Signals and Systems (EKT230)
Tutorial IV
1. Compute the Fourier transform of each of the following signals :
(a) x[n] = u[n-2] – u[n-6]
5
X (e j )   x[n]e  jn
n2
 e  j 2  e  j 3  e  j 4  e  j 5
 e 2 j  e 3 j  e 4 j  e 5 j
1
(b) x[n] = ( )  n u[n  1]
2
X ( e j ) 

 x[n]e
 jn
n  

1
1
 (2)
n
e  jn
n  

1
  ( ) n e jn
n 1 2

1
1
 ( e j )  ( e j ) n
2
n 0 2
1
1
 ( e j )(
)
1 j
2
1 ( e )
2
j
e
(
)
2  e j
(c) x[n] = 2n sin(
X (e j ) 

n)u[n]
4

2
n
sin(
n  

0
2
n
sin(
n  

  2  n sin(
n 0

4

4

4
n)u[n]e  jn
n ) e  j n
(n))e jn
e
 e j / 4 n  jn
e
  2 n 
2
j
n 0




1 
1
1


  ( ) n e  j / 4 n e jn   ( ) n e j / 4 n e jn 
2 j  n 0 2
n 0 2


 j / 4 n
-2-
Tutorial IV




1 
1
1



2 j   1  j ( / 4 ) 
 1 j (  / 4  )  
 1  e

1  e

2

 2
(d) x[n] = a|n|
X ( e j ) 

 x[n]e
 jn
n  
1

 a  n e  jn   a n e  j n

n  
n 0


  (ae j ) n   (ae  j ) n
m 1
n 0

1
1  ae  j
m 0
1
1
 (ae j )(
)
j
1  ae
1  ae  j
(ae j )(1  ae  j )  (1  ae j )

(1  ae j )(1  ae  j )
 (ae j )  (ae j ) n 
(ae j )(1  ae  j )  (1  ae j )
(1  ae j )(1  ae  j )
1 a2

e j  e  j
1  2a (
)  a2
2
1 a2

1  2a cos   a 2

2. The following are the Fourier Transform of discrete-time signals. Determine the signal
corresponding to each transform.
(a)
X(ejω) =
1,
0,

3
4

|  |
3
4
4
3

|  |  ,0 |  |
4
4


4
0
3

4
4
-3-
x[n] 
1
2
 X (e
j
Tutorial IV
)e jn d
2
 / 4
 3 / 4 jn

  e d   e jn d 


3 / 4
  /4

3

/
4
 / 4

1  e j (3 / 4) n
e j (3 / 4) n




2
jn  / 4
jn 3 / 4 


1   e j (3 / 4) n e j ( / 4 ) n   e  j ( / 4) n e  j (3 / 4) n





2   jn
jn  
jn
jn
1

2
  e j (3 / 4 ) n e  j (3 / 4) n   e j ( / 4 ) n


 2 j 
2 j   2 j

1 
3
 

n)  sin( n) 
 sin(
n 
4
4 

(b)
1
n
1
1  ( )e  j
3
X (e j ) 
1
1
1  e  j  e  2 j
4
8

A
B
A
B

1  j
1  j
(1  e ) (1  e )
2
4
1  13 (2)
2

1
9
(1  (2))
4
1  13 (4)
7

1
9
(1  (4))
2
X ( e j ) 
2/9
7/9

1
1
(1  e  j ) (1  e  j )
2
4
Take inverse Fourier transform,
n
n

7 1 
2  1 

x[n]         u[n]
9 4 

9  2 


 


e  j ( / 4 ) n  


2 j  
-4Tutorial IV
3. A signal x(t) has the indicated Laplace
transform X(s). Plot the poles and zeros in the
s-plane and determine the Fourier Transform of x(t) without inverting X(s).
s2 1
(a) X ( s)  2
s  5s  6
(b) X ( s) 
s2 1
s2  s 1
-5-
Tutorial IV
4. Use the basic Laplace transforms and the Laplace transform properties given in Tables to
determine the unilateral Laplace transform of the following signals :
d
(a) x(t )  {te t u (t )}
dt
(b) x(t )  tu(t ) * cos( 2t )u (t )
(c) x(t )  u (t  1) * e 2t u (t  1)
-6-
(d) x(t )  t
Tutorial IV
d t
{e cos(t )u (t )}
dt
5. Use the basic Laplace transforms and the Laplace transform properties given in Tables
to determine the time signals corresponding to the following unilateral Laplace
transforms:
(a) X ( s )  (
1
1
)(
)
s2 s3
(b) X ( s)  e  2 s
(c) X ( s ) 
d
1
(
)
ds ( s  1) 2
1
(2 s  1) 2  4
-7-
Tutorial IV
6. Use the method of partial fractions to find the time signals corresponding to the following
unilateral Laplace transforms :
(a) X ( s ) 
s3
s  3s  2
(b) X ( s ) 
5s  4
s  3s 2  2 s
(c) X ( s) 
s2  3
( s  2)( s 2  2s  1)
2
3
7. Use Laplace transform to determine the transfer function and impulse response of the system.
-8(a)
d
y (t )  10 y (t )  10 x(t )
dt
(b)
d2
d
d
y (t )  y (t )  2 y (t )  4 x(t )  5 x(t )
2
dt
dt
dt
Tutorial IV
8. Use the tables of z-transforms and the z-transform properties to determine the z-transforms of
the following signals :
1
(a) x[n]  ( ) n u[n] * 2 n u[n  1]
2
1
1
(b) x[n]  n( ) n u[n] * ( ) n u[n  2]
2
4
(c) x[n]  u[n]
-9-
Tutorial IV
9. Use the method of partial fractions to obtain the time-domain signals corresponding to the
following z-transforms:
7
1  z 1
1
6
(a) X ( z ) 
, | z |
1
1
2
(1  z 1 )(1  z 1 )
2
3
1
(b) X ( z ) 
(1 
7 1
z
6
1 1
1
z )(1  z 1 )
2
3
, | z |
1
3
- 10 -
Tutorial IV
1
z
4
(c) X ( z )  2
, | z | 4
z  16
3z 2 
(d) X ( z ) 
2z 4  2z 3  2z 2
, | z | 1
z 2 1
10. Determine the impulse response corresponding to the following transfer functions if (i) the
system is stable or (ii) the system is causal:
3
2  ( ) z 1
2
(a) H ( z ) 
1
(1  2 z 1 )(1  z 1 )
2
- 11 -
(b) H ( z ) 
Tutorial IV
4z
1
1
z2  z 
4
16
11. Determine (i) transfer function and (ii) difference equation representation of the causal
systems described by the following difference equations:
1
(a) y[n]  y[n  1]  2 x[n  1]
2
- 12 -
Tutorial IV
(b) y[n] = x[n] –x[n-2] + x[n-4] – x[n-6]
12. Determine the zero-input response, zero-state response and the complete response of the
systems by using the unilateral z-transform. Given are the inputs and the initial conditions of
the following difference equations.
(a) y[n] 
1
1
y[n  1]  2 x[n], y[1]  1, x[n]  ( ) n u[n]
3
2
Take z-transform of X(z) and Y(z),
1
1
1  z 1
2
1
Y ( z )  ( z 1Y ( z )  1)  2 X ( z )
3
X ( z) 
i) For zero-input response(ZIR),
- 13 -
Tutorial IV
1
Y ( z )  ( z 1Y ( z )  1)  0
3
1
1
Y ( z )  z 1Y ( z )   0
3
3
1
1
Y ( z )(1  z 1 ) 
3
3
1
3
Y ( z ) ZIR 
1
1  z 1
3
Take inverse z-transform,
ZIR is
n
y[n] ZIR
11
   u[n]
33
ii) For zero-state response(ZSR),
1
Y ( z )  ( z 1Y ( z ))  2 X ( z )
3
1
1
Y ( z )(1  z 1 )  2(
)
1
3
1
1 z
2
1
1
Y ( z ) ZSR  2(
)(
)
1 1
1 1
1 z
(1  z )
2
3
By using partial fraction method,
A
B

1
1
1  z 1 1  z 1
2
3
2
2
A

1
5
1  (2)
3
3
6
A
5
Y ( z) 
6
Y ( z ) ZSR
4
5
5


1 1
1 1
1 z
1 z
2
3
2
2

1
5
1  (3)
2
2
4
B
5
B
- 14 -
Take inverse z-transform,
6  1 n 4  1 n 
y[n] ZSR         u[n]
5  3  
 5  2 
1
y[n  2]  x[n  1], y[1]  1, y[2]  0, x[n]  2u[n]
9
Take z-transform of X(z) and Y(z),
(b) y[n] 
2
1  z 1
1
Y ( z )  ( z  2Y ( z )  z 1 )  z 1 X ( z )
9
X ( z) 
i) For zero-input response(ZIR),
1
Y ( z )  ( z  2Y ( z )  z 1 )  0
9
1
1
Y ( z )  z  2Y ( z )  z 1
9
9
1 2
1 1
Y ( z )(1  z )  z
9
9
1 1
z
9
Y ( z ) ZIR 
1
1  z 2
9
Use partial fraction method
Tutorial IV
- 15 Az
Bz

1
1
z
z
3
3
Y ( z)
z
A
B



1
1
1
1
z
z ( z  )( z  ) z 
z
3
3
3
3
1
1
1
( z  )Y ( z )
(z  )
B( z  )
3
3
3

 A
1
1
1
z
( z  )( z  )
z
3
3
3
Y ( z) 
A
1
1
(z  )
3

z
1
3
1
B
(z 
1
3
3
2

z 
3
2
1
3
(1 / 9)(3 / 2) (1 / 9)(3 / 2)

1
1
z
z
3
3




1
1
1



1 1 
6  1 1
1 z 
1 z
3
3


Y ( z ) ZIR 
Y ( z ) ZIR
Take the inverse z-transform of Y(z)ZIR
1  1   1 
      
6  3   3 
n
y[n] ZIR
n

u[n]

ii) For zero-state response(ZSR),
Tutorial IV
- 16 1
Y ( z )  ( z  2Y ( z ))  z 1 X ( z )
9
1
2 z 1
Y ( z )(1  z  2 ) 
9
1  z 1
2 z 1
Y ( z ) ZSR  (
)
1 2
1
(1  z )(1  z )
9
Use partial fraction method,
Az
Bz
Cz


1
1 z 1
z
z
3
3
3
A
2
3
B
4
9
C
4
9/4
3/ 4
3/ 2
Y ( z ) ZIR 


1
1
1
1 z
1  z 1 1  z 1
3
3
Y ( z) 
Take inverse z-transform,
9 3  1 n 3  1 n 
y[n] ZSR          u[n]
2  3  
 4 4  3 
Tutorial IV
Signals and Systems (EKT230)
Tutorial IV
Appendix 1
Lampiran 1
FOURIER TRANSFORM
Signal
Transform
LAPLACE TRANSFORM
Signal
Transform
1
u(t)
s
1
tu(t)
s2
Z-TRANSFORM
Signal
Transform
 [n]
(t)
1
1
2()
u(t)
1
  ( )
j
(t - )
e-s
 n u[n]
e-atu(t)
1
a  j
e-atu(t)
1
sa
n n u[n]
1
( s  a) 2
[cos(1n)]u[n]
1  z 1 cos 1
1  z 1 2 cos 1  z 2
1
u[n]
1
1
1  z 1
1
1  z  1
z 1
(1  z 1 ) 2
te-atu(t)
a  j 2
te-atu(t)
e-a|t|
2a
2
a 2
[cos(1t )]u(t )
s
2
s  12
[sin( 1n)]u[n]
z 1 sin 1
1  z 1 2 cos 1  z 2
[sin( 1t )]u(t )
1
2
s  12
[r n cos(1n)]u[n]
1  z 1 r cos 1
1  z 1 2r cos 1  r 2 z 2
sa
( s  a ) 2  12
[r sin( 1n)]u[n]
z 1 r sin 1
1  z 1 2r cos 1  r 2 z  2
1
2
e t
2
/2
e
 2 / 2
[e
 at
cos(1t )]u(t )
[e  at sin( 1t )]u (t )
1
( s  a) 2  12
n