Chapter 5 the Laplace Transform

EE 422G Notes: Chapter 5 Instructor: Zhang

Chapter 5 The Laplace Transform

5-1 Introduction

(1) System analysis

static system: y(t) = ax(t) => easy (simple processing)

dynamic system:

Can we determine y(t) for given u(t) easily?

Easier solution method

(1) we have systematic way to obtain H(f) based on the differential equation

(2) we can obtain X(f)

è Fourier transform: an easier way

(2) Problem

Fourier transform of the input signal:

if x(t) does not go to zero when and

ð X(f) typically does not exist! (the existence of X(f) if not guaranteed)

A very strong condition, can not be satisfied by many signals!

(3) Solution:

Why should we care about t<0 for system analysis? We do not care!

x(t) does not go to zero (when ), but may!

(Will be much easier. )

ð use “single-sided” Fourier transform of , instead of “double-sided” Fourier transform of x(t).

very useful, let’s use a new name for it: Laplace transform.

(4) Laplace Transform

Definition: Laplace transform of x(t)

What is a Laplace transform of x(t)?

A time function? No, t has been eliminated by the integral with respect to t!

A function of s ( s is complex variable)

(5) System analysis using Laplace transform

(6) How is the inverse transform defined?

(7) Will we often use the definition of the inverse transform to find time

function?

No!

What will we do?

Express X(s) as sum of terms for which we know the inverse transforms!

5-2 Examples of Evaluating Laplace Transforms using the definition

(1) x(t)=1 and step function x(t)=u(t)

(2)

Define a new complex variable

we know

(3)

No constraint on s.

5-2B Discussion: Convergence of the Laplace Transform

(1) To assure converge, must be psotive enough such that goes to zero when t goes to positive infinite

(2) Region of absolute convergence and pole

(3) How to obtain Fourier transform form Laplace transform:

Important: why introduce Laplace transform; definition of Laplace transform as a modification of Fourier transform; find the Laplace transforms of the three basic functions based on the (mathematical) definition of Laplace transform.

Chapter structure

Part one: Definition (5.1)

Part two: Direct evaluation of Laplace transforms of simple time-

domain function.

Easy? Not at all!

ð complex (not simple) functions: even harder

ð tools for application of Laplace transform in system analysis

Part three: Rest of the chapter

What tools: tools for easier Laplace transform evaluation

tools for easy inverse transform

5-3 Some Laplace Transform theorems

(Tools for evaluating Laplace transform based on the Laplace

transforms of the basic functions)

5.3.1 Linearity

Assume ( and are time independent)

then

HW#2-1: Assume , and .

(a) Is equal to ?

(Answer: Yes or No)

(b) If , is it true

(Answer: Yes or No)

Example 5.1

(1) Find

Key to solution : express as linear combination of , ,

and/or :

let

Can we use and to express ?

(2) Find

5-3-2 Transforms of Derivatives

Assume

Then

Proof:

(1) Definition

(2) Integration by parts:

General equation:

(3) Use the above equation

Why? If we assume

must go to zero. Otherwise,

does not exist !

use as lower limit =>

HW#2-2: Assume . Prove

HW#2-3: Express using

Example 5-2

Find i(t) using Laplace transform method for t>0

Solution:

(1) Before switched from 1 to 2 at t=0

(2) System equation (t>0)

KVL:

(3) Solve system equation using Laplace transform

5-3-3 Laplace Transform of an integral

Assume

Then

where

Proof :

(1)

(2)

(3)

Proved!

Example 5.3:

Find I(s) = L(i(t))

Solution:

(1) Differential equation

KVL :

(2) Laplace transform

5-3-4. Complex Frequency shift (s-shift) Theorem

Assume

Then

Example 5-4 Find

Solution:

5-3-4 Delay Theorem

· question: How to express delayed function?

· Assume

Then

(If , it will not be a delay!)

Proof :

Question: will be true if

No! (it will not be a delay)

Example 5-5: Square wave beginning at t = 0

5-3-5 Convolution

Signal 1: Signal 2 :

Therefore, if

Look at

Then

5-3-7 Product

5-3-8 Initial Value Theorem

Example: A demonstration where x(0) is obvious

It is evident:

Using Laplace transform

5-3-9 Final Value Theorem: if and are Laplace transformable, then

(condition: has no poles on or in the right-half s-plan or exists)

5-3-10 Scaling

a>0: x(at) ¬ a times fast (if a>1)

or slow (if a<1) as x(t)

What do we expect on ?

5-4 Inversion of Rational Functions

(1) Ways to find from

(1) (Contour Integral)

(2) Transform pair

Therefore

(2) All kinds of Laplace Transform ? No!

We almost only see

rational function X(s) Delay

ð Consider Rational Functions only!

(3) Non proper Rational Function

proper Rational Function

Non proper m>=n

proper m<n

Non proper => proper + Polynomial (using long division)

How to find inverse Laplace transform for polynomials?

èconsider proper rational functions only!

(4) Proper Rational Functions: Partial Fraction Expansion

è sum of

Let’s look at examples, and then summarize!

Techniques:

Common Denominator Factorize first!

Specific value of s Expand second!

Heaviside’s Expansion Find coefficients third!

Matlab

Example 5-9: Simple Factors

Solution:

(1) Factorize and expand

(2) Common Denominator Methods

specific values of s

Can you solve for A and B?

Heaviside Expansion

and

Example 5-10 Imaginary Roots

Solution: what do we have:

A1+A2 must be real number

(-A1+A2)j must be real number

Heaviside Expansion => A3=1 and A4= - 2.

s=1 =>

s=2 =>

Can we solve for c1 and c2?

c1=1 c2=1

Too complex: use MATLAB

Example 5-11 Repeated linear Factors

Example 5-12

Example 5-13 Complex - Conjugate Factors

Example 5-14 Repeated Quadratic Factors

* Summary of Partial–Fraction Expansion

(1) Expansion Structure:

Simple Roots (including complex conjugate)

=> could be complex.

Repeated Roots: m multiplicity

real number or complex number

(2) Avoid complex number

For complex conjugates:

(3) Inverse Laplace transform

Matlab use for Partial – Fraction Expansion.

Have to be memorized:

(1) Table 5-2 all except for No. 7

(2) Table 5-3 No. 1-No. 7
Appendix: Partial-Faction Expansion with MatLab

1. Command (MatLab Help)

2. An Introduction

3. An Example: Complex Conjugate

1. Command (MatLab Help)

» help residue

RESIDUE Partial-fraction expansion (residues).

[R,P,K] = RESIDUE(B,A) finds the residues, poles and direct term of

a partial fraction expansion of the ratio of two polynomials B(s)/A(s).

If there are no multiple roots,

B(s) R(1) R(2) R(n)

---- = ------+ ------+ ... + ------+ K(s)

A(s) s - P(1) s - P(2) s - P(n)

Vectors B and A specify the coefficients of the numerator and

denominator polynomials in descending powers of s. The residues

are returned in the column vector R, the pole locations in column

vector P, and the direct terms in row vector K. The number of

poles is n = length(A)-1 = length(R) = length(P). The direct term

coefficient vector is empty if length(B) < length(A), otherwise

length(K) = length(B)-length(A)+1.

If P(j) = ... = P(j+m-1) is a pole of multplicity m, then the

expansion includes terms of the form

R(j) R(j+1) R(j+m-1)

------+ ------+ ... + ------

s - P(j) (s - P(j))^2 (s - P(j))^m

[B,A] = RESIDUE(R,P,K), with 3 input arguments and 2 output arguments,

converts the partial fraction expansion back to the polynomials with

coefficients in B and A.

Warning: Numerically, the partial fraction expansion of a ratio of

polynomials represents an ill-posed problem. If the denominator

polynomial, A(s), is near a polynomial with multiple roots, then

small changes in the data, including roundoff errors, can make

arbitrarily large changes in the resulting poles and residues.

Problem formulations making use of state-space or zero-pole

representations are preferable.