) = z − 0.9 Z Using cos(Ω0 n ) u[ n ] ← →
Y( z) =
[
]
z z − cos(Ω0 ) , z >1 z 2 − 2 z cos(Ω0 ) + 1
[
]
0.3232 0.6768 z + 0.3591 z z − cos(Ω0 ) z z = + z − 0.9 z 2 − 2 z cos(Ω0 ) + 1 z − 0.9 z 2 − 2 z cos(Ω0 ) + 1
Using the result derived in the text, N ( z) y[ n ] = Z −1 z 1 + H( p1 ) cos(Ω0 n + ∠ H( p1 )) u[ n ] D( z) where p1 = e
j
π 4
π j 4 e e −1 0.3232 u[ n ] cos Ω0 n + ∠ π y[ n ] = Z z + π z − 0.9 j j e 4 − 0.9 e 4 − 0.9 j
π 4
π n y[ n ] = 0.3232(0.9) u[ n ] + 1.3644 cos n − 1.0517 u[ n ] 4 Using the DTFT,
π π Ω − Ω + 1 e 4 + comb 4 Y( jΩ) = jΩ comb 2π e − 0.9 2 2π jΩ
12-5
M. J. Roberts - 7/12/03 π π π π j −j Ω− Ω+ 4 4 e e 1 4 4 Y( jΩ) = comb + comb π π j −j 2 2 π 2 π 4 4 e − 0.9 e − 0.9
π π π π π Ω− Ω+ jπ − j − j j 4 e 4 − 0.9 + e 4 comb 4 e 4 − 0.9 e 4 comb 2π 2π 1 Y( jΩ) = π π j4 − j 4 2 e e 0 9 0 9 − . − .
π π π π π π Ω− Ω+ Ω− Ω+ j −j 4 + comb 4 − 0.9 e 4 comb 4 + e 4 comb 4 comb 2π 2π 2π 2π 1 Y( jΩ) = π π −j 2 j 1 − 0.9 e 4 + e 4 + 0.81 π π π π Ω− Ω+ Ω− Ω+ 0 9 . 4 + comb 4 − 4 + (1 − j ) comb 4 1 comb + comb j ( ) 2π 2π 2π 2 2π 1 Y( jΩ) = 2 π 1.81 − 1.8 cos 4
π Ω− 4 comb 2π Y( jΩ) =
1 2
π π Ω− Ω+ 4 + comb 4 comb 2π π 2 π Ω+ . 0 9 4 − + comb 2 π π 2π Ω Ω − + 4 − comb 4 + j comb 2π 2π π 1.81 − 1.8 cos 4
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M. J. Roberts - 7/12/03
π π π π Ω− Ω+ Ω− Ω+ 0 9 . 0.9 4 + comb 4 −j 4 − comb 4 comb 1 − comb 2π 2π 2π 2 2 2π 1 Y( jΩ) = 2 0.5372 2πn 2πn Y( jΩ) = 0.6768 cos + 1.1847 sin 8 8 2πn Y( jΩ) = 1.3644 cos − 1.0518 . Check. 8 (b)
z2 H( z) = 2 z − 1.6 z + 0.63
[
]
0.03482 1.454 z z − cos(Ω0 ) z2 0.489 z − 1.666 z = + − Y( z) = 2 z − 1.6 z + 0.63 z 2 − 2 z cos(Ω0 ) + 1 z − 0.7 z − 0.9 z 2 − 2 z cos(Ω0 ) + 1 Using the result derived in the text, N ( z) y[ n ] = Z −1 z 1 + H( p1 ) cos(Ω0 n + ∠ H( p1 )) u[ n ] D( z) where p1 = e
j
π 4
j π4 e
2
j π4 j j H e = = = 2 π π j 1 + j j j − 1.6 + 0.63 −0.5014 − j 0.1314 4 4 e − 1.6e + 0.63 2 j j n n y[ n ] = 0.03482(0.7) + 1.454 (0.9) + − cos Ω0 n + ∠ − u[ n ] j j 0 . 5014 + 0 . 1314 0 . 5014 + 0 . 1314 2πn n n y[ n ] = 0.03482(0.7) + 1.454 (0.9) + 1.9293 cos − 1.3145 u[ n ] 8
Y( jΩ) =
j 2Ω
e
j 2Ω
e − 1.6e jΩ
π π Ω − Ω + 1 4 + comb 4 comb 2π + 0.63 2 2π
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M. J. Roberts - 7/12/03
π π Ω− Ω+ 4 4 comb comb 2π 2π j Y( jΩ) = + 2 j − 1.6 1 + j + 0.63 j + 1.6 1 − j − 0.63 2 2 π π Ω− Ω+ 4 4 comb comb 2π 2π j Y( jΩ) = − − 2 0.5013 + j 0.1314 0.5013 − j 0.1314 π π Ω− Ω+ 4 (0.5013 − j 0.1314 ) − comb 4 (0.5013 + j 0.1314 ) comb 2π 2π j Y( jΩ) = − 2 (0.5013 + j 0.1314)(0.5013 − j 0.1314) π π π π Ω− Ω+ Ω− Ω+ 4 − comb 4 − j 0.1314 comb 4 + comb 4 0.5013comb 2π 2π 2π 2π j Y( jΩ) = − 2 0.2686 2πn 2πn y[ n ] = 1.866 sin − 0.4892 cos 8 8 2πn y[ n ] = 1.9291cos − 1.3144 . Check. 8 7. Sketch the magnitude frequency response of these systems from their pole-zero diagrams.
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M. J. Roberts - 7/12/03
Im(z)
|H(e
[z]
jΩ
)|
2
(a)
Re(z)
-π
π
Ω
0.5
Im(z)
[z]
(b)
|H(e
Re(z) 0.5
1
jΩ
)|
2
-π
Im(z)
π
Ω
[z]
0.5
|H(e jΩ)| 3
(c)
Re(z) 0.5
-π
π
Ω
-0.5
8. Use the Jury stability test to determine which of these transfer functions are for unstable systems. z2 − z (a) H( z) = 3 z − 0.25 z 2 − 0.6528 z + 0.2083 The Jury array is 1 0.2083 −0.6528 −0.25 1 2 1 −0.25 −0.6528 0.2083 . 3 −0.9566 0.114 0.6007 D(1) = 1 − 0.25 − 0.6528 + 0.2083 = 0.3055 > 0
(−1) D(−1) = (−1) 3 (−1 − 0.25 + 0.6528 + 0.2083) = 0.3889 > 0 3
−0.9566 > 0.6007 Stable
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M. J. Roberts - 7/12/03
(b) H( z) =
z −1 z − 0.9 z − 0.65 z 2 + 0.873z 4
3
1 0 0.873 −0.65 −0.9 1 2 1 −0.9 −0.65 0.873 0 3 1 0.9 0.65 −0.873 4 −0.873 0.65 0.9 1 5 0.2379 1.4675 1.4357 0.2379 < 1.4357 Unstable (c) H( z) =
z z − 1.5 z + 0.5 z 2 + 0.25 z − 0.25 4
3
1 −0.25 0.25 0.5 −1.5 1 2 1 −1.5 0.5 0.25 −0.25 3 −0.9375 1.4375 −0.625 0.125 4 0.125 −0.625 1.4375 −0.9375 5 0.8633 −1.2395 0.40625 D(1) = 1 − 1.5 + 0.5 + 0.25 − .25 = 0 −1) D(−1) = 1 + 1.5 + 0.5 − 0.25 − 0.25 = 2.5 > 0 ({ 4
1
1 > −0.25 −0.9375 > 0.125 0.8633 > 0.40625 Marginally Stable
9. Draw a root locus for each system with the given forward and feedback path transfer functions. z −1 4z , H 2 ( z) = 1 z − 0.8 z+ 2 z( z − 1) z −1 z = 4K T( z ) = 4 K 1 1 z + z − 0.8 z + ( z − 0.8) 2 2
(a) H1 ( z) = K
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M. J. Roberts - 7/12/03
Im(z)
Re(z)
z −1 4 , H 2 ( z) = 1 z − 0.8 z+ 2 z −1 z −1 1 = 4K T( z ) = 4 K 1 z − 0.8 1 z+ z + ( z − 0.8) 2 2 (b) H1 ( z) = K
Im(z)
Re(z)
1 z 5 (c) H1 ( z) = K , H 2 ( z) = 3 1 z− z− 4 4 z+
1 z z + 5 T( z ) = K 1 z − z − 4
3 4
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M. J. Roberts - 7/12/03
Im(z)
Re(z)
(d) H1 ( z) = K
z
, H 2 ( z) =
1 4 z( z + 2) T( z ) = K 1 3 z − z − 4 4 z−
z+2 3 z− 4
Im(z)
Re(z)
1 , H 2 ( z) = 1 1 2 2 z − z− 3 9 1 T( z ) = K 2 1 z − z + 3 3
(e) H1 ( z) = K
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M. J. Roberts - 7/12/03
Im(z)
Re(z)
10. Using the impulse-invariant design method, design a DT system to approximate the CT systems with these transfer functions at the sampling rates specified. Compare the impulse and unit step (or sequence) responses of the CT and DT systems. (a)
H( s) =
6 , f s = 4 Hz s+6
h( t) = 6e
−6 t
u( t) ⇒ h[ n ] = 6e
Unit step response:
3 − n 2
H −1 ( s) =
u[ n ] ⇒ H( z) =
6z z−e
3 − 2
=
6z z − 0.2231
1 6 1 1 = − ⇒ h −1 ( t) = (1 − e −6 t ) u( t) s s+6 s s+6
Unit sequence response: H −1 ( z) =
z 6z 1.723 7.723 = z − z − 1 z − 0.2231 z − 1 z − 0.2231
[
]
h −1[ n ] = 7.723 − 1.723(0.2231) u[ n ] n
Unit Step Response
Unit Sequence Response
h (t)
h [n]
-1
-1
1
8
1
(b)
H( s) =
t
6 , f s = 20 Hz s+6
12-13
-5
10
n
M. J. Roberts - 7/12/03
h( t) = 6e −6 t u( t) ⇒ h[ n ] = 6e
Unit step response:
−
3 n 10
H −1 ( s) =
u[ n ] ⇒ H( z) =
6z z−e
−
3 10
=
6z z − 0.7408
1 6 1 1 = − ⇒ h −1 ( t) = (1 − e −6 t ) u( t) s s+6 s s+6
Unit sequence response: H −1 ( z) =
z 6z 17.148 23.148 = z − z −1 z − 1 z − 0.7408 z − 0.7408
[
]
h −1[ n ] = 23.148 − 17.148(0.7408) u[ n ] n
Unit Step Response h- 1(t)
Unit Sequence Response h- 1[n]
1
25
1
t
-5
30
n
11. Using the impulse-invariant and step-invariant design methods, design digital filters to approximate analog filters with these transfer functions. In each case choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the “s” plane. Graphically compare the step responses of the digital and analog filters. (a)
H( s) =
2 s + 3s + 2 2
CT Impulse Response:
h( t) = 2(e − t − e −2 t ) u( t)
Sampling Rate:
fs =
10 = 3.183 ⇒ Ts = 0.31416 π
Impulse Invariant: h[ n ] = 2(e −0.31416 n − e −0.62832 n ) u[ n ]
DT Impulse Response:
[
]
h[ n ] = 2 (0.7304 ) − (0.5335) u[ n ] z-Domain Transfer Function: H( z) = 0.3938
z z − 1.264 z + 0.3897 2
12-14
n
n
M. J. Roberts - 7/12/03
z Transform of Step Response: z z H −1 ( z) = 0.3938 2 z − 1 z − 1.264 z + 0.3897 10.05 3.092 7.955 H −1 ( z) = 0.3938 − + z − 1 z − 0.7306 z − 0.5334 DT Step Response:
[ [n] = [3.1312 − 5.4183(0.7304)
h −1[ n ] = 3.1312 − 3.9575(0.7304 ) h −1
n −1
n
+ 1.22(0.53348)
n −1
]u[n − 1]
]
+ 2.287(0.53348) u[ n ] n
(Last form is correct because h −1[0] = 0 .) Step Invariant: Laplace Transform of Step Response: H −1 ( s) =
1 2 1 2 1 2 1 = = − + 2 s s + 3s + 2 s ( s + 1)( s + 2) s s + 1 s + 2
CT Step Response:
h −1 ( t) = (1 − 2e − t + e −2 t ) u( t)
Sampling Rate:
fs =
DT Step Response:
h −1[ n ] = (1 − 2e −0.31416 n + e −0.6283 n ) u[ n ]
10 = 3.183 ⇒ Ts = 0.31416 π
(
)
h −1[ n ] = 1 − 2(0.7304 ) + (0.5335) u[ n ] n
z Transform of Step Response: H −1 ( z) =
z z z −2 + z −1 z − 0.7304 z − 0.5335
Transfer Function: H( z) =
z − 1 z z z −2 + z z −1 z − 0.7304 z − 0.5335
H( z) = 1 − 2
z −1 z −1 + z − 0.7304 z − 0.5335
H( z) = 0.0726
z + 0.7314 z − 1.2639 z + 0.3897 2
12-15
n
M. J. Roberts - 7/12/03
h- 1(t)
Unit Step Response
1
Impulse-Invariant Unit Sequence Response h- 1[n]
t
3π
3.1309
-5
Step-Invariant Unit Sequence Response h- 1[n]
n
30
1
-5
(b)
H( s) =
n
30
6s 6s 10 16 = =− + s + 13s + 40 ( s + 5)( s + 8) s+5 s+8 2
CT Impulse Response:
h( t) = (16e −8 t − 10e −5 t ) u( t)
Sampling Rate:
fs =
40 = 12.732 ⇒ Ts = 0.07854 π
Impulse Invariant: DT Impulse Response: h[ n ] = (16e −0.6283 n − 10e −0.3927 n ) u[ n ]
[
]
h[ n ] = 16(0.5335) − 10(0.6752) u[ n ] n
n
H( z) =
Transfer Function:
16 z 10 z − z − 0.5335 z − 0.6752
z 2 − 0.9114 z H( z) = 6 2 z − 1.2087 z + 0.3602 z Transform of Step Response: H −1 ( z) = 6
z z 2 − 0.9114 z z − 1 z 2 − 1.2087 z + 0.3602
3.459 z 3.044 z 0.5848 z H −1 ( z) = 6 + − z −1 z − 0.6753 z − 0.5334
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M. J. Roberts - 7/12/03
H( z) = 6
z 3 − 0.9098 z 2 z 3 − 2.209 z 2 + 1.569 z − 0.3602
DT Step Response:
(
)
h −1[ n ] = 6 0.5848 + 3.459(0.6753) − 3.044 (0.5334 ) u[ n ] n
n
Step Invariant: Laplace Transform of Step Response: H −1 ( s) =
1 6s 6 2 2 = = − 2 s s + 13s + 40 ( s + 5)( s + 8) s + 5 s + 8
CT Step Response:
h −1 ( t) = (2e −5 t − 2e −8 t ) u( t)
Sampling Rate:
fs =
DT Step Response:
h −1[ n ] = 2 (0.6752) − (0.5335) u[ n ]
40 = 12.732 ⇒ Ts = 0.07854 π
(
n
n
z Transform of DT Step Response: H −1 ( z) =
0.2834 z z − 1.2087 z + 0.3602 2
Transfer Function: H( z) = h- 1(t)
0.2835 z − 0.2835 z − 1.209 z + 0.3602 2
Unit Step Response
0.34094
Impulse-Invariant Unit Sequence Response h- 1[n]
2.3562
t
7.7833
-5
Step-Invariant Unit Sequence Response h- 1[n]
30
n
0.34265
-5
30
12-17
n
)
M. J. Roberts - 7/12/03
(c)
H( s) =
250 s + 10 s + 250 2
CT Impulse Response:
h( t) = 16.667e −5 t sin(15 t) u( t)
Sampling Rate:
f s = 10
250 = 25.165 ⇒ Ts = 0.03974 2π
Impulse Invariant: DT Impulse Response: h[ n ] = 16.667e −0.1987 n sin(0.5961n ) u[ n ] h[ n ] = 16.667(0.8198) sin(0.5961n ) u[ n ] n
H( z) = 7.671
Transfer Function:
z z − 1.357 z + 0.6721 2
z Transform of Step Response: H −1 ( z) = 7.671
z z 2 z − 1 z − 1.357 z + 0.6721
2.174 z − 2.133 3.174 H −1 ( z) = 7.671 − 2 z − 1 z − 1.357 z + 0.6721 DT Step Response: 24.348 u[ n − 1] − 36.245(0.8198) n sin(0.5959 n ) u[ n ] h −1[ n ] = n −1 +35.55(0.8198) sin(0.5959( n − 1)) u[ n − 1] Alternate solution: z 3.174 z − 2.133 1 H −1 ( z) 3.174 = 7.671 = 7.671 − 2 2 z − 1 z − 1.357 z + 0.6721 z z − 1 z − 1.357 z + 0.6721 3.174 z 3.174 z z 2 − 0.672 z 3.174 z 2 − 2.133z z 3 174 7 . 671 . H −1 ( ) = 7.671 − 2 − = 2 z − 1.357 z + 0.6721 z −1 z − 1.357 z + 0.6721 z −1 3.174 z z 2 − 0.6785 z 0.0065 z + 2 H −1 ( z) = 7.671 − 3.174 2 z − 1.357 z + 0.6721 z − 1.357 z + 0.6721 z −1 α = 0.8198 , Ω0 = 0.5959
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M. J. Roberts - 7/12/03
3.174 z z 2 − 0.6785 z 0.4601z + 0.01413 2 H −1 ( z) = 7.671 − 3.174 2 z − 1.357 z + 0.6721 z − 1.357 z + 0.6721 z −1
{
}
h −1[ n ] = 24.348 − 24.348(0.81982) [cos(0.59594 n ) + 0.01413 sin(0.59594 n )] u[ n ] n
Step Invariant: CT Step Response:
sin(15 t) h −1 ( t) = 1 − e −5 t cos(15 t) + u( t) 3
Sampling Rate:
f s = 10
250 = 25.165 ⇒ Ts = 0.03974 2π
DT Step Response: sin(0.5961n ) n h −1[ n ] = 1 − (0.8198) cos(0.5961n ) + u[ n ] 3 z Transform of Step Response: H −1 ( z) = 0.3279 Transfer Function: h- 1(t)
z 3 − 0.4873z 2 + 0.4486 z z 3 − 2.357 z 2 + 2.029 z − 0.6721
z 2 − 0.4873z + 0.4486 H( z) = 0.3279 2 z − 1.357 z + 0.6721
Unit Step Response
1.3498
1.1922
t
Impulse-Invariant Unit Sequence Response h- 1[n] 33.2209
-5
Step-Invariant Unit Sequence Response h- 1[n]
30
n
1.3457
-5
30
n
12. Using the difference-equation method and all backward differences, design digital filters to approximate analog filters with these transfer functions. . In each case, if a sampling frequency is not specified choose a sampling frequency which is 10 times the 12-19
M. J. Roberts - 7/12/03
magnitude of the distance of the farthest pole or zero from the origin of the “s” plane. Graphically compare the step responses of the digital and analog filters. (a)
H( s) = s , f s = 1MHz x( t) − x( t − Ts ) d Y( s) = s ⇒ y( t) = ( x( t)) ≅ dt Ts X( s) x[ n ] − x[ n − 1] X( z) − z −1 X( z) y[ n ] = ⇒ Y( z) = Ts Ts H( z) =
z −1 Y( z) 1 − z −1 z − 1 = = = 10 6 Ts zTs z X( z)
H −1 ( s) = 1 ⇒ h −1 ( t) = δ ( t) H −1 ( z) = 10 6 ⇒ h −1[ n ] = 10 6 δ [ n ]
h-1(t) 1 t h-1[n] 6
10
n (b)
H( s) =
1 , f s = 1kHz s
d Y( s) 1 = ⇒ ( y( t)) = x( t) dt X( s) s y( t) − y( t − Ts ) y[ n ] − y[ n − 1] ⇒ x[ n ] = Ts Ts −1 Ts zTs z Y( z) − z Y( z) X( z) = ⇒ H( z) = = 10 −3 −1 = z −1 z −1 Ts 1− z x( t) ≅
H −1 ( s) =
1 ⇒ h −1 ( t) = ramp( t) s2
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M. J. Roberts - 7/12/03
2
z z −3 H −1 ( z) = 10 = 10 z z − 1 (z − 1) 2 −3
Z Using ramp[ n ] ← →
z (z − 1) 2
z Z ramp[ n + 1] ← → z − 0 2 ( z − 1) Therefore z H −1 ( z) = 10 −3 z ⇒ h −1[ n ] = 10 −3 ramp[ n + 1] (z − 1) 2
h-1(t) 1
h-1[n]
10 (c)
H( s) =
1
t
-3
n
2 s + 3s + 2 2
h( t) = 2(e − t − e −2 t ) u( t) H −1 ( s) =
1 2 1 2 1 = − + 2 s s + 3s + 2 s s + 1 s + 2
h −1 ( t) = (1 − 2e − t + e −2 t ) u( t) f s = 3.1831 ⇒ Ts = 0.31416 z2 H( z) = 0.09225 2 z − 1.375 z + 0.4673 H −1 ( z) = 0.09225
z z2 z − 1 z 2 − 1.375 z + 0.4673
16.518 z 6.682 z 10.834 z H −1 ( z) = 0.09225 − + z −1 z − 0.7607 z − 0.6143
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M. J. Roberts - 7/12/03
(
)
h −1[ n ] = 0.09225 10.834 − 16.518(0.7607) + 6.682(0.6143) u[ n ] h- 1(t)
n
n
Unit Step Response
1
3π
t
Difference-Equation Unit Sequence Response h- 1[n] 1
-5
30
n
13. Using the matched-z-transform method, design digital filters to approximate analog filters with these transfer functions. In each case, if a sampling frequency is not specified choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the “s” plane (unless all poles or zeros are at the origin, in which case the sampling rate will not matter, in this method). Graphically compare the step responses of the digital and analog filters. (a)
H( s) = s Zero at s = 0. Transformation is s − a → 1 − e aT z −1 . Therefore H( z) = 1 − z −1 =
z −1 z
1 H −1 ( s) = s = 1 ⇒ h −1 ( t) = δ ( t) s H −1 ( z) =
z z −1 = 1 ⇒ h −1[ n ] = δ [ n ] z −1 z
h-1(t) 1 t h-1[n] 1 n (b)
H( s) =
1 s 12-22
M. J. Roberts - 7/12/03
Pole at s = 0. Transformation is s − a → 1 − e aT z −1 . Therefore H( z) =
1 z −1 = 1− z z −1
H −1 ( s) =
11 1 = ⇒ h −1 ( t) = ramp( t) s s s2
H −1 ( z) =
z z z =z ⇒ h −1[ n ] = ramp[ n + 1] z −1 z −1 (z − 1) 2
h-1(t) 1
h-1[n]
1 (c)
1
t
n
H( s) =
2s 2s = , Double pole at s = −5. s + 10 s + 25 ( s + 5) 2
H( s) =
2 10 − s + 5 ( s + 5) 2
2
h( t) = 2(e −5 t − 5 te −5 t ) u( t) f s = 7.9577 ⇒ Ts = 0.12566 H −1 ( s) =
1 2s 2 = 2 s s + 10 s + 25 ( s + 5) 2
h −1 ( t) = 2 te −5 t u( t) Using s − a → 1 − e aTs z −1 , s → 1 − z −1 and s − 5 → 1 − 0.5335 z −1 Then H( s) =
2(1 − z −1 ) 2s 2 z( z − 1) = 2 2 → H( z) = −1 2 (s + 5) (1 − 0.5335z ) (z − 0.5335)
12-23
M. J. Roberts - 7/12/03
z 2 z( z − 1) 2z 2 z H −1 ( z) = 2 = 2 = 2z z − 1 ( z − 0.5335) (z − 0.5335) (z − 0.5335) 2 Z → Using nα n u[ n ] ←
(n + 1)α
n +1
z (z − α )2
z z2 u[ n + 1] ←→ z 2 − 0 = 2 (z − α ) (z − α ) Z
and h −1[ n ] = 2( n + 1)(0.5335) h (t)
n +1
u[ n + 1]
(c) Unit Step Response
-1
0.14599
3.7699
t
(c) Matched-z Unit Sequence Response h- 1[n] 2.134
-5
30
n
14. Using the bilinear-z-transform method, design digital filters to approximate analog filters with these transfer functions In each case choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the “s” plane. Graphically compare the step responses of the digital and analog filters. (a)
H( s) =
s − 10 s + 10
[
]
h( t) = δ ( t) − 20e −10 t u( t) H −1 ( s) =
1 s − 10 1 2 =− + ⇒ h −1 ( t) = (2e −10 t − 1) u( t) s s + 10 s s + 10
H( z) = 0.5219
z − 1.9161 z − 0.5219
H −1 ( z) = 0.5219
z − 1.9161 z z − 1.9161 = 0.5219 z 2 z − 1.522 z + 0.5219 z − 1 z − 0.5219
1.914 z 2.914 z H −1 ( z) = 0.5219 − z − 0.5218 z −1 12-24
M. J. Roberts - 7/12/03
(
)
h −1[ n ] = 1.521(0.5218) − 1 u[ n ] n
Unit Step Response
h- 1(t)
1.885
t
-1
Bilinear-z Unit Sequence Response h- 1[n] 30
n
-1
(b)
H( s) =
10 s + 11s + 10 2
h( t) = (1.1111e − t − 1.1111e −10 t ) u( t) 10 1 1 10 1 1 9 10 1 H −1 ( s) = 2 = − 9 + 9 = − + s s + 11s + 10 s s + 1 s + 10 9 s s + 1 s + 10 h −1 ( t) =
9 − 10e − t + e −10 t u( t) 9
H( z) = 0.007281
z 2 + 2z + 1 z 2 − 1.461z + 0.4901
H −1 ( z) = 0.007281z
z 2 + 2z + 1 z 3 − 2.461z 2 + 1.951z − 0.4901
z 1.077 z 0.08455 z − + z − 1 z − 0.9391 z − 0.5219
H −1 ( z) =
(
)
h −1[ n ] = 1 − 1.077(0.9391) + 0.08455(0.5219) u[ n ] n
12-25
n
M. J. Roberts - 7/12/03
Unit Step Response
h (t) -1
0.8313
1.885
t
Bilinear-z Unit Sequence Response h- 1[n] 0.83648
30
(c)
H( s) = h( t) =
n
3s s + 11s + 10 2
10 −10 t e − e − t ) u( t) ( 3 1 3s 3 1 1 1 = 2 = − 2 s s + 11s + 10 s + 11s + 10 3 s + 1 s + 10
H −1 ( s) = h −1 ( t) =
e − t − e −10 t u( t) 3
H( z) = 0.06953
z2 − 1 z 2 − 1.461z + 0.4901
H −1 ( z) = 0.06953
z +1 z z2 − 1 = 0.06953z 2 2 z − 1.461z + 0.4901 z − 1 z − 1.461z + 0.4901
0.3232 z 0.2536 z − z − 0.9391 z − 0.5219
H −1 ( z) =
[
]
h −1[ n ] = 0.3232(0.9391) − 0.2536(0.5219) u[ n ] n
h- 1(t)
n
Unit Step Response
0.2322
1.885
t
Bilinear-z Unit Sequence Response h- 1[n] 0.23256
30
12-26
n
M. J. Roberts - 7/12/03
15. Design a digital-filter approximation to each of these ideal analog filters by sampling a truncated version of the impulse response and using the specified window. In each case choose a sampling frequency which is 10 times the highest frequency passed by the analog filter. Choose the delays and truncation times such that no more than 1% of the signal energy of the impulse response is truncated. Graphically compare the magnitude frequency responses of the digital and ideal analog filters using a dB magnitude scale versus linear frequency. fc = 1 ; type = 'LP' ; fs = 10*fc ; % % Lowpass, Rectangular Window % h = FIRDF(type,fc,fs,'RE',0.01) ; N = length(h) ; Ts = 1/fs ; n = [0:N-1]' ; F = [0:0.001:1/2]' ; [H,F] = DTFT(n,h,F) ; subplot(2,2,1) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/2,-120,0],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Lowpass - Rectangular Window','Times',24,'n','c') ; subplot(2,2,2) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/10,-5,0],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Lowpass - Rectangular Window','Times',24,'n','c') ; % % Lowpass, von Hann Window % h = FIRDF(type,fc,fs,'VH',0.01) ; N = length(h) ; Ts = 1/fs ; n = [0:N-1]' ; F = [0:0.001:1/2]' ; [H,F] = DTFT(n,h,F) ; subplot(2,2,3) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/2,-120,20],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Lowpass - Von Hann Window','Times',24,'n','c') ; subplot(2,2,4) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/10,-5,0],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Lowpass - Von Hann Window','Times',24,'n','c') ; % % % % % % % % % % % % % % %
Function to design a digital filter using truncation of the impulse response to approximate the ideal impulse response. user specifies the type of ideal filter, LP BP
-
The
lowpass bandpass
the cutoff frequency(s) fcs (a scalar for LP and HP a 2-vector for BP and BS), the sampling rate, fs, the type of window, RE VH BA HA BL
-
rectangular von Hann Bartlett Hamming Blackman
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M. J. Roberts - 7/12/03
% % and the allowable truncation error, % err, as a fraction of the total impulse response signal energy. % % The function returns the filter coefficients as a vector. % % function h = FIRDF(type,fcs,fs,window,err) % function h = FIRDF(type,fcs,fs,window,err) if fs == 0 | fs == inf | fs == -inf, disp('Sampling rate is unusable') ; else fs Ts = 1/fs ; type = upper(type) ; window = upper(window) ; switch type case 'LP', fc = fcs(1) ; zc1 = 1/(2*fc) ; N = 200*zc1/Ts ; n = [0:N]' ; Etotal = sum(sinc(2*fc*n*Ts).^2) E = 0 ; N = 1 ; while abs(E-Etotal) > Etotal*err, N = 2*N ; n = [0:N]' ; E = sum(sinc(2*fc*n*Ts).^2) ; end delN = floor(N/10) ; while abs(E-Etotal)1, N = N-delN ; n = [0:N]' ; E = sum(sinc(2*fc*n*Ts).^2) ; delN = floor(delN/2) ; end N = ceil(N/2)*2 ; % Make number of pts even nmid = (N/2-1) ; w = makeWindow(window,n,N) ; h = w.*sinc(2*fc*(n-nmid)*Ts) ; h = h/sum(h) ; case 'BP' fl = fcs(1) ; fh = fcs(2) ; fmid = (fh + fl)/2 ; df = abs(fh-fl) ; zc1 = 1/df ; N = 200*zc1/Ts ; n = [0:N]' ; Etotal = sum((2*df*sinc(df*n*Ts).*... cos(2*pi*fmid*n*Ts)).^2) ; E = 0 ; N = 1 ; while abs(E-Etotal) > Etotal*err, N = 2*N ; n = [0:N]' ; E = sum((2*df*sinc(df*n*Ts).*... cos(2*pi*fmid*n*Ts)).^2) ; end delN = floor(N/10) ; while abs(E-Etotal)1, N = N-delN ; n = [0:N]' ; E = sum((2*df*sinc(df*n*Ts).*... cos(2*pi*fmid*n*Ts)).^2) ;
12-28
M. J. Roberts - 7/12/03
delN = floor(delN/2) ; end N = ceil(N/2)*2 ; % Make number of pts even nmid = (N/2-1) ; n = [0:N-1]' ; w = makeWindow(window,n,N) ; h = w.*2.*df.*sinc(df*(n-nmid)*Ts).*... cos(2*pi*fmid*(n-nmid)*Ts) ; h = h/sum(h.*cos(2*pi*fmid*(n-nmid)*Ts)) ; end end function w = makeWindow(window,n,N) switch window case 'RE' w = ones(N,1) ; case 'VH' w = (1-cos(2*pi*n/(N-1)))/2 ; case 'BA' w = 2*n/(N-1).*(0<=n & n<=(N-1)/2) + ... (2-2*n/(N-1)).*((N-1)/2<=n & n
(a)
Lowpass, f c = 1 Hz , Rectangular window |H(ej2πfTs )| dB
|H(ej2πfTs )| dB 5
f
1
-120
(b)
f
-5
Lowpass, f c = 1 Hz , von Hann window |H(ej2πfTs )| dB
|H(ej2πfTs )| dB
20 5
1
f
-120
f
-5
16. Draw a canonical-form block diagram for each of these system transfer functions. (a)
H( z) =
z( z − 1) z + 1.5 z + 0.8 2
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M. J. Roberts - 7/12/03
Y(z) 1 z
X(z)
1 z
1.5 0.8 (b) H( z) =
z 2 − 2z + 4 1 2 z − (2 z + z + 1) 2 1 2 z −z+2 z 2 − 2z + 4 = 2 H( z) = 1 1 1 1 z3 + z − 2z 3 + z − 4 4 2 2 1 2
2 1 z
X(z)
1 z
Y(z)
1 z
1 4
1 4
17. Draw a cascade-form block diagram for each of these system transfer functions. (a)
H( z) =
z 1 3 z + z − 3 4
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M. J. Roberts - 7/12/03
X(z)
(b) H( z) =
z −1 4 z + 2z 2 + 2z + 3
1 z
1 z
1 3
3 4
Y(z)
3
1 z −1 4 H( z) = z + 0.888 z 2 − 0.388 z + 0.8446
1 z
1 z
0.888
0.388
X(z)
1 4
1 z
Y(z)
0.8446 18. Draw a parallel-form block diagram for each of these system transfer functions. (a)
H( z) =
z 1 3 z + z − 3 4 36 12 H( z) = 39 + 52 3 1 z+ z− 4 3
1 z
12 39
1 3 X(z)
1 z 3 4
(b) H( z) =
8z 3 − 4 z 2 + 5z + 9 7z 3 + 4 z 2 + z + 2
12-31
36 52
Y(z)
M. J. Roberts - 7/12/03
H( z) = 1.143 −
0.2599 0.9646 z − 1.278 − 2 z + 0.8212 z − 0.2497 z + 0.3479 1.143
X(z)
1 z
Y(z)
0.2599
0.8212 0.9646 1.278 1 z
1 z
0.2497 0.3479 19. For the system in Figure E19 write state equations and output equations.
4 -2 x[n]
D
D
y[n]
D
2 3 1 5
1 2
Figure E19 A 3-state DT system Assigning states to the responses of delay elements in the order 1,2,3 right-to-left,
12-32
M. J. Roberts - 7/12/03
q1[ n + 1] = q 2 [ n ]
q 2 [ n + 1] = q 3 [ n ] 2 1 1 q 3 [ n + 1] = x[ n ] − q 3 [ n ] − q 2 [ n ] − q1[ n ] 3 5 2 y[ n ] = −2 q 2 [ n ] + 4 q 3 [ n ] Writing these equations in standard state-space form, q1[ n + 1] 0 1 0 q1[ n ] 0 q 2 [ n + 1] = 01 01 12 q 2 [ n ] + 0 x[ n ] q 3 [ n + 1] − − − q 3 [ n ] 1 2 5 3 , q1[ n ] y[ n ] = [0 −2 4 ]q 2 [ n ] q 3 [ n ] 20. Write a set of state equations and output equations corresponding to these transfer functions. (a) H( z) =
0.9 z z − 1.65 z + 0.9 2
We can write a resursion relation directly from the transfer function. y[ n + 2] = 0.9 x[ n + 1] + 1.65 y[ n + 1] − 0.9 y[ n ]
or
y[ n + 1] = 0.9 x[ n ] + 1.65 y[ n ] − 0.9 y[ n − 1]
Let q1[ n ] = y[ n − 1] and let q 2 [ n ] = y[ n ] . Then q1[ n + 1] = q 2 [ n ]
q 2 [ n + 1] = 0.9 x[ n ] + 1.65 q 2 [ n ] − 0.9 q1[ n ] . y[ n ] = q 2 [ n ]
Writing the state equations in standard form, 1 q1[ n ] 0 q1[ n + 1] 0 x[ n ] = −0 9 1 65 + . q 2 [ n ] 0.9 q 2 [ n + 1] . q1[ n ] y[ n ] = [0 1] q 2 [ n ]
(b) H( z) =
4 ( z − 1) (z − 0.9)(z − 0.7) 12-33
M. J. Roberts - 7/12/03
This system can be modeled by a canonical block diagram. 4 4 X(z)
1 z
1 z
Y(z)
Y1 (z)
1.6 0.63 Let q1[ n ] be the response of the right delay element and let q 2 [ n ] be the response of the left delay element. Then q1[ n + 1] = q 2 [ n ] q 2 [ n + 1] = x[ n ] + 1.6 q 2 [ n ] − 0.63 q1[ n ] . y[ n ] = 4 q 2 [ n ] − 4 q1[ n ]
Writing the state equations in standard form, 1 q1[ n ] 0 q1[ n + 1] 0 x[ n ] = −0 63 1 6 + . q 2 [ n ] 1 q 2 [ n + 1] . y[ n ] = [−4
q1[ n ] 4 ] q 2 [ n ]
21. Convert the difference equation, 2πn 10 y[ n ] + 4 y[ n − 1] + y[ n − 2] + 2 y[ n − 3] = cos u[ n ] 16 into a set of state equations and output equations. The difference equation can be written in the form, 2πn y[ n ] = 0.1cos u[ n ] − 0.4 y[ n − 1] − 0.1 y[ n − 2] − 0.2 y[ n − 3] 16 or
2πn y[ n + 1] = 0.1cos u[ n + 1] − 0.4 y[ n ] − 0.1 y[ n − 1] − 0.2 y[ n − 2] 16 Assign the state variables, q1[ n ] = y[ n ] , q 2 [ n ] = y[ n − 1] , q 3 [ n ] = y[ n − 2]. Then
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M. J. Roberts - 7/12/03
2πn q1[ n + 1] −0.4 −0.1 −0.2 q1[ n ] 1 0 0 0.1cos 16 u[ n + 1] 0 0 0 q 2 [ n ] + 0 0 0 q 2 [ n + 1] = 1 q 3 [ n + 1] 0 0 1 0 q 3 [ n ] 0 0 0 q1[ n ] y[ n ] = [1 0 0]q 2 [ n ] q 3 [ n ] 22. Convert the state equations and output equation, n q1[ n + 1] −2 −5 q1[ n ] 1 0 1 u[ n ] = + 3 q 2 [ n + 1] 1 0 q 2 [ n ] 0 0 0 q [n] y[ n ] = [1 0] 1 q 2 [ n ]
into a single difference equation. Using the relationships between the q’s and y, we can write n y[ n + 1] −2 −5 y[ n ] 1 0 1 u[ n ] = + 3 y[ n ] 1 0 y[ n − 1] 0 0 0 Mulitplying matrices and using only the top equation that results, n
1 y[ n + 1] = −2 y[ n ] − 5 y[ n − 1] + u[ n ] 3 or 1 y[ n ] + 2 y[ n − 1] + 5 y[ n − 2] = 3
n −1
u[ n − 1] .
23. Find the responses of the system described by this set of state equations and output equations. (Assume the system is initially at rest.) q1[ n + 1] 3 1 q1[ n ] 4 = + u[ n ] q1[ n + 1] 0 −2 q1[ n ] 3 y1[ n ] 1 −1q1[ n ] = y1[ n ] 2 0 q1[ n ] The transfer function is
12-35
M. J. Roberts - 7/12/03
H( z) = C[ zI − A ]
−1
−1
1 −1z − 3 −1 4 0 B+D= + z + 2 3 0 2 0 0
1 z + 2 4 z + 11 z − 3 4 1 −1 3z − 9 1 −1 0 H( z ) = = = 2 2 2 0 z − z − 6 3 2 0 z − z − 6
z + 20 8 z + 22 z2 − z − 6
z + 20 2 H( z ) = z − z − 6 8 z + 22 2 z − z − 6 The z transform of the excitation vector (in this case, a scalar) is X( z) =
z . z −1
Therefore the z-domain response vector is z + 20 z + 20 2 z2 − z − 6 z (z − 1)(z − z − 6) =z Y( z) = H( z) X( z) = 8 z + 22 z − 1 8 z + 22 2 2 z − z − 6 ( z − 1)( z − z − 6) 2.3 z − 3 + Y( z) = z 4.6 + z − 3
1.2 − z+2 0.4 − z+2
3.5 z − 1 5 z − 1
2.3( 3) n + 1.2(−2) n − 3.5 y[ n ] = u[ n ] n n 4.6( 3) + 0.4 (−2) − 5 24. Find the overall transfer functions of these systems in the form of a single ratio of polynomials in z. (a)
12-36
M. J. Roberts - 7/12/03
X(z)
Y(z) 1 z 0.6 1 z 0.8
H( z) =
z2 1 1 = = 1 + z −1 (0.6 + 0.8 z −1 ) 1 + 0.6 z −1 + 0.8 z −2 z 2 + 0.6 z + 0.8
(b)
1 z Y(z)
X(z)
0.6
1 z 0.8 1 1 2 + 1.4 z −1 z 2 + 0.7 z = 2 H( z) = + = 1 + 0.6 z −1 1 + 0.8 z −1 1 + 1.4 z −1 + 0.48 z −2 z 2 + 1.4 z + 0.48 25. Find the DT-domain responses, y[ n ] , of the systems with these transfer functions to the unit sequence excitation, x[ n ] = u[ n ] . (a)
H( z) =
z z − 1.8 z + 0.82
Y( z) =
z z 50 49 z − 41 = − 2 2 z − 1 z − 1.8 z + 0.82 z − 1 z − 1.8 z + 0.82
2
Z Using α n sin(Ω0 n ) u[ n ] ← →
zα sin(Ω0 ) , z>α z − 2αz cos(Ω0 ) + α 2 2
where α = 0.9055 and Ω0 = 0.1102 ,
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M. J. Roberts - 7/12/03
Y( z) =
0.1z 0.1z 50 49 z − 41 50 = − 490 2 − 410 z −1 2 − 2 z − 1.8 z + 0.82 z − 1 z − 1.8 z + 0.82 z − 1 z − 1.8 z + 0.82
y[ n ] = 50 u[ n − 1] − 490(0.9055) sin(0.1102 n ) u[ n ] + 410(0.9055) n
(b)
H( z) =
z 2 − 1.932 z + 1 z( z − 0.95)
Y( z) =
z z 2 − 1.932 z + 1 z 2 − 1.932 z + 1 = z − 1 z( z − 0.95) (z − 1)(z − 0.95)
Y( z) = 1 +
n −1
sin(0.1102( n − 1)) u[ n − 1]
1.36 1.342 − z − 1 z − 0.95
[
y[ n ] = δ [ n ] + 1.36 − 1.342(0.95)
n −1
]u[n − 1]
26. Sketch the magnitude frequency response of these systems from their pole-zero diagrams. Im(z)
[z]
0.5
(a)
|H(e jΩ)|
Re(z) 0.2
0.866
3
-0.5
-π
Im(z)
Ω
|H(e jΩ)|
[z]
25
0.45
(b)
π
-π
Re(z)
π
Ω
0.866 1
-0.45
27. Using the impulse-invariant design method, design a DT system to approximate the CT systems with these transfer functions at the sampling rates specified. Compare the impulse and unit step (or sequence) responses of the CT and DT systems.
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M. J. Roberts - 7/12/03
(a)
H( s) =
712 s , f s = 20 Hz s + 46 s + 240
H( s) =
712 s 712 s 1.1765 0.1765 = = 712 − s + 40 s + 46 s + 240 ( s + 40)( s + 6) s+6
2
2
h( t) = 712(1.1765e
−40 t
3 − n −2 n 10 − 0.1765e ) u( t) ⇒ h[ n ] = 7121.1765e − 0.1765e u[ n ] −6 t
z z z z = 7121.1765 H( z) = 7121.1765 − 0.1765 3 −2 − 0.1765 − z−e z − 0.1353 z − 0.7408 10 z−e Unit step response: 1 712 s 712 0.02941 0.02941 = = 712 − + 2 s + 40 s s + 46 s + 240 ( s + 40)( s + 6) s+6 h −1 ( t) = 20.941(e −6 t − e −40 t ) u( t) H −1 ( s) =
Unit sequence response: H −1 ( z) = 712
z z z − 0.1765 1.1765 z −1 z − 0.1353 z − 0.7408
2.858 0.1565 1.1565 3.858 H −1 ( z) = 712 1.1765 z − − − 0.1765 z z −1 z − 0.1353 z − 1 z − 0.7408 0.1841z 0.5044 z 0.6796 z H −1 ( z) = 712 − + z −1 z − 0.1353 z − 0.7408
[
]
h −1[ n ] = 712 0.6796 − 0.1841(0.1353) + 0.5044 (0.7408) u[ n ] (b)
n
n
H( s) =
712 s , f s = 200 Hz s + 46 s + 240
H( s) =
712 s 712 s 1.1765 0.1765 = = 712 − s + 40 s + 46 s + 240 ( s + 40)( s + 6) s+6
2
2
h( t) = 712(1.1765e −40 t − 0.1765e −6 t ) u( t) ⇒ h[ n ] = 712(1.1765e −0.2 n − 0.1765e −0.03 n ) u[ n ] z z z z H( z) = 7121.1765 − 0.1765 −0.2 − 0.1765 −0.03 = 7121.1765 z−e z−e z − 0.8187 z − 0.9704 Unit step response:
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M. J. Roberts - 7/12/03
1 712 s 712 0.02941 0.02941 = = 712 − + 2 s + 40 s s + 46 s + 240 ( s + 40)( s + 6) s+6
H −1 ( s) =
h −1 ( t) = 20.941(e −6 t − e −40 t ) u( t) Unit sequence response: H −1 ( z) = 712
z z z − 0.1765 1.1765 z − 1 z − 0.8187 z − 0.9704
32.784 4.5157 5.5157 33.784 H −1 ( z) = 712 1.1765 z − − − 0.1765 z z −1 z −1 z − 0.8187 z − 0.9704 5.3127 z 5.7864 z 0.5263z H −1 ( z) = 712 − + z −1 z − 0.8187 z − 0.9704
[
]
h −1[ n ] = 712 0.5263 − 5.317(0.8187) + 5.7864 (0.9704 ) u[ n ] n
n
(c) Unit Step Response h (t)
(c) Unit Sequence Response h [n]
-1
-1
12
750
1
t
-5
(d) Unit Step Response h- 1(t)
30
n
(d) Unit Sequence Response h- 1[n]
12
3000
1
t
-5
200
n
28. Using the impulse-invariant and step-invariant design methods, design digital filters to approximate analog filters with these transfer functions. In each case choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the “s” plane. Graphically compare the step responses of the digital and analog filters. (a)
H( s) =
16 s s + 10 s + 250 2
CT Impulse Response:
sin(15 t) h( t) = 16e −5 t cos(15 t) − u( t) 3
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M. J. Roberts - 7/12/03
f s = 10
Sampling Rate:
250 = 25.165 ⇒ Ts = 0.03974 2π
Impulse Invariant: DT Impulse Response: sin(0.5961n ) n h[ n ] = 16(0.8198) cos(0.5961n ) − u[ n ] 3 z 2 − 1.238 z z 2 − 1.357 z + 0.6721 (Checked by hand. Correct.) H( z) = 10.75
Transfer Function:
z Transform of Step Response: H −1 ( z) = 10.75
z 3 − 1.238 z 2 z 3 − 2.357 z 2 + 2.029 z − 0.6721
−0.7538 z 1.754 z 2 − 0.5063z + 2 H −1 ( z) = 10.75 z − 1.357 z + 0.6719 z −1 0.7538 z z 2 − 0.6785 z 0.46 z + 0.8746 2 H −1 ( z) = 10.75 − + 1.754 2 z −1 z − 1.357 z + 0.6719 z − 1.357 z + 0.6719 α = 0.8197 Ω0 = 0.5957 DT Step Response:
{
}
h −1[ n ] = 10.75 −0.7538 + 1.754 (0.8197) [cos(0.5957 n ) + 0.8746 sin(0.5957 n )] u[ n ] n
Step Invariant: CT Step Response:
h −1 ( t) = 1.0667e −5 t sin(15 t) u( t)
Sampling Rate:
f s = 10
DT Step Response:
h −1[ n ] = 1.0667(0.8198) sin(0.5961n ) u[ n ]
250 = 25.165 ⇒ Ts = 0.03974 2π n
z Transform of Step Response: H −1 ( z) = 1.0667
0.8198 z z − 1.357 z + 0.6721 2
12-41
M. J. Roberts - 7/12/03
Transfer Function: H( z) = 1.0667
z −1 0.8198 z 2 z z − 1.357 z + 0.6721
H( z) = 1.0667
0.8198 z( z − 1) z − 1.357 z 2 + 0.6721z
h- 1(t)
3
Unit Step Response
0.65563
1.1922
t
Impulse-Invariant Unit Sequence Response h- 1[n] 12.2749 30 -5
n
-15.3933
Step-Invariant Unit Sequence Response h- 1[n] 0.66613
-5
(b)
30
H( s) =
s+4 s + 12 s + 32
H( s) =
s+4 1 = (s + 4)(s + 8) s + 8
n
2
Pole-zero cancellation.
CT Impulse Response:
h( t) = e −8 t u( t)
Sampling Rate:
fs =
40 = 12.732 ⇒ Ts = 0.07854 π
Impulse Invariant: DT Impulse Response:
h[ n ] = e −0.6283 n u[ n ] = (0.5335) u[ n ]
Transfer Function:
H( z) =
n
z z − 0.5335
z Transform of Step Response: H −1 ( z) =
z z 2.1436 z 1.1436 z = − z − 1 z − 0.5335 z −1 z − 0.5335
DT Step Response:
[
]
h −1[ n ] = 2.1436 − 1.1436(0.5335) u[ n ]
12-42
n
M. J. Roberts - 7/12/03
Step Invariant: CT Step Response:
h −1 ( t) =
Sampling Rate:
fs =
DT Step Response:
h −1[ n ] =
1 − e −8 t u( t) 8
40 = 12.732 ⇒ Ts = 0.07854 π 1 − (0.5335) u[ n ] 8 n
z Transform of Step Response: H −1 ( z) = 0.05831
z z − 1.533z + 0.5335 2
H( z) = 0.05831
Transfer Function: h- 1(t)
1 z − 0.5335
Unit Step Response
0.125
2.3562
t
Impulse-Invariant Unit Sequence Response h- 1[n] 2.1436
Step-Invariant Unit Sequence Response h [n]
30
n
-1
0.125
30
(c)
n
s2 + 4 0.125 2.125 1.25 = + − 2 s s+8 s+4 s( s + 12 s + 32) 1 + 17e −8 t − 10e −4 t CT Impulse Response: h( t) = u( t) 8 H( s) =
fs =
Sampling Rate: Impulse Invariant: DT Impulse Response: 12-43
40 = 12.732 ⇒ Ts = 0.07854 π
M. J. Roberts - 7/12/03
1 + 17(0.5335) − 10(0.7304 ) h[ n ] = u[ n ] 8 n
n
Transfer Function: H( z) =
z z 1 z + 17 − 10 z − 0.5335 z − 0.7304 8 z −1
z 2 − 1.9182 z + 0.9334 H( z) = z ( z − 1)( z − 0.5335)( z − 0.7304 ) z 3 − 1.9182 z 2 + 0.9334 z H( z) = 3 z − 2.2639 z 2 + 1.6536 z − 0.3897 z Transform of Step Response: H −1 ( z) =
z z 3 − 1.9182 z 2 + 0.9334 z z − 1 z 3 − 2.2639 z 2 + 1.6536 z − 0.3897
H −1 ( z) =
z 4 − 1.918 z 3 + 0.934 z 2 z 4 − 3.264 z 3 + 3.917 z 2 − 2.043z + 0.3897
H −1 ( z) =
0.04354 z 0.125 z 3.387 z 2.43z + − 2 + z −1 (z − 1) z − 0.7304 z − 0.5335
DT Step Response:
[
]
h −1[ n ] = 0.04354 + 0.125 n + 3.387(0.7304 ) − 2.43(0.5335) u[ n ] n
n
Step Invariant: Laplace Transform of Step Response: H −1 ( s) =
s2 + 4 0.125 0.04688 0.3125 0.2656 = 2 − + − 2 2 s s s+4 s+8 s ( s + 12 s + 32)
CT Step Response: t h −1 ( t) = − 0.04688 + 0.3125e −4 t − 0.2656e −8 t u( t) 8 Sampling Rate:
fs =
DT Step Response:
12-44
40 = 12.732 ⇒ Ts = 0.07854 π
M. J. Roberts - 7/12/03
h −1[ n ] = (0.009818 n − 0.04688 − 0.26563e −0.62832 n + 0.3125e −0.31416 n ) u[ n ]
(
)
h −1[ n ] = 0.009818 n − 0.04688 − 0.26563(0.5335) + 0.3125(0.7304 ) u[ n
z Transform of Step Response: H −1 ( z) = 0.009818
z z z z − 0.26563 + 0.3125 2 − 0.04688 z −1 z − 0.5335 z − 0.7304 (z − 1)
Transfer Function: H( z) = 0.009818 H( z) =
z −1 z −1 z −1 + 0.3125 2 − 0.04688 − 0.26563 z − 0.5335 z − 0.7304 (z − 1)
0.04952 z 3 − 0.1472 z 2 + 0.147 z − 0.4939 z 4 − 3.264 z 3 + 3.917 z 2 − 2.043z + 0.3897
h- 1(t)
Unit Step Response
0.24767
Impulse-Invariant Unit Sequence Response h- 1[n]
2.3562
t
3.7938
-5
Step-Invariant Unit Sequence Response h- 1[n]
30
n
0.24769
-5
30
n
29. Using the difference-equation method and all backward differences, design digital filters to approximate analog filters with these transfer functions. . In each case choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the “s” plane. Graphically compare the step responses of the digital and analog filters. (a)
H( s) =
[
s2 s2 + 3s + 2
]
h( t) = δ ( t) − 4 e −2 t + e − t u( t)
12-45
n
M. J. Roberts - 7/12/03
H( z) = 0.4673
z 2 − 2z + 1 z 2 − 1.375 z + 0.4673
s 1 2 =− + ⇒ h −1 ( t) = (2e −2 t − e − t ) u( t) s + 3s + 2 s +1 s + 2
H −1 ( s) =
2
z z 2 − 2z + 1 z( z − 1) = 0.4673 2 2 z − 1 z − 1.375 z + 0.4673 z − 1.375 z + 0.4673 (Pole-zero cancellation) H −1 ( z) = 0.4673
1.228 z 0.7609 z − z − 0.6141 z − 0.7609
H −1 ( z) =
[
]
h −1[ n ] = 1.228(0.6141) − 0.7609(0.7609) u[ n ] n
n
Unit Step Response
h (t) -1
1
3π
t
Difference-Equation Unit Sequence Response h [n] -1
0.4671
30
(b)
H( s) = h( t) =
n
s + 60 s + 120 s + 2000 2
e −100 t + e −20 t u( t) 2
H −1 ( s) = h −1 ( t) =
1 s + 60 1 6 5 1 = − − 2 s s + 120 s + 2000 200 s s + 20 s + 100
6 − 5e −20 t − e −100 t u( t) 200
f s = 159.15 ⇒ Ts = 0.0062832 H( z) = 0.0047211 H −1 ( z) = 0.00472
z 2 − 0.7263z z 2 − 1.5025 z + 0.5456
z 3 − 0.7263z 2 z 3 − 2.502 z 2 + 2.048 z − 0.5456
12-46
M. J. Roberts - 7/12/03
4.706 0.651 6.356 H −1 ( z) = 0.00472 − − z − 1 z − 0.8884 z − 0.6141
(
)
h −1[ n ] = 0.03 − 0.02221(0.8884 ) − 0.00307(0.6141) u[ n ] n
h (t)
n
Unit Step Response
-1
0.029424
0.1885
t
Difference-Equation Unit Sequence Response h- 1[n] 0.029363
-5
(c)
H( s) =
30
n
16 s s + 10 s + 250 2
sin(15 t) h( t) = 16e −5 t cos(15 t) − u( t) 3 H −1 ( s) =
16 16 15 = s + 10 s + 250 15 ( s + 5) 2 + 225 2
h −1 ( t) = 1.067e −5 t sin(15 t) u( t) H( z) = 0.3548
z( z − 1) z − 1.338 z + 0.558 2
z z( z − 1) z2 = H −1 ( z) = 0.3548 0.3548 2 z − 1 z 2 − 1.338 z + 0.558 z − 1.338 z + 0.558 z 2 − 0.669 z 0.3323z z + 2.0132 2 H −1 ( ) = 0.3548 2 z − 1.338 z + 0.558 z − 1.338 z + 0.558
α = 0.747 , Ω0 = 0.461 h −1[ n ] = 0.3548(0.747) [cos(0.461n ) + 2.013 sin(0.461n )] u[ n ] n
12-47
M. J. Roberts - 7/12/03
h- 1(t)
Unit Step Response
0.65563
1.1922
t
Difference-Equation Unit Sequence Response h- 1[n] 0.4747
30
n
30. Using the matched-z-transform method, design digital filters to approximate analog filters with these transfer functions. In each case choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the “s” plane (unless all poles or zeros are at the origin, in which case the sampling rate will not matter, in this method). Graphically compare the step responses of the digital and analog filters. s2 H( s) = 2 (a) s + 1100 s + 10 5
[
]
h( t) = δ ( t) − 1111e −1000 t + 11.111e −100 t u( t) H −1 ( s) =
1 s2 s 2 5 = 2 s s + 1100 s + 10 s + 1100 s + 10 5
H −1 ( s) =
s = (s + 100)(s + 1000)
h −1 ( t) =
1 10 1 − 9 s + 1000 s + 100
10e −1000 t − e −100 t u( t) 9
f s = 1591.55 Hz and Ts = 628.32 µs z 2 − 2z + 1 (z − 1) = H( z) = 2 z − 1.473z + 0.501 ( z − 0.9391)( z − 0.5335) 2
H −1 ( z) =
z z( z − 1) (z − 1) = z − 1 ( z − 0.9391)( z − 0.5335) ( z − 0.9391)( z − 0.5335)
H −1 ( z) =
1.1474 z 0.1474 z − z − 0.5335 z − 0.9391
[
2
]
h −1[ n ] = 1.1474 (0.5335) − 0.1474 (0.9391) u[ n ] n
n
12-48
M. J. Roberts - 7/12/03
h- 1(t)
Unit Step Response
1
0.01885
h- 1[n]
t
Matched-z Unit Sequence Response
1
-5
(b)
n
30
s2 + 100 s + 5000 H( s) = 2 s + 120 s + 2000
[
]
h( t) = δ ( t) − 62.5e −100 t + 42.5e −20 t u( t) 1 s2 + 100 s + 5000 1 20 5 17 = + − H −1 ( s) = 2 s s + 120 s + 2000 8 s s + 100 s + 20 20 + 5e −100 t − 17e −20 t h −1 ( t) = u( t) 8 H( z) =
z 2 − 1.389 z + 0.5335 z 2 − 1.415 z + 0.4705
H −1 ( z) =
z z 2 − 1.389 z + 0.5335 z 3 − 1.389 z 2 + 0.5335 z = z − 1 z 2 − 1.415 z + 0.4705 z 3 − 2.415 z 2 + 1.886 z − 0.4705
H −1 ( z) =
2.617 z 2.09 z 0.4732 z − + z −1 z − 0.8819 z − 0.5335
[
]
h −1[ n ] = 2.617 − 2.09(0.8819) + 0.4732(0.5335) u[ n ] n
h- 1(t)
n
Unit Step Response
2.451
0.1885
t
Matched-z Unit Sequence Response h- 1[n] 2.5688
-5
(c)
H( s) =
30
s +4 s( s + 12 s + 32) 2
2
12-49
n
M. J. Roberts - 7/12/03
1 + 17e −8 t − 10e −4 t h( t) = u( t) 8 1 s2 + 4 0.125 0.04688 0.3215 0.2656 = 2 − + − 2 s s( s + 12 s + 32) s s s+4 s+8
H −1 ( s) =
[
]
h −1 ( t) = 0.125 t − 0.04688 + 0.3215e −4 t − 0.2656e −8 t u( t) z 3 − 1.975 z 2 + z z 3 − 2.264 z 2 + 1.654 z − 0.3897
H( z) = H −1 ( z) =
z z 3 − 1.975 z 2 + z z 4 − 1.975 z 3 + z 2 = z − 1 z 3 − 2.264 z 2 + 1.654 z − 0.3897 z 4 − 3.264 z 3 + 3.917 z 2 − 2.043z + 0.3897
H −1 ( z) =
0.1958 z 0.7543z 4.627 z 2.873 + − 2 − z −1 z − 0.7304 z − 0.5335 (z − 1)
[
]
h −1[ n ] = 0.1958 n − 0.7543 + 4.627(0.7304 ) − 2.873(0.5335) u[ n ] n
h- 1(t)
n
Unit Step Response
0.24767
2.3562
t
Matched-z Unit Sequence Response h [n] -1
5.1201
-5
30
n
31. Using the bilinear-z-transform method, design digital filters to approximate analog filters with these transfer functions In each case choose a sampling frequency which is 10 times the magnitude of the distance of the farthest pole or zero from the origin of the “s” plane. Graphically compare the step responses of the digital and analog filters. (a)
H( s) =
s2 s2 + 100 s + 250000
{
}
h( t) = δ ( t) − 100e −50 t [cos( 497.49 t) + 4.9245 sin( 497.49 t)] u( t) H −1 ( s) =
1 s2 s s = 2 = 2 2 s s + 100 s + 250000 s + 100 s + 250000 ( s + 50) + 247500 12-50
M. J. Roberts - 7/12/03
H −1 ( s) =
s + 50 497.49 − 0.1005 2 (s + 50) + 247500 (s + 50) 2 + 247500
h −1 ( t) = e −50 t [cos( 497.49 t) − 0.1005 sin( 497.49 t)] u( t) z 2 − 2z + 1 H( z) = 0.8609 2 z − 1.552 z + 0.8918 H −1 ( z) = 0.8609
z z 2 − 2z + 1 z( z − 1) = 0.8609 2 2 z − 1 z − 1.552 z + 0.8918 z − 1.552 z + 0.8918
α = 0.9444 , Ω0 = 0.6064 z 2 − 0.776 z 0.5382 z − 0.4162 2 H −1 ( z) = 0.8609 2 z − 1.552 z + 0.8918 z − 1.552 z + 0.8918 h −1[ n ] = 0.8609(0.9444 ) [cos(0.6064 n ) − 0.4162 sin(0.6064 n )] u[ n ] n
h- 1(t)
Unit Step Response
1
0.037699
t
Bilinear-z Unit Sequence Response h [n] -1
0.8609 30
(b)
H( s) =
n
s2 + 100 s + 5000 s2 + 120 s + 2000
[
]
h( t) = δ ( t) − 62.5e −100 t + 42.5e −20 t u( t) H −1 ( s) = h −1 ( t) =
1 s2 + 100 s + 5000 = s s2 + 120 s + 2000
1 20 5 17 − + 8 s s + 100 s + 20
20 + 5e −100 t − 17e −20 t u( t) 8
H( z) = 0.9762
z 2 − 1.3942 z + 0.5392 z 2 − 1.404 z + 0.4602
12-51
M. J. Roberts - 7/12/03
z 2 − 1.3942 z + 0.5392 z z 2 − 1.3942 z + 0.5392 z . = 0 9762 z 3 − 2.404 z 2 + 1.864 z − 0.4602 z − 1 z 2 − 1.404 z + 0.4602
H −1 ( z) = 0.9762
2.5 z 1.999 z 0.4756 z − + z − 1 z − 0.8818 z − 0.5219
H −1 ( z) =
[
]
h −1[ n ] = 2.5 − 1.999(0.8818) + 0.4756(0.5219) u[ n ] n
h- 1(t)
n
Unit Step Response
2.451
0.1885
t
Bilinear-z Unit Sequence Response h [n] -1
2.4541
30
(c)
n
s2 + 4 H( s) = 2 s + 12 s + 32
[
]
h( t) = δ ( t) − 17e −8 t + 5e −4 t u( t) 1 s2 + 4 0.125 0.04688 0.3215 0.2656 = 2 − + − 2 s s( s + 12 s + 32) s s s+4 s+8
H −1 ( s) =
[
]
h −1 ( t) = 0.125 t − 0.04688 + 0.3215e −4 t − 0.2656e −8 t u( t) H( z) = 0.6617
z 2 − 1.9752 z + 1 z 2 − 1.25 z + 0.3802
z 2 − 1.9752 z + 1 z z 2 − 1.9752 z + 1 z . = 0 6617 z 3 − 2.25 z 2 + 1.631z − 0.3802 z − 1 z 2 − 1.25 z + 0.3802
H −1 ( z) = 0.6617 H −1 ( z) =
[
0.125 z 1.08 z 1.617 z − + z −1 z − 0.7285 z − 0.5219
]
h −1[ n ] = 0.125 − 1.08(0.7285) + 1.617(0.5219) u[ n ] n
12-52
n
M. J. Roberts - 7/12/03
h- 1(t)
Unit Step Response
1
2.3562
t
Bilinear-z Unit Sequence Response h- 1[n] 0.662
30
n
32. Design a digital-filter approximation to each of these ideal analog filters by sampling a truncated version of the impulse response and using the specified window. In each case choose a sampling frequency which is 10 times the highest frequency passed by the analog filter. Choose the delays and truncation times such that no more than 1% of the signal energy of the impulse response is truncated. Graphically compare the magnitude frequency responses of the digital and ideal analog filters using a dB magnitude scale versus linear frequency. % Program to generate the FIR impulse responses of some approximations to % ideal filters and plot a comparison of the magnitude frequency responses of % the approximate and ideal filters close all ; % % Bandpass, Rectangular Window % fc = [10 20] ; type = 'BP' ; fs = 10*fc(2) ; h = FIRDF(type,fc,fs,'RE',0.01) ; N = length(h) ; Ts = 1/fs ; n = [0:N-1]' ; F = [0:0.001:1/2]' ; [H,F] = DTFT(n,h,F) ; subplot(2,2,1) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/2,-140,0],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Bandpass - Rectangular Window','Times',24,'n','c') ; subplot(2,2,2) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/10,-5,0],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Bandpass - Rectangular Window','Times',24,'n','c') ; % % Bandpass, Blackman Window % h = FIRDF(type,fc,fs,'BL',0.01) ; N = length(h) ; Ts = 1/fs ; n = [0:N-1]' ; F = [0:0.001:1/2]' ; [H,F] = DTFT(n,h,F) ; subplot(2,2,3) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/2,-140,0],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Bandpass - Blackman Window','Times',24,'n','c') ;
12-53
M. J. Roberts - 7/12/03
subplot(2,2,4) ; p = xyplot(F*fs,20*log10(abs(H)),[0,fs/10,-5,0],'\itf ',... '|H(\ite^{{\itj}2{\pi}{\itf}{\itT}_s} )|',... 'Times',18,'Times',14,... 'Bandpass - Blackman Window','Times',24,'n','c') ; % Function to design a digital filter using truncation of the % impulse response to approximate the ideal impulse response. The % user specifies the type of ideal filter, % % LP lowpass % BP bandpass % % the cutoff frequency(s) fcs (a scalar for LP and HP a 2-vector % for BP and BS), the sampling rate, fs, the type of window, % % RE rectangular % VH von Hann % BA Bartlett % HA Hamming % BL Blackman % % and the allowable truncation error, % err, as a fraction of the total impulse response signal energy. % % The function returns the filter coefficients as a vector. % % function h = FIRDF(type,fcs,fs,window,err) % function h = FIRDF(type,fcs,fs,window,err) if fs == 0 | fs == inf | fs == -inf, disp('Sampling rate is unusable') ; else fs Ts = 1/fs ; type = upper(type) ; window = upper(window) ; switch type case 'LP', fc = fcs(1) ; zc1 = 1/(2*fc) ; N = 200*zc1/Ts ; n = [0:N]' ; Etotal = sum(sinc(2*fc*n*Ts).^2) E = 0 ; N = 1 ; while abs(E-Etotal) > Etotal*err, N = 2*N ; n = [0:N]' ; E = sum(sinc(2*fc*n*Ts).^2) ; end delN = floor(N/10) ; while abs(E-Etotal)1, N = N-delN ; n = [0:N]' ; E = sum(sinc(2*fc*n*Ts).^2) ; delN = floor(delN/2) ; end N = ceil(N/2)*2 ; % Make number of pts even nmid = (N/2-1) ;
12-54
M. J. Roberts - 7/12/03
w = makeWindow(window,n,N) ; h = w.*sinc(2*fc*(n-nmid)*Ts) ; h = h/sum(h) ; case 'BP' fl = fcs(1) ; fh = fcs(2) ; fmid = (fh + fl)/2 ; df = abs(fh-fl) ; zc1 = 1/df ; N = 200*zc1/Ts ; n = [0:N]' ; Etotal = sum((2*df*sinc(df*n*Ts).*... cos(2*pi*fmid*n*Ts)).^2) ; E = 0 ; N = 1 ; while abs(E-Etotal) > Etotal*err, N = 2*N ; n = [0:N]' ; E = sum((2*df*sinc(df*n*Ts).*... cos(2*pi*fmid*n*Ts)).^2) ; end delN = floor(N/10) ; while abs(E-Etotal)1, N = N-delN ; n = [0:N]' ; E = sum((2*df*sinc(df*n*Ts).*... cos(2*pi*fmid*n*Ts)).^2) ; delN = floor(delN/2) ; end N = ceil(N/2)*2 ; % Make number of pts even nmid = (N/2-1) ; n = [0:N-1]' ; w = makeWindow(window,n,N) ; h = w.*2.*df.*sinc(df*(n-nmid)*Ts).*... cos(2*pi*fmid*(n-nmid)*Ts) ; h = h/sum(h.*cos(2*pi*fmid*(n-nmid)*Ts)) ; end end function w = makeWindow(window,n,N) switch window case 'RE' w = ones(N,1) ; case 'VH' w = (1-cos(2*pi*n/(N-1)))/2 ; case 'BA' w = 2*n/(N-1).*(0<=n & n<=(N-1)/2) + ... (2-2*n/(N-1)).*((N-1)/2<=n & n
(a)
Bandpass, f low = 10 Hz , f high = 20 Hz, Rectangular window
(b)
Bandpass, f low = 10 Hz , f high = 20 Hz, Blackman window
12-55
M. J. Roberts - 7/12/03
|H(ej2πfTs )|
Bandpass - Rectangular Window |H(ej2πfTs )|
100
f
20
-140
f
-5
|H(ej2πfTs )|
Bandpass - Blackman Window |H(ej2πfTs )|
100
f
-140
20
f
-5
33. Draw a canonical-form block diagram for each of these system transfer functions. z2 H( z) = 4 2 z + 1.2 z 3 − 1.06 z 2 + 0.08 z − 0.02
(a)
z2 1 H( z) = 2 z 4 + 0.6 z 3 − 0.53z 2 + 0.04 z − 0.01 1 2 1 z
X(z)
1 z
1 z
Y(z) 1 z
0.6 0.53
0.04 0.01
(b)
H( z) =
(2z
z 2 ( z 2 + 0.8 z + 0.2)
2
+ 2 z + 1)( z 2 + 1.2 z + 0.5) H( z) =
z 4 + 0.8 z 3 + 0.2 z 2 2 z 4 + +4.4 z 3 + 4.4 z 2 + 2.2 z + 0.5 12-56
Y1 (z)
M. J. Roberts - 7/12/03
H( z) =
z 4 + 0.8 z 3 + 0.2 z 2 1 2 z 4 + 2.2 z 3 + 2.2 z 2 + 1.1z + 0.25
0.8 Y(z)
0.2 1 z
X(z)
1 z
1 z
1 z
Y1 (z)
2.2 2.2 1.1 0.25
34. Draw a cascade-form block diagram for each of these system transfer functions. (a)
z2 z H( z) = 2 + z − 0.1z − 0.12 z − 1 H( z) = z
2 z 2 − 1.1z − 0.12 z − 0.6433 z + 0.0933 z × × = 2× 2 z + 0.3 z −1 z − 0.4 (z − 0.1z − 0.12)(z − 1)
2 X(z)
1 z
0.6433
1 z
0.3
0.0933
1 z 0.4
z z −1 (b) H( z) = z z2 1+ z − 1 z2 − 1 2
12-57
Y(z)
M. J. Roberts - 7/12/03
1 z 2 + 0.7071z 1 z − 0.7071 2 z z × H( ) = = × z 1 2 z + 0.6028 z 2 − 1.103z + 0.4147 2z 3 − z 2 − + 2 2 z2 −
1 2
0.7071
X(z)
1 z
1 z
0.7071
0.6028
1 z
Y1 (z)
1.103 0.4147
35. Draw a parallel-form block diagram for each of these system transfer functions. (a)
H( z) = (1 + z −1 ) H( z) = 18
18 (z − 0.1)(z + 0.7)
z +1 257.1 247.5 9.643 =− + + z( z − 0.1)( z + 0.7) z z − 0.1 z + 0.7
X(z)
1 z
257.1
1 z
247.5
0.1
1 z
9.643
0.7
z z −1 (b) H( z) = z z2 1+ z − 1 z2 − 1 2
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Y(z)
Y(z)
M. J. Roberts - 7/12/03
1 1 0.5474 z 2 − 0.3822 z 0.04735 z 2 z z − H( ) = = 2 3 z 2 z 1 z 2 − 1.103z + 0.4147 z + 0.6028 z − − + 2 4 4 z2 −
0.04735
1 z 0.6028 0.5474
Y(z)
0.3822
1 z
X(z)
1 z
Y1 (z)
1.103 0.4147
36. Write a set of state equations and output equations corresponding to these transfer functions (which are for DT Butterworth filters). 0.06746 z 2 + 0.1349 z + 0.06746 H( z) = z 2 − 1.143z + 0.4128
(a)
0.06746 0.1349
0.06746
X(z)
1 z
Q2(z)
1 z
Q1(z)
1.143 0.4128
From the block diagram we can immediately write the state equations,
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Y(z)
M. J. Roberts - 7/12/03
1 Q1 ( z) 0 z Q1 ( z) 0 = − . + X( z) z Q 2 ( z) 0 4128 1.143Q 2 ( z) 1 The response can be written in terms of the states as Y( z) = 0.06746 z Q 2 ( z) + 0.1349 Q 2 ( z) + 0.06746 Q1 ( z) . Then using one of the state equations and substituting, Y( z) = 0.06746[−0.4128 Q1 ( z) + 1.143 Q 2 ( z) + X( z)] + 0.1349 Q 2 ( z) + 0.06746 Q1 ( z) Then, putting the output equation in standard form, Q ( z) Y( z) = [0.0396 0.212] 1 + 0.06746 X( z) . Q 2 ( z)
(b)
H( z) =
0.0201z 4 − 0.0402 z 2 + 0.0201 z 4 − 2.5494 z 3 + 3.2024 z 2 − 2.0359 z + 0.6414
0.0201
0.0402
0.0201
1 z
X(z)
Q4(z)
1 z
Q3(z)
1 z
Q2(z)
1 z
Q1(z)
2.5494
3.2024
2.0359 0.6414
We can immediately write the state equations, 1 0 0 Q1 ( z) 0 z Q1 ( z) 0 0 1 0 Q 2 ( z) 0 z Q 2 ( z) = 0 + X( z) . z Q 3 ( z) 0 0 0 1 Q 3 ( z) 0 z Q 4 ( z) −0.6414 2.0359 −3.2024 2.5494 Q 4 ( z) 1 The response is
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Y(z)
M. J. Roberts - 7/12/03
Y( z) = 0.0201Q1 ( z) − 0.0402 Q 3 ( z) + 0.0201z Q 4 ( z) which, using the state equations can be rewritten as
or or
−0.6414 Q1 ( z) + 2.0359 Q 2 ( z) Y( z) = 0.0201Q1 ( z) − 0.0402 Q 3 ( z) + 0.0201−3.2024 Q 3 ( z) + 2.5494 Q 4 ( z) + X( z) Y( z) = 0.0721Q1 ( z) − 0.0409 Q 2 ( z) − 0.1046 Q 3 ( z) + 0.0512 Q 4 ( z) + 0.0201 X( z) Q1 ( z) Q 2 ( z) + 0.0201 X( z) Y( z) = [0.0721 −0.0409 −0.1046 0.0512] Q 3 ( z ) Q 4 ( z)
37. For the system in Figure E37 write state equations and response equations.
D
x[n]
10
y[n]
2 3
D
1 5
1 2
Figure E37 A 2-state DT system Let the response of the left delay element be q1[ n ] and let the response of the right delay element be q 2 [ n ]. Then 1 1 q 2 [ n + 1] = x[ n ] − q1[ n ] + q1[ n + 1] + q 2 [ n ] 2 5 2 q1[ n + 1] = ( x[ n ] − q1[ n ]) 3 y[ n ] = 10 q 2 [ n ] Using the second equation, the first equation can be rewritten as 1 2 1 q 2 [ n + 1] = x[ n ] − q1[ n ] + ( x[ n ] − q1[ n ]) + q 2 [ n ] . 5 3 2 Then the state equations can be written as
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M. J. Roberts - 7/12/03
2 2 q1[ n + 1] − 3 0 q1[ n ] 3 = + x[ n ] q 2 [ n + 1] − 17 1 q 2 [ n ] 17 15 2 15 q1[ n ] y[ n ] = [0 10] q 2 [ n ] 38. Find the response of the system in Exercise E37 to the excitation, x[ n ] = u[ n ] . (Assume that the system is initially at rest.) From Exercise 37 the state equations are 2 2 q1[ n + 1] − 3 0 q1[ n ] 3 = + x[ n ] q 2 [ n + 1] − 17 1 q 2 [ n ] 17 15 2 15 q1[ n ] y[ n ] = [0 10] q 2 [ n ] 2 z + 3 −1 H( z) = C[ zI − A ] B + D = [0 10] 17 15
0 1 z− 2
−1
2 3 17 + 0 15
1 1 2 0 z − 2 3 z − 3 17 17 2 2 17 − z + z z 10 17 z 3 3 = 0 10 15 = 10 15 = H( z) = [0 10] 15 [ ] 1 1 15 2 1 1 1 1 17 1 1 z2 + z − z2 + z − z2 + z − z + z− 6 3 6 3 6 3 15 6 3 Y( z) =
z 10 17 z 13.6 2.59 4.857 = + − 15 z 2 + 1 z − 1 z − 1 z − 1 z + 2 z − 1 6 3 3 2
n −1 n −1 2 1 y[ n ] = 13.6 + 2.59 − − 4.857 u[ n − 1] 3 2
39. A DT system is excited by a unit sequence and the response is n −1 n −1 1 3 y[ n ] = 8 + 2 − 9 u[ n − 1] . 2 4
Write state equations and output equations for this system.
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M. J. Roberts - 7/12/03
X( z) = Y( z) =
z z −1
1 1 8 +2 −9 1 3 z −1 z− z− 2 4
The transfer function is
1 1 8 +2 −9 1 3 z −1 z− z− Y( z) 2 4 = H( z) = = z X( z) z −1
1 8 z − z − 2
1 3 3 + 2( z − 1) z − − 9( z − 1) z − 2 4 4 1 3 (z − 1) z − z − 2 4 z z −1
Simplifying, H( z) =
z . 5 3 z2 − z + 4 8
Y(z) X(z)
1 z
1 z
Y1 (z)
5 4 3 8 Let q1[ n ] be the response of the right delay element and let q 2 [ n ] be the response of the left delay element. Then q1[ n + 1] = q 2 [ n ] q 2 [ n + 1] = x[ n ] + y[ n ] = q 2 [ n ]
5 3 q 2 [ n ] − q1[ n ] . 4 8
Writing the state equations in standard form,
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M. J. Roberts - 7/12/03
q1[ n + 1] 0 = − 3 q n + 1 [ ] 2 8
q1[ n ] y[ n ] = [0 1] q 2 [ n ]
1 q [ n ] 0 5 1 + x[ n ] 1 4 q 2 [ n ]
40. Define new states which transform this set of state equations and output equations into a set of diagonalized state equations and output equations and write the new state equations and output equations. 2πn q1[ n + 1] −0.4 −0.1 −0.2 q1[ n ] 2 −0.5 0.1cos u[ n ] 16 −0.2 q 2 [ n ] + 1 0 0 n q 2 [ n + 1] = 0.3 3 q 3 [ n + 1] 1 u[ n ] −1.3q 3 [ n ] 0 0 3 4 q [n] y1[ n + 1] 1 0 −1 1 = q 2 [ n ] y 2 [ n + 1] 0 0.3 0.7 q 3 [ n ] The eigenvalue matrix is 0 0 −0.8365 + j 0.0717 0 0 . Λ= −0.8365 − j 0.0717 0 0 −0.027 The solution of the equation, ΛT = TA , for the transformation matrix, T, is 0.8908 + j 0.1086 0.1046 + j 0.0219 −0.4280 + j 0.0098 T = 0.8908 − j 0.1086 0.1046 − j 0.0219 −0.4280 − j 0.0098 . 0.1891 −0.2556 −0.9481 The new state-variable vector is
q2 [ n ] = Tq1[ n ]
and the new diagonalized state equations are 2πn q1[ n + 1] −0.8365 + j 0.0717 0 0 q1[ n ] 1.8862 + j 0.2391 −1.7294 − j 0.0248 0.1cos u[ n ] 16 −0.8365 − j 0.0717 0 0 q 2 [ n ] + 1.8862 − j 0.2391 −1.7294 + j 0.0248 n q 2 [ n + 1] = 3 u n q 3 [ n + 1] [ ] −0.027 q 3 [ n ] 0 0 −1.4592 0.6951 4 q [n] y1[ n + 1] 1.3188 + j 0.0988 1.3188 − j 0.0988 −0.4447 1 = q 2 [ n ] y 2 [ n + 1] −0.2682 + j 0.0135 −0.2682 − j 0.0135 −0.1521 n q 3 [ ]
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M. J. Roberts - 7/12/03
41. Find the response of the system described by this set of state equations and output equations. (Assume the system is initially at rest.) 1 1 u[ n ] − − q1[ n + 1] 2 q1[ n ] 2 −3 n 5 = + 1 1 3 u n 7 [ ] n n q + 1 q [ ] [ ] 4 2 0 2 10 u[ n ] q1[ n ] n y[ n ] = [ 4 −1] + [1 0] 3 u n [ ] n q 2 [ ] 4 1 z + 2 −1 H( z) = C[ zI − A ] B + D = [ 4 −1] 0
−1
2 −3 + [1 0] 7 z − 1 1 10 1 5
19 8 7 1 z − 10 − 5 2 z − 5 −3z + 10 1 1 1 0 z+ z + 2 −3 z+ 2 2 + 1 0 2 1 0 4 1 H( z) = [ 4 −1] = − + ] [ ] 2 1 7 [ ] 1 7 1 1 [ z2 − z − z − z− 5 20 5 20 71 69 71 69 −13z + 7 z − 10 −13z + 10 7 z − 10 10 H( z ) = +1 + [1 0] = 1 7 1 7 1 7 z2 − z − z2 − z − z2 − z − 5 20 5 20 5 20 71 145 2 34 13z − z + 5 z − 20 10 H( z ) = − 1 7 1 7 z2 − z − z2 − z − 5 20 5 20 The z-domain vector of excitations is z z −1 X( z) = z . z − 3 4 The response is 145 2 34 z + 5 z − 20 Y( z) = H( z) X( z) = 1 7 z2 − z − 5 20
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z 71 13z − z − 1 10 z − 1 7 3 z 2 − z − z − 5 20 4
M. J. Roberts - 7/12/03
34 145 z− 5 20 −z Y( z) = z 7 2 1 (z − 1) z − z − z − 5 20 z2 +
Y( z) = 1 +
or
71 10 3 2 1 7 z − z − 4 5 20 13z −
1.2222 3.889 2.889 31.8 23.33 4.533 + + − − + z −1 z − 0.7 z + 0.5 z − 0.75 z − 0.7 z + 0.5 Y( z) = 1 +
1.2222 31.8 27.219 1.644 − + − z −1 z − 0.75 z − 0.7 z + 0.5
n −1 n −1 n −1 3 7 1 y[ n ] = δ [ n ] + 1.2222 − 31.8 + 27.219 − 1.644 − u[ n − 1] 10 2 4
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