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7/31/2019 EECE 301 Note Set 34 DT Using ZT to Solve
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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
7/31/2019 EECE 301 Note Set 34 DT Using ZT to Solve
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Ch. 1 Intro
C-T Signal Model
Functions on Real Line
D-T Signal Model
Functions on Integers
System Properties
LTICausal
Etc
Ch. 2 Diff EqsC-T System Model
Differential Equations
D-T Signal ModelDifference Equations
Zero-State Response
Zero-Input Response
Characteristic Eq.
Ch. 2 Convolution
C-T System Model
Convolution Integral
D-T System Model
Convolution Sum
Ch. 3: CT FourierSignalModels
Fourier Series
Periodic Signals
Fourier Transform (CTFT)
Non-Periodic Signals
New System Model
New Signal
Models
Ch. 5: CT FourierSystem Models
Frequency Response
Based on Fourier Transform
New System Model
Ch. 4: DT Fourier
SignalModels
DTFT
(for Hand Analysis)DFT & FFT
(for Computer Analysis)
New Signal
Model
Powerful
Analysis Tool
Ch. 6 & 8: LaplaceModels for CT
Signals & Systems
Transfer Function
New System Model
Ch. 7: Z Trans.
Models for DT
Signals & Systems
Transfer Function
New System
Model
Ch. 5: DT Fourier
System Models
Freq. Response for DT
Based on DTFT
New System Model
Course Flow DiagramThe 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).
7/31/2019 EECE 301 Note Set 34 DT Using ZT to Solve
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ZT For Difference Eqs.
Given a difference equation that models a D-T system we may want to solve it:
-with ICs
-with ICs 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.
Well consider the ZT/Difference Eq. approach first
7/31/2019 EECE 301 Note Set 34 DT Using ZT to Solve
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, ...,,nnx
y
nbxnayny
210for][
]1[IC
][]1[][:Given
=
=
=+
Solve for:y[n] for n = 0, 1, 2,
Solving a First-order Difference Equation using the ZT
Take ZT of differential equation: { } { }][]1[][ nbxZnaynyZ =+
{ } { } { }][]1[][ nxbZnyaZnyZ =+
Use Linearity
of ZT
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:
{ } ]1[)(]1[1
+=
yzYznyZ
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[ ] )(]1[)()( 1 zbXyzYzazY =++ Using these results gives:
Which is an algebraic equation that can be solved for Y(z):
)(11
]1[)(
11
zXaz
b
az
ayzY
+
+
+
=
Not the best form for doing
Inverse ZT we want things
in terms ofz notz-1
)(]1[)( zXaz
bz
az
zayzY +++=
Multiply each term byz/z
az
bz
zH +=)(
On ZT Table
{ })()(][)](1[][ 1 zXzHZnuaayny n +=
Part due to input signal
modified by Transfer Function
If |a| < 1 this dies out as n ,its an IC-driven transient
If the ICs are zero,
this is all we have!!!
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Ex.: Solving a Difference Equation using ZT: 1st-Order System w/ Step Input
1)(][][For
==
z
zzXnunx
Then using our general results we just derived we get:
++
+
=
1
]1[)(
z
z
az
bz
az
zayzY
For now assume that a -1 so we dont have a repeated root.
Then doing Partial Fraction Expansion we get (and we have to do the PFE by
hand because we dont know a but it is not that hard!!!)
( ) ( )1
]1[)( 11
+
++
+
= ++
z
z
az
z
az
zayzY a
baab
[ ] ,...2,1,0)1()(1
)](1[][ =++
+= naaa
baayny
nnn
IC-Driven Transient:decays if system is stable
Input-Driven Output 2 Terms:
1st term decays (Transient)
2nd term persists (Steady State)
Now using
ZT Table
we get:
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]1[][]2[]1[][ 1021 +=++ nxbnxbnyanyany
Solving a Second-order Difference Equation using the ZT
The Given Difference Equation:
Assume that the input is causal
Assume you are given ICs:y[-1] & y[-2]
Find the system responsey[n] forn = 0, 1, 2, 3,
) )
)()(
]2[]1[)(]1[)()(
1
10
12
2
1
1
zXzbzXb
yyzzYzayzYzazY
+=
+++++
Take the ZT using the non-causal right-shift property:
)(11
]2[]1[]1[)(
2
2
1
1
110
2
2
1
1
21
21 zXzaza
zbb
zaza
yazyayazY
++++
++=
Due to ICs decays
if system is stable H(z)
Due to input will have
transient part andsteady-state part
7/31/2019 EECE 301 Note Set 34 DT Using ZT to Solve
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22
11
1
22
11
2
1
21
11
]2[]1[]1[
)(
++
=++
= zaza
BzA
zaza
yazyaya
zYzi
Lets take a look at the IC-Driven transient part:
21
2
2
)(
azaz
BzAzzYzi
++
+=Multiply top and bottom byz2:
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:
( )22
12
2
1
2
1
)cos(2
)cos()sin(
][)sin(
][)cos(
azaz
zCCazC
nunaC
nunaC
o
oo
o
n
o
n
+
+
+
)sin()cos(
)sin(
2cos
0
01
0
21
2
11
02
==
==
Ca
BCAC
a
aaa
Compare
&Identify
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Finally, by a trig ID we know that
][)cos(][)sin(][)cos( 21 nunCanunaCnunaC on
o
n
o
n +=+
So all of this machinery leads to the insight that the IC-Driven transient
of a second-order system will look like this:
][)cos(][ nunCany on
zi +=
where:
1. The frequency 0 and exponential aare set by the Characteristic Eq.
2. The amplitude Cand the phase areset by the ICs
==
2
11
022
cosa
aaa
Note: If |a2| < 1 then
we get a decaying
response!!
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== =+M
i
i
N
i
i nxbinyany01
]1[][][
Containsx[n],x[n-1],
If this system is causal, we wonthavex[n+1],x[n+2], etc. here
sIC'theondepends)(
...)(
...)(1
10
1
1
1
=
+++=++++=
zC
zbzbzbzB
azazazzAMN
M
NN
NN
NN
Solving a Nth-order Difference Equation using the ZT
Transforming gives:
)()(
)(
)(
)()( zX
zA
zB
zA
zCzY += Transient and steady
state part due to input
H(z) transfer function
Transient part due to ICs
7/31/2019 EECE 301 Note Set 34 DT Using ZT to Solve
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DifferenceEquation
DifferenceEquation
TransferFunction
TransferFunction
Frequency
Response
Frequency
ResponseImpulse
Response
Impulse
Response
Pole/Zero
Diagram
Pole/Zero
Diagram
DTFT
ZT
ZT (Theory)Inspect (Practice)
Roots
Unit Circle
Time Domain Z / Freq Domain
Discrete-Time System Relationships
h[k]
][)]1([]1[][
][)]1([]1[][
011
011
nkfbnkfbkfbkfb
nkyankyakyaky
nn
n
++++
=++++
"
"
011
1
011
1)(azazazbzbzbzbzH
nn
n
nn
nn
++++++++=
"
"
== jezzHH |)()(
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Difference
Equation
Difference
Equation
Time Domain
)()1()( nxnyny =Input-Output Form
Transfer
Function
Transfer
Function
ZT (Theory)
Inspect (Practice)
)1()()( += nynxny Recursion Form
Z / Freq Domain
11)(
)()( == zzX
zYzH
Example System Relationships
)(1)( 1 zXzzY =
= zzzH )(
7/31/2019 EECE 301 Note Set 34 DT Using ZT to Solve
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Transfer
Function
Transfer
Function
Frequency
Response
Frequency
Response
Unit Circle
Z / Freq Domain
11)(
=
zzH
==
j
j
ez
ez1
On
Unit
Circle
],(1
)(
= j
e
H
=jez 1
Example System (cont.)
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[ ]
[ ]
[ ] [ ]
=
+=
+=
=
=
)cos(1
)sin(tan)(
)sin()cos(1)(
)sin()cos(1
)sin()cos(11)(
1
22
H
H
j
jeH
j
Plotting Transfer Function
Eulers
Equation
Group into
Real & Imag
StandardEqs for
Mag. & Angle
Example System (cont.)
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Transfer
Function
Transfer
Function
Pole/Zero
Diagram
Pole/Zero
Diagram
Roots
Z/Freq Domain
=
=
z
z
zzH
11)(
Unit
Circle
Pole
Den. = 0When z =
Zero
Num. = 0When z = 0
Re{z}
Im{z}
If < 1: Inside UC
If > 1: Outside UC
Example System (cont.)
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Transfer
Function
Transfer
Function
Frequency
Response
Frequency
ResponseImpulse
Response
Impulse
Response
DTFT
ZTUnit Circle
Time Domain Z / Freq Domain
Example System (cont.)
11)(
=
zzH
],(
1)(
=
je
H
n
nh =)(
Inv. ZT
Table
=
denh j1)(
Inv. DTFT
Use Table if Possible