136Lecture 15 Thursday March 12 Fourier Series Continued; Impulse Response; Dirac Delta function, Greens functionsand Convolutions.Response xp to a periodic force....We consider a forcing F(t) that is periodic. It may be written in its Fourier Series:F(t) = ao/2 + Σ an cos(n ωt) + Σ bn sin(n ωt)for which the coefficients a and b can be found from whatever the original description of F(t) is,using the integral formulas for a and b. E.g. if a F(t) was a square wave we found F(t) = (4/π)[ sin ω t + (1/3) sin 3ω t + (1/5) sin(5ω t) + . . . ]This force is a sum of harmonic forcings, so the steady state response is thecorresponding sum of harmonic responses:xp(t) = ao/2k + Σ (an) G(nω) cos(n ωt-φ(nω)) + Σ (bn ) G(nω) sin(n ωt-φ(nω))where the G and the φ are found from the formulas developed for the harmonic forcingsG(nω) =1 / k(1 − (nω/ωn)2)2+ (2ζnω/ωn)2φ(nω) = tan−1[2ζnωωn/ (ωn2− n2ω2)]( ζ being c/2mωn and ωn = √k/m ) This form for xp is established by appeal to the known formfor the particular solutions for each individual harmonic force composing the Fourier series forF, and to the principle of superposition: {the xp for a sum of forces is the sum of the individualxp}The above expression for the particular response is ugly, especially if the expressions for G and φwere inserted. Let us therefore focus on the meaning of it. What is the essential physics? Wenote that this xp isi) steady state; that is, it has no component which decays with time.ii) periodic with the same period as F(t). It can be seen that if t increases by T = 2π/ωthen xp is unchanged. By employing trig formulas for the cosine and sine of the sum of twoangles, the above could, if you wished, be explicitly rewritten as a Fourier series ( cosines andsines without the phases φ) . That is rarely a useful thing to do.iii) looks like the original Fourier series, but with each term amplified by a differentfactor G depending on the frequency nω of that term, and phase delayed by a different amountφ(nω) depending on the frequency nω of that term.137Resonant responses to periodic forcings.The above expressionxp(t) = ao/2k + Σ (an) G(nω) cos(n ωt-φ(nω)) + Σ (bn ) G(nω) sin(n ωt-φ(nω))will be resonant if any of the terms are resonant. That is, if for some n, nω ~ ωnatural , then theG factor for that term will be large and that term might dominate the sum. (assuming thecorresponding a and b do not both vanish. ) A forcing might be a square wave, but the responsecan be dominated by a single harmonic sinusoid if some multiple n of the frequency 2π/ω of theforcing is close to the natural frequency of the mass-spring system.e.g. consider the square wave F(t) with period 2π so that ω = 1 (indicated here with a bold line)acting on a system with small ζ and a natural frequency ωnat of about 3. The response is periodicwith period T =1 but is dominated – though not exclusively - by a term that goes like sin(3t-π/2).A forcing with period T contains in it frequencies ω that are multiples of the fundamentalω= 2π/T. The nth term of the Fourier series will be resonant if ωnatural is equal to 2nπ/T (orclose to it) for some n. Resonance occurs if the frequency of forcing f = 1/T equalsωnatural /2π, or ωnatural /4π or ωnatural /6π etc… (assuming the corresponding an or bn does notvanish)Schematically we might plot thetime-averaged energy of the steadystate response versus the frequency f =ω/2π of the periodic forcing. It willshow peaks at subharmonics of thenatural frequency ω = ωnat / nThis plot of E-steady-state versusforcing frequency f may be slightlycounter-intuitive. Why, if the138frequency of forcing is half, or a third, of the natural frequency, do we still get resonance? ( asopposed to twice or three times it) But a little thought establishes its reasonableness. Thinkof the swing set again; if you push only every other swing, you still get a good response.Complex Fourier SeriesAnother version of the Fourier series, the Complex Fourier Series, is mathematically moreabstract, but simpler too. ConsiderF(t) = Σ cn exp( inωt) where the sum is now over all n from -infinity to +infinity and the c are complex.The reality of F; F = F* impliesc -n = cn*By expansion of the exp(inωt)'s using the Euler identities one can make the followingidentifications between the coefficients of the regular Fourier Series and the complex FourierSeries:an = cn + c -n and bn = i (cn – c -n)The integral formula for the Fourier coefficients cm can be obtained very easily. Multiply bothsides of the above complex Fourier series for F(t) by exp(-i m ω t) and integrate over a fullcontiguous period T.∫F(t)exp(-imω)dt = [ Σn=−∞∞cnexp(inωt ) ]∫exp(-imω)dt On the right hand side all terms will vanish upon integration except for the one term where nhappens to equal m. This term will consist of an integral of cm over the period T. Thereforecm= (1 / T ) F(t )exp(−imωt )dt∫which , BTW, does obey c–n = cn*.Response to a complex Fourier series… Given an F(t) represented as above as a sum of complexexponentials, we recall that each such term cnexp(inωt ) gives rise to a particular solution cn%G(nω) exp(inωt ) where the complex amplification factor %G was given by %G(ω) = (1/k) / [ 1 - (ω/ωnat)2 + 2i ζ ω/ωnnat ]( see lecture 13). Thus the response to the complex Fourier series F(t) isxp(t) = Σ cn %G(nω) exp( inωt)139The notation for complex Fourier series is more compact than it is for the cosines and sines, atthe expense of having to deal with complex quantities.Query: This xp looks to be complex? Is it? Answer: No, because the terms in negative n arecomplex conjugates of the terms in positive n. Because c-n= cn* and %G(-nω) = %G(nω)*.So far we've discussed power series forces and their particular solutions, harmonic, and periodic.Another classic is the impulse response, which provides an introduction to the important topic ofGreen's functions:The impulse response G(t)G(t) is defined as the particular solution specific to a specified special initial condition. It is thex(t) that follows quiescent initial conditions (quiescent Initial conditions means xo = 0, and vo =0) upon forcing by a unit "impulse".m d2x/dt2 + c dx/dt + k x = δ(t)where δ(t) is the "Dirac