Fourier Transform Properties

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

Max Woesner

1. Find ${\mathcal {F}}[cos(w_{0}t)g(t)]\!$

Recall $w_{0}=2\pi f_{0}\!$, so ${\mathcal {F}}[cos(w_{0}t)g(t)]={\mathcal {F}}[cos(2\pi f_{0}t)g(t)]=\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt\!$

Also recall $cos(\theta )={\frac {1}{2}}(e^{j\theta }+e^{-j\theta })\!$,so $\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }{\frac {1}{2}}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!$

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

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

reviewed by Joshua Sarris

2. Find ${\mathcal {F}}{\bigg [}\int _{-\infty }^{\infty }g(t)h^{*}(t)dt{\bigg ]}\!$

Recall $g(t)={\mathcal {F}}^{-1}[G(f)]=\int _{-\infty }^{\infty }G(f)e^{j2\pi ft}df\!$
Similarly, $h(t)={\mathcal {F}}^{-1}[H(f)]=\int _{-\infty }^{\infty }H(f)e^{j2\pi ft}df\!$
So ${\mathcal {F}}{\bigg [}\int _{-\infty }^{\infty }g(t)h^{*}(t)dt{\bigg ]}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }G(f^{'})e^{j2\pi f^{'}t}df^{'}{\Bigg (}\int _{-\infty }^{\infty }H(f^{''})e^{j2\pi f^{''}t}df^{''}{\Bigg )}^{*}dt\!$
Now $\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }G(f^{'})e^{j2\pi f^{'}t}df^{'}{\Bigg (}\int _{-\infty }^{\infty }H(f^{''})e^{j2\pi f^{''}t}df^{''}{\Bigg )}^{*}dt=\int _{-\infty }^{\infty }G(f^{'})\int _{-\infty }^{\infty }H^{*}(f^{''})\int _{-\infty }^{\infty }e^{j2\pi (f^{''}-f^{'})t}dtdf^{''}df^{'}\!$

Note that $\int _{-\infty }^{\infty }e^{j2\pi (f^{''}-f^{'})t}dt=\delta (f^{''}-f^{'})\!$

Added step per Nick's suggestion

Substituting gives us $\int _{-\infty }^{\infty }G(f^{'})\int _{-\infty }^{\infty }H^{*}(f^{''})\int _{-\infty }^{\infty }e^{j2\pi (f^{''}-f^{'})t}dtdf^{''}df^{'}=\int _{-\infty }^{\infty }G(f^{'})\int _{-\infty }^{\infty }H^{*}(f^{''})\delta (f^{''}-f^{'})df^{''}df^{'}\!$

And $\int _{-\infty }^{\infty }G(f^{'})\int _{-\infty }^{\infty }H^{*}(f^{''})\delta (f^{''}-f^{'})df^{''}df^{'}=\int _{-\infty }^{\infty }G(f^{'})H^{*}(f^{'})df^{'}\!$

Since $f^{'}\!$ is a simply a dummy variable, we can conclude that:

${\mathcal {F}}{\bigg [}\int _{-\infty }^{\infty }g(t)h^{*}(t)dt{\bigg ]}=\int _{-\infty }^{\infty }G(f)H^{*}(f)df\!$

"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: $\int _{a1}^{a2}\int _{b1}^{b2}X(s')Y(s'')\delta (s''-s')\,ds'\,ds''=\int _{a1}^{a2}X(s')Y(s')\,ds'$

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 ${\mathcal {F}}\left[g(t)e^{j2\pi f_{0}t}\right]$

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

${\mathcal {F}}\left[g(t)e^{j2\pi f_{0}t}\right]=\int _{-\infty }^{\infty }\left[g(t)e^{j2\pi f_{0}t}\right]e^{-j2\pi ft}\,dt$

Combining terms, we get:

$\int _{-\infty }^{\infty }\left[g(t)e^{j2\pi f_{0}t}\right]e^{-j2\pi ft}\,dt=\int _{-\infty }^{\infty }g(t)e^{-j2\pi (f-f_{0})t}\,dt$

Now let's make the following substitution $\displaystyle \theta =f-f_{0}$

This now gives us a surprisingly familiar function:

$\int _{-\infty }^{\infty }g(t)e^{-j2\pi (f-f_{0})t}\,dt=\int _{-\infty }^{\infty }g(t)e^{-j2\pi \theta t}\,dt$

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

We can now conclude that:

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

PLEASE ENTER PEER REVIEW HERE

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

-- 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:

${\mathcal {F}}\left[g(t-t_{0})e^{j2\pi f_{0}t}\right]=\int _{-\infty }^{\infty }\left[g(t-t_{0})e^{j2\pi f_{0}t}\right]e^{-j2\pi ft}\,dt$

Rearranging terms we get:

$\int _{-\infty }^{\infty }\left[g(t-t_{0})e^{j2\pi f_{0}t}\right]e^{-j2\pi ft}\,dt=\int _{-\infty }^{\infty }g(t-t_{0})e^{-j2\pi (f-f_{0})t}\,dt$

Now lets make the substitution $\lambda =t-t_{0}\rightarrow t=\lambda +t_{0}$.
This leads us to:

$\int _{-\infty }^{\infty }g(t-t_{0})e^{-j2\pi (f-f_{0})t}\,dt=\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})(\lambda +t_{0})}\,dt$

After some simplification and rearranging terms, we get:

$\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})(\lambda +t_{0})}\,dt=\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})\lambda }e^{-j2\pi (f-f_{0})t_{0}}\,dt$

Rearranging the terms yet again, we get:

$\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})\lambda }e^{-j2\pi (f-f_{0})t_{0}}\,dt=e^{-j2\pi (f-f_{0})t_{0}}\left[\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})\lambda }\,dt\right]$

We know that the exponential in terms of $\displaystyle t_{0}$ is simply a constant and because of the Fourier Property of complex modualtion, we finally get:

${\mathcal {F}}\left[g(t)e^{j2\pi f_{0}t}\right]=G(f-f_{0})e^{-j2\pi (f-f_{0})t_{0}}$

PLEASE ENTER PEER REVIEW HERE

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Joshua Sarris

Find ${\mathcal {F}}[sin(w_{0}t)g(t)]\!$

Recall $w_{0}=2\pi f_{0}\!$,

so expanding we have,

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

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

so we can convert to exponentials.

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

Now integrating gives us, ($**$ I believe you are missing 'j' in the denominator of the second term)

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

So we now have the identity,

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

or rather

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

Reviewed by Max

Also reviewed by Nick Christman -- Looks good. I found one typo (I think), see $**$. Good job Josh!