EECE 301
Signals & Systems
Prof. Mark Fowler
Note Set #34
• D-T Systems: Z-Transform … Solving Difference Eqs. & Transfer Func.
• Reading Assignment: Sections 7.4 – 7.5 of Kamen and Heck
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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 2/16
ZT For Difference Eqs.
Given a difference equation that models a D-T system we may want to solve it:
-with IC’s
-with IC’s of zero
Apply ZT to the Difference Equation
Use the Transfer Function Approach
Note… the ideas here are very much like what we did with the Laplace
Transform for CT systems.
We’ll consider the ZT/Difference Eq. approach first…
3/16
Solving a First-order Difference Equation using the ZT
Given : y[n ] + ay[n − 1] = bx[n ]
Solve for: y[n] for n = 0, 1, 2,…
IC = y[ −1]
x[n ] for n = 0, 1, 2 , ...
Take ZT of differential equation: Z {y[n ] + ay[n − 1]} = Z {bx[n ]} Use Linearity
of ZT
Z {y[n ]}+ aZ {y[n − 1]} = bZ {x[n ]}
Y(z)
X(z)
Need Right-Shift Property…
but which one???
Because of the non-zero IC we need to use the non-causal form:
Z {y[n − 1]} = z −1Y ( z ) + y[−1]
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[
]
Using these results gives: Y ( z ) + a z −1Y ( z ) + y[ −1] = bX ( z )
Which is an algebraic equation that can be solved for Y(z):
Y ( z) =
b
− ay[ −1]
X ( z)
+
−1
−1
1 + az
1 + az
Not the best form for doing
Inverse ZT… we want things
in terms of z not z-1
Multiply each term by z/z
Y ( z ) = −ay[ −1]
z
bz
+
X ( z)
z+a z+a
On ZT Table
bz
Part due to input signal
H ( z) =
modified by Transfer Function
z+a
y[n ] = − ay[ −1]( −a ) n u[n ] + Z −1{H ( z ) X ( z )}
If |a| < 1 this dies out as n ↑,
its an IC-driven transient
If the ICs are zero,
this is all we have!!!
5/16
Ex.: Solving a Difference Equation using ZT: 1st-Order System w/ Step Input
z
For x[n ] = u[n ] ↔ X ( z ) =
z −1
Then using our general results we just derived we get:
Y ( z) =
− ay[−1]z ⎛ bz ⎞⎛ z ⎞
+⎜
⎟
⎟⎜
z+a
⎝ z + a ⎠⎝ z − 1 ⎠
For now assume that a ≠ -1 so we don’t have a repeated root.
Then doing Partial Fraction Expansion we get (and we have to do the PFE by
hand because we don’t know a… but it is not that hard!!!)
− ay[ −1]z ( aab+1 )z ( ab+1 )z
Y ( z) =
+
+
z+a
z+a
z −1
[
b
y[n ] = −ay[ −1]( −a ) +
a ( −a )n + (1) n
a +1
n
IC-Driven Transient:
decays if system is stable
]
Now using
ZT Table
we get:
n = 0, 1, 2,...
Input-Driven Output… 2 Terms:
1st term decays (Transient)
2nd term persists (Steady State)
6/16
Solving a Second-order Difference Equation using the ZT
The Given Difference Equation:
y[n] + a1 y[n − 1] + a2 y[n − 2] = b0 x[n] + b1 x[n − 1]
Assume that the input is causal
Assume you are given ICs: y[-1] & y[-2]
Find the system response y[n] for n = 0, 1, 2, 3, …
Take the ZT using the non-causal right-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 )
− a1 y[ −1] − a2 y[ −1]z −1 − a2 y[ −2]
b0 + b1 z −1
+
Y ( z) =
X ( z)
−1
−2
−1
−2
1 + a1 z + a2 z
1 + a1 z + a2 z
Due to IC’s… decays
if system is stable
H(z)
Due to input – will have
transient part and
steady-state part
7/16
Let’s take a look at the IC-Driven transient part:
− a1 y[ −1] − a2 y[ −1]z −1 − a2 y[ −2]
A − Bz −1
Yzi ( z ) =
=
−1
−2
1 + a1 z + a2 z
1 + a1 z −1 + a2 z −2
Multiply top and bottom by z2:
Az 2 + Bz
Yzi ( z ) = 2
z + a1 z + a2
Now to do an inverse ZT on this requires a bit of trickery…
Take the bottom two entries on the ZT table and form a linear combination:
C1a n cos(Ωo n )u[n ]
+ C2 a n sin(Ωo n )u[n ]
a = a2
C1 = A
↔
C1 z 2 + a (C2 sin(Ωo ) − C1 cos(Ωo ) )z
z 2 − 2a cos(Ωo ) z + a 2
⎡ − a1 ⎤
Ω0 = cos ⎢
⎥
⎢⎣ 2 a2 ⎥⎦
cos(Ω0 )
B
− C1
C2 =
sin(Ω0 )
a sin(Ω0 )
−1
Compare
&
Identify
8/16
Finally, by a trig ID we know that
C1a n cos(Ωo n )u[n ] + C2a n sin(Ωo n )u[n ] = Ca n cos(Ωo n + θ )u[n ]
So… all of this machinery leads to the insight that the IC-Driven transient
of a second-order system will look like this:
y zi [n ] = Ca n cos(Ωo n + θ )u[n ]
…where:
1. The frequency Ω0 and exponential a
are set by the Characteristic Eq.
a = a2
⎡ − a1 ⎤
Ω0 = cos−1 ⎢
⎥
⎢⎣ 2 a2 ⎥⎦
2. The amplitude C and the phase θ are
set by the ICs
Note: If |a2| < 1 then
we get a decaying
response!!
9/16
Solving a Nth-order Difference Equation using the ZT
N
M
Contains x[n], x[n-1],…
i =1
i =0
If this system is causal, we won’t
have x[n+1], x[n+2], etc. here
y[n] + ∑ ai y[n − i ] = ∑ bi x[n − 1]
A( z ) = z N + a1 z N −1 + ... + a N −1 z + a N
B( z ) = b0 z N + b1 z N −1 + ... + bM z N − M
C ( z ) = depends on the IC' s
Transforming gives:
Y ( z) =
C ( z ) B( z )
+
X ( z)
A( z ) A( z )
Transient and steady
state part due to input
H(z) – transfer function
Transient part due to ICs
10/16
Discrete-Time System Relationships
Time Domain
Z / Freq Domain
Pole/Zero
Pole/Zero
Diagram
Diagram
Roots
Difference
Difference
Equation
Equation
ZT (Theory)
Inspect (Practice)
y[ k ] + an −1 y[k − 1] + " + a1 y[k − (n − 1)] + a0 y[k − n] =
bn f [k ] + bn −1 f [k − 1] + " + b1 f [k − (n − 1)] + b0 f [k − n]
h[k]
bn z n + bn −1z n −1 + " + b1 z + b0
H ( z) = n
z + an −1 z n −1 + " + a1z + a0
Unit Circle
ZT
DTFT
Impulse
Impulse
Response
Response
Transfer
Transfer
Function
Function
Frequency
Frequency
Response
Response
H ( Ω ) = H ( z ) | z = e jΩ
11/16
Example System Relationships
Time Domain
Difference
Difference
Equation
Equation
Z / Freq Domain
ZT (Theory)
Inspect (Practice)
Input-Output Form
y (n) − α y (n − 1) = β x(n)
Transfer
Transfer
Function
Function
[
]
Y (z) 1−α z = β X (z)
−1
Recursion Form
y (n) = β x(n) + α y (n − 1)
β
Y ( z)
H ( z) =
=
−1
X ( z) 1 − α z
βz
H ( z) =
z −α
12/16
Example System (cont.)
Z / Freq Domain
z = e jΩ
z −1 = e − jΩ
On
Unit
Circle
H ( z) =
β
1 − α z −1
Transfer
Transfer
Function
Function
z −1 = e − jΩ
β
H (Ω ) =
1 − α e − jΩ
Ω ∈ (−π , π ]
Unit Circle
Frequency
Frequency
Response
Response
13/16
Example System (cont.)
Plotting Transfer Function
β
β
=
1 − α e − jΩ 1 − [α cos(Ω) − jα sin(Ω)]
β
=
[1 − α cos(Ω)] + jα sin(Ω)
Euler’s
Equation
H (Ω ) =
H (Ω) =
β
[1 − α cos(Ω)] + [α sin(Ω)]
2
⎡ α sin(Ω) ⎤
∠H (Ω) = − tan −1 ⎢
⎥
1
α
cos(
)
−
Ω
⎦
⎣
2
Group into
Real & Imag
Standard
Eqs for
Mag. & Angle
14/16
Example System (cont.)
Z/Freq Domain
Im{z}
Unit
Circle
Pole/Zero
Pole/Zero
Diagram
Diagram
Re{z}
Roots
Transfer
Transfer
Function
Function
β
βz
H ( z) =
=
−1
1−α z
z −α
Zero
Num. = 0
When z = 0
Pole
Den. = 0
When z = α
If α < 1: Inside UC
If α > 1: Outside UC
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Example System (cont.)
Time Domain
Z / Freq Domain
H ( z) =
Inv. ZT
Table
β
1 − α z −1
Transfer
Transfer
Function
Function
h(n) = βα n
Impulse
Impulse
Response
Response
Inv. DTFT
π
β
h( n) = ∫
dΩ
−π 1 − α e − jΩ
Use Table if Possible
ZT
Unit Circle
DTFT
Frequency
Frequency
Response
Response
β
H (Ω ) =
1 − α e − jΩ
Ω ∈ (−π , π ]
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