Computing Fourier transform for a real odd signal

Question 1

Given a real and odd signal$x(t)$, such that$\vert X(\omega)\vert = e^{-\vert \omega\vert}$ is the magnitude of Fourier transform.

Question: Find the Fourier Transform$X(w)$.

My attempt:

We know,$x_{o}(t) = \frac{x(t) - x(-t)}{2}$ where$x_{o}(t)$ denotes odd part of signal. Since$x(t) $ is odd, we have$x(-t)=-x(t).$ Therefore$x_{o}(t)=x(t).$

Therefore$\mathscr{F}(x(t)) = \mathscr{F}(x_{o}(t))$hence

$$X(\omega) = j\cdot \text{Im}(X(w))\; \text{using properties of Fourier Transform}$$

Since$\mathbb{Re}X(\omega)=0$, we get$X(\omega) = j e^{-\vert \omega \vert}.$

But$x(t)$ being real, we shd have this property also satisfied;$X(\omega) = X^{*}(-\omega)$, for all$\omega \in \mathbb{R}$. This is not being satisfied by the$X(\omega)$(due to the$j$ term), I calculated above.

Can I get some feedback if my$X(\omega)$ is properly calculated? Thanks

Question 2

$X(\omega) = |X(\omega)| e^{i \Phi(\omega)} = |X(\omega)| (\cos(\Phi(\omega))+i \sin(\Phi(\omega))$

Since we know that the signal is real and odd, its phase spectrum is odd and purely imaginary.

So for purely imaginary phase spectrum,$\cos(\Phi(\omega)) = 0$. But It should also be odd function of$\omega$. So the only function that will satisfy this is$\Phi(\omega) = \frac{\pi}{2}\operatorname{sgn}(\omega)$

Plugging back,$X(\omega) = |X(\omega)| i \sin\left(\frac{\pi}{2} \operatorname{sgn}(\omega)\right) = i |X(\omega)| \operatorname{sgn}(\omega) = i e^{-|\omega|} \operatorname{sgn}(\omega)$

$x(t) = \mathcal{F}^{-1}(i e^{-|\omega|} \operatorname{sgn}(\omega)) = \frac{4\pi t}{1+4\pi^2 t^2}$, where$\mathcal{F}^{-1}(F(\omega)) = \int\limits_{-\infty}^{\infty}F(\omega) e^{i 2 \pi \omega t} d\omega$

Question 3

thanks for the help @Srini. I wasn't aware that the phase spectrum needs to be odd for a real signal. Appreciate it

Question 4

Real and odd signal has odd and imaginary phase spectrum. Real and even signal has real and even phase spectrum

Question 5

To be accurate, the spectrum is imaginary, not phase spectrum

Srini 2,3651 gold badge15 silver badges25 bronze badges · Accepted Answer · 2025-11-21 23:42:19Z

$X(\omega) = |X(\omega)| e^{i \Phi(\omega)} = |X(\omega)| (\cos(\Phi(\omega))+i \sin(\Phi(\omega))$

Since we know that the signal is real and odd, its phase spectrum is odd and purely imaginary.

So for purely imaginary phase spectrum,$\cos(\Phi(\omega)) = 0$. But It should also be odd function of$\omega$. So the only function that will satisfy this is$\Phi(\omega) = \frac{\pi}{2}\operatorname{sgn}(\omega)$

Plugging back,$X(\omega) = |X(\omega)| i \sin\left(\frac{\pi}{2} \operatorname{sgn}(\omega)\right) = i |X(\omega)| \operatorname{sgn}(\omega) = i e^{-|\omega|} \operatorname{sgn}(\omega)$

$x(t) = \mathcal{F}^{-1}(i e^{-|\omega|} \operatorname{sgn}(\omega)) = \frac{4\pi t}{1+4\pi^2 t^2}$, where$\mathcal{F}^{-1}(F(\omega)) = \int\limits_{-\infty}^{\infty}F(\omega) e^{i 2 \pi \omega t} d\omega$

thanks for the help @Srini. I wasn't aware that the phase spectrum needs to be odd for a real signal. Appreciate it
Real and odd signal has odd and imaginary phase spectrum. Real and even signal has real and even phase spectrum
To be accurate, the spectrum is imaginary, not phase spectrum

Movatterモバイル変換

Stack Exchange Network

Computing Fourier transform for a real odd signal

1 Answer1

You mustlog in to answer this question.

Related

Hot Network Questions

Subscribe to RSS