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

reviewed by Joshua Sarris

2. Find ${ℱ}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)h^{*}(t)dt]$
Recall $g(t)={{ℱ}}^{-1}[G(f)]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}G(f)e^{j2\pi ft}df$
Similarly, $h(t)={{ℱ}}^{-1}[H(f)]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}H(f)e^{j2\pi ft}df$
So ${ℱ}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)h^{*}(t)dt]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}G(f^{\text{'}})e^{j2\pi f^{\text{'}}t}d{f}^{\text{'}}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}H(f^{{\text{'}}^{\prime }})e^{j2\pi f^{{\text{'}}^{\prime }}t}d{f}^{{\text{'}}^{\prime }}{)}^{*}dt$
Now $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}G(f^{\text{'}})e^{j2\pi f^{\text{'}}t}d{f}^{\text{'}}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}H(f^{{\text{'}}^{\prime }})e^{j2\pi f^{{\text{'}}^{\prime }}t}d{f}^{{\text{'}}^{\prime }}{)}^{*}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}G(f^{\text{'}})\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{H}^{*}(f^{{\text{'}}^{\prime }})\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi (f^{{\text{'}}^{\prime }}-f^{\text{'}})t}dtd{f}^{{\text{'}}^{\prime }}d{f}^{\text{'}}$

Note that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi (f^{{\text{'}}^{\prime }}-f^{\text{'}})t}dt=\delta (f^{{\text{'}}^{\prime }}-f^{\text{'}})$

So ${ℱ}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)h^{*}(t)dt]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}G(f)H^{*}(f)df$

"I was going to make a comment on the delta identity, but after looking at it closer I think it is fine. One comment I have is that you might consider adding one more step, showing the delta function in the integral and pulling the integrands together to make it look like a double integral -- it isn't necessary and I understood the transition, but it helps the proof/identity look a little more complete. Good job!"

Example: $\underset{a1}{\overset{a2}{\int }}\underset{b1}{\overset{b2}{\int }}X(s^{\prime })Y(s^{″})\delta (s^{″}-s^{\prime })d{s}^{\prime }d{s}^{″}=\underset{a1}{\overset{a2}{\int }}X(s^{\prime })Y(s^{\prime })d{s}^{\prime }$

Reviewed by Nick Christman

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Nick Christman

Note: After scratching my head for a couple of hours, I decided that I would try a different Fourier Property. In fact, I chose a property that would need to be defined in order to show my second property.

1. Find ${ℱ}[g(t)e^{j2\pi f_{0}t}]$

This is a fairly straightforward property and is known as complex modulation

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

Combining terms, we get:

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

Now let's make the following substitution $\theta =f-f_0$

This now gives us a surprisingly familiar function:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)e^{-j2\pi (f-f_0)t}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)e^{-j2\pi \theta t}dt$

This looks just like $G(\theta )$!

We can now conclude that:

${ℱ}[g(t)e^{j2\pi f_{0}t}]=G(\theta )=G(f-f_0)$

PLEASE ENTER PEER REVIEW HERE

2. Find ${ℱ}[g(t-t_0)e^{j2\pi f_{0}t}]$

-- Using the above definition of complex modulation and the definition from class of a time delay (a.k.a "the slacker function"), I will attempt to show a hybrid of the two...

By definition we know that:

${ℱ}[g(t-t_0)e^{j2\pi f_{0}t}]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[g(t-t_0)e^{j2\pi f_{0}t}]e^{-j2\pi ft}dt$

Rearranging terms we get:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[g(t-t_0)e^{j2\pi f_{0}t}]e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t-t_0)e^{-j2\pi (f-f_0)t}dt$

Now lets make the substitution $\lambda =t-t_0\to t=\lambda +t_0$.
This leads us to:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t-t_0)e^{-j2\pi (f-f_0)t}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)(\lambda +t_0)}dt$

After some simplification and rearranging terms, we get:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)(\lambda +t_0)}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }{e}^{-j2\pi (f-f_0)t_0}dt$

Rearranging the terms yet again, we get:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }{e}^{-j2\pi (f-f_0)t_0}dt=e^{-j2\pi (f-f_0)t_0}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }dt]$

We know that the exponential in terms of $t_0$ is simply a constant and because of the Fourier Property of complex modualtion, we finally get:

${ℱ}[g(t)e^{j2\pi f_{0}t}]=G(f-f_0)e^{-j2\pi (f-f_0)t_0}$

PLEASE ENTER PEER REVIEW HERE

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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,

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

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,

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

or rather

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

Reviewed by Max