We know the below,

$$ \mathscr{F}\big\{x(t)\big\}=X(f) \tag{1} $$ $$ \mathscr{F}\big\{x(-t)\big\}=X(-f) \tag{2} $$ $$ \mathscr{F}\big\{x^*(t)\big\}=X^*(-f) \tag{3} $$

Now, if for some signal

$$ x(-t)=x^*(t) \tag{4} $$

Then, is it safe to assume the following?

$$ X(-f)=X^*(-f) \tag{5} $$

or does it depend on the type of signal?

  • $\begingroup$ You can still validate the most appropriate answer $\endgroup$ Commented May 21, 2020 at 19:46

3 Answers 3


You are correct. Your last equation is simply a fancy way of saying that $X(f)$ is real valued.

In general: if it's real in one domain, it's conjugate symmetric in the other.


Yes, if eqs. (2) and (3) hold for any "type of signal" (which they do), then (5) must hold.

Inserting (4) into (2) we get $$ \mathscr{F}\big\{x^*(t)\big\} = X(-f) $$ and using (3) $$ X(-f) = X^*(-f) $$

If we substitute $f = - g$ we get $$ X(g) = X^*(g) $$ which, as Hilmar has already observed, means that $X(f)$ is real-valued. This is to be expected as, according to (4), $x(t)$ exhibits the conjugate complex symmetry.


The answers by @Deve and @Hilmar are technically perfect. I would like to provide some additional insights, with a few questions.

First, do you know of a signal satisfying this reversed-time/conjugate identity:


A first obvious idea is to choose among real and symmetric signals. A natural one in the Fourier framework is the cosine.

Now, let us get a little more complex (pun intented).

So second, what about the real sine? It is anti-symmetric. But if you remember that $i^*=-i$, the function $t\to i.\sin t$ also becomes a solution as well. So, by additivity, the function

$$t\to e^{i t}$$

(called complex exponential or cisoid) is also a solution. And its Fourier transform (as a generalized function) is indeed real (albeit somehow "infinite"). Going further, any linear combination of cisoids with real coefficients will do it.

Your question illustrates how Fourier duality is important, and how using it can simplify some issues. As seen in SYMMETRY OF THE DTFT FOR REAL SIGNALS:

In other terms, if a signal $ x(n)$ is real, then its spectrum is Hermitian (``conjugate symmetric'').

Here, your base signal $x$ is Hermitian, and the Fourier version is real. So to understand it better, just imagine $t$ is a frequency variable, and $f$ is its time dual. The standard representation is provided in Digital Analysis of Geophysical Signals and Waves/Complex Symmetry Properties.

Complex Symmetry Properties

It is also called the Heyser corkscrew/spiral.


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