- #1

schniefen

- 178

- 4

- Homework Statement
- Find the Fourier transform of ##f(t)=e^{-a |t|}\cos{(bt)}##, where ##a## and ##b## are positive constants.

- Relevant Equations
- I use the convention ##\tilde{f}(\omega)= \int_{-\infty}^{\infty}f(t)e^{-i\omega t} \mathrm{d}t##. Also, I can use the fact that ##f(t)=-i\Theta(t)e^{-i\omega_0 t-\gamma t}## has Fourier transform ##\tilde{f}(\omega)=\frac{1}{\omega-\omega_0+i\gamma}##, although this is not explicitly hinted at.

First,

We can get rid of the absolute value by splitting the integral up

My strategy then is to rewrite the two integrands on the form ##-i\Theta(t)e^{-i\omega_0 t-\gamma t}## and use the known formula, as given under Relevant equations. Is this an approach you would have taken?

##\tilde{f}(\omega)=\int_{-\infty}^{\infty}e^{a|t|}\cos(bt)e^{-i\omega t} \mathrm{d}t##

We can get rid of the absolute value by splitting the integral up

##\int_{-\infty}^{0}e^{at}\cos(bt)e^{-i\omega t} \mathrm{d}t+ \int_{0}^{\infty}e^{-at}\cos(bt)e^{-i\omega t} \mathrm{d}t##

Using ##\cos(x)=\frac12(e^{ix}+e^{-ix})##, the first integral reduces to

##\frac12\int_{-\infty}^{0}e^{at}(e^{ibt}+e^{-ibt})e^{-i\omega t}\mathrm{d}t=\frac12\int_{-\infty}^{0}e^{t(a+i(b-\omega))}\mathrm{d}t+\frac12\int_{-\infty}^{0}e^{t(a+i(-b-\omega))}\mathrm{d}t ##

My strategy then is to rewrite the two integrands on the form ##-i\Theta(t)e^{-i\omega_0 t-\gamma t}## and use the known formula, as given under Relevant equations. Is this an approach you would have taken?