Fourier Transform Property review

Max Woesner

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Homework #5 - Reveiew a Fourier Transform Property

Joshua Sarris derived the following Fourier Transform Property here for Homework #4.

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

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

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My Review

Josh, Josh, it appears that you copied my code and forgot to change some necessary elements so it is works for your equation, such as the cosines on lines 4, 11, and 13.
Also, you forgot a $j\!$ in the second term of the second equation on line 9, and your identity for $sin(\theta )\!$ has a sign error. It should be:
$sin(\theta )={\frac {1}{j2}}(e^{j\theta }-e^{-j\theta })\!$
The derivation should be as follows.

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,
$\int _{-\infty }^{\infty }{\frac {1}{j2}}[e^{j2\pi f_{0}t}-e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt={\frac {1}{j2}}\int _{-\infty }^{\infty }e^{-j2\pi (f-f_{0})t}g(t)dt-{\frac {1}{j2}}\int _{-\infty }^{\infty }e^{-j2\pi (f+f_{0})t}g(t)dt={\frac {1}{j2}}G(f-f_{0})-{\frac {1}{j2}}G(f+f_{0})\!$
So we now have the identity, ${\mathcal {F}}[sin(w_{0}t)g(t)]={\frac {1}{j2}}[G(f-f_{0})-G(f+f_{0})]\!$
Your answer looks correct, but I don't know how you got it from the equation above it. Your operator between the $G\!$ terms changed from a $+\!$ to a $-\!$