Fourier Transform Properties

Some properties to choose from if you are having difficulty....

Max Woesner

1. Find ${ℱ}[cos(w_{0}t)g(t)]$
Recall $w_0=2\pi f_0$, so ${ℱ}[cos(w_{0}t)g(t)]={ℱ}[cos(2\pi f_{0}t)g(t)]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt$
Also recall $cos(\theta )=\frac{1}{2}(e^{j\theta }+e^{-j\theta })$,so $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt$
Now $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt=\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi (f-f_0)t}g(t)dt+\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi (f+f_0)t}g(t)dt=\frac{1}{2}G(f-f_0)+\frac{1}{2}G(f+f_0)$
So ${ℱ}[cos(w_{0}t)g(t)]=\frac{1}{2}[G(f-f_0)+G(f+f_0)]$

2. Find ${ℱ}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{g}^{*}(t)h(t)dt]$

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Nick Christman

Find ${ℱ}[10^{t}g(t)e^{j2\pi f{t}_0}]$

To begin, we know that

${ℱ}[10^{t}g(t)e^{j2\pi f{t}_0}]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}10^{t}g(t)e^{j2\pi f{t}_0}{e}^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}10^{t}g(t)e^{j2\pi f(t_0-t)}dt$

But recall that $e^{j2\pi f(t_0-t)}\equiv \delta (t_0-t)\text{ or }\delta (t-t_0)$

Because of this definition, our problem has now been simplified significantly:

${ℱ}[10^{t}g(t)e^{j2\pi f{t}_0}]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}10^{t}g(t)\delta (t-t_0)dt=10^{t_0}g(t_0)$

Therefore,

${ℱ}[10^{t}g(t)e^{j2\pi f{t}_0}]=10^{t_0}g(t_0)$

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Joshua Sarris

Find ${ℱ}[sin(w_{0}t)g(t)]$

Recall $w_0=2\pi f_0$,

so expanding we have,

${ℱ}[sin(w_{0}t)g(t)]={ℱ}[sin(2\pi f_{0}t)g(t)]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt$

Also recall $sin(\theta )=\frac{1}{j2}(e^{j\theta }+e^{-j\theta })$,

so we can convert to exponentials.

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}sin(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{j2}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt$

Now integrating gives us,

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{j2}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt=\frac{1}{j2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi (f-f_0)t}g(t)dt+\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi (f+f_0)t}g(t)dt=\frac{1}{j2}G(f-f_0)+\frac{1}{j2}G(f+f_0)$

So we now have the identity,

${ℱ}[cos(w_{0}t)g(t)]=\frac{1}{j2}[G(f-f_0)+G(f+f_0)]$

orr rather

${ℱ}[cos(w_{0}t)g(t)]=\frac{1}{2}j[G(f-f_0)-G(f+f_0)]$

Reviewed by Max