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I would like to calculate the inverse Fourier transform of the following

$$H(f) = \frac{1-e^{-2\pi i f t}}{2\pi i f}$$

Can anyone tell me and explain to me how to do that? I don't want just an answer, I would like to understand the method. Thanks a lot in advance.

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    $\begingroup$ In the future, please use LaTeX formatting. $\endgroup$ May 31, 2015 at 13:31

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I multiply by $e^{i 2 \pi ft}$ and integrate

$$\int^\infty _ {-\infty} \frac{e^{i 2 \pi ft}}{i 2 \pi f}-\frac{1}{i 2 \pi f}df$$

Second term is odd, so it will be zero. I am using euler's identity for the first term:

$$\int^\infty _ {-\infty} \frac{\cos (2 \pi ft)}{i 2 \pi f}+i\frac{\sin (2 \pi ft)}{i 2 \pi f}df$$

First term is even function over odd function, so it is odd overall. It will be zero. We have the second term. After cancelling the $i$'s:

$$\int^\infty _ {-\infty} \frac{\sin (2 \pi ft)}{ 2 \pi f}df$$

Rewrite as:

$$t \int^\infty _ {-\infty} \frac{\sin (2 \pi ft)}{ 2 \pi f t}df$$

The inner integral is a famous integral. I will not derive it here. Look here: enter link description here Here is the tricky part:

$$\frac{1}{2\pi} \text{sign}(t) \int^\infty _ {-\infty} \frac{\sin (2 \pi ft)}{ 2 \pi f t}d( 2 \pi f t)$$

You need to have the $sign(t)$ because if $t$ is negative. This is something that comes up very often we you change the variable of integration such that it absorbs a variable which may take negative values. The bounds of the integral must reverse. So we get:

$$\frac{\text{sign} (t)}{2}$$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ May 31, 2015 at 19:18

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