Fourier transform

[logistic_guy]

Evaluate the Fourier transform of the damped sinusoidal wave \(\displaystyle g(t ) = e^{-t}\sin( 2\pi f_c t )u(t )\) where \(\displaystyle u(t)\) is the unit step function.

 

[khansaheb]

Evaluate the Fourier transform of the damped sinusoidal wave \(\displaystyle g(t ) = e^{-t}\sin( 2\pi f_c t )u(t )\) where \(\displaystyle u(t)\) is the unit step function.

show us your effort/s to solve this problem.

 

[logistic_guy]

The Fourier transform of \(\displaystyle g(t)\) is:

\(\displaystyle \mathcal{F}\{g(t)\} = G(f) = \int_{-\infty}^{\infty}g(t)e^{-j\omega t} \ dt = \int_{-\infty}^{\infty}e^{-t}\sin( 2\pi f_c t )u(t )e^{-j2\pi f t} \ dt = \int_{0}^{\infty}e^{-t}\sin( 2\pi f_c t )e^{-j2\pi f t} \ dt\)

It is very difficult to solve this integral, so we do this trick:

Let \(\displaystyle f(t) = e^{-t}u(t)\) and \(\displaystyle z(t) = \sin( 2\pi f_c t )\)

Then

\(\displaystyle G(f) = F(f)*Z(f)\)

where \(\displaystyle *\) denotes convolution.

From basic Fourier transforms, we know that:

\(\displaystyle F(f) = \frac{1}{1 + j2\pi f}\)

\(\displaystyle Z(f) = \frac{1}{2j}\big[\delta(f - f_c) - \delta(f + f_c)\big]\)

Then,

\(\displaystyle G(f) = F(f)*Z(f) = \frac{1}{1 + j2\pi f}*\left(\frac{1}{2j}\big[\delta(f - f_c) - \delta(f + f_c)\big]\right)\)

Properties of Dirac delta function with convolution give us:

\(\displaystyle G(f) = \frac{1}{2j}\left(\frac{1}{1 + j2\pi(f - f_c)} - \frac{1}{1 + j2\pi(f + f_c)}\right)\)