ELEC 3004: Signals, Systems &...

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LTI & Laplace Transforms ELEC 3004: Signals, Systems & Control Dr. Surya Singh, Prof. Brian Lovell & Dr. Paul Pounds Lecture # 4 March 12, 2012 [email protected] http://courses.itee.uq.edu.au/elec3004/2012s1/ © 2012 School of Information Technology and Electrical Engineering at the University of Queensland 2 Schedule of Events Week Date Lecture Title 1 1-Mar Overview 5-Mar Signals & Systems 8-Mar Sampling 12-Mar LTI & Laplace Transforms 15-Mar Convolution 19-Mar Discrete Fourier Series 22-Mar Fourier Transform 26-Mar Fourier Transform Operations 29-Mar Applications: DFFT and DCT 2-Apr Exam 1 (10%) 5-Apr (Guest Lecture from Industry) 16-Apr Data Acquisition & Interpolation 19-Apr Noise 23-Apr Filters & IIR Filters 26-Apr FIR Filters 30-Apr Multirate Filters 3-May Filter Selection 7-May Holiday 10-May Quiz (10%) 14-May z-Transform 17-May Introduction to Digital Control 21-May Stability of Digital Systems 24-May Estimation 28-May Kalman Filters & GPS 31-May Applications in Industry 6 2 3 4 5 13 7 8 9 10 11 12

Transcript of ELEC 3004: Signals, Systems &...

LTI & Laplace Transforms

ELEC 3004: Signals, Systems & Control

Dr. Surya Singh, Prof. Brian Lovell & Dr. Paul Pounds

Lecture # 4 March 12, 2012

[email protected]

http://courses.itee.uq.edu.au/elec3004/2012s1/© 2012 School of Information Technology and Electrical Engineering at the University of Queensland

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Schedule of EventsWeek Date Lecture Title

1 1-Mar Overview5-Mar Signals & Systems8-Mar Sampling

12-Mar LTI & Laplace Transforms15-Mar Convolution19-Mar Discrete Fourier Series22-Mar Fourier Transform26-Mar Fourier Transform Operations29-Mar Applications: DFFT and DCT

2-Apr Exam 1 (10%)5-Apr (Guest Lecture from Industry)

16-Apr Data Acquisition & Interpolation19-Apr Noise23-Apr Filters & IIR Filters26-Apr FIR Filters30-Apr Multirate Filters3-May Filter Selection7-May Holiday

10-May Quiz (10%)14-May z-Transform17-May Introduction to Digital Control21-May Stability of Digital Systems24-May Estimation28-May Kalman Filters & GPS31-May Applications in Industry

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Overview

• Laplace transform– Finite power signals

1. Unilateral Laplace transform

2. Bilateral Laplace transform

• Transform Analysis of Linear systems– Circuit Analysis

– Transfer functions

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Laplace Transform

• Problem: FT of a signal may not always exist!– finite power (and not periodic), – e.g., x(t) = u(t)exp(-at)) with a < 0– Or x(t) = u(t)cos(5t)!

• Solution: Force signal to have finite energy– Multiply by convergence factor (exp(-σt))– i.e., new signal xσ(t) = exp(-σt)x(t)– Therefore, FT of xσ(t) exists

∫∞

∞−

∞−

+−=

−=

dttjwtxwX

dtjwttxwX

))(exp()()(

)exp()()(

σσ

σσ Fourier

Laplace

⇓Rearranging…

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Bilateral Laplace Transform• For compactness we write

– s = σ+jw

∫∞+

∞−

∞−

=

−=

j

j

dsstsXj

tx

dtsttxsX

σ

σπ)exp()(

2

1)(

)exp()()( Laplace transformX(s) = L{x (t )}

Inverse Laplace transformx(t) = L

-1{X (s )}

Note:

i.e., F-1{Xσ(ω)}

i.e., × by inverse of converge factor

Finally re-arrange

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Unilateral Laplace Transform

• Problem with two-sided (bilateral) transform

– choice of σ can cause ambiguities in L-1{X (s )}

• i.e., as different x(t)’s have same X(s)!• e.g., L{exp(at)u(t)} = L{-exp(at)u(-t)}

• Solution: assume x(t) to be causal (one-sided)– i.e., x(t) = 0 when t < 0

– This is termed the unilateral Laplace Transform

– Integration is now from 0 ≤ t < ∞• This solution works for most practical signals

– Bilateral required for (non-deterministic) random signals • (see later)

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Unilateral Laplace Transform

• One-sided Laplace transform

• Laplace transform

– X(s) = L{x(t)}

• Inverse Laplace transform

– x(t) = L-1{X(s)}

∫∞

−=0

)exp()()( dtsttxsX

0- indicates origin is included in integration 0 ≤ t < ∞

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Convergence of Laplace Transform

• Consider signal

• Convergence dependent on both σand a

– Note: ℜ{s} = σ• Region of Convergence (ROC)

– Finite integral (energy)

)()exp()( tuattx −=

a=-1

a=1a=2

( )( )[ ]

{ } asas

sX

ajwa

tasas

dtjwtta

dttassX

−>ℜ+

=

>+++

=

+−+−=

−+−=

+−=

,1

)(

0,)(

1

exp1

)exp())(exp(

))(exp()(

0

0

0

σσ

σEffectively same as

� Fourier Transform

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Laplace Examples

( ){ } ( ) ( )

( )

( )0,

1exp

exp

exp

0

0

0

>=

−−=

−=

−=

σss

st

dtst

dtsttutuLUnit step function:

Impulse function: ( ){ } ( ) ( )

( ) ( )

( ) ( ) 10exp

exp

exp

0

0

0

0

=−=

−=

−=

+

+

∞−

dtts

dtstt

dtstttL

δ

δ

δδ

( ) ( ) ( ) ( ) ( ) ( ) ( )000

0

0

0

fdttfdtttfdtttf === ∫∫∫+

+

+

+

∞−

δδδ

Remember:

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Interpretation of Laplace Transform

• Represents signals, x(t), as sum of

– growing/decaying cosine waves

• at all frequencies (continuous), X(s)

– exp(σt )|X(s)|dw/2π is amplitude of growing/decaying cosine wave

• In frequency band w to w + dw

– ∠X(s) is phase shift of cosine wave

• parameter σ (ℜ{s}) determines rate of growth or decay

– Note: σ = 0 is the Fourier Transform ☺

• Constant amplitude cosine waves

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Complex Phasors

Decaying σ < 0

Growing σ > 0

constant magnitude σ = 0As per Fourier Transform

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Linear Transforms

• So far, we have looked at

– Fourier series

• Trigonometrical & Complex

– Fourier transform

– Laplace transform

• All represent signals as a

– Weighted sum (or integration) of

– Complex exponentials (that are orthogonal)

– e.g., complex FS, x(t) = Σ Xn exp(jnw0t)

• This relates directly to linear systems

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Linear Transforms & Linear Systems

Useful, due to two properties of linear systems

1. Superposition principle

– Inputs are complex exponentials (sinusoids)

– Output is same exponentials, but with different weights, i.e., delay and amplify/attenuate input

( ) ( ){ } ( ){ } ( ){ }1 2 1 2H ax t bx t aH x t bH x t+ = +

In other words, each phasor can be considered individuallyand output calculated by summation of the individual phasors

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Linear Transforms & Linear Systems

2. Ordinary differential equations (ODE)– Output y(t) related to input x(t) by ODE

m

m

mn

n

n dt

xdb

dt

dxbxb

dt

yda

dt

dyaya +++=+++ LL 1010

0 0 0

22

0 0 02

(exp( )) exp( )

(exp( )) ( ) exp( )

djn t jn jn t

dt

djn t jn jn t

dt

ω ω ω

ω ω ω

=

=

Differentiation is simple: • Frequency unchanged • Magnitude changes (complex)

Note: initially consider harmonically related sinusoids nω0 as per Fourier Series

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Example RC Circuit

0 1 0 0

0

( ) exp( )

( ) exp( )

dya y t a b jn t

dtdy

y t RC jn tdt

ω

ω

+ =

+ =

x(t) (input)

R

C y(t) (output)

i

Specifically:a0 = 1 a1 = RCb0 = 1derivation shortly…

Using Kirchhoff’s laws, system described (modelled) by ODE

x(t) sinusoidal input(1 energy store ∴ 1st order)

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Example RC Circuit

• If input single sinusoid xn(t) = exp(jnw0t)

– and system linear, (steady-state) output has form:

• Substituting (assumed) solution into ODE0( ) exp( )ny t K jn tω=

where K is complex constant (weight)

( ) ( )0 1 0 0 0

00

0 0 1

0

0

0 0 1 0

( ) ( ) exp

( ) exp( )( )

( ) exp( )

1

( ) 1 ( )

n n

n

n n

n

a y t a jn y t b jn t

by t jn t

a jn a

y t H jn t

bH

a jn a jn RC

ω ω

ωω

ω

ω ω

+ =

=+

=

= =+ +

Where Hn is the system transfer

function to thissingle phasor (K)

Same frequencyDifferent amp. &/or phase

( ) ( ) ( )0n

n

dy tjn y t

dxω=Note:

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Response of Linear System

• Input represented as Fourier series

X1exp(jw0t)

X2exp(j2w0t)

Xnexp(jnw0t)

Σ Linear Systemy(t) x(t)

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Response of Linear System

• Applying superposition principle– Applying individual phasors & shifting Xn terms

Σ

X1

H1 exp(jw0t)Linear System

y(t)

X2

H2 exp(j2w0t)Linear System

Xn

Hn exp(jnw0t)Linear System

exp(jw0t)

exp(j2w0t)

exp(jnw0t)

Each phasor is orthogonal and ∴ can be evaluated separately

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Response of Linear System

• Output represented as Fourier series– output FS coefficients Yn = Hn Xn

Σ y(t)

H1X1exp(jw0t)

H2X2exp(j2w0t)

HnXnexp(jnw0t)∑

−∞=

=n

nn tjnwXHty )exp()( 0

Input |.| & ∠ at this frequency

System |.| & ∠ at this frequency

Output |.| & ∠ at this frequency

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General Approach

• Represent input as weighted sum of – Complex exponential basis functionsbasis functionsbasis functionsbasis functions,

– e.g., FS: exp(jnω0t): sinusoids at harmonic frequencies

• Basis functions are orthogonal– Amplitude (& phase) at each freq. evaluated separately

• System described by ODE– Response to exponentials (differentials) easy to calculate

• Frequency remains constant, Amp. & Phase change

– e.g., if response 2nd harmonic is 2ω0exp(2jω0)

• System is linear– Output is sum of responses of individual exponentials

• i.e., sum response at each frequency

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Laplace Transforms

• General approach was illustrated with

– Periodic input (i.e., we used Fourier series)

– Steady state response

• But, in general interested in

– non-periodic input and

– Both transient & steady state response

• So, we use the Laplace transform

– Basis functions exp(st), where s = σ+jw– output response is H(s) exp(st)

– where H(s) is system transfer functiontransfer functiontransfer functiontransfer functionNote: still orthogonal

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Laplace System Analysis

Steps in Laplace system Analysis,

1. Laplace transform input signal– X(s) = L{x(t)}

2. Calculate system transfer function– H(s)

3. Calculate (Laplace) output using multiplication– Y(s) = H(s)X(s)

4. Inverse Laplace transform– y(t) = L-1 {Y(s)}

Key: Calculating system transfer function: H(s)

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Laplace Transfer Function• H(s) completely defines the system

• Defined as:

)(

)(

)}({

)}({)(

sX

sY

txL

tyLsH ==

R

ty

dt

dyC

R

tx

txdt

dyC

R

tytx

)()(

)( of in terms Rearanging

)()(

+=

=−

x(t)

R

C y(t)

iA

Apply Kirchoff’s current law at node A:

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Laplace Transfer Function

{ }

)1(

1

)(

)()(

:LPF)order (first function Transfer

)(1)(

0)0( conditions initial zero Assume

)()0()(

)(

Transforms Laplace Taking

RCssX

sYsH

sYR

CsR

sX

yR

sYyssYC

R

sX

+==

+=

=

+−=

Example 1.9MGT

( ) ( ) ( )+−=

0fssFtf

dt

dLNote:

Note: similarity to Hn for FS

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Laplace Circuit Analysis

• Transform circuit elements– then apply Kirchoff’s current law

• Impedance of Capacitor = 1/Cs

– (Note: impedance of inductor = Ls)

X(s)

R

1/Cs Y(s)

A

)1(

1)(

)(1

)()()(

RCssH

sCsYCs

sY

R

sYsX

+=

==−

This is quick and easy way of doing circuit analysis

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Laplace Circuit Analysis

• Impedance of Inductor = Ls

X(s)

R

Ls Y(s)

A

)()1(

1)(

)(1)()(

)(1)()(

RLs

Ls

Ls

RsH

sYsLR

sYsX

dttyLR

tytx

+=

+=

=−

=−∫By Kirchoff’s current law:

First order HPF

( ) ( )s

sFdfL

t

=

∫0

ξξNote:

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Summary

• Laplace transform similar to Fourier– Applicable to broad range of signals

– Most one-sided (causal) finite energy and power

• Particularly useful for– solving ODE’s i.e., analysing linear, time-invariant systems (e.g., circuit

analysis)

• Based on summation (integration) of– (orthogonal) exponential (basis) functions

• like Fourier series and transform

– System response scaled versions of these• amplitude and phase change (only) at each frequency