EECE 301 Note Set 28a CT Partial Fractions

7/31/2019 EECE 301 Note Set 28a CT Partial Fractions

1/8

1/8

EECE 301

Signals & Systems

Prof. Mark Fowler

Note Set #28a

C-T Systems: Laplace Transform Partial Fractions Reading Assignment: Notes 28a Reading on Partial Fractions

(On my website)


2/8

2/8

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).


3/8

3/8

There is a set of notes on-line called . Reading Notes on Partial

Fraction Expansion

You should read these on your own

They lead you through several examples of using partial fractions to take a

ratio of two high-order polynomial and expand it into a sum of ratios of

simple polynomials.

You should be able to:

Perform a simple partial fraction by hand (e.g., during an

exam)

Use matlab to compute more complicated partial fractions

Here Ill cover the main ideas


4/84/8

When trying to find the inverse Laplace transform it is helpful to be able to break a

complicated ratio of two polynomials into forms that are on the Laplace Transformtable.

Partial Fraction Expansion and Inverse LT

+=

+ RCsssRCs

RC --

/1

11

)/1(

/1 11LL

+

=

RCss

--

/1

11 11LL

Linearity

of LT

)(tu= )()/( tue RCt=

Here the factoring could be done by inspection in general, this factoringcan be done using the method of partial fraction expansion (PFE)

Factor LT

into simpler

forms

An Example:


5/85/8

Motivation Going the Other Way

( ) ( )2

5

4

3

2)( +

++

+=

sssY

Suppose we had this:

We want to go from

here to here

This has poles at

s = -3 and s =-5

A term like thisis called a

Direct Term

We could find the common denominator and combine:

( )( )( )

( )( )( )

( )( )( )( )

158

47222

53

532

53

34

53

52)(

2

2

++

++=

++

+++

++

++

++

+=

ss

ss

ss

ss

ss

s

ss

ssY


6/86/8

Ex. #1: No Direct Terms, No Repeated Roots, No Complex Roots

Well illustrate with an example like you should be able to do on an exam:

If the highest power in the numerator is less than the highest power in the

denominator then there will be no direct terms.

By using the quadratic formula to find the roots of the denominator we can

verify that there are no repeated or complex roots.

23

13)(

2 ++=ss

ssY

If there are no repeated roots and no direct terms we can

always write it as

)2()1()( 21

++

+=

s

r

s

rsY

The numbers r1 and r2 are called the residues we need to find them!

)2)(1(13)(++ = ss

ssYThe roots are: s = -2 and s = -1 so we can write:


7/87/8

Now we exploit what we know:

)2()1()2)(1(

13 21

++

+=

++

s

r

s

r

ss

s

Multiply each side by (s+1) gives:

)2(

)1(

)1(

)1(

)2)(1(

)1)(13( 21

+

++

+

+=

++

+

s

sr

s

sr

ss

ss

Canceling (s+1) where we can gives:

)2(

)1(

)2(

)13( 21

+

++=

+

s

srr

s

s

Setting s = 1 gives:

4

)21(

)13(11 ==

+

rr

11)1)((

=+=

sssYr

All of this is

summarized

by this


8/8

8/8

Similarly we find the other residue using: 7)2)((22=+=

=sssYr

)2(

7

)1(

4)(+

++=

sssY

Then we have:

Each of these terms is on the LT Table, so we get

)(7)(4

)2(

7

)1(

4)(

2

11

tuetue

ssty

tt

+=

++

+

= LL

See the set of notes on my website called . Reading Notes on Partial

Fraction Expansion for details on how to use MATLAB to find PartialFraction Expansions for the more complicated cases.

Documents

EECE 301 Note Set 28a CT Partial Fractions