Fourier Transform question F o sin(10t)

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priscared
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Fourier Transform of sin(10t)

Hi all,
Can some1 explain how to get the complex Fourier transform of sin(10t)

I understand how to steal it off a Fourier transform table, but i have no idea how to do it manually.

Any help anyone?
Cheers
 
Attachments
  • Exam question with red hand-drawn circle around part (c): h(t) = e^(-2t) for t >= 0 and 0 for t < 0; asks (i) Fourier transform H(f) and (ii) transform of sin(10t) h(t).
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The Fourier transform of a pure tone (the sine function) is two delta function, one at [tex]+\frac{10}{2\pi}[/tex] and the other at [tex]-\frac{10}{2\pi}[/tex] (I could have my [tex]{2\pi}[/tex]'s wrong). The last part of this question should be a simple sampling in the frequency domain of the transform of the function in part A.
 
thanks for your reply.
yeah i was way off. Thanks for the clarification
 
Hi,
When answering question above, you stated the need "The last part of this question should be a simple sampling in the frequency domain of the transform of the function in part A."

I believe i have the right rule attached.

e^(j 2pi fc t)g(t) <-> G(f-fc) (this formula is attached in jpeg)

So, once i find the Fourier transform of the first function. I simply need to shift it by fc.

I have had a go at it, i think its right. (answer attached)

If you could point me in the right direction, that'd be great. cheers
 
Attachments
  • Modulation property of the Fourier transform: F[e^(j2πf_c t) g(t)] = G(f - f_c).
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  • Handwritten working: g(t) = sin(10t)h(t) = (1/(2j)·e^(j10t) − 1/(2j)·e^(−j10t))h(t), then fc = 10/(2π) = 5/π, then the transform = (1/(2j))·H(f − 5/π) − (1/(2j))·H(f + 5/π).
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