# Inverse Laplace Transform-Theory of Signals And Systems-Lecture Slides, Slides for Signals and Systems Theory. Birla Institute of Technology and Science

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This lecture is part of lecture series for Signals and Systems course. Dr. Aishwarya Vyasa delivered this lecture at Birla Institute of Technology and Science. It includes: Inverse, Laplace, Transform, Shift, Property, R...
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Lecture #18

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How Do We Perform Inverse Laplace Transform?

• In 6.003, we will only deal with Laplace transform that are: 1) Rational, i.e.X(s) = N(s)/D(s) ;

And/or: 2) exponential, i.e.X(s) = e-sT .

• For case 2), use shift property

• For case 1), use PFE.

• For case 1) & 2), i.e.

x(t ! T )" # $e! sTX(s) (Similar to the FT property x(t ! T )" #$ e ! j%T

X( j%)

X(s) = e !sT

N(s) / D(s)( ) X 1 (s)"x

1 (t )

1 2 4 3 4 4

Then b

x(t) = x 1 ( t)

t!t"T = x1 (t " T )

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Inverse Laplace Transforms Via Partial Fraction Expansion and Properties

Example:

Three possible ROC’s — corresponding to three different signals

Recall

X(s) = s + 3

s +1( ) s ! 2( ) =

A

s +1 +

B

s ! 2

A = ! 2

3 , B =

5

3

1

s + a , Re(s) < !a " # $! e !at u(!t) left - sided and 1 s + a , Re(s) > !a " #$ e

!at u(t) right - sided

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Example (cont.) ROC I: — Left-sided signal.

x(t) =

ROC II: — Two-sided signal, has Fourier Transform.

x(t) =

ROC III: — Right-sided signal.

x(t) =

! Ae !t u(!t) ! Be

2t u(!t)

! 2

3 e ! t u(t) +

5

3 e

2t u(!t)"

# $% & ' ( 0 as t( ± ) Ae ! t u(t) + Be 2t u(t) = 2 3 e ! t ! 5 3 e 2t" #$

% & ' u(!t) Diverges as t( !)

= ! 2

3 e !t

+ 5

3 e

2t" # $% & ' u(t) Diverges as t( + ) 6 Properties of Laplace Transforms • Many parallel properties of the CTFT, but for Laplace transforms we need to determine implications for the ROC • For example: Linearity ROC at least the intersection of ROC’s of X1(s) and X2(s) ROC can be bigger (due to pole-zero cancellation) E.g. x1(t) = x2(t) and a = -b Then a x1(t) + b x2(t) = 0 → X(s) = 0 ⇒ ROC entire s-plane ax 1 ( t) + bx 2 (t) ! " # aX 1 (s) + bX 2 (s) docsity.com 4 7 Time Shift x(t ! T ) " #$ e !sT

X(s) , same ROC as X(s)

e ! sT

s + 2 , Re(s) > !2 " # $e !2 t u(t) t# t!T ! T = " 3 e 3s s + 2 , Re(s) > "2 #$ % e"2(t+3)u(t + 3)

E.g.,

e3s

s + 2 , Re(s) > !2 " # $? Causal or not? 8 Time-Domain Differentiation x(t) = 1 2!j X(s)e st ds " # j$

"+ j$% , dx(t) dt = 1 2!j sX(s)e st ds " # j$

" + j$% ROC could be bigger than the ROC of X(s), if there is pole-zero cancellation. E.g. x(t) = u(t) ! " # 1 s , Re(s) > 0 s-Domain Differentiation !tx(t ) " #$ dX(s)

ds , with same ROC as X(s) (Derivation is

similar to d

dt

! s)

!

dx(t) dt

" # $sX(s), with ROC containing the ROC of X(s) dx(t) dt = !(t ) " #$ 1 = s %

1

s ROC = entire s & plane

E.g. te !at u( t) " # $! d ds 1 s + a % & ' ( ) * = 1 s + a( )2 , Re(s) > !a docsity.com 5 9 Convolution Property x(t) h(t) y(t) = h( t)! x(t) For x(t)! " # X(s), y(t)! " # Y(s), h(t)! " # H(s) • This is the main reason why Laplace transform is useful in solving problems involving LTI systems, just like the Fourier transform. However, because Laplace transform and its inverse transform are not symmetric, the reverse relation x(t)•h(t) X(s)∗H(s) is not true, different from Fourier transform. Thus, L-transform is not useful in modulation and sampling. × Then Y (s) = H(s) ! X(s) 10 Convolution Property (cont.) ROC of Y(s) = H(s)X(s), at least the overlap of the ROC’s of H(s) & X(s) a) ROC could be empty if there is no overlap between the two ROCs E.g. x(t) = etu(t) Re(s) > 1 and h(t) = -e-tu(-t) Re(s) < -1 b) ROC could be larger than the overlap of the two. E.g. x(t)∗h(t) = δ(t) ⇒ ROC entire s-plane. docsity.com 6 11 The System Function of an LTI System h(t) ! " # H(s) — the system function The system function characterizes the system ⇓ System dynamics correspond to properties of H(s) and its ROC A first example: System is stable ⇒ The frequency response H(jω), which is the Fourier transform of h(t), is well defined. ! " # h(t) dt$%

+%

& < % ' ROC of H (s)

includes the j( axis

x(t) h(t) y(t)

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Geometric Evaluation of Rational Laplace Transforms

Example #1: X1(s) = s - a A first-order zero

Graphic evaluation of X1(s1) = s1 - a

⇒|X1(s1)| = |s1 - a| & X1(s1) = ∠(s1 - a)

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Example #2: A first-order pole

X 2 (s) =

1

s ! a =

1

X 1 (s)

Example #3: A higher-order rational Laplace transform

and !X(s) = !M + ! i=1

R

" s # $i( ) # ! j=1 P " s #% j( ) ! X2 (s) = 1 X1(s) log X2 (s) = " logX1(s)( ) and !X 2 (s) = "!X 1 (s) X(s) = M s ! "i( )i=1 R # s !$ j( )j=1

P

#

X(s) = M s ! "ii=1

R

# s ! $jj=1 P # 14 First-Order System H(s) = 1 s! +1 = 1 / ! s +1 / ! , Re(s) > " 1 ! Graphical evaluation of H(jω): H( j! ) = 1/ " j! + 1 / " h(t) = 1 ! e " t! u(t) s(t) = [1 " e " t! ] u(t ) !H( j") = #$ = # tan#1 "

1 / %

& ' (

) * + = # tan#1("% )

| H( j! ) | = 1 / "

! 2 + (1 / " )2 =

1 ! = 0

1 / 2 ! = 1 / "

~ 1 /!" ! >>1 / "

#

$% & % = 0 ! = 0 "# / 4 ! =1 /$

"# / 2 ! >>1 / $% & ' ( ' docsity.com 8 15 Bode Plot of the first-order system H( j! ) = 1/ " j! + 1 / " 20 dB/decade changes by -π/2 = 0 ! = 0 "# / 4 ! =1 /$

"# / 2 ! >>1 / $% & ' ( ' | H( j! ) | = 1 / " ! 2 + (1 / " )2 = 1 ! = 0 1/ 2 ! =1 / " ~1 / !" ! >>1 / " #$ %

& %

!H( j") = #$= # tan #1 ("% ) 16 Second-Order System (causal) E.g.LCR circuit, a mechanical system with a spring and friction H(s) = ! n 2 s 2 + 2"!ns +!n 2 ROC Re(s) > Re(pole) 0 < ! <1 " complex poles ! =1 " Underdamped double pole at s = #$n

! >1 "

Critically damped

2 poles on negative real axis

Overdamped

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Demo Pole-zero diagrams, frequency response, and step response of first-order and second-order CT causal systems

Important point: Dynamics of a rational system is completely characterized by its pole-zero diagram (except for a prefactor).

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Bode Plot of a Second-order System

40 dB/decade Top is flat when ζ= 1/√ 2 = 0.707 ⇒ a LPF for ω < ωn

changes by -π

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Unit-impulse and Unit-step Response of a Second-order System

No oscillations when ζ ≥ 1

⇒ Critically (=) and over (>) damped.

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Next lecture covers O&W pp. 698-720

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First-order All-Pass System

H(s) = s ! a

s + a , Re(s) > !a (a > 0)

(a) Two vectors have

the same length

= (! "# 2 ) "#

2

(b) !H( j" ) = #1 $#2 = ! " 2# 2 = ! " = 0 ! / 2 " = a ~ 0 " >> a #$ %

& %

⇒ |H(jω)| = 1

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