EE313 Linear Systems and Signals
Fall 2010
Frequency Response of
Discrete-Time Systems
Prof. Brian L. Evans
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
Initial conversion of content to PowerPoint
by Dr. Wade C. Schwartzkopf
Complex Exponentials
• Two-sided complex
exponential zn when
input into LTI systems
Output will be same complex
exponential weighted by H(z)
Provided that z is in region of
convergence for H(z)
yn  hn xn  hn z n


nm


h
m
z

m  
z

n
 h[m] z
m
m  
 z n H ( z)
• When we specialize the z-domain to frequency
domain, the magnitude of H(z) will control
which frequencies are attenuated or passed
23 - 2
Frequency Response for LTI Systems
• Continuous
time
• Discrete
time
H  j e j  t
e jt
cos t 
h(t )
 
H e  cos k  H e 
H e j e j k
e j k
cos k 
H  j cos t  H  j
h[k ]
j
j
• Real-valued impulse response: H(e-j ω) = H*(ej ω)
Input
e  j  k  e j  k  2 cos( k )
Output H e j   e j  k  H e j   e j  k  H * e j   e j  k  H e j   e j  k
 e
e
 H e  e
 2 H e  cos ω k  H e 
 He
j
j
 jH ( e j )  j  k
j
j
jH ( e j ) j  k
e
23 - 3
Response to Sampled Sinusoids
• Start with a continuous-time sinusoid
x(t )  cos t 
• Sample it every Ts seconds (see slide 7-6)
x[n]  x(t ) t  nT  cos n Ts 
s
• We show discrete-time sinusoid with
cos n   cos Ts n 
   Ts
• Resulting in
• Discrete-time frequency is equal to continuoustime frequency multiplied by sampling period
23 - 4
Example
• Calculate the frequency response of the system
given as a difference equation as
y[n  1]  0.8 y[n]  x[n  1]
• Assuming zero initial conditions we take the ztransform of both sides
Y ( z )z  0.8  X ( z ) z
Y ( z)
z
1
H ( z) 


X ( z ) z  0.8 1  0.8 z 1
• Since pole is inside unit circle,
 
He
j
1
1


 j
1  0.8e
1  0.8cos   j sin  
23 - 5
Example
• Group real and
imaginary parts
• Absolute value
(magnitude
response)
  1  0.8 cos 1  j0.8 sin 
1
H e  
1  0.8 cos    j 0.8 sin 
H ej 
j

a  j b  a 2 b 2
b
a  j b  tan 1  
a
• Find angle
(phase response)

1
1  0.8 cos  2  0.8 sin  2
1
1.64  1.6 cos 
 0.8 sin  
H e   0  tan 

 1  0.8 cos  
j
1
23 - 6
Magnitude and Phase Responses
• Output of system for input cos(0 k) is simply




H e j 0 cos  0 k  H e j 0
 
H e j

5
Magnitude
response
 
H e j
Phase
response

2
3

2
3

53.13

23 - 7
Discrete-time Frequency Response
• As in previous example, frequency response of
a discrete-time system is periodic with 2
Why? Frequency response is function of complex
exponential which has period of 2 : e j   e j   2  m 
• Absolute value of discrete-time frequency
response is even and angle is odd symmetric
Discrete-time sinusoid is symmetric around 
cos  2m n   cosn  2mn  cos n
cos   x n   cos x n  cosn   sin  x n sin n 
 cos x n  cosn 
 cos  x n   cos  x n
23 - 8
Aliasing and Sampling Rate
• Continuous-time sinusoid can have a frequency
from 0 to infinity
• By sampling a continuous-time sinusoid,
sample
cos t   cos n T   cos n 
t  nT
• Discrete-time frequency  unique from 0 to 
0  T    0   

 f s  0  2f  f s  0  f  f s / 2
T
We only can represent frequencies up to half of the
sampling frequency.
Higher frequencies exist would be “wrapped” to some
other frequency in the range.
23 - 9
Effect of Poles and Zeros of H(z)
• The z-transform of a difference equation can be
written in general form as
H ( z )  bn
z  z1 z  z2  z  zm 
z   1 z   2  z   m 
• Complex number as vector in complex plane
z and zi are both complex numbers
Their difference is also a complex
number (vector in complex plane)
Im
zi
z  zi
z
Re
23 - 10
Effect of Poles and Zeros of H(z)
• Each difference term in H(z) may be represented
as a complex number in polar form
Magnitude is distance of
pole/zero to chosen point
(frequency) on unit circle
Angle is angle of vector with
respect to horizontal axis
jm
j1
j2
r
e
r
e

r
e
H e j   bn 1 j1 2 j 2 m j m
d1e d 2e  d m e
Im
1
1 x
r
z2 1 2
o
2
2 x
r1r2  rm j 1 2 m 1  2  m 
 bn
e
d1d 2  d m
d1
d2
r2

z1 1
o
23 - 11
Re
Digital Filter Design
• Poles near unit circle indicate filter’s passband(s)
• Zeros on/near unit circle indicate stopband(s)
• Biquad with zeros z0 and z1, and poles p0 and p1
Transfer function
Magnitude response
|a – b| is distance
between complex
numbers a and b
H ( z)  C
H (e
j
z  z0 z  z1 
z  p0 z  p1 
e
)C
e
H ( e j )  C
Distance from point on unit
circle ej and pole location p0
j




 z0 e j  z1
j
 p0 e j  p1
e j  z0 e j  z1
e j  p0 e j  p1
23 - 12
Digital Filter Design Examples


z  z0 z  z1 
1  z0 z 1 1  z1 z 1 
C
• Transfer function H ( z)  C 
1  p0 z 1 1  p1z 1 
z  p0 z  p1 
• Poles (X) & zeros (O) in conjugate symmetric pairs
For coefficients in unfactored transfer function to be real
• Filters below have what magnitude responses?
Im(z)
O
O
Im(z)
X
X
Im(z)
O
X
Re(z)
O
X
Re(z)
X
O
X
O
Zeros are on the unit circle
lowpass
highpass
bandpass
Re(z)
bandstop
allpass
notch?
Poles have radius r
Zeros have radius 1/r
23 - 13
DSP First Demonstrations
• IIR Filters (Chapter 8)
• Three-domain demonstrations
IIR filter with one pole
IIR filter with one pole and one zero
Radial movement of poles (second-order section)
• Z-to-Freq demonstration
Figure 4: Movie takes points on unit circle to create
magnitude response (double click on plot to start
animation)
Last figure: Movie takes slice of 3-D view of the zdomain to create magnitude response (double click on
plot to start animation)
23 - 14
DSP First Demonstrations
• BIBO stable causal systems
All poles inside unit circle
• Filter amplitude response at 
Enhance it by placing a pole close to ej
Suppress it by placing a zero close to or at ej
• Poles/zeros at origin
Do not affect amplitude response
Add a phase of (–  T ) which is a pure delay
• Poles and zeros can cancel each other’s effect if
placed close together
23 - 15
Effect of Poles/Zeros
H
H
T
x

o
x

-
-/2
H
H
H
H
T
o
T
x

o
T
x
x
x

-
-
H
H
23 - 16