Fourier Transform of product of heaviside step function and another function

  • Thread starter Dextrine
  • Start date
  • #1
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Homework Statement


A damped harmonic oscillator is driven by a force of the form f(t)=h(t) t^2 Exp(-t), where h(t) is a Heaviside step function. The Oscillator satisfies the equation x''+2x'+4x=f(t). Use pencil-and-paper methods involving Fourier transforms and inverse transforms to find the
response of the oscillator, x(t), assuming that x(0)=0 and x'(0)=1.

Homework Equations


The Fourier Transform F[f(t)]
The Inverse Fourier Transform F^(-1)[f(ω)]
Integration by parts

The Attempt at a Solution


First thing I did was take the Fourier transform of the left hand side, which I'm sure I got correct. The part I'm stuck on is taking the fourier transform of f(t). I used integration by parts using
u=t^2/e^t dv=h(t)(Exp[i ω t])
du=(2t-t^2)/e^t v=πδ(ω)+i/ω

but now when I try to complete the integration, i get an integral that does not converge because of the i/ω. There have been multiple heaviside problems that I have been working on that all end up the same, and I'm not sure where to go from here.
 
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Answers and Replies

  • #2
102
7
In general actually, how would one go about solving the fourier transform of a function with the form f(t)H(t)

for example, how could I do the fourier transform of tH(t)? Again, I've tried doing it by parts, but nothing good comes of it
 
  • #3
vela
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The way you split the integrand up is unusual. The step function allows you to write the integral as
$$\int_{-\infty}^\infty t^2e^{-t}h(t) e^{i\omega t}\,dt = \int_0^\infty t^2e^{(i\omega-1)t}\,dt,$$ which you can integrate by parts. A trick you could use to avoid integration by parts is to note that ##t^2 e^{i\omega t} = -\frac{\partial^2}{\partial \omega^2}e^{i\omega t}##, so you can say
$$\int_0^\infty t^2e^{(i\omega-1)t}\,dt = -\frac{\partial^2}{\partial \omega^2}\int_0^\infty e^{(i\omega-1)t}\,dt$$
 
  • #4
102
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Thank you so much you are a life saver. My homework became super easy after your post!
 

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