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 $X^*(-j\omega)$?


  • 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 $X(j\omega)$. $\endgroup$
    – jojeck
    May 28, 2015 at 12:34

1 Answer 1


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






where $X_R(\omega)$ and $X_I(\omega)$ refer to the real and imaginary parts of $X(j\omega)$, respectively. [Note that they are not the respective Fourier transforms of the real and imaginary parts of the time domain signal.]

Also, the notation $X^*(j\omega)$ is to be understood as $\left[X(j\omega)\right]^*$, i.e., the whole expression for $X(j\omega)$ is conjugated. Similarly for $X^*(-j\omega)$, i.e., $X^*(-j\omega)=\left[X(-j\omega)\right]^*$.

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

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


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

as claimed in the quote of your question.

  • $\begingroup$ 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? $\endgroup$ May 28, 2015 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, 2015 at 13:04
  • $\begingroup$ @David: The order doesn't matter. $\endgroup$
    – Matt L.
    May 28, 2015 at 13:24
  • $\begingroup$ @DilipSarwate: I added some more information to my answer, hopefully clearing up any misconceptions. $\endgroup$
    – Matt L.
    May 28, 2015 at 13:29
  • 1
    $\begingroup$ @jason: No, why? $\endgroup$
    – Matt L.
    May 28, 2015 at 13:51

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