Fourier transform of a shifted sine wave

  • #1
durandal
5
2
Homework Statement
Find the Fourier transform of sin(4t-4)
Relevant Equations
Regular Fourier transforms
This is my attempt at a solution. I have used Eulers formula to rewrite the sine function and then used the Fourier transform of complex exponentials. My solution is not correct and I don't understand if I have approached this problem correctly. Please help.

$$ \mathcal{F}\{\sin (4t-4) \} = \frac{1}{2j} \mathcal{F} \{ e^{j(4t-4)} - e^{-j(4t-4)} \} $$
$$ \frac{1}{2j}(e^{-4j} \mathcal{F}\{ e^{4jt}\} - e^{4j} \mathcal{F}\{ e^{-4jt} \}) = \frac{1}{2j}(e^{-4j} 2 \pi \delta(\omega - 4)-e^{4j} 2 \pi \delta(\omega + 4))$$
$$ \mathcal{F(\omega)} = \frac{\pi}{j}(e^{-4j} \delta(\omega - 4) - e^{4j}\delta(\omega + 4)) $$



IMG_81A2E3B3A67B-1.jpeg
 
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  • #2
Hi,
:welcome:

Well, you are not very far off when compared to this one
Perhaps a factor ##\sqrt{2\pi}## can be found somewhere ?

##\ ##
 
  • #3
BvU said:
Hi,
:welcome:

Well, you are not very far off when compared to this one
Perhaps a factor ##\sqrt{2\pi}## can be found somewhere ?

##\ ##
I tried inputting this solution but it was incorrect. I can't see where a factor of ##\sqrt{2\pi}## would come from?
 
  • #4
I figured it out, I made a mistake when I assumed that the time shift was ##(t-4)##, but the function can be rewritten as ##\sin(4t-4) = \sin(4(t-1))##, with time shift ##(t-1)##. The transform then becomes ##\frac{\pi}{j}e^{-j \omega}(\delta(\omega -4) - \delta(\omega + 4))##, which is correct.
 
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