Transcript The z-Transform

The z-Transform
The representation for a sampled function was
shown to be

x s (t ) 
 x[ n ] ( t  nT ).
n  
Taking the Laplace transform of this function we
have

X s (s) 
 x [ n ]e
n  
 snT
.
If we let z=esT, we have

X s (s) 
 x[ n ] z
n
 X ( z ).
n  
The samples in the sum are multiplied by z-n. Each
factor z-n corresponds to a factor of e-snT in Laplace
domain. Multiplying by e-snT in Laplace domain
corresponds to a delay in time domain:

e
 snT
X (s)  e
 snT
e
 st
x ( t ) dt
0


e

 s ( t  nT )
x ( t ) dt 

e
 su
x ( u  nT ) du

 L  x ( t  nT ) .
Here, the bilateral Laplace tranformation is used.
Example: Find the z-transform of the discrete-time
step function
1
x[ n ]  u [ n ]  
0
( n  0 ),
( n  0 ).
u[n]
n
1 2 3 4 5
By simply inserting this discrete-time function into
the definition of the z-transform,

X (z) 
 x[ n ] z
n
n  
we have

X (z) 
 u[n ]z
n  


z
n0
n
.
n
.
To evaluate this sum, we use the formula for an
infinite geometric series:

a
n0
n

1
1 a
.
This summation converges to the above expression
if |a|<1. A derivation of this expression is on the
following slide.
N
SN 
a
n
.
n0
S N 1  S N  a
N 1
S N [1  a ]  1  a
SN 
S 
1 a
1 a
N 1
N 1
1 a
1 0
 1  aS N .

.
1
1 a
.
.
Using the formula for the infinite geometric series
the z-transform of the unit step function is

X (z) 
z
n
n0

1
1 z
1
.
The convergence criterion is
|z
1
|  1,
or,
| z |  1.
Since z is a complex number, this last statement is
equivalent to those values of z whose magnitude is
greater than one:
2
2
zr  zi  1,
or,
2
2
z r  zi  1.
The region
2
2
zr  zi  1
corresponds to a circle in the complex plane. (zr is
like x and zi is like y.) Thus, the region
2
2
zr  zi  1
is the region exterior to the unit circle.
Im{z}
2
2
zr  zi  1
Re{z}
Im{z}
2
2
zr  zi  1
Re{z}
Example: Find the z-transform of
 1
x[ n ]   u [  n  1]  
 0
(   n  0 ).
( n  0 ).
-u[-n-1]
-3 -2 -1
n
1 2 3 4 5
In this case, we have an anticausal function. Strictly
speaking, it is impossible to generate this function,
but it may arise as a solution to a design. Such
functions correspond to nonrealizable systems.
The z-transform of this function will be a sum like the
z-transform for the step function, but the index will
be negative:
1
X (z)  
z
n
n  

  z
n 1
n
 1

 
 1
1  z

1  1 
 
 1 z



z
1 z
z
z 1
1
1 z
1
.
z


We see that the z-transform of the anticausal step
function is the same as that of the ordinary step
function. This z-transform converges for |z|<1 (look
at the last summation before the geometric series
formula is applied).
The region |z|<1 corresponds to
2
2
zr  zi  1.
Im{z}
2
2
zr  zi  1
Re{z}
Example: Find the z-transform of a discrete-time
exponential function
n

a
n
x[ n ]  a u [ n ]  
0
( n  0 ),
( n  0 ),
where |a| < 1.
anu[n]
n
1 2 3 4 5
To find the z-transform, we proceed in much the
same fashion as we did before:

X (z) 
a
n
z
n
n0


 az 
1
n0

1
1  az
1
.
n
The region of convergence is |az-1 | < 1, or |z|>a.
Notice that the regions of convergence do not
contain any of the poles of the z-transform. The
poles are where the denominator is zero. For the
unit step function z-transform, the pole was at z=1.
For anun, the pole is at z=a. Poles are marked with
an ‘x’.
Im{z}
2
2
2
zr  zi  a
2
2
zr  zi  a
2
Re{z}
2
2
zr  zi  1
Example: Find the z-transform of a discrete-time
ramp function
n
x[ n ]  r [ n ]  nu [ n ]  
0
( n  0 ),
( n  0 ).
r[n]
n
1 2 3 4 5
The first step in finding this z-transform is the same
as with other z-transforms:

X (z) 
 nz
n
.
n0
To proceed from here, we need to use a small trick:
d
dz
z 
1 n
1
 
n z
1 n 1
 nz
 n 1
.
The last expression is the term inside the sum of the
z-transform of the ramp multiplied by z. So if we
multiply the transform by z-1z, we have
X (z)  z
1

 nz
 n 1
n0
 z
1


n0
d
dz
z 
1 n
1
 z
1
dz
 z
1
 z
1
 z 
1 n
n0
d
dz
1

d
1
1
1 z
1
1  z 
1 2
1
.
The region of convergence is the same as that for
the unit step function: |z|>1.
Example: Find the z-transform of a discrete-time
impulse function
( n  0 ),
1
x[ n ]   [ n ]  
0
( n  0 ).
[n]
n
-2 -1
1
2
3
When evaluating the z-transform for this function, we
see that only the n=0 term is non-zero
X (z)  z
0
 1.
A summary of the z-transforms of the causal signals
is shown in the table on the following slide.
x[n ]
 [n ]
X (z)
1
1
u [n ]
n
a u [n ]
r [n ]
1 z
1
1
1  az
z
1
1
1  z 
1 2
There are certain general characteristics that apply
to all z-transforms. For example, if we multiply a ztransform by z-1, we achieve a delay in time-domain:
1
z X (z)  z

1
 x[ n ] z
n
n  


 x[ n ] z
 ( n 1)
n  


 x [ n  1] z
n  
n
.
(In the last step, we substituted n-1 for n.)
So,
Z { x[ n  1]} 
1
z X ( z ).
This property is perhaps the most important property
of z-transforms.
In (continuous-time) linear system theory, we
described the input/output relationship of a system
by the following diagram:
x (t )
* h (t )
L
y (t )
L
H (s)
X (s)
Y (s)
-1
We can find the output y(t) from the input x(t) by
either using convolution with the impulse response
or by multiplication by the transfer function in
Laplace domain. Does a similar diagram exist for
discrete-time functions?
x[n ]
* h[ n ]
Z
y [n ]
Z
H (z)
X (z)
Y (z)
-1
To see if such relationships exist, let’s look at the
bottom half of this diagram where we multiply X(z)
by H(z) to get Y(z).
Y ( z )  H ( z ) X ( z ).
By applying the definition of the z-transform, we can
see what the equivalent (discrete-) time relationship
would be.
Y (z)  H (z) X (z)


 h[ n ] z  x[ m ] z
n
n  



m
m  

  h[ n ] x[ m ] z
(n m )
n   m  



  h[ n  m ] x[ m ] z
n
n   m  
In the last step, we substituted n with n-m.
Since

Y (z) 
 y[ n ] z
n
,
n  
we must necessarily have

y[ n ] 
 x[ m ]h[ n  m ] .
m  
This last expression is that of a discrete-time
convolution:

y [ n ]  x[ n ] * h[ n ] 
 x[ m ]h[ n  m ] .
m  
Example: Find the discrete-time convolution of the
following two functions:
x[n]
2
1
n
-2 -1
h[n]
1
2
3
4
1
n
-2 -1
1
2
3
4
We start by evaluating the terms in the summation:
x[m] and h[n-m] for various values of n.
x[m]
2
1
m
-2 -1
h[-m]
1
2
3
4
n=0
1
m
-2 -1
1
2
3
4
h[1-m]
n=1
1
m
-2 -1
h[2-m]
1
2
3
4
n=2
1
m
-2 -1
h[3-m]
1
2
3
4
n=3
1
m
-2 -1
1
2
3
4
We then find the sum of the products of the two
functions.
x[m]
2
1
m
-2 -1
h[-m]
1
2
3
4
n=0
1
m
-2 -1
1
2
3
4
The values marked in red indicate values in both
functions whose product is not zero. Non-red values
will have a zero product. So from the red values we
have
y[ 0 ] 
 x[ m ]h [  m ]  (1)( 1)  1 .
Proceeding for n=1,2,…, we have
n=1
x[m]
2
1
m
-2 -1
h[1-m]
1
2
3
4
1
m
-2 -1
y [1] 
1
2
3
4
 x[ m ]h[1  m ]  (1)( 1)  ( 2 )(1)  3 .
n=2
x[m]
2
1
m
-2 -1
h[2-m]
1
2
3
4
1
m
-2 -1
y[ 2 ] 
1
2
3
4
 x[ m ]h[ 2  m ]  ( 2 )( 1)  (1)( 1)  3 .
x[m]
n=3
2
1
m
-2 -1
h[3-m]
1
2
3
4
1
m
-2 -1
y[3] 
1
2
3
4
 x[ m ]h[ 3  m ]  (1)( 1)  1 .
So we have the following values for y[n]:
1

3
y[ n ]  
3
1

( n  0 ),
( n  1),
( n  2 ),
( n  3 ).
The values for y[n] are zero for other values of n
(n<0, n>3).
y[n]
3
2
1
n
-2 -1
1
2
3
4
The same result could have been achieved using ztransforms:
X (z)  1  2z
1
2
z .
1
H (z)  1  z .
Y (z)  X (z)H (z)

 1 2z
 1  3z
1
1
z
2
 3z
2
1  z 
1
3
z .
Taking the inverse z-transform, we have
y [ n ]   [ n ]  3 [ n  1]  3 [ n  2 ]   [ n  3 ].
In MATLAB, we carry out the calculations for the
previous example using the conv and filt functions:
>> x = [1 2 1];
>> h = [1 1];
>> y = conv(x,h)
y =
1
3
3
1
>>X = filt(x,1)
Transfer function:
1 + 2 z^-1 + z^-2
>>H = filt(h,1)
Transfer function:
1 + z^-1
>>Y = X*H
Transfer function:
1 + 3 z^-1 + 3 z^-2 + z^-3
We can also plot the functions as follows:
>> plot(0:2,x,'or')
>> axis([-1 4 -1 4])
>> plot(0:1,h,'ob')
>> axis([-1 4 -1 4])
>> plot(0:3,y,'om')
>> axis([-1 4 -1 4])
The plots are shown on the following pages.
Input
4
3
xn
2
1
0
-1
-1
0
1
2
n
3
4
Impulse Response
4
3
hn
2
1
0
-1
-1
0
1
2
n
3
4
Output
4
3
yn
2
1
0
-1
-1
0
1
2
n
3
4