Linear EE313 Systems and Signals
Spring 2009
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
Complex exponentials have special property when they are input into LTI systems. Output will be same complex exponential weighted by H[z]
y[ n] = h[ n] f [ n] = h[ n] z n =
m = n
h[ m] z n m
m =
=z
h[m]z 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
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Frequency Response
For continuous-time systems, response to complex e j t H ( j ) e j t sinusoid is
frequency response cos( t ) H ( j ) cos( t + H ( j ) )
frequency response
For discrete-time systems, response to complex sinusoid is e j k H (e j ) e j k
cos( k ) H ( e j ) cos( k + H ( e j ) )
For real-valued impulse response, H(e-j ) = H*(ej )
e j k + e j k H e j e j k + H e j e j k
2 cos( k ) 2 H (e j ) cos( k )
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()
()
Response to Sampled Sinusoids
Start with a continuous-time sinusoid Sample it every T seconds (substitute t = n T) We show discrete-time sinusoid with
cos( k ) = cos( n T ) cos( n T ) cos( t )
Resulting in = T Discrete-time frequency is equal to continuoustime frequency multiplied by sampling period
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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 can take the z-transform of this difference equation
Y [ z ]( z 0.8) = X [ z ]z Y [ z] z 1 H [ z] = = = X [ z ] z 0.8 1 0.8 z 1 j
Since z = e
He
[]
j
1 1 = = j 1 0.8e 1 0.8( cos j sin )
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Example
Group real and imaginary parts
He
[]
j j
1 = (1 0.8 cos ) + j 0.8 sin 1
The absolute value (magnitude response) is
He
[ ] = (1 0.8 cos ) + j 0.8 sin
= 1
(1 0.8 cos ) 2 + ( 0.8 sin ) 2
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1 = 1.64 1.6 cos
Example
The angle (phase response) is
H e
()
j
0.8 sin = 0 tan 1 0.8 cos
1
where 0 comes from the angle of the nominator and the term after comes from the denominator of H ( e j )
Reminder: Given a complex number a + j b the absolute value and angle is given as
a + j b = a 2 +b 2 1 b a + j b = tan a
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Example
We can calculate the output of this system for a sinusoid at any frequency by substituting with the frequency of the input sinusoid.
H ( e j )
5
H ( e j )
53.13
2
3
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2
3
Discrete-time Frequency Response
As in previous example, frequency response of a discrete-time system is periodic 2
with Why? Frequency response is function of the complex j j ( + 2 m) exponential which is periodic with 2 : e = e
Absolute value of discrete-time frequency response is even and angle is odd symmetric.
Discrete-time sinusoid is symmetric around cos( ( + 2m ) n ) = cos( n + 2mn ) = cos n cos( ( x ) n ) = cos( x n ) cos( n ) sin ( x n ) sin ( n ) = cos( x n ) cos( n ) cos( + x ) = cos( x )
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Aliasing and Sampling Rate
Continuous-time sinusoid can have a frequency from 0 to infinity By sampling a continuous-time sinusoid,
cos( t )
sample t = nT
0 T < 0 < = f s 0 2f < 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.
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Discrete-time frequency unique from 0 to
cos( n T ) = cos( n )
Effect of Poles and Zeros of H[z]
The z-transform of a difference equation can be written in a general form as
H [ z ] = bn
( z z1 )( z z2 ) ( z zm ) ( z 1 )( z 2 ) ( z m )
We can think of complex number as a vector in the complex plane. Im
Since z and zi are both complex numbers the difference is again a complex number thus a vector in the complex plane.
zi z zi z
Re
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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 the distance of the pole/zero to the chosen point (frequency) on unit circle. Angle is the angle of vector with the horizontal axis. r1e j1 r2e j2 rm e jm H ( e j ) = bn d1e j1 d 2e j 2 d m e j m
Im
1
d1 d2 r...