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LECTURE 14: LAPLACE TRANSFORM PROPERTIES

5 Laplace transform (3 lectures): Laplace transform as Fourier transform with

convergence factor. Properties of the Laplace transform

Specific objectives for today:• Linearity and time shift properties• Convolution property• Time domain differentiation & integration

property• Transforms table

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LECTURE 14: RESOURCES

Core material SaS, O&W, Chapter 9.5&9.6

Recommended material MIT, Lecture 18

Laplace transform properties are very similar to the properties of a Fourier transform, when s=j

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REMINDER: LAPLACE TRANSFORMS Equivalent to the Fourier transform when s=j

Associated region of convergence for which the integral is finite

Used to understand the frequency characteristics of a signal (system)

Used to solve ODEs because of their convenient calculus and convolution properties (today)

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dtetxsX st)()(

)()( sXtxL

j

j

stdsesXj

tx

)(

2

1)(

Laplace transform

Inverse Laplace transform

LINEARITY OF THE LAPLACE TRANSFORM

If

and

Then

This follows directly from the definition of the Laplace transform (as the integral operator is linear). It is easily extended to a linear combination of an arbitrary number of signals

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)()( 11 sXtxL

)()( 22 sXtxL

)()()()( 2121 sbXsaXtbxtaxL

ROC=R1

ROC=R2

ROC= R1R2

TIME SHIFTING & LAPLACE TRANSFORMS If

Then

Proof Now replacing t by t-t0

Recognising this as

A signal which is shifted in time may have both the magnitude and the phase of the Laplace transform altered.

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)()( sXtxL

)()( 00 sXettx st

L

j

j

stj dsesXtx

)()( 21

)()}({ 00 sXettxL st

j

j

ststj

j

j

ttsj

dsesXe

dsesXttx

))((

)()(

0

0

21

)(21

0

ROC=R

ROC=R

EXAMPLE: LINEAR AND TIME SHIFT Consider the signal (linear sum of two time shifted sinusoids)

where x1(t) = sin(0t)u(t). Using the sin() Laplace

transform example

Then using the linearity and time shift Laplace transform properties

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)4(5.0)5.2(2)( 11 txtxtx

0}Re{)(20

20

1

ss

sX

0}Re{5.02)(20

2045.2

s

seesX ss

CONVOLUTION The Laplace transform also has the multiplication property, i.e.

Proof is “identical” to the Fourier transform convolution

Note that pole-zero cancellation may occur between H(s) and X(s) which extends the ROC

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ROC=R1

ROC=R2

ROCR1R2

)()( sXtxL

)()( sHthL

)()()(*)( sHsXthtx

L

}{1)()(

1}{1

2)(

2}{2

1)(

ssHsX

ss

ssH

ss

ssX

EXAMPLE 1: 1ST ORDER INPUT & FIRST ORDER SYSTEM IMPULSE RESPONSE

Consider the Laplace transform of the output of a first order system when the input is an exponential (decay?)

Taking Laplace transforms

Laplace transform of the output is

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asas

sX

}Re{1

)(

)()(

)()(

tueth

tuetxbt

at

bsbs

sH

}Re{,1

)(

},max{}Re{11

)( basbsas

sY

Solved with Fourier transforms when a,b>0

EXAMPLE 1: CONTINUED … Splitting into partial fractions

and using the inverse Laplace transform

Note that this is the same as was obtained earlier, expect it is valid for all a & b, i.e. we can use the Laplace transforms to solve ODEs of LTI systems, using the system’s impulse response

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},max{}Re{111

)( basbsasab

sY

)()()( 1 tuetuety btatab

)()( sHthL

EXAMPLE 2: SINUSOIDAL INPUT Consider the 1st order (possible unstable) system response with input x(t)

Taking Laplace transforms

The Laplace transform of the output of the system is therefore

and the inverse Laplace transform is

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)()cos()(

)()(

0 tuttx

tueth at

asas

sH

}Re{1

)(

0}Re{)(20

2

ss

ssX

asa

a

s

as

a

asass

ssY

11

},0max{}Re{1

)(

20

220

2

20

20

2

20

2

ataettaa

tuty

)cos()sin(

)()( 0002

02

DIFFERENTIATION IN THE TIME DOMAIN Consider the Laplace transform derivative in the

time domain

sX(s) has an extra zero at 0, and may cancel out a corresponding pole of X(s), so ROC may be larger

Widely used to solve when the system is described by LTI differential equations

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ROC=R)()( sXtxL

j

j

stdsesXj

tx

)(

2

1)(

)()(

ssXdt

tdx L

ROCR

j

j

stdsessXjdt

tdx

)(

2

1)(

EXAMPLE: SYSTEM IMPULSE RESPONSE Consider trying to find the system response (potentially

unstable) for a second order system with an impulse input x(t)=(t), y(t)=h(t)

Taking Laplace transforms of both sides and using the linearity property

where r1 and r2 are distinct roots, and calculating the inverse transform

The general solution to a second order system can be expressed as the sum of two complex (possibly real) exponentials

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)()()()(

2

2

txtcydt

tdyb

dt

tyda

)()())((

11)()}({

1))}(({

)}({)()()(

2

2

1

1

212

2

2

2

rs

k

rs

k

rsrsacbsassHtyL

cbsastyL

tLtycLdt

tdybL

dt

tydaL

)()()( 2121 tuektuekty trtr

LECTURE 14: SUMMARY Like the Fourier transform, the Laplace transform is linear and represents time shifts (t-T) by multiplying by e-sT

Convolution

Convolution in the time domain is equivalent to multiplying the Laplace transforms

Laplace transform of the system’s impulse response is very important H(s) = h(t)e-stdt. Known as the transfer function.

Differentiation

Very important for solving ordinary differential equations

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)()()(*)( sHsXthtxL

)()(

ssXdt

tdx L

ROCR

ROCR1R2

QUESTIONS Theory SaS, O&W, Q9.29-9.32 Work through slide 12 for the first order system

Where the aim is to calculate the Laplace transform of the impulse response as well as the actual impulse response

Matlab Implement the systems on slides 10 & 12 in

Simulink and verify their responses by exact calculation.

Note that roots() is a Matlab function that will calculate the roots of a polynomial expression

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)()()(

ttbydt

tdya