3
$\begingroup$

I have a very simple question.

In Oppenheim book, it says that:

If CT Fourier transform of $x(t)$ is $X(j\omega)$ then, CT Fourier transform of $x^*(t)$ is $X^*(-j\omega)$.

What I can't understand is what is $X^*(-j\omega)$? Or what is $X^*(j\omega)$? Can you tell me what I should understand when I see $X^*(-j\omega)$ or $X^*(j\omega)$?

For example $x(t) = (a + bj)t$ and $X(j\omega) = (c + dj)\omega$ what should be $X^*(-j\omega)$?

Thanks

$\endgroup$
  • 1
    $\begingroup$ That is the general notation of the Fourier Transform Pairs. $x(t)$ is any time domain function, whereas $X(j\omega)$ (or sometimes $X(\omega)$) is the Fourier Transform of that function. Additionally $^*$ denotes the complex conjugate. What you are asking for is the Conjugation Property of the FT. It means that whenever you take the complex conjugate of your time signal, then it is equivalent to taking the complex conjugate and frequency reversal of your $X(j\omega)$. $\endgroup$ – jojek May 28 '15 at 12:34
9
$\begingroup$

$X^*(j\omega)$ is the complex conjugate of $X(j\omega)$. So if

$$X(j\omega)=X_R(\omega)+jX_I(\omega)$$

then

$$X^*(j\omega)=X_R(\omega)-jX_I(\omega)$$

and

$$X^*(-j\omega)=X_R(-\omega)-jX_I(-\omega)$$

To gain some more understanding, consider the definition of the Fourier transform:

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$$

So we have

$$X(-j\omega)=\int_{-\infty}^{\infty}x(t)e^{j\omega t}dt$$

and

$$X^*(-j\omega)=\int_{-\infty}^{\infty}x^*(t)e^{-j\omega t}dt=\mathcal{F}\{x^*(t)\}$$

as claimed in the quote of your question.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ You might want to add the interpretation of $X^*(-j\omega)$ to your answer. Do we replace $j\omega$ by $-j\omega$ wherever we find $\omega$ in the formula for $X(j\omega)$ while leaving $\omega$ (by itself) unchanged and also replace $j$ by $-j$ to account for the conjugation or what? $\endgroup$ – Dilip Sarwate May 28 '15 at 12:52
  • $\begingroup$ That's one of the reasons I don't like the $X(j\omega)$ notation. The other is - what's the order of operations? Is is flipping the frequency variable and then conjugating the result or conjugating and then flipping the frequency. It isn't clear from the notation. $\endgroup$ – David May 28 '15 at 13:04
  • $\begingroup$ @David: The order doesn't matter. $\endgroup$ – Matt L. May 28 '15 at 13:24
  • $\begingroup$ @DilipSarwate: I added some more information to my answer, hopefully clearing up any misconceptions. $\endgroup$ – Matt L. May 28 '15 at 13:29
  • 1
    $\begingroup$ @jason: No, why? $\endgroup$ – Matt L. May 28 '15 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.