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

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

-- I was going to make a comment on the delta identity, but after looking at it closer I think it is fine. Good job!

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:

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

But this is simply $G(f'){\mbox{ where }}f'=f-f_{0}$. Therefore,

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

• • • • PLEASE ENTER PEER REVIEW HERE ****

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

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

Rearranging terms we get:

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

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

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

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

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:

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

• • • • 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,

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

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