EE 102 spring 2001-2002 Handout #5Lecture 3The Laplace transform• definition & examples• properties & formulas– linearity– the inverse Laplace transform– time scaling– exponential scaling– time delay– derivative– integral– multiplication by t– convolution3–1Ideathe Laplace transform converts integral and differential equations intoalgebraic equationsthis is like phasors, but• applies to general signals, not just sinusoids• handles non-steady-state conditionsallows us to analyze• LCCODEs• complicated circuits with sources, Ls, Rs, and Cs• complicated systems with integrators, differentiators, gainsThe Laplace transform 3–2Complex numberscomplex number in Cartesian form: z = x + jy• x = z,thereal part of z• y = z,theimaginary part of z• j =√−1 (engineering notation); i =√−1 is polite term in mixedcompanycomplex number in polar form: z = rejφ• r is the modulus or magnitude of z• φ is the angle or phase of z• exp(jφ)=cosφ + j sin φcomplex exponential of z = x + jy:ez= ex+jy= exejy= ex(cos y + j sin y)The Laplace transform 3–3The Laplace transformwe’ll be interested in signals defined for t ≥ 0the Laplace transform of a signal (function) f is the function F = L(f)defined byF (s)=∞0f(t)e−stdtfor those s ∈ C for which the integral makes sense• F is a complex-valued function of complex numbers• s is called the (complex) frequency variable,withunits sec−1; t is calledthe time variable (in sec); st is unitless• for now,weassume f contains no impulses at t =0common n otation convention: lower case letter denotes signal; capitalletter denotes its Laplace transform, e.g., U denotes L(u), VindenotesL(vin),etc.The Laplace transform 3–4Examplelet’s find Laplace transform of f(t)=et:F (s)=∞0ete−stdt =∞0e(1−s)tdt =11 − se(1−s)t∞0=1s − 1provided we can say e(1−s)t→ 0 as t →∞,whichistrue for s>1:e(1−s)t=e−j(s)t  =1e(1− s)t= e(1− s)t• the integral defining F makes sense for all s ∈ C with s>1 (the‘region of convergence’ of F )• but the resulting formula for F makes sense for all s ∈ C except s =1we’ll ignore these (sometimes important) details and just say thatL(et)=1s − 1The Laplace transform 3–5More examplesconstant: (or unit step) f(t)=1(for t ≥ 0)F (s)=∞0e−stdt = −1se−st∞0=1sprovided we can say e−st→ 0 as t →∞,whichistrue for s>0 sincee−st=e−j(s)t  =1e−( s)t= e−( s)t• the integral defining F makes sense for all s with s>0• but the resulting formula for F makes sense for all s except s =0The Laplace transform 3–6sinusoid: first express f(t)=cosωt asf(t)=(1/2)ejωt+(1/2)e−jωtnow we can find F asF (s)=∞0e−st(1/2)ejωt+(1/2)e−jωtdt=(1/2)∞0e(−s+jω)tdt +(1/2)∞0e(−s−jω)tdt=(1/2)1s − jω+(1/2)1s + jω=ss2+ ω2(valid for s>0;finalformulaOKfors = ±jω)The Laplace transform 3–7powers of t: f(t)=tn(n ≥ 1)we’ll integrate by parts, i.e.,usebau(t)v(t) dt = u(t)v(t)ba−bav(t)u(t) dtwith u(t)=tn, v(t)=e−st, a =0, b = ∞F (s)=∞0tne−stdt = tn−e−sts∞0+ns∞0tn−1e−stdt=nsL(tn−1)provided tne−st→ 0 if t →∞,whichistrue for s>0applying the formula recusively, we obtainF (s)=n!sn+1valid for s>0;finalformulaOKforalls =0The Laplace transform 3–8Impulses at t =0if f contains impulses at t =0we choose to include them in the integraldefining F :F (s)=∞0−f(t)e−stdt(you can also choose to not include them, but this changes some formulaswe’ll see &use)example: impulse function, f = δF (s)=∞0−δ(t)e−stdt = e−stt=0=1similarly for f = δ(k)we haveF (s)=∞0−δ(k)(t)e−stdt =(−1)kdkdtke−stt=0= ske−stt=0= skThe Laplace transform 3–9Linearitythe Laplace transform is linear:iff and g are any signals, and a is anyscalar, we haveL(af)=aF, L(f + g)=F + Gi.e.,homogeneity & superposition holdexample:L3δ(t) − 2et=3L(δ(t)) − 2L(et)=3−2s − 1=3s − 5s − 1The Laplace transform 3–10One-to-one propertythe Laplace transform is one-to-one:ifL(f)=L(g) then f = g(well, almost; see below)• F determines f• inverse Laplace transform L−1is well defined(not easy to show)example (previous page):L−13s − 5s − 1=3δ(t) − 2etin other words, the only function f such thatF (s)=3s − 5s − 1is f(t)=3δ(t) − 2etThe Laplace transform 3–11what ‘almost’ means: if f and g differ only at a finite number of points(where there aren’t impulses) then F = Gexamples:• f defined asf(t)=1 t =20 t =2has F =0• f defined asf(t)=1/2 t =01 t>0has F =1/s (same as unit step)The Laplace transform 3–12Inverse Laplace transformin principle we can recover f from F viaf(t)=12πjσ+j∞σ−j∞F (s)estdswhere σ is large enough that F (s) is defined for s ≥ σsurprisingly, this formula isn’t really useful!The Laplace transform 3–13Time scalingdefine signal g by g(t)=f(at),wherea>0;thenG(s)=(1/a)F (s/a)makes sense: times are scaled by a,frequencies by 1/alet’s check:G(s)=∞0f(at)e−stdt =(1/a)∞0f(τ)e−(s/a)τdτ =(1/a)F (s/a)where τ = atexample: L(et)=1/(s − 1) soL(eat)=(1/a)1(s/a) − 1=1s − aThe Laplace transform 3–14Exponential scalinglet f be a signal and a ascalar,and define g(t)=eatf(t);thenG(s)=F (s − a)let’s check:G(s)=∞0e−steatf(t) dt =∞0e−(s−a)tf(t) dt = F (s − a)example: L(cos t)=s/(s2+1),andhenceL(e−tcos t)=s +1(s +1)2+1=s +1s2+2s +2The Laplace transform 3–15Time delaylet f be a signal and T>0;definethesignalg asg(t)=00≤ t<Tf(t − T ) t ≥ T(g is f,delayed by T seconds & ‘zero-padded’ up to T )ttt = Tf(t) g(t)The Laplace transform 3–16then we have G(s)=e−sTF (s)derivation:G(s)=∞0e−stg(t) dt =∞Te−stf(t − T ) dt=∞0e−s(τ+T )f(τ) dτ= e−sTF (s)The Laplace transform 3–17example: let’s find the Laplace transform of a rectangular pulse signalf(t)=1 if a ≤ t ≤ b0 otherwisewhere 0 <a<bwe can write f as f = f1− f2wheref1(t)=1 t ≥ a0 t<af2(t)=1 t ≥ b0 t<bi.e., f is a unit step delayed a seconds, minus a unit step delayed b secondshenceF (s)=L(f1) −L(f2)=e−as− e−bss(can check by direct integration)The Laplace transform