Wednesday, June 29, 2011
CHAPTER 16
Consider the circuit shown below.
P.P.16.1
s Io 4/s
10/s
Using current division, 4 10 s I o 4 s4 s s V o ( s ) 4 I o
160 s ( s 2)
2
A s
4
+ V o (s)
40 s (s
2
4s 4)
160 s ( s 2)
B s2
2
C
( s 2) 2
160 A (s 2 4s 4) B ( s 2 2s) Cs Equating coefficients : 80 4 A s0 : s1 : s2 :
A 40 0 4A 2B C 0 A B B -A -40
Hence, 0 4 A 2 B C V o ( s )
40 s
40 s2
C -80
80 ( s 2) 2
v o ( t ) 40 (1
e
-2t
2 t e -2t ) u( t ) V
The circuit in the s-domain is shown below.
P.P.16.2
1
+
75/(s+2)
V o (s)
2s
i(0)/s
2
At node o, V o
75
s 2 V o V o i (0) 0 s 1 2s 2 1 1 1 V 75 o s2 2 2 s 150 s 50 s V o ( s 2)(3s 1) ( s 2)( s 1 3)
where i(0) 0A
A s2
B s 1 3
Solving for A and B we get, A = [50(–2)]/(–2+1/3) = 300/5 = 60, B = [50(–1/3)]/[(–1/3)+2] = –150/15 = –10 V o
60 s2
10 s 1 3
Hence,
v o ( t ) 60 e -2t
10e -t 3 u( t )V
v(0) V0 is incorporated as shown below.
P.P.16.3
V(s) + I 0 /s
1/sC
R
CV 0
V
We apply KCL to the top node. I0 V 1 CV0 sCV sC V s R R
V V V where A
I0 s (sC 1 R ) V0 s 1 RC V0 s 1 RC
I0 C
CV0 sC 1 R I0 C
s (s 1 RC) A s
B s 1 RC
I 0 R ,
1 RC
V(s)
V0 s 1 RC
v( t ) ((V 0
B
I 0 R s
I 0 R ) e
-t
I0 C - 1 RC
- I 0 R
I 0 R s 1 RC I 0 R ),
t
0,
where
RC
P.P.16.4 We solve this problem the same as we did in Example 16.4 up to the point where we find V1 . Once we have V1 , all we need to do is to divide V1 by 5s to and add in the contribution from i(0)/s to find I L .
I L = V 1 /5s – i(0)/s = 7/(s(s+1)) – 6/(s(s+2)) – 1/s = 7/s – 7/(s+1) – 3/s + 3/(s+2) – 1/s = 3/s – 7/(s+1) + 3/(s+2) Which leads to i L (t) = (3 – 7e –t + 3e –2t)u(t)A
We can use the same solution as found in Example 16.5 to find iL .
P.P.16.5
All we need to do is divide each voltage by 5s and then add in the contribution from i(0). Start by letting iL = i 1 + i 2 + i 3 . I 1 = V 1 /5s – 0/s = 6/(s(s+1)) – 6/(s(s+2)) = 6/s – 6/(s+1) – 3/s + 3/(s+2) –t
or
–2t
i 1 = (3 – 6e + 3e )u(t)A
I 2 = V 2 /5s – 1/s = 2/(s(s+1)) – 2/(s(s+2)) – 1/s = 2/s – 2/(s+1) – 1/s + 1 /(s+2) –1/s –t
or
–2t
i 2 = (–2e + e )u(t)A
I 3 = V 3 /5s – 0/s = –1/(s(s+1)) + 2/(s(s+2)) = –1/s + 1/(s+1) + 1/s – 1/(s+2) –t
or
–2t
i 3 = (e – e )u(t)A
–t –2t This leads to i L (t) = i 1 + i 2 + i 3 = (3 – 7e + 3e )u(t) A
P.P.16.6
Ix
1/s
1 +
+
30/s
Vo
+
2
4I x
(a) Take out the 2 Ω and find the Thevenin equivalent circuit. V Th = Ix
1/s
1 +
30/s
+
V Th
+
4I x
Using mesh analysis we get, –30/s +1Ix +I x /s + 4I x = 0 or (1 + 1/s + 4)Ix = 30/s or Ix = 30/(5s+1) V Th = 30/s – 30/(5s+1) = (150s+30–30s)/(s(5s+1)) = 30(4s+1)/(s(5s+1)) = 24(s+0.25)/(s(s+0.2))
Ix
1/s
1
+
30/s
+
I sc
4I x
I x = (30/s)/1 = 30/s Isc = 30/s + 4(30/s)/(1/s) = 30/s + 120 = (120s+30)/s = 120(s+0.25)/s Z Th = V Th /I sc = {24(s+0.25)/(s(s+0.2))}/{120(s+0.25)/s} = 1/(5(s+0.2))
1 5( s 0.2) +
24( s 0.25)
+
s ( s 0. 2)
Vo
2
24( s 0.25) V o =
s ( s 0.2) 1
5( s 0.2) (b)
2
2
24( s 0.25) s (0.2 2 s 0.4)
2 =
24( s s( s
0.25) 0.3)
or
60(4 s
1)
s(10 s
3)
+
Initial value:
v o (0 ) = Lim sV o = 24 V s ∞
Final value:
vo (∞) = Lim sV o = 24(0+0.25)/(0+0.3) = 20 V s 0
(c)
Partial fraction expansion leads to Vo = 20/s + 4/(s+0.3) Taking the inverse Laplace transform we get, v o (t) = (20 + 4e –0.3t)u(t)V -3t If x(t ) 10e u (t ) , then X ( s)
P.P.16.7
Y ( s ) H ( s ) X ( s )
A Y ( s ) ( s 3)
20 s ( s 3)( s 6)
A s3
10 s3
B s6
-20 B Y ( s ) ( s 6) s -6 40 Y ( s )
- 20 s3
y( t ) (-20 e H(s)
s -3
40 s6
-3t
2s (s 6)
40 e -6t )u( t )
2 (s 6 6)
2
s6
12 s6
h ( t ) 2 (t ) 12 e -6t u(t ) P.P.16.8
I1
By current division, 2 1 2s I s 4 2 1 2s 0
H (s)
I1 I0
2 1 2s s 4 2 1 2s
4s 2s
2
1 12s
1
P.P.16.9
2s
(a)
Vo Vi
1 || 2 s 1 1 || 2 s
1 2 s 2s
1
1 2 s
.
2 s4
H (s)
Vo Vi
2 s
4
(b)
h ( t ) 2 e -4t u(t )
(c)
Vo (s) H (s) Vi (s)
A s Vo (s) Vo (s)
(d)
s (s 4)
1
, s 0
A s
B s4
B (s 4) Vo (s)
2
s -4
-1 2
1 1
1 2 s s 4
1
v o (t )
2
2
(1 e -4t ) u(t ) V
v i ( t ) 8 cos(2 t )
Vo (s) H (s) Vi (s)
A (s 4) Vo (s)
s -4
Vi (s)
8s s2
16s (s 4)(s
2
4)
4 A (s 4)
Bs C (s 2
4)
- 16 5
Multiplying both sides by (s 4)(s 2
4) gives 16s A (s 4) B (s 2 4s) C (s 4)
Equating coefficients : s2 :
0 AB
s1 :
16 4B C
s0 :
0 4 A 4C
Vo (s)
B -A C
16
5 C -A
16 5
16
(1) (2) (3)
1 s 1 16 1 s 1 2 2 2 2 5 s 4 s 4 5 s 4 s 4 2 s 4
v o ( t ) 3.2
e
-4t
cos(2 t ) 0.5 sin(2 t ) u( t ) V
P.P. 16.10 Consider the circuit below.
i R
R 1
i
L +
vL
-
+ C
+ _
vs
R 2
v -
i R
i C
vo
R2i
But
dv dt
(1)
v
vs
vs
v
i R
R1
Hence, R1
i C
dv dt
or
v
v R1C
i
vs
C
R1C
(2)
Also, v L
v + v L + v o =0
L
di dt
v vo
But v o = iR 2 . Hence
+ vo
i v / L vo / L
v
iR2
L L Putting (1) to (3) into the standard form
(3)
1 1 1 v v R C C 1 R 1C vs I R2 i i 0 L L v vo 0 R2 i
If we let R 1 = 1, R 2 = 2, C = ½, L = 1/5, then 2 2 2 , A B 0 , C 0 2 5 10
s 2 5
s 10 2
sI A
s 10 2 5 s 2 1 ( sI A) 2 s 12s 30
H (s) C(sI A ) 1 B
=
P.P. 16.11
0
s 10 2 2 2 s 2 0 5 s2
12s 30
20 s 2 12s 30
20 s
2
12s
30
Consider the circuit below. i 1
L
vo
2 io
i1
R 1
+ v -
C
R 2
i2
At node 1, i1
v R1
or
v
C v i 1
R1C
v
1 C
i
i1 C
(1)
This is one state equation. At node 2, io i i2
(2)
Applying KVL around the loop containing C, L, and R 2 , we get
v L i io R2 0 or
i
v
R2
io L L Substituting (2) into (3) gives v R R i 2 i 2 i2 (4) L L L vo = v (5) From (1), (3), (4), and (5), we obtain the state model as 1 1 1 0 v i1 R1C C v C
1 i L
R2 i2 R2 i 0 L L
vo 1 0 v 0 0 i1 i 0 1 i 0 1 i 2 o Substituting R 1 = 1, R 2 =2, C = ½, L = ¼ yields
v 2 2 v 2 0 i1 4 8 i 0 8 i2 i vo 1 0 v 0 0 i1 i 0 1 i 0 1 i 2 o P.P. 16.12
Let so that
x1 = y
(1)
(3)
Let
x1
y
x2
x1 y
x3
x2 y
(2) (3)
Finally, let
(4)
then (5) From (1) to (5), we obtain,
x1 y (t ) 1 0 0 x2 x3 P.P.16.13
The circuit in the s-domain is equivalent to the one shown below. Vo
+ Z
- Vo
Z
(Vo ) Z Z R || 1 sC
Thus, - 1
R 1 sRC
or
- 1 Z ,
where
R 1 sRC - (1 sRC) R
For stability,
R -1
or
From another viewpoint, Vo -(Vo ) Z
-1 R
(1 Z) Vo
Vo
0
R 1 V 0 1 sRC o (sRC R 1) Vo
0
R 1 s V 0 RC o For stability
R 1 RC
must be positive, i.e.
R 1 0
P.P.16.14 (a)
-1
or
R
Following Example 15.24, the circuit is stable when 25 0 or α > –25
(b)
For oscillation, 25 0
α = –25
or
P.P.16.15
Vo Vi
s
R R sL
1 sC
s2
s
R
L R L
1 LC
Comparing this with the given transfer function, R 1 4 20 and L LC If we select R 2 , then 2 L 500 mH 4
and
C
1 20L
1 10
100 mF
Consider the circuit shown below.
P.P.16.16
Y3 Y4 Y1
Y2
+
V in
Clearly,
V1
V2
+
Vo
0
V2
At node 1,
V1 ) Y1 (V1 Vo ) Y3 (V1 0) Y2 Vin Y1 V1 (Y1 Y2 Y3 ) Vo Y3
(Vin or
(1)
At node 2,
(V1 0) Y2 or
V1
- Y4 Y2
(0 Vo ) Y4
Vo
(2)
Substituting (2) into (1),
Vin Y1 or
If we select Y1
Vo Vin
1
R 1
Vo Vin
- Y4 Y2
Vo (Y1 Y2 - Y1Y2
Y4 (Y1 Y2 , Y2
Y3 ) Y2 Y3
sC1 , Y3 sC 2 , and -s
Y3 ) Vo Y3
Y4
C1 R 1
1 sC1 sC 2 s 2 C1C 2 R 2 R 1 1
1 R 2
, then
-s
Vo
Vin
s2
s
1 R 1C 2
1 1 1 R 2 C1 C 2 R 1 R 2 C1C 2 1
Comparing this with the given transfer function shows that 1 R 1C 2
If R 1
2,
1 1 6, R 2 C1 C 2 1
1 R 1 R 2 C1C 2
10
10 k , then C2
1 R 2 C1
1 2 10 3
0.5 mF 1
5
R 2
5C1
1 1 C1 C 6 51 6 C1 2 0.1 mF R 2 C1 C 2 5 C 2 1
R 2
1 5C1
1 (5)(0.1 10 -3 )
2 k
Therefore, C 1 = 100 µF, C 2 = 500 µF, R 2 = 2 k Ω.
