How to find the inverse Fourier transform of $u(\omega) e^{-j \frac{\pi}{2}} + u(-\omega) e^{j \frac{\pi}{2}}$?

Question

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.

Answer 1

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.