Transcript Document

ECE
8443
– PatternContinuous
Recognition
EE
3512
– Signals:
and Discrete
LECTURE 34: PROPERTIES OF THE Z-TRANSFORM
AND THE INVERSE Z-TRANSFORM
• Objectives:
Modulation, Summation, Convolution
Initial Value and Final Value Theorems
Inverse z-Transform by Long Division
Inverse z-Transform by Partial Fractions
Difference Equations
• Resources:
MIT 6.003: Lecture 23
Wiki: Inverse Z-Transform
CNX: Inverse Z-Transform
Arslan: The Inverse Z-Transform
CNX: Properties
ISIP: Pole/Zero Demo
URL:
Audio:
Properties of the z-Transform
ax1[n]  bx2 [n]  aX1[ z]  bX 2 [ z]
• Linearity:
x[n  n0 ]  z n0 X [ z]
dX [ z ]
nx[n]   z
dz
• Time-shift:
• Multiplication by n:
Proof: X(z) 

n
x
[
n
]
z

n  


dX ( z )
dX ( z )
 n 1
 n  x[n]z
 z
 n  x[n]z n  Znx[n]
dz
dz
n  
n  
z
a n x[n]  X  
a
• Multiplication by an:

  (a
Proof: Z a n u(n) 

n
x[n])z n
n  
• Multiplication by ejn:

z
  x[n] 
a
n  
n
z
 X 
a

e jn x[n]  X e  jn z

   
( j / 2)X e z   X e

z 
• Multiplication by cosn: cos(n) x[n]  (1/ 2) X e jn z  X e  jn z
• Multiplication by sinn: sin(n) x[n] 
• Summation:
n
v[n]   x[i]  V ( z ) 
i 0
EE 3512: Lecture 34, Slide 1
jn
1
X ( z)
1  z 1
 jn
Convolution
• Convolution: x[n] * h[n] 

 x[k ]h[n  k ]
 X ( z) H ( z)
k  

 

 

Proof: Z x[n] * h[n]  Z   x[k ]h[n  k ]     x[k ]h[n  k ]z  n
 k  
 n    k  

 

  x[k ]  h[n  k ]z  n 
k  
 n  

Change of index on the second sum: m  n  k


 
 
( m  i ) 
k )  
m 
Zx[n] * h[n]   x[k ]  h[m]z

x
[
k
]
z
h
[
m
]
z
 
 

k  
m
 k 
 m

 X ( z) H ( z)

The ROC is at least the intersection of the ROCs of x[n] and h[n], but can be a
larger region if there is pole/zero cancellation.
• The system transfer function is completely analogous to the CT case:

h[n]  H ( z ) 
 h[n]z
n
n  
• Causality: h[n]  0 n  0
Implies the ROC must be the exterior of a circle and include z = .
EE 3512: Lecture 34, Slide 2
Initial-Value and Final-Value Theorems (One-Sided ZT)
• Initial Value Theorem: x[0]  lim X ( z )

z 
Proof: lim X ( z )  lim  x[n]z n  lim x[0]  x[1]z 1  ...  x[0]
z 
z 
n 0
z 
x[n]  lim( z  1) X ( z )
• Final Value Theorem: lim
n 
z 1
• Example:
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
 2
z  z  0.5

z 1
5
 10
.5
• Tables 7.2 and 7.3 in the textbook contain a summary of the z-Transform
properties and common transform pairs.
EE 3512: Lecture 34, Slide 3
Inverse Laplace Transform
• Recall the definition of the inverse Laplace transform via contour integration:
  j
1
1
st


x(t ) 
X
s
e
ds

X ( s )e st ds


2j   j
2j C
• The inverse z-transform follows from this:
1
n 1
x[n] 
X
(
z
)
z
dz

2j C
Evaluation of this integral is beyond the scope of this course. Instead, as with
the Laplace transform, we will restrict our interest in the inverse transform to
rational forms (ratio of polynomials). We will see shortly that this is convenient
since linear constant-coefficient difference equations can be converted to
polynomials using the z-transform.
• As with the Laplace transform, there are two common approaches:
 Long Division
 Partial Fractions Expansion
• Expansion by long division Is also known as the power series expansion
approach and can be easily demonstrated by an example.
EE 3512: Lecture 34, Slide 4
Long Division
z 2 1
• Consider: X ( z )  3
z  2z  4
Solution:
z 1
z 1  0 z 2  3z 3  4 z 4
z 3  2z  4 z 2  1
z 3  2z  4 z 2  1
z 2  2  4 z 1
z 2  2  4 z 1
 3  4 z 1
 3  4 z 1
 6 z  2  12z 3
3
z 1  0 z 2  3 z 3
z 3  2z  4 z 2  1
 4 z 1  6 z  2  12z 3
 4 z 1
z 2  2  4 z 1
 8 z 3  16z  4
6 z  2  20z 3  16z  4
 3  4 z 1
 6 z  2  12z 3
3
 4 z 1  6 z  2  12z 3
 X ( z )  z 1  0 z 2  3z 3  4 z 4  ...
 x[n]  0 [n]  1 [n  1]  3 [n  3]  4 [n  4]  ...
EE 3512: Lecture 34, Slide 5
Implications of stability?
Inverse z-Transform Using MATLAB
8z 3  2 z 2  5z
• Consider: X ( z )  3
z  1.75z  .75
MATLAB:
Syms X x z
X = (8*z^3+2*z^2-5*z)/(z^3-1.75*z+.75);
x = iztrans(X)
x = 2*(1/2)^n+2*(-3/2)^n+4
Evaluate numerically:
num = [8 2 -5 0];
den = [1 0 -1.75 .75];
x = filter(num, den, [1 zeros(1,9)])
Output:
8 2 9 -2.5 14.25 -11.125 26.8125 -30.1563 55.2656
EE 3512: Lecture 34, Slide 6
Inverse z-Transform Using Partial Fractions
• Rational transforms can be factored using the same partial fractions approach
we used for the Laplace transforms.
• The partial fractions approach is preferred if we want a closed-form solution
rather than the numerical solution long division provides.
z3 1
• Example: X ( z )  3
z  z2  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:
A( z )  z 3  z 2  z  2  ( z  2)(z  0.5  j 0.866)(z  0.5  j 0.866)
c
c1
c1
X ( z ) c0



 3
z
z z  0.5  j 0.866 z  0.5  j 0.866 z  2
1
 X ( z) 
c0  
( z ) 
 0.5
z

2

 z 0
 X ( z)
z  0.5  j 0.866
c1  
 0.429  j 0.0825
 z
 z  0.5 j 0.866
 X ( z)
z  2  0.643
c3  
 z
 z 2
EE 3512: Lecture 34, Slide 7
Inverse z-Transform (Cont.)
We can compute the inverse using our table of common transforms:
c z
c1 z
c1 z
X ( z )  c0 

 3
z  0.5  j 0.866 z  0.5  j 0.866 z  2
c3
c1
c1
 c0 


1  0.5  j 0.866z 1 1  0.5  j 0.866z 1 1  2 z 1
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]
The exponential terms can be converted to a single cosine using a
magnitude/phase conversion:
p1  (0.5) 2  (0.866) 2  1
p1    tan1
0.866 4

rad
0.5
3
c1  (0.429) 2  (0.0825) 2  0.437
0.0825
 0.19 rad (10.89)
0.429
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]
c1  tan1
 c0 [n]  2 c1 p1 cos(p1 n  c1 )  c3 (2) n u[n]
 0.5 [n]  0.874cos(
EE 3512: Lecture 34, Slide 8
4
n  0.19)  0.643(2) n u[n]
3
Inverse z-Transform (Cont.)
This can be verified using MATLAB:
z3 1
X ( z)  3
z  z2  z  2
num = [1 0 0 1];
den = [1 -1 -1 -2 0];
[r, p] = residue(num, den)
r=
p=
0.6429
2.0000
0.4286 – 0.825i
-0.5000 + 0.8660i
0.4286 + 0.825i
-0.5000 – 0.8660i
-0.5000
0
The first 20 samples of the output can be computed numerically using:
num = [1 0 0 1];
den = [1 -1 -1 -2 0];
x = filter(num, den, [1 zeros(1,19)]);
• Using MATLAB as a resource for solving homework problems can greatly
reduce the time you spend doing busywork.
EE 3512: Lecture 34, Slide 9
First-Order Difference Equations
• Consider a first-order difference equation:
y[n]  ay[n  1]  bx[n]
• We can apply the time-shift property:


Y ( z)  a z 1Y ( z)  y[1]  bX ( z)
• We can solve for Y(z):
Y ( z)  
ay[1]
b

X ( z)
1
1
1  az
1  az
• The response is again a function of two things: the response due to the initial
condition and the response due to the input.
• If the initial condition is zero:
b
Y ( z)
b
Y ( z) 
X
(
z
)

H
(
z
)


X ( z ) 1  az 1
1  az 1
• Applying the inverse z-Transform:
 b 
n
h[n]  Z 1 

b
(

a
)
u[n]
1 
1  az 
• Is this system causal? Why?
• Is this system stable? Why?
• Suppose the input was a sinusoid. How would you compute the output?
EE 3512: Lecture 34, Slide 10
Example of a First-Order System
• Consider the unit-step response of this system:
1
z

1  z 1 z  a
ay[1]
b
ay[1]
b  1 
Y ( z)  

X
(
z
)



1
1
1
1 
1 
1  az
1  az
1  az
1  az  1  z 
x[n]  u[n]

X ( z) 
ay[1]z
bz 2


za
( z  a)(z  1)
• Use the (1/z) approach for the inverse transform:
V ( z)  1 
bz 2
ab /(1  a) b /(1  a)
 


z
za
z 1
 z  ( z  a)(z  1)
ay[1]z ab /(1  a) z b /(1  a) z
ay[1]z
b  az
z 
Y ( z)  







za
za
z 1
za
1  a  z  a z 1
b
y[n]  ay[1](a) n 
[a(a) n  (1) n ]
1 a
b
 ay[1](a) n 
[(a) n 1  1], n  0, 1, 2, ...
1 a
• The output consists of a DC term, an exponential term due to the I.C., and an
exponential term due to the input. Under what conditions is the output stable?
EE 3512: Lecture 34, Slide 11
Second-Order Difference Equations
• Consider a second-order difference equation:
y[n]  a1 y[n  1]  a2 y[n  2]  b0 x[n]  b1 x[n  1]
• We can apply the time-shift property:




Y ( z)  a1 z 1Y ( z)  y[1]  a2 z 2Y ( z)  z 1 y[1]  y[2]  b0 X ( z)  b1 z 1 X ( z)
• Assume x[-1] = 0 and solve for Y(z):
b0  b1 z 1
a2 y[2]  a1 y[1]  a2 y[1]z 1
Y ( z) 

X ( z)
1
2
1
2
1  a1 z  a2 z
1  a1 z  a2 z
• Multiplying z2/z2:
b0 z 2  b1 z
 a1 y[1]  a2 y[2]z 2  a2 y[1]z
Y ( z) 
 2
X ( z)
2
z  a1 z  a2
z  a1 z  a2
• Assuming the initial conditions are zero:
b0 z 2  b1 z
Y ( z)  2
z  a1 z  a2
• Note that the impulse response is of the form:
z 2  (a cos ) z
h[n]  a cosn u[n]  H ( z )  2
z  (2a cos ) z  a 2
This can be visualized as a complex pole pair with a center frequency and
bandwidth (see Java applet).

n

EE 3512: Lecture 34, Slide 12
Example of a Second-Order System
• Consider the unit-step response of this system:
1
z
x[n]  u[n]  X ( z ) 

1  z 1 z  a
y[n]  1.5 y[n  1]  0.5 y[n  2]  x[n]  x[n  1] where y[1]  2, y[2]  1
b0 z 2  b1 z
 a1 y[1]  a 2 y[2]z 2  a 2 y[1]z
Y ( z) 
 2
X ( z)
2
z  a1 z  a 2
z  a1 z  a 2
 (1.5)(2)  (0.5)(1) z 2  (0.5)(2) z
z2  z
 z 




z 2  1 . 5 z  0. 5
z 2  1 . 5 z  0 .5  z  1 
 3.5 z 2  z
z2
 2
 2
z  1.5 z  0.5 z  1.5 z  0.5
• We can further simplify this:
 2.5 z 2  z
Y ( z)  2
z  1.5 z  0.5
0.5 z
3z


z  0.5 z  1
• The inverse z-transform gives:
y[n]  0.5(0.5) n  3(1) n , n  0, 1, 2, ...
EE 3512: Lecture 34, Slide 13
[not e: ( z 2  z ) z  z 2 ( z  1)]
MATLAB:
num = [1 -1 0];
den = [1 1.5 .5];
n = 0:20;
x = ones(1, length(n));
zi = [-1.5*2-0.5*1, -0.5*2];
y = filter(num, den, x, zi);
Nth-Order Difference Equations
• Consider a general difference equation:
N
M
i 1
i 1
y[n]   ai y[n  i]   bi x[n  i]
• We can apply the time-shift property once again:
N
M
Y ( z )   ai z Y ( z )   bi z i X ( z )
i
i 1
(assuming zero initialconditions)
i 1
N

M

i 
Y ( z ) 1   ai z   X ( z )  bi z i 
 i 1

 i 1

M
Y ( z)
H ( z) 

X ( z)
b z
i 1
N
i
i
1   a i z i
b0  b1 z 1  b2 z  2  ...  bM z  M

1  a0  a1 z 1  a 2 z  2  ...  a N z  N
i 1
• We can again see the important of poles in the stability and overall frequency
response of the system. (See Java applet).
• Since the coefficients of the denominator are most often real, the transfer
function can be factored into a product of complex conjugate poles, which in
turn means the impulse response can be computed as the sum of damped
sinusoids. Why?
• The frequency response of the system can be found by setting z = ej.
EE 3512: Lecture 34, Slide 14
Transfer Functions
• In addition to our normal transfer function
components, such as summation and
multiplication, we use one important additional
component: delay.
• This is often denoted by its z-transform equivalent.
• We can illustrate this with an example (assume
initial conditions are zero):
EE 3512: Lecture 34, Slide 15
x[n]
D
x[n]
z-1
y[n]  x[n  1]
y[n]  x[n  1]
Y ( z)  z 1 X ( z)
Transfer Function Example
• Redraw using z-transform:
• Write equations for the behavior at
each of the summation nodes:
zQ1 ( z )  Q2 ( z )  X ( z )
zQ2 ( z )  Q1 ( z )  3Y ( z )
Y ( z )  2Q1 ( z )  Q2 ( z )
• Three equations and three unknowns: solve the first for Q1(z) and substitute
into the other two equations.
Q1 ( z )  z 1Q2 ( z )  z 1 X ( z )


zQ2 ( z )  z 1Q2 ( z )  z 1 X ( z )  3Y ( z )
Q2 ( z )  z  2 Q2 ( z )  z  2 X ( z )  3z 1Y ( z )


1
z  2 X ( z )  3 z 1Y ( z )
2
1 z
1
 1

2
1
1
2
1
Y ( z )  2 z 1 
z
X
(
z
)

3
z
Y
(
z
)

2
z
X
(
z
)

z
X
(
z
)

3
z
Y ( z)
2
2

1 z
1  z

Sim plify...
Y ( z)
2z  1
H ( z) 
 2
X ( z ) z  3z  5
Q2 ( z ) 

EE 3512: Lecture 34, Slide 16



Basic Interconnections of Transfer Functions
EE 3512: Lecture 34, Slide 17
Summary
• Introduced additional properties of the z-transform.
• Derived the convolution property for DT LTI systems.
• Introduced two practical ways to compute the inverse z-transform: long
division and partial fractions expansion.
• Worked examples of each and demonstrated how to solve these problems
using MATLAB.
• Demonstration: Frequency response using a Java applet that allows you to
visualize poles and zeros in the complex plane.
• Demonstrated the solution of Nth-order difference equations using the
z-transform: general response is an exponential.
• Demonstrated how to develop and decompose signal flow graphs using the
z-transform: introduced a component, the delay, which is equivalent to
differentiation in the s-plane.
EE 3512: Lecture 34, Slide 18