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I have been trying to find the following inverse Fourier transform but without success:

$$ H(\omega) = \begin{cases} e^{-j \frac{\pi}{2}} & \omega \gt 0 \\ e^{j \frac{\pi}{2}} & \omega \lt 0 \\ \end{cases} $$

I have tried using the inverse Fourier transform but of course the $e^{j \omega t}$ won't converge.

I have also tried using the Fourier's transform property of duality given that $H(\omega)$ can be expressed as a sum of 2 unit steps multiplied by the constant $e^{\pm j \frac{\pi}{2}}$. That lead me to $\frac{1}{\pi t}$ which doesn't seem correct to me.

Restrictions on how to solve it:

  1. Don't use any kind of sorcery math that's not taught in an engineering/CS school.
  2. If you use the definition of Fourier Transform I would be glad if you used this one:

$X(\omega ) = \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt}$ and $x(t) = {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {X(\omega ){e^{j\omega t}}d\omega }$

Thanks.

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  • $\begingroup$ Why doesn't $1/t$ seem correct to you? $\endgroup$
    – Matt L.
    Sep 19 at 8:49
  • $\begingroup$ My mistake. It lead me to $ \frac{1}{\pi t} $. But I don't know, it's just a feeling. Can I show that it is correct? I don't know how to calculate the Fourier transform of that and get back to $H(\omega)$. $\endgroup$ Sep 19 at 8:58
  • $\begingroup$ Your result is correct. Take a look at my answer. $\endgroup$
    – Matt L.
    Sep 19 at 9:09
  • $\begingroup$ Might want to lead with $e^{j \frac \pi 2} = j$. I think they put those in as exponentials to throw you off. $\endgroup$
    – TimWescott
    Sep 20 at 14:47

1 Answer 1

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I think you did the right thing, namely use known transform pairs to evaluate the inverse Fourier transform of the given function. Solving the corresponding integral will get you into mathematical difficulties the type of which an engineer usually tries to avoid.

I would look at it as follows. The given function can be written as

$$H(\omega)=-j\textrm{sgn}(\omega)\tag{1}$$

where $\textrm{sgn}(\omega)$ is the signum function. In good Fourier transform tables you'll find that

$$\mathcal{F}\big\{\textrm{sgn}(t)\big\}=\frac{2}{j\omega}\tag{2}$$

If you don't have such a table, just derive $(2)$ from the more commonly known Fourier transform of the step function $u(t)$, using $\textrm{sgn}(t)=2u(t)-1$.

We can easily derive the duality relationship

$$\mathcal{F}\big\{x(t)\big\}=X(\omega)\Longrightarrow \mathcal{F}\big\{X(-t)\big\}=2\pi x(\omega)\tag{3}$$

Combining $(1)$, $(2)$, and $(3)$ results in

$$h(t)=\frac{1}{\pi t}\tag{4}$$

A system with impulse response $h(t)$ given by $(4)$ is a Hilbert transformer. You'll find some mathematical background in the link. Note that all improper integrals involving $h(t)$ are defined as principal values.

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  • $\begingroup$ Thanks. This is actually one sub-sub-question from previous exams in my uni. Now I am trying for the exams. One think that lead me to think I am wrong are the following 2 sub-questions: 1. Find $y(t)$ given an arbitrary $x(t)$. Should I give them the convolution improper integral as a result? 2. What would the output of the system be for input $x(t) = sin(\Omega t)$. Either the convolution improper integral or (most likely) multiply the Fourier transform of $x(t)$ with $H(\omega)$ and then inverse it. What do you think about these sub-questions. I am not asking you to solve them. $\endgroup$ Sep 19 at 9:20
  • $\begingroup$ Miss-typed many things. Can't edit again my comment... Sorry. $\endgroup$ Sep 19 at 9:25
  • $\begingroup$ In general, you'd get the convolution integral, but for a sinusoidal input, the answer is much simpler. Think in the frequency domain! $\endgroup$
    – Matt L.
    Sep 19 at 9:28
  • $\begingroup$ Yes. It just would be a shift by $\frac{\pi}{2}$. So we have $y(t) = sin(\Omega t + \frac{pi}{2})$ $\endgroup$ Sep 19 at 9:30
  • $\begingroup$ Should be $-\pi/2$ instead of $+\pi/2$. $\endgroup$
    – Matt L.
    Sep 19 at 9:32

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