EECE 301
Signals & Systems
Prof. Mark Fowler
Note Set #35
• D-T Systems: Z-Transform … Stability of Systems, Frequency Response
• Reading Assignment: Section 7.5 of Kamen and Heck
1/14
Course Flow Diagram
The arrows here show conceptual flow between ideas. Note the parallel structure between
the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis).
New Signal
Models
Ch. 1 Intro
C-T Signal Model
Functions on Real Line
System Properties
LTI
Causal
Etc
D-T Signal Model
Functions on Integers
New Signal
Model
Powerful
Analysis Tool
Ch. 3: CT Fourier
Signal Models
Ch. 5: CT Fourier
System Models
Ch. 6 & 8: Laplace
Models for CT
Signals & Systems
Fourier Series
Periodic Signals
Fourier Transform (CTFT)
Non-Periodic Signals
Frequency Response
Based on Fourier Transform
Transfer Function
New System Model
New System Model
Ch. 2 Diff Eqs
C-T System Model
Differential Equations
D-T Signal Model
Difference Equations
Ch. 2 Convolution
Zero-State Response
C-T System Model
Convolution Integral
Zero-Input Response
Characteristic Eq.
D-T System Model
Convolution Sum
Ch. 4: DT Fourier
Signal Models
DTFT
(for “Hand” Analysis)
DFT & FFT
(for Computer Analysis)
Ch. 5: DT Fourier
System Models
Freq. Response for DT
Based on DTFT
New System Model
New System Model
Ch. 7: Z Trans.
Models for DT
Signals & Systems
Transfer Function
New System
Model 2/14
Stability of DT Systems
For systems with rational H(z):
∞
It is " BIBO" stable if
∑ h[n] < ∞
n =0
B( z )
Recall : H ( z ) =
A( z )
Where A( z ) = z N + a1 z N −1 + ... + a N −1 z + a N
Any common roots in B(z) and A(z) are assumed to have been cancelled.
Let A( z ) have roots of p1 , p2 ,..., p N
poles of H(z)
Then H ( z ) =
B( z )
( z − p1 )( z − p2 )...( z − p N )
and
h[ n] = h1[n] + h2 [n] + ... + hN [n]
Note : each h1[n ] will have ( pi ) n u[n ]
decays if pi < 1
3/14
Result:
∞
∑ h[n] < ∞
is equivalent to
n =0
all pi < 1
i.e., poles are inside
the unit circle
⇓
⇓
System is stable
System is stable
j
Im{z}
“unit circle”
Re{z}
−1
1
For a Stable System
−j
• Poles must be “inside unit circle”
• Zeros can be anywhere
Aside: Complex poles and complex zeros must occur as conjugate pairs
4/14
Frequency Response
All the same as for the CT case! (e.g. how sinusoids go through, how general
signals go through)
∞
H (Ω) = ∑ h[n ]e − jΩn = H ( z ) z =e jΩ
n =0
Using Matlab to Compute Frequency Response:
Some bi may be 0
b0 + b1 z −1 + b2 z −2 + ... + bN z − N
H ( z) =
a0 + a1 z −1 + a2 z − 2 + ... + a N z − N
must put any zero bi into the vector
>> num = [b0 b1 ... bN ]
>> den = [a0 a1 ... a N ]
>> omega = -pi : ? : pi
Some ai may be 0
must put any zero ai into the vector
Pick appropriate spacing
>> H = freqz(num, denom, omega)
>> plot(omega/pi, abs(H))
>> plot(omega/pi, angle(H))
5/14
Relationship between the ZT and the DTFT
Recall: H (Ω) = H ( z ) jΩ
z =e
Let’s explore this idea with some
pictures for an explicit case…
This causes H(z) = 0
for z = 0
Consider the Z-Transform given by:
H ( z) =
(1 − 0.8e
1
j 0.3π
z
−1
)(1 − 0.8e
− j 0.3π
z
−1
)
=
(z − 0.8e
z
j 0.3π
)(z − 0.8e
− j 0.3π
)
These cause H(z) = ∞
for z = 0.8e±j0.3π
Pole-Zero Plot For This H(z)
Im{z}
0.3π
0.8
Re{z}
6/14
So… from this pole-zero plot
we can then imagine that the
plot of the |H(z)| might look
something like this:
Pole-Zero Plot For This H(z)
Im{z}
0.3π
0.8
Re{z}
|H(z)|
And we know that the
Frequency Response
is just the Transfer
Function evaluated on
the Unit Circle.
H (Ω) = H ( z ) z =e jΩ
7/14
Now… plot just those values on the unit circle:
Now…“Cut” here…
and unwrap
This shows the Frequency Response H(Ω) where Ω is the angle around the
unit circle… this explains why H(Ω) is a periodic function of Ω
8/14
This shows the previous plot “cut and unwrapped”…
and plotted on the Ω axis:
Normalized for convenience
9/14
Effect of Poles & Zeros on Frequency Response of DT filters
Im{z}
Note: Including a
pole or zero at the
origin …
ℜ{z}
Im{z}
ℜ{z}
Placing a
zero at ±π…
Ω
Ω
…doesn’t change
the magnitude but
does change the
phase
Im{z}
ℜ{z}
…makes
|H(π)| = 0
Ω
Im{z}
Placing more
zeros/poles…
ℜ{z}
Ω
… gives sharper
transitions.
Im{z}
ℜ{z}
Ω
Figure from B.P. Lathi, Signal Processing and Linear Systems
10/14
So… from these plots and ideas we can see that we could design simple DT
filters by deciding where to put poles and zeros.
This is not a very good design approach…
… but this insight is crucial to understanding transfer functions.
The following charts in this set of notes shows a filter designed not by
placing poles and zeros but rather by using one really good computer-based
design method for designing DT filters.
11/14
A practical DT filter (Designed using MATLAB’s remez)
(See Digital Signal Processing course to learn the design process)
Here is the impulse response h[n]… it is assumed to be zero where not shown…
Note that it has only finite many non-zero samples
Called a “Finiteimpulse Response”
(FIR) filter
12/14
All these zeros, right on the unit circle, pull the frequency response
down to create the stop band
13/14
The effect of
the zeros on
the unit circle
Note, this filter has linear phase in the passband… this is the ideal phase
response (as we saw back in Ch. 5 for CT filters)
FIR DT filters are well-suited to getting linear phase and are therefore very
widely used.
14/14