Why Laplace Transforms?

Why Laplace transforms?First-order RC cctKVL vS (t ) v R (t ) vC (t ) 0instantaneous for each tt 0VA vS-R vR -i(t)v c-CSubstitute element relationsvS (t ) V Au (t ), vR (t ) Ri (t ), i (t ) CdvC (t )dtOrdinary differential equation in terms of capacitor voltagedvC (t )RC vC (t ) V Au (t )dt1Laplace transform RC[ sVC ( s ) vC (0)] VC ( s ) V AsvC (0)V A / RC Solve VC ( s ) s ( s 1 / RC ) s 1 / RC& ,# t t)RCRC'' vC (0)eInvert LT vC (t ) V A **1 e!u (t ) Volts(% "MAE40 Linear Circuits133

Time domain (t domain)LinearcctDifferentialequationLaplace transformsComplex frequency domain(s domain)Laplacetransform nAlgebraictechniquesInverse Laplacetransform L-1ResponsetransformThe diagram commutesSame answer whichever way you goMAE40 Linear Circuits134

Laplace Transform - definitionFunction f(t) of timePiecewise continuous and exponential orderf (t ) Kebt F ( s) ò f (t )e st dt0-0- limit is used to capture transients and discontinuities at t 0s is a complex variable (s jw)There is a need to worry about regions of convergence ofthe integralUnits ofs are sec-1 HzA frequencyIff(t) is volts (amps) then F(s) is volt-seconds (amp-seconds)MAE40 Linear Circuits135

Laplace transform examplesStep function – unit Heavyside Functionì0, for t 0After Oliver Heavyside (1850-1925)u (t ) íî1, for t ³ 0Exponential functionAfter Oliver Exponential (1176 BC- 1066 BC)Delta (impulse) function d(t)MAE40 Linear Circuits136

Laplace transform examplesStep function – unit Heavyside Functionì0, for t 0After Oliver Heavyside (1850-1925)u (t ) íî1, for t ³ 0 F ( s ) u (t )e st dt e st dt 0 0 e st s 0e (σ jω )t σ jω 01if σ 0sExponential functionAfter Oliver Exponential (1176 BC- 1066 BC)Delta (impulse) function d(t)MAE40 Linear Circuits137

Laplace transform examplesStep function – unit Heavyside Functionì0, for t 0After Oliver Heavyside (1850-1925)u (t ) íî1, for t ³ 0 F ( s ) u (t )e st dt e st dt 0 0 e st s 0e (σ jω )t σ jω 01if σ 0sExponential functionAfter Oliver Exponential (1176 BC- 1066 BC) F(s) e ( s α )t αt ste dt 0 e0 (s α )tedt s α01 if σ αs αDelta (impulse) function d(t)MAE40 Linear Circuits138

Laplace transform examplesStep function – unit Heavyside Functionì0, for t 0After Oliver Heavyside (1850-1925)u (t ) íî1, for t ³ 0 F ( s ) u (t )e st dt e st dt 0 0 e st s 0e (σ jω )t σ jω 01if σ 0sExponential functionAfter Oliver Exponential (1176 BC- 1066 BC) F(s) e ( s α )t αt ste dt 0 e0 (s α )tedt s α01 if σ αs αDelta (impulse) function d(t) F ( s ) δ (t )e st dt 1 for all s0 MAE40 Linear Circuits139

Laplace Transform Pair TablesSignalWaveformimpulseδ (t )stepu (t )ramptu (t )exponentiale αt u (t )damped rampte αt u (t )sinesin ( βt ) u (t )cosinecos( βt )u (t )damped sinee αt sin ( βt )u (t )damped cosinee αt cos( βt )u (t )MAE40 Linear CircuitsTransform11s1s21s α1( s α ) 2βs2 β 2ss2 β 2β( s α ) 2 β 2s α( s α ) 2 β 2140

Laplace Transform PropertiesLinearity – absolutely critical propertyFollows from the integral definitionL{Af1 (t ) Bf 2 (t )} AL{f1 (t )} BL{f 2 (t )} AF1 ( s ) BF2 ( s )ExampleL(Acos(βt)) MAE40 Linear Circuits141

Laplace Transform PropertiesLinearity – absolutely critical propertyFollows from the integral definitionL{Af1 (t ) Bf 2 (t )} AL{f1 (t )} BL{f 2 (t )} AF1 ( s ) BF2 ( s )Example A' AAL(A cos(βt)) L& (e e )) L (e ) L (e )%2( 22A 1A 1 2 s jβ 2 s jβAs s βjβt2MAE40 Linear Circuits jβtjβt jβt2142

Laplace Transform PropertiesIntegration propertyProof&t# F (s )L % f (τ )dτ " s 0!/t, )t& stL . f (τ ) dτ ' f (τ ) dτ e dt-0* 0 (0%MAE40 Linear Circuits143

Laplace Transform PropertiesIntegration propertyProofDenoteso&t# F (s )L % f (τ )dτ " s 0!/t, )t& stL . f (τ ) dτ ' f (τ ) dτ e dt-0* 0 (0%t e stx , and y f (τ ) dτs0dxdy st e , and f (t )dtdt )t& ) e st t&1 st dt f(τ)dτ f(t)eIntegrate by parts L ' f (τ ) dτ ' s'(0 % '( % 0 s 00MAE40 Linear Circuits144

Laplace Transform PropertiesDifferentiation Propertydf (t ) 'L&# sF ( s ) f (0 )% dt "Proof via integration by parts again " df (t)% df (t) st stL#e dt f (t)e s f (t)e st dt& dt '0 0 0 dt sF(s) f (0 )[]Second derivative (/ d 2 f (t ) %/df (t ) % df( d . df (t ) %(L' L' , sL ' (0 ) )& dt # dt& dt - dt * #/& dt 2 /# s 2 F ( s ) sf (0 ) f "(0 )MAE40 Linear Circuits145

Laplace Transform PropertiesGeneral derivative formula" d m f (t) % mm 1m 2(m 1))L# sF(s) sf(0 ) sf(0 ) f(0 )&m dt ' Translation propertiess-domain translationL{e αt f (t )} F ( s α )t-domain translationL{f (t a )u (t a )} e as F ( s ) for a 0MAE40 Linear Circuits146

Laplace Transform PropertiesInitial Value PropertyFinal Value Propertylim f (t ) lim sF ( s )t 0 s lim f (t ) lim sF ( s )t s 0Caveats:Laplace transform pairs do not always handlediscontinuities properlyOften get the average valueInitial value property no good with impulsesFinal value property no good with cos, sin etcMAE40 Linear Circuits147

Rational FunctionsWe shall mostly be dealing with LTs which arerational functions – ratios of polynomials in sbm s m bm 1s m 1 b1s b0F (s) an s n an 1s n 1 a1s a0( s z1 )( s z2 ) ( s zm ) K( s p1 )( s p2 ) ( s pn )pi are the poles and zi are the zeros of the functionK is the scale factor or (sometimes) gainA proper rational function has n³mA strictly proper rational function has n mAn improper rational function has n mMAE40 Linear Circuits148

A Little Complex AnalysisWe are dealing with linear cctsOur Laplace Transforms will consist of rational functions(ratios of polynomials in s) and exponentials like e-stThese arise from discrete component relations of capacitors and inductors the kinds of input signals we apply– Steps, impulses, exponentials, sinusoids, delayedversions of functionsRational functions have a finite set of discrete polese-st is an entire function and has no poles anywhereTo understand linear cct responses you need to look atthe poles – they determine the exponential modes inthe response circuit variables.Two sources of poles: the cct – seen in the response to Icsthe input signal LT poles – seen in the forced responseMAE40 Linear Circuits149

Residues at polesFunctions of a complex variable with isolated, finite order poleshave residues at the polesSimple pole: residue Multiple pole: residue lim ( s a) F ( s)s a1dlim((s a) F(s))(m 1)!dsm 1ms am 1The residue is the c-1 term in the Laurent Series Bundle complex conjugatepole pairs into second-order terms ifyou want( s α jβ )( s α jβ ) s 2 2αs α 2 β 2[()]but you will need to be carefulInverse Laplace Transform is a sum of complex exponentialsFor circuits the answers will be realMAE40 Linear Circuits150

Inverting Laplace Transforms in PracticeWe have a table of inverse LTsWrite F(s) as a partial fraction expansionF(s) bm sm bm 1sm 1 b1s b0an sn an 1sn 1 a1s a0(s z1 )(s z2 ) (s zm ) K(s p1 )(s p2 ) (s pn )αqα1α2α 31α 32α 33 . 23(s p1 ) (s p2 ) (s p3 ) (s p3 ) (s p3 )s pq()Now appeal to linearity to invert via the table Surprise!Computing the partial fraction expansion is best done bycalculating the residuesMAE40 Linear Circuits151

Inverting Laplace TransformsCompute residues at the poles1d m 1 &lim( s a) m F ( s)#!"(m 1)! s a ds m 1 %lim ( s a ) F ( s )s a2 s 2 5s 2( s 1) 2 ( s 1) 3213 Example332s 1 (s 1) (s 1)3(s 1)(s 1)MAE40 Linear Circuits152

Inverting Laplace TransformsCompute residues at the poles1d m 1 &lim( s a) m F ( s)#!"(m 1)! s a ds m 1 %lim ( s a ) F ( s )s a2 s 2 5s 2( s 1) 2 ( s 1) 3213 Example332s 1 (s 1) (s 1)3(s 1)(s 1)lims -1( s 1)3 (2 s 2 5s )( s 1)3 -3d é ( s 1)3 ( 2 s 2 5s ) ùêú 1lim3s -1 ds êëúû( s 1)1d 2 é ( s 1)3 ( 2 s 2 5s ) ùêú 2lim2! s -1 ds 2 êëúû( s 1)32-1 é 2 s 5s ùú e t (2 t - 3t 2)u (t )L êê 3úë ( s 1) ûMAE40 Linear Circuits153

T&R, 5th ed, Example 9-12 3)20(sFind the inverse LT of F ( s) ( s 1)( s 2 2s 5)MAE40 Linear Circuits154

T&R, 5th ed, Example 9-12 3)20(sFind the inverse LT of F ( s) ( s 1)( s 2 2s 5)k1k2k 2*F (s) s 1 s 1 j2 s 1 j220( s 3)k1 lim ( s 1) F ( s ) 102s 1s 2 s 5 s 1k2 20( s 3)lim(s 1 2 j ) F (s) ( s 1)( s 1 2 j ) s 1 2 js 1 2 j5j π 5 5 j 5 2e 455&( 1 j 2)t j π( 1 j 2)t j π #4 5 2e4 !u (t )f (t ) 10e t 5 2e !%"5π #& t t 10e 10 2e cos(2t )!u (t )4 "%MAE40 Linear Circuits155

Not Strictly Proper Laplace Transforms3 2 8s6s12sFind the inverse LT of F ( s) s 2 4s 3MAE40 Linear Circuits156

Not Strictly Proper Laplace Transforms3 2 8s6s12sFind the inverse LT of F ( s) s 2 4s 3Convert to polynomial plus strictly proper rational functionUse polynomial divisions 2F (s) s 2 2s 4s 30.5 0.5 s 2 s 1 s 3Invert as normaldδ (t )&f (t ) 2δ (t ) 0.5e t 0.5e 3t #!u (t )% dt"MAE40 Linear Circuits157

Multiple PolesLook for partial fraction decompositionK ( s z1 )k1k 21k 22F (s) 2 s p2( s p1 )( s p2 )1 s p2 ( s p2 )Ks Kz1 k1 ( s p2 ) 2 k 21 ( s p1 )( s p2 ) k 22 ( s p1 )Equate like powers of s to find coefficientsk1 k 21 0 2k1 p2 2k 21 ( p1 p2 ) k 22 Kk1 p 2 k12 p1 p2 k 22 p1 Kz12SolveMAE40 Linear Circuits158

Recall motivating example for LTFirst-order RC cctKVL vS (t ) v R (t ) vC (t ) 0instantaneous for each tt 0VA vS-R vR -i(t)v c-CSubstitute element relationsvS (t ) V Au (t ), vR (t ) Ri (t ), i (t ) CdvC (t )dtOrdinary differential equation in terms of capacitor voltagedvC (t )RC vC (t ) V Au (t )dt1Laplace transform RC[ sVC ( s ) vC (0)] VC ( s ) V AsvC (0)V A / RC Solve VC ( s ) s ( s 1 / RC ) s 1 / RC& ,# t t)RCRC'' vC (0)eInvert LT vC (t ) V A **1 e!u (t ) Volts(% "MAE40 Linear Circuits159

An Alternative s-Domain Approacht 0VA vS-R vR -Ri(t)v c-C1VA sI(s)1sC Vc(s)- vC ( 0)sTransform the cct element relationsWork in s-domain directlyOK since L is linearvC (0)1Impedance sourceVC ( s ) IC (s) CssAdmittance sourceI C ( s ) sCVC ( s ) CvC (0)KVL in s-DomainMAE40 Linear Circuits1sCRVC ( s ) CRvC (0) VC ( s ) V As160

Time-varying inputsSuppose vS(t) VAcos(bt), what happens?t 0 vS-i(t)R vR -V cos(βt)v c-CsV A 22s βI(s)RVc(s)A-KVL as before1sC vC ( 0)ssV A( RCs 1)VC ( s ) RCvC (0) 2s β2sV AvC (0)RCVC ( s ) 2 2( s β )( s 1 ) s 1 RCRC t t&#VAVASolve vC (t ) cos( βt θ ) e RC vC (0)e RC ! u (t )221 (βRC)1 (βRC)%"161MAE40 Linear Circuits

135 MAE40 Linear Circuits Laplace Transform -definition Function f(t)of time Piecewise continuous and exponential order 0-limit is used to capture transients and discontinuities at t 0sis a complex variable (s jw)