# Fourier transform of ##e^{-a |t|}\cos{(bt)}##

• schniefen
In summary, the first equation is simplified by reversing the order of the integrals, and the second equation is simplified by taking the derivative with respect to time of the real part of the first equation.
schniefen
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,

##\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?

I think you are completely justified in carrying out the integration just as you would if you were to integrate ## \int e^{at} \ dt ##, without resorting to a formula for the F.T. Consider the Euler formula ## e^{ix}=\cos{x}+i \sin{x} ## for the details for making the integration valid when complex numbers are involved. Edit: e.g. You can use the complex numbers to do the integration, and then verify that it gives the correct result for ## \int e^{-at} \cos(bt) \, dt ## by taking the derivative of the real part of the result you get for ## \int e^{-at} e^{ibt} \, dt ##, and see that you do indeed get ## e^{-at} \cos(bt) ##.

for a recent homework posting that is somewhat related.

Last edited:
schniefen
schniefen said:
##\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?
Starting from two integrals in the last line, change intgral variables from ##t## to ##-t## so that integrand become ##[0,+\infty)##.
Then you may write it
$$\int_0^{+\infty} g_k(t)dt = \int_{-\infty}^{+\infty} g_k(t)\theta (t) dt$$
where
$$g_1(t)=e^{-ta-i(b-\omega)t}$$
$$g_2(t)=e^{-ta-i(-b-\omega)t}$$
so that you may make use of what you prepared.

schniefen

## What is the Fourier transform of $$e^{-a |t|}\cos{(bt)}$$?

The Fourier transform of $$e^{-a |t|}\cos{(bt)}$$ is given by $$\frac{a}{a^2 + (b - \omega)^2} + \frac{a}{a^2 + (b + \omega)^2}$$, where $$\omega$$ is the frequency variable in the Fourier domain.

## How do you derive the Fourier transform of $$e^{-a |t|}\cos{(bt)}$$?

To derive the Fourier transform, you can use the property that the Fourier transform of $$e^{-a|t|}$$ is $$\frac{2a}{a^2 + \omega^2}$$ and apply the modulation property. The cosine term can be expressed using Euler's formula as a sum of exponentials, and the linearity of the Fourier transform can be used to find the result.

## What are the conditions on $$a$$ and $$b$$ for the Fourier transform to exist?

The parameter $$a$$ must be a positive real number to ensure that $$e^{-a|t|}$$ is integrable over all $$t$$. The parameter $$b$$ can be any real number, as it represents the frequency of the cosine function, which does not affect the integrability of the overall expression.

## Why is the absolute value function used in $$e^{-a |t|}$$?

The absolute value function in $$e^{-a |t|}$$ ensures that the function is even and exponentially decays for both positive and negative values of $$t$$. This symmetry simplifies the Fourier transform calculation and ensures the function is integrable over the entire real line.

## Can the Fourier transform of $$e^{-a |t|}\cos{(bt)}$$ be used in practical applications?

Yes, the Fourier transform of $$e^{-a |t|}\cos{(bt)}$$ is frequently used in signal processing and communications. It can model damped oscillatory signals and is useful in analyzing systems with exponential decay and harmonic components. Applications include filter design, spectral analysis, and solving differential equations.

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