Pendulum with Dirac Comb excitationlθ¨+bθ˙+gθ=G,∑−∞∞δ(t−nT)where l, b, g, G are constants and T=2πω.Show that the resulting motion is given byθ(t)=GTlω2+2Gcos⁡(ωt−π2)Tbω+ [terms with frequencies ≥2ω]and explain why the higher frequency terms are supressed.My first take was to rearrange toθ¨+blθ˙+ω2θ=Gl,∑−∞∞δ(t−nT)where ω=glThen, taking the Laplace transform of both sides I gotΘ(s)=Gl,ω2−(b2l)2,ω2−(b2l)2(s+b2l)2−((b2l)2−ω2),11−e−sTwhich, as far as I'm concerned transforms toθ(t)=Gl,ω2−(b2l)2∑n=0∞,H(t−nT),e−b2l(t−nT)sin⁡(ω2−(b2l)2(t−nT))And, assuming that this is a correct form of the solution, I can't see how that is equivalent with the function given in the question. I reckon it has something to do with using Fourier series/transform instead? If so, I'm not sure how to do that. Or, is there a way to convert my solution into the given one?I've been struggling with this for a good few days now, so any help would be much appreciated.

Question

Pendulum with Dirac Comb excitationlθ¨+bθ˙+gθ=G,∑−∞∞δ(t−nT)where l, b, g, G are constants and T=2πω.Show that the resulting motion is given byθ(t)=GTlω2+2Gcos⁡(ωt−π2)Tbω+ [terms with frequencies ≥2ω]and explain why the higher frequency terms are supressed.My first take was to rearrange toθ¨+blθ˙+ω2θ=Gl,∑−∞∞δ(t−nT)where ω=glThen, taking the Laplace transform of both sides I gotΘ(s)=Gl,ω2−(b2l)2,ω2−(b2l)2(s+b2l)2−((b2l)2−ω2),11−e−sTwhich, as far as I&#039;m concerned transforms toθ(t)=Gl,ω2−(b2l)2∑n=0∞,H(t−nT),e−b2l(t−nT)sin⁡(ω2−(b2l)2(t−nT))And, assuming that this is a correct form of the solution, I can&#039;t see how that is equivalent with the function given in the question. I reckon it has something to do with using Fourier series/transform instead? If so, I&#039;m not sure how to do that. Or, is there a way to convert my solution into the given one?I&#039;ve been struggling with this for a good few days now, so any help would be much appreciated.

Ezakhenitne · Accepted Answer

Step 1The distinction between apparent solutions via Laplace transform and via Fourier transform depends meaningfully on the function-space that the &quot;target&quot; function is in, and the desired solution is in.Suppressing some of the constants to reduce clutter, consider the equation&amp;nbsp;θ&amp;nbsp;″+aθ′+θ=∑n∈ZδnUnder various external context assumptions, we might suppose that the solution θ is at worst a tempered distribution (the right-hand side is such). This excludes exponentially-growing (either forward or backward) solutions&amp;nbsp;θ. In particular, unless the constants a,b on the left-hand side are such that solutions to the characteristic equation&amp;nbsp;λ2+aλ+b=0, the homogeneous equation has no tempered-distributions solutions (except 0).Step 2The Poisson summation formula is the assertion that&amp;nbsp;∑nδn&amp;nbsp;is its own Fourier transform. Thus, again up to essentially irrelevant constants,-x2θ^+bixθ^+cθ^&amp;nbsp;=&amp;nbsp;∑nδnwhich suggest thatθ^&amp;nbsp;=&amp;nbsp;∑nδn-x2+bix+c&amp;nbsp;=&amp;nbsp;∑nδn-n2+bin+cTransforming back, again suppressing constants,&amp;nbsp;θ(t)&amp;nbsp;=&amp;nbsp;∑ne2πint-n2+bin+c.

Pendulum with Dirac Comb excitation l \ddot\theta+b\dot \theta+g\theta=G\,\sum_{-\infty}^\inf

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