DSP3108 DSP Applications
04: Z-Transform, LTI, Causality, BIBO
As there are quite a number of complex issues in this area, the following summary emphasises the important
concepts:
The Z-Transform is defined:
x( n )  X  z  

 x( n ) z
n
n 
z  esT
T
e
jT
 e j
1
fs
N 1
In practice X  z    x( n) z  n 1 sided transform  is used
n 0
The Unit Impulse
1, n  0
 ( n)  
0, n  0

  z     ( n) z  n  1  z 0 0  z 1  0  z 2  .......  1
n 0
The Unit Step
1, n  0
u( n )  
0, n  0

U  z    u( n) z  n  1  z 0 1  z 1  1  z 2  .......
n 0
 1  z 1  z 2  z 3  ....... 
z
z 1
Let S  1  r  r 2  r 3  .......
rS  r  r 2  r 3  .......
S  rS  1  r  r 2  r 3  .......  r  r 2  r 3  .......  1
S 1  r   1
S
1
1 r
U  z   1  z 1  z 2  z 3  ....... 
1
1
z

1
1 z
z 1
DSP3108 DSP Applications
Delay of a Signal
Let y  n   x(n  k )  Y  z   z  k X  z 
The Unit Delay
Let x  n    (n  k )  X  z   z k
Table of Z-Transforms
Sequence
Z-Transform
1
1
2
3
4
n
5
6
n
7
8
n
9
10
11
12
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DSP3108 DSP Applications
Linear time invariance (LTI)
If an input to a system is x1(n) and its output is y1(n) then it is linear if:
x1 ( n )  y1 ( n )
a2 x2 ( n )  y2 ( n)
a1 x1 ( n )  a2 x2 ( n )  a1 y1 ( n )  a2 y2 ( n )
The system is time invariant if
x( n  k )  y ( n  k )
A system is LTI if both properties are met.
Bounded input bounded output (BIBO)
A system is BIBO if a bounded input (limited) produces a bounded output.
Causality
A system is causal if the output is zero for n<0.
Problems:
Find the z-transform of the following sequences:
x( n )   ( n )
x( n )  4 ( n  1)
x( n )  8 ( n  2)  4 ( n  3)  2 ( n  3)
x( n )  u( n )  u( n  6)
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DSP3108 DSP Applications
A more advanced treatment
The Z-Transform is defined:
x( n )  X  z  

 x( n ) z
n
n 
ze e
1
T
fs
sT
jT
 e j
N 1
In practice X  z    x( n ) z  n 1 sided transform  is used
n 0
You can say that this is just a definition and take it as verbatim. If you
have a problem with the comprehension, then just accept the definition.
However, when you go back to the Laplace transform, multiplying a signal
by exp(-sT) delays the signal by T. i.e.

x(t )  X  s    x(t )exp   st dt
0

x(t  T )   x(t  T )exp   st dt
0


 x(u)exp   s  u  T  du
T


T
0
 exp   sT   x(u )exp   su du  exp   sT   x(u )exp   su du  exp   sT  X  s 
This is why we are defining the z transform in this way as it effectively creates delays. Note that the integral
lower limit can be set to 0 since the signal is delayed to T anyway.
The Unit Impulse
This signal only exists at n=0.
1, n  0
 ( n)  
0, n  0

  z     ( n) z  n  1  z 0 0  z 1  0  z 2  .......  1
n 0
The Unit Step
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DSP3108 DSP Applications
This signal is 1 for all n≥0.
1, n  0
u( n )  
0, n  0

U  z    u( n) z  n  1  z 0 1  z 1  1  z 2  .......
n 0
1
 1  z  z 2  z 3  .......
Consider the arithmetic series:
S  1  r  r 2  r 3  .......
rS  r  r 2  r 3  .......
S  rS  1  r  r 2  r 3  .......  r  r 2  r 3  .......  1
S 1  r   1
S
1
1 r
Therefore we can express U(z) un closed form by letting r=z-1:
1, n  0
u( n )  
0, n  0

  u( n) z  n  1  z 0 1  z 1  1  z 2  .......
n 0
U  z   1  z 1  z 2  z 3  ....... 
1
z

1
1 z
z 1
The Unit Delay
1, n  k
Let x  n    ( n  k )  
0, n  k

X  z    (n  k ) z  n
n 0
 0  z  0  z 1  0  z 2  .......  1  z  k  .......  z  k
0
Delay of a Signal
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DSP3108 DSP Applications
Let y  n   x( n  k )

Y  z    x( n  k ) z  n
n 0
Let u  n  k , or n  u  k

Y  z    x(u ) z
uk
 u  k 


uk
u 0
z  k  x(u ) z  u  z  k  x(u ) z  u
 z X  z
k
Note that you can sum from u=0 since the signal is delayed until k anyway.
Also the change of variable from n to u in the final sum has no effect on
the result. It is still the standard z transform. So the result is quite simple.
Any time a signal is delayed, you simply multiply its z-transform by z to the
power of minus that delay.
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DSP3108 DSP Applications
Table of Z-Transforms
Sequence
Z-Transform
1
1
2
3
4
n
5
6
n
7
8
n
9
10
11
12
Consider one of the trigonometric examples.
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DSP3108 DSP Applications
Let y  n   b n cos  na 

Y  z    b n cos  na  z  n
n 0


1
b n exp  jna   exp   jna  z  n

2 n 0

1 
1 
1 n
1 n




b
exp
ja
z

b
exp

ja
z






 2 
2 n 0 
n 0
1  n 1  n
r1   r2 , r1  b exp  ja  z 1 , r2  b exp   ja  z 1

2 n 0
2 n 0
1
1
1
1




1
2 1  r 1 2 1  r2  2 1  b exp  ja  z  2 1  b exp   ja  z 1 



2 1  b exp   ja  z 1  1  b exp  ja  z 1 
4 1  b exp  ja  z 1  1  b exp   ja  z 1 

2  b  exp  ja   exp   ja   z 1
2 1  b  exp  ja   exp   ja   z 1  b 2 z 2
2  2b cos  a  z 1
2 1  2b cos  a  z 1  b 2 z 2 

z
z 2  bz cos  a 
2
 2bz cos  a   b 2 


z

1  b cos  a  z 1
1  2b cos  a  z
1
 b 2 z 2 
z  z  b cos  a  
2
 2bz cos  a   b 2 
All of the other trigonometric examples can be derived in a similar way. Also note:
b n cos  na  
z
z  z  b cos  a  
2
 2bz cos  a   b2 
Let b  1:
cos  na  
z
z  z  cos  a  
2
 2 z cos  a   1
Note: It is not expected that the trigonometric definitions would have to be derived. The above example is
shown just for reference purposes.
Linear time invariance (LTI)
If an input to a system is x1(n) and its output is y1(n) then it is linear if:
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DSP3108 DSP Applications
x1 ( n )  y1 ( n )
a2 x2 ( n )  y2 ( n)
a1 x1 ( n )  a2 x2 ( n )  a1 y1 ( n )  a2 y2 ( n )
The system is time invariant if
x( n  k )  y ( n  k )
A system is LTI if both properties are met.
Bounded input bounded output (BIBO)
A system is BIBO if a bounded input (limited) produces a bounded output.
Causality
A system is causal if the output is zero for n<0.
Problem:
Find the z-transform of the following sequences:
x( n )   ( n )
x( n )  4 ( n  1)
x( n )  8 ( n  2)  4 ( n  3)  2 ( n  3)
x( n )  u( n )  u( n  6)
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