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

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

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Nick Christman

Find ${\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]$

To begin, we know that

${\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}e^{-j2\pi ft}\,dt=\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi f(t_{0}-t)}\,dt$

But recall that $e^{j2\pi f(t_{0}-t)}\equiv \delta (t_{0}-t){\mbox{ or }}\delta (t-t_{0})$

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

${\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=\int _{-\infty }^{\infty }10^{t}g(t)\delta (t-t_{0})\,dt=10^{t_{0}}g(t_{0})$

Therefore,

${\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=10^{t_{0}}g(t_{0})$

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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,

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

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

so we can convert to exponentials.

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

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,

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

orr rather

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

Reviewed by Max