UNIVERSITY OF SOUTH AFRICA
Assignment THREE Solutions (2015)
Question 1
1.1 By definition definition
∞
L (f (x)) =
e−st f (t)dt.
0
Therefore 1.1.1
∞
L (cos(αx)) =
e−st cos(αt)dt =
0
1.1.2
s s2
∞
L (sin(αx)) =
e−st sin(αt)dt =
0
1.1.3
+ α2 α
s2
+ α2
∞
L (cosh(αx)) =
e−st cosh(αt)dt =
0
1.1.3
∞
αx
L (10
)=
e−st 10αt dt =
0
,
,
s s2
− α2
,
1 . s − α ln10
1.2 Suppose y = y (x),
with y (0) = y 0 , y (0) = y 0 and the Laplace transform L (y ) = Y . The Laplace transforms for y , y , y (3) , y (4) and the constant C are
1.2.1
∞
L (y ) =
e−st y (t)dt = sY − y0 ,
0
1.2.2
∞
L (y ) =
0
1.2.3
e−st y (t)dt = s 2 Y − sy0 − y0 ,
∞
=
L y(3)
0
e−st y (3) dt = s 3 Y − s2 y0 − sy0 − y0 ,
1
1.2.3
∞
=
L y (4)
0
(3)
e−st y (4) dt = s 4 Y − s3 y0 − s2 y0 − sy0 − y0 .
1.2.4
∞
L (C ) =
e−st Cdt =
0
C . s
Question 2
Suppose we are given a system described by the differential equation (0) = 1 and and y (0) = 1. y − y = t, where y (0) Here t is the independent variable and y the dependent variable.
2.1 Solve Solve the problem problem using Laplace transform transforms. s. That is, 2.1.1 first apply the Laplace transform transform to the equation, with L(y) = Y , L (y − y − t) = L (y ) − L (y ) − L (t) = 0 .
That is, L (y − y − t) = p 2 Y − py0 − y0 − Y −
1 p2
= 0.
Since y0 = y 0 = 1, then That is, p2 Y − p − 1 − Y −
That is, 2
( p − 1)Y − That is, Y =
That is, Y =
( p + 1) p2 p2
1 p2
−
= 0.
1 p2
= 0.
( p + 1) p2 1 + 2 2 . 2 2 p ( p − 1) p ( p − 1) 3 1 1 − 2− . 2( p − 1) p 2(1 + p)
2.1.2 G( p) =
3 1 1 − 2 − . 2( p − 1) p 2(1 + p)
2.1.3 This implies implies y =
3et e −t − − t. 2 2
[TURN OVER]
3
EMT4806 Assignment 3 Solutions
2.2 Solve Solve the same problem using the reduction reduction of order method. Details Details on this method can b e found in chapter three of your textbook(Duffy). Let the equation (0) = 1 and and y (0) = 1, y − y = t, where y (0) be solvable through y + y = v
and v − v = t.
The solution of the latter is v = − x − 1 + Ce t .
Substituting it into the first equation, together with the initial conditions, give y =
3et e −t − − t. 2 2
2.3 The approach approach through through Laplace transforms transforms is easy and less complicate complicated. d.
[TURN OVER]
4
EMT4806 Assignment 3 Solutions
Question 3
Consider a system which is initially in a quiescent state and is described by the differential equation x(4) + 2x − x + 2x = 2u + 3u
Applying the Laplace transforms:
L x(4) + 2x − x + 2x = L (2u + 3u)
L x(4) + 2L (x ) − L (x ) + 2 L (x) = 2 L (u ) + 3 L (u)
+ 2 p2 X − + 2X = 2 sU + + 3U p4 X + − pX +
p4 + 2 p2 − p + 2 X = (2 p + 3) U
2 p + 3 X = 4 U p + 2 p2 − p + 2
G =
p4
2 p + 3 + 2 p2 − p + 2
Question 4
Consider the periodic function f ( t) in the graph, where f 1 be the function which agrees with f on [0, 4] , and is zero elsewhere:
4.1 Determinini Determinining ng f 1 in terms of Heaviside unit step functions:
()= −
f 1 t
Hence,
t, t < 2 , t, t > 2
.
f 1 (t) = t [H (t + 2) − H (t − 2)].
[TURN OVER]
5
EMT4806 Assignment 3 Solutions
4.2 Finding Finding the Laplace transform transform F 1 ( p) = L {f 1 (t)}:
L[f 1 (t)] = L [t[H (t + 2) − H (t − 2)]] = L [t[H (t + 2)]] − L[t[H (t − 2)]]
That is, e2s − e−2s L[f 1 (t)] = L [t[H (t + 2) − H (t − 2)]] = s2
4.3 Using the answer answer to 4.2 to compute compute F ( p) = L {f ( ( t)}:
e2s − e−2s e 6s − e−6s e10s − e−10s + + + · · · F ( p) = s2 s2 s2
Question 5
L
˙ ˙
x1 x2
= −7010 = −7010 L
sX 1 sX 2
0 1 + 400 40 0 − 0 + 400 0 −40 x1
L
x2
.
1
X 1
s
.
X 2
or
0 0 = −7010 s
X 1
s
X 2
0 + 400 0 −40 1
X 1
s
X 2
,
or
s + 70
−10 s
0 = 400 + 40 0 1
X 1
s
,
X 2
or
40 + 1 = 2800 + 110 + 10 X 1 X 2
s
s
s2
0 0 400 70 + 1 s
s
,
or [TURN OVER]
6
1 = 2800 + 110 + X 1
s
X 2
s2
EMT4806 Assignment 3 Solutions
40+s s
400
,
10 s
or
= X 1 X 2
400(40+ s) s(2800+110 s+s2 ) 4000 s(2800+110 s+s2 )
,
or
=
40 7s 10 7s
=
40 7 10 7
X 1 X 2
− −
40 7(70+s) 10 40 + 21(70+ s) 3(40+s)
,
40 −70t e 7 10 −40t 3 e
.
or
x1 x2
− −
+
40 −70t 21 e
Question 6
Suppose we are given a system initially in a quiescent state (so all initial conditions are 0) described by the difference equation 6yk+2 − yk+1 − yk = 6uk+1 + 12 uk .
6.1 Applying Applying the Z transforms: transforms:
6Z [yk+2 ] − Z [yk+1 ] − Z [yk ] = 6Z [uk+1 ] + 12Z [uk ],
6z (Z [yk+1 ] − y1 ) − z (Z [( [(yk ] − y0 ) − Z [yk ] = 6z (Z [uk ] − u0 ) + 12Z [uk ],
6z (z {Z [yk ] − y0 } − y1 ) − z (Z [( [(yk ] − y0 ) − Z [yk ] = 6z (Z [uk ] − u0 ) + 12Z [uk ],
6z (zZ }) − zZ − = 6zU + 12U, − Z =
6
( 6z + 12) U. z 2 − z − 1 Z = (6
[TURN OVER]
7
Transfer equation G =
EMT4806 Assignment 3 Solutions
6z + 12 U = 2 . 6z − z − 1 Z
The singularities follow from 6z 2 − z − 1 = 0 , giving
1 3
z1 = − , z1 =
1 . 2
Both fall within the unit circle, therefore the system is stable.
6
−1
−1
xk yk zk
= 12 6 uk vk
Question 7
Solving the difference equation yn+2 + 4 yn+1 + 3 yn = u n+1 − 2un
if the initial conditions are given as y0 = 0, y1 = 0, and the input is u k = 5 for all k : The Z transform transform generates z 2 + 4z + 3 Y = (z − 2) Z [un ].
That is, z − 2 z − 2 Y = 2 Z [un ] = 2 Z [un ] = z + 4z + 3 z + 4z + 3
Since Z [un ] =
then Y = −
Reversing the Z transforms:
5 3 − 2(3 + z ) 2(1 + z )
Z [un ].
5z , z − 1
5z 15z 25z + − . 8(−1 + z ) 4(1 + z ) 8(3 + z )
5 8
yn = − +
15 25 (−1)n − (−3)n . 4 8
Question 8
[TURN OVER]
8
EMT4806 Assignment 3 Solutions
Solving the following state–space equation by taking a Z –transform –transform and using an inverse matrix, given that x1 (0) = x 2 (0) = 0 and uk = 2:
( + 1) ( + 1)
x1 k
−0
=
x2 k
1 ( ) 1 + ( ) 0 − ( ) − 2] ( ) x1 k
1 8
3 4
uk
x1 k
= [1
yk
x2 k
x2 k
After application of the Z- transforms, the first expression can be expressed in the form
z
−1 + X 1
1 8
z
3 4
X 2
=
10
U
Since the inverse of the matrix
z 1 8
z
−1 + 3 4
is 1 z (z + 34 ) +
1 8
1
z + 34 − 18
z
,
we get
10 01 X 1
1
=
X 2
z (z + 34 ) +
1 8
1 10
z + 34 − 18
z
U.
Hence,
X 1
=
X 2
1 z (z +
3 1 4) + 8
z + 34 − 18
2z . z − 1
That is,
X 1 X 2
=
−
56z 15(−1+z ) 2z 15(−1+z )
+
−
16z 128z 3(1+2z ) − 5(1+4z ) 4z 16z + 5(1+4 z) 3(1+2z )
.
Taking the anti Z- transforms:
x1 x2
=
− − − + − − − + − 56 15
16 6 2 4 15 6
1 n 2 1 n 2
128 20 16 20
1 4 1 4
n n
.
[TURN OVER]
9
EMT4806 Assignment 3 Solutions
Finally, from the second expression, the solution follows:
yk
= [1
+ − − − − 2] + − − − − 1 n 2 1 n 2
56 15
16 6 2 4 15 6
128 20 16 20
1 4 1 4
That is, yk
=
1
4+4 −
2
n
1
− 8 −
c UNISA 2015
4
n
.
n n
.
