eece 301 note set 34 dt using zt to solve

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


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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).


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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


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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


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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


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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 )(


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13/16

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



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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



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Transfer

Function

Transfer

Function

Frequency

Response

Frequency

ResponseImpulse

Response

Impulse

Response

DTFT

ZTUnit Circle

Time Domain Z / Freq Domain


11)(

=

zzH

],(

1)(

=

je

H

n

nh =)(

Inv. ZT

Table

=

denh j1)(

Inv. DTFT

Use Table if Possible

eece 301 note set 34 dt using zt to solve

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