Chapter Five
Z-Transform and Applications
5.1 Introduction to Z-transform
As you recall, we talked first about differential equations, then difference
equations. The methods of solving difference equations was in very many
respects parallel to the methods used to solve differential equations. We then
learned about the Laplace transform, which is a useful tool for solving
differential equations and for doing system analysis on continuous-time
systems. Our development now continues to the Z-transform. This is a
transform technique used for discrete time signals and systems. As you
might expect, many of the tools and techniques that we developed using
Laplace transforms will transfer over to the Z-transform techniques. The Ztransform is simply a power series representation of a discrete-time
sequence. For example, if we have the sequence x[0]; x[1]; x[2]; x[3], the Ztransform simply multiplies each coe_cient in the sequence by a power of z
corresponding to its index. In this example
X(z) = x[0] + x[1]z-1+ x[2]z-2+ x[3]z-3:
Note that negative powers of z are used for positive time indexes.
Z-transform converts a discrete-time signal into a complex frequency-domain
representation. It is similar to the Laplace transform for continuous signals
Generally we can write the Z-transform as
X ( z )= x n z n
n
X ( z )= x [n] zn
n
n is an integer time index;
frequency
z=e
-j
is a complex number; - angular
5.2 Region of Convergence ROC
The set of values of z for which the z-transform converges
x ( n) z n
| X ( z ) |
| x(n) || z |
Give a sequence, the set of values of z for
which the z-transform converges, i.e., |X(z)|<, is called the region of
convergence.
x n r
-n
For certain choices of r the sum maybe made finite
Each value of r represents a circle of radius r
The region of convergence is made of circles
Example 1: Let xn = anun a causal sequence
X ( z )= x n z n
n
For
|a z1|<1
X ( z )=
gives that
(a z1)n
n=0
= 1a z1
|z|>|a| ROC is circle
N ( z)
z
=
z a D( z)
N(z) = 0
D(z) = 0
z = 0; zero(s).. Roots of numerator: X(z) = 0
z = a; pole(s).. Roots of denominator: X(z)
2
n 0
n 1
1 a n z n a n z n a n z n X ( z ) a nu (n 1)z n
n
-a u(-n-1) anti causal sequence
| a
z|
n 0
For convergence of X(z), we require that
| z || a | | a 1 z | 1
X ( z ) 1 (a 1 z ) n 1
n 0
x ( n) a n ,
1
z
1
1 a z z a
0 n N 1
Example 3:
1 z a
N 1
z
za
N
1 N
1 ( az )
1 az 1
N 1
N 1
n 0
n0
X ( z ) a n z n (az 1 ) n
ROC: 0 < z < doesnt include any pole
Example 2: x(n) =
x(n) ( 12 ) n u ( n) ( 13 ) n u (n)
Example 4:
Conclusion:
Z-transform exists only within the ROC!
z-transform and ROC uniquely specify the signal
poles cannot exist in the ROC; only on its boundary
Note: if the ROC contains the unit circle (|z| = 1), the system is stable
Properties of ROC
A ring or disk in the z-plane centered at the origin.
The Fourier Transform of x(n) is converge absolutely iff the ROC
includes the unit circle.
The ROC cannot include any poles
Finite Duration Sequences: The ROC is the entire z-plane except
possibly z=0 or z=.
Right sided sequences: The ROC extends outward from the outermost
finite pole in X(z) to z=.
Left sided sequences: The ROC extends inward from the innermost
nonzero pole in X(z) to z=0.
5.3 Z Transform Pairs
5.4 Z Transform Properties
1. Linearity: suppose we x1[k] and x2[k] are signals and a & b are complex
constants. Then
Z{ax1[k] + bx2[k]} = a Z{ x1[k]} + b Z{ Xx2[k]}
2. Time Advance: let x[k] be a discrete time signal with Z-transform X(z)
so,
Z{x[k + N]} = ZNX(z) ZNx[0] ZN-1x[1] - . Z2x[N 2] Zx[N 1]
N 1
X [k + N ]= ZN X ( z ) x [k ]Zk
k=0
3. Time Delay: it is useful for solving linear differential equation with
constant coefficients.
N 1
X [k N ]= Z-N X (z )+ x [ j 1]Z j
j=0
4. Multiplication by ak:
Z {ak x [k ]}=X ( z /a)
5. Multplication by k:
kx[k] = - z
dX (z)
dz
Exercise:
i. Z{ak sin(kwt)}
ii.
Z{k ak}
iii.
Z{k2 ak}
6. Convolution: suppose that x[k] and y[k] are signals with ZT of X(z) and
Y(z) so
Z{x[k]*y[k]} = X(z)Y(z)ROC is their intersection
7. Final Value theorem:
x [ k ] = lim X ( z ) ( z1)
k
lim
k
8. Initial value theorem:
x [ 0 ]=lim X ( z )
z
9. Time reversal:
X[-k] = X(1/z)
Time Shifting:
10.
x [ k k 0 ] =zk X (z)
0
11.
z 0 x [k ]=X (
Scaling in Z-domain:
z
)
z0
5.5 Inverse Z-Transform
A signal is called the inverse Z-transform of X(z) and denoted by Z-{X(z)}
For signal x[n] for
n0
Sequence
1. d[n]
2. u[n]
3. -u[-n-1]
Transform
1
z/(z-1)
z/(z-1)
4. d[n-m]
z-m
m<0
5. anu[n]
a|
6. -anu[-n-1]
a|
7. nanu[n]
a|
8. -nanu[-n-1]
a|
9. [cosw0n]u[n]
10. [sinw0n]u[n]
11. [rncosw0n]u[n]
12. [rnsinw0n]u[n]
13. anu[n] - anu[n-N]
ROC
all z
|z|>1
|z|<1
all z except 0 if m>0 or if
z/(z-a)
|z|>|
z/(z-a)
|z|<|
az/(z-a)2
|z|>|
az/(z-a)2
|z|<|
(z2-[cosw0]z)/(z2-[2cosw0]z+1)
[sinw0]z)/(z2-[2cosw0]z+1)
(z2-[rcosw0]z)/(z2-[2rcosw0]z+r2)
[rsinw0]z)/(z2-[2rcosw0]z+r2)
(zN-aN)/zN-1(z-a)
|z|>1
|z|>1
|z|>r
|z|>r
|z|>0
Inverse ZT can be determined using a Partial Fraction
Let the Z-domain Transfer Function is given by
a + a z 1 +a z2+ +a p z p
H ( z )= 0 1 1 1 2
b0 + b1 z +b1 z + + bq zq
We can follow the following steps to find the Inverse ZT
Step 1: if p q use long division to convert H(z) in to sum of polynomials
then go to step 3.
H ( z )=a0 +c 1 z 1 + ..+c p1 z (p 1 )
Next is let
Else if
p<q
R ( z) =
d 0 +d 1 z 1 + ..+d q1 z(q1)
b0 + b1 z1 + .+bq zq
d0 + d1 z1+ ..+ d q1 z(q1)
b 0 +b1 z1 + .+b q z q
go to step 2
Step 2: write R(z) only in the positive power of z i.e,
d0 z q +d 1 z q 1 + ..+d q 1 z
R ( z) =
b 0 z q +b 1 z q1 + .+b q
Step 3: Multiply both side of step 2 by
z1
and factor denominator
q1
z1 R ( z )=
d 0 z +
( za 1 )( za2 ) .( zeq 1)(z eq 2)2
Step 4: write in partial fraction form
c1
c0
A
B
z1 R ( z )=
+
+ ..
+
za 1 za 2
zeq 1 ( zeq 2)2
Then find the value of constants A, B c1...
Step 5: Substitute the constants and multiply both side by Z
c z
c0 z
Az
Bz
R ( z) =
+
+ .. 1
+
za 1 za 2
zeq 1 ( zeq 2)2
8
Step 6: put all terms in the power of
z1
c1
c0 z
A
B
+
+ ..
+
1
1
1
1a 1 z
za 2 z
zeq 1 z
( z eq 2 z1 )2
R ( z) =
So we can write the transfer function as
H ( z )=a0 +c 1 z 1 +c 2 z 2 + .+ R( z )
Step 7: Finally the inverse ZT is given by
h [ n ] =c 0 [ n ] +c 2 [ n1 ] + .+ c pq [ n( pq) ] + A e 1n u [ n ] + B e2n u [ n ] + +c 1 eq1n u [ n ] + c0
n
/(e+1) n eq1 u [ n ]
X ( z)
z3 1
z3 z2 z 2
Example 5:
A( z ) z 3 z 2 z 2 ( z 2)( z 0.5 j 0.866)( z 0.5 j 0.866)
X ( z ) c0
c1
c1
c
3
z
z z 0.5 j 0.866 z 0.5 j 0.866 z 2
X ( z)
( z )
z
c0
z 0
1
0 .5
2
X ( z)
z 0.5 j 0.866
z
c1
X ( z)
z 2
z
c3
0.429 j 0.0825
z 0.5 j 0.866
0.643
z 2
In this example, the
order of the numerator and denominator are the same. For this case, we can
use a trick of factoring X(z)/z:
c1 z
c1 z
cz
3
z 0.5 j 0.866 z 0.5 j 0.866 z 2
c1
c1
c3
c0
1
1
1 0.5 j 0.866 z
1 0.5 j 0.866 z
1 2 z 1
X ( z ) c0
x[n] c0 [n] c1 (0.5 j 0.866) n u[n] c1 ( 0.5 j 0.866) n u[n] c3 2 n u[n]
Example 6: Find the Inverse ZT
5
4( ) z1
2
a) H ( z )=
5
1( )z1
6
21 1
)z
5
H ( z )=
1 1 18 2
1+
z
z
10
25
1(
b)
c)
( ) ( )
H ( z )=
1+ z1
2
z2 z
3
( )
5.6 Initial-Value and Final-Value Theorems
x[0] lim X ( z )
z
Initial Value Theorem:
lim X ( z ) lim
z
x[n]z
n 0
lim x[0] x[1]z 1 ... x[0]
z
lim x[n] lim ( z 1) X ( z )
n
z 1
Final Value Theorem:
10
3z 2 2 z 4
3z 2 2 z 4
X (z) 3
z 2 z 2 1.5 z 0.5 ( z 1)( z 2 z 0.5)
lim x[ n] [( z 1) X ( z )] z 1
n
3z 2 2 z 4
5
10
2
z z 0.5 z 1 .5
Example:
5.7 Zero i/p and Zero state response
Example 6: use ZT to determine the ZIR and ZSR of the system
2 y [ k +2]+3 y [k +1]+ y [k ]=x [k +2]+ x [k +1]x [k ]
Assume that y [1]=2, y [2]=1 and
x [k ]
is discrete time unit step
function.
Solution: it is Advanced that is Non Causal so to make Causal let
the system will be
2 y [k ]+3 y [k1]+ y [k2]=x [k ]+x [k1]x [k2]
Z { y [ k1 ] } =z1 Y ( z ) + y [1]
Z { y [ k2 ] } =z2 Y ( z ) + z1 y [ 1 ] + y [2]
Z { x [ k 1 ] }=z1 X ( z ) + x [1]
Z { x [ k 2 ] }=z 2 X ( z ) + z1 x [ 1 ] + x [2]
x [ 1 ] =x [ 2 ] =0
Because it is unit step function from 0 to
3 { z Y ( z )+ y [ 1 ] }+ 2Y ( z )+ z Y ( z ) + z y [ 1 ] + y [2 ] = X ( z ) + z X ( z ) + z X ( z )
1
1
2
1
1+ z z
5+ 2 z
(
)
(
)
Y z=
X z
2
2
1
2
2+3 z + z
2+3 z + z
1+ z1z2 ( )
(
)
Y zs z =
X z
2+3 z2 + z2
Y zi (z )=
5+2 z1
2+3 z 1 + z2
and
X (z )=
z
z1
11
k =k 2
Then
y zs [ k ] =z {Y zs ( z ) }
y z i [ k ] =z {Y zi ( z ) }
and
Note: ZT has the same definition for Transfer function and
Frequency Response s= jw so replace
response
12
z=e
jw
for frequency
