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 ![{\displaystyle {\mathcal {F}}[sin(w_{0}t)g(t)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/982c9e181460db5faafc47902bc3dd11e92fb7f8)
Recall
,
so expanding we have,
![{\displaystyle {\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\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21e5b0a54b4a33ce953b9c0b8926cc6954272b6c)
Also recall
,
so we can convert to exponentials.
![{\displaystyle \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\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c884a2c617a28b78d035b777debd2cdd6db5836d)
Now integrating gives us,
![{\displaystyle \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})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b61e8ba351acb869a7b02b195639e7c650beaf3c)
So we now have the identity,
![{\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]={\frac {1}{j2}}[G(f-f_{0})+G(f+f_{0})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92e71d5106bc977b2bb2ae0d16b2f94317ccfccb)
orr rather
![{\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]={\frac {1}{2}}j[G(f-f_{0})-G(f+f_{0})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44387854d8ef3e82c40a05710160b0e340679d8b)
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 
in the second term of the second equation on line 9, and your identity for 
has a sign error. It should be:
The derivation should be as follows.
Find
Recall ,
so expanding we have,
![{\displaystyle {\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\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6225ec01a20e83a776ded489642cdb29d069d33c)
Also recall ,
so we can convert to exponentials.
![{\displaystyle \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\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f88b7b79b8a8cc5dcc6a94c71ad76dd9feda0422)
Now integrating gives us,
![{\displaystyle \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})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac8d866fae65506a3008381940532ee85a4eaae1)
So we now have the identity,
![{\displaystyle {\mathcal {F}}[sin(w_{0}t)g(t)]={\frac {1}{j2}}[G(f-f_{0})-G(f+f_{0})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6b41814c468a20025b44e90e113ba5897671c85)
Your answer looks correct, but I don't know how you got it from the equation above it. Your operator between the 
terms changed from a 
to a 