Suppose we have the Fourier transform pair $x(t)$ and $X(\omega)$ such that $$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t$$

The duality property states that $X(t)$ and $2\pi x(-\omega)$ constitute a Fourier transform pair. I was trying to prove this statement when I ran into the following problem: the Fourier transform of a signal is sometimes denoted by $X(j\omega)$, such that $$X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t$$

Using this notation, how do we even state the duality property? Note that evaluating $X(t)$ in this case would effectively replace $j\omega$ (as opposed to just $\omega$ in the previous case) by $t$ which is clearly wrong. Would we have to write $X(jt)$ for consistency?

Returning to the question of deriving the duality property and using the notation I used in the beginning, what is the flaw in the following approach?

Since $$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t$$ we let $t$ be dummy variable $u$, and evaluate the function $X(\cdot)$ at $t$. We then have $$X(t) = \int_{-\infty}^{\infty} x(u) e^{-j ut} \mathrm{d}u$$ With the change of variables $\omega = -u$ we get $$X(t) = -\int_{-\infty}^{\infty} x(-\omega) e^{j\omega t} \mathrm{d}\omega$$ $$X(t) = -\frac{1}{2\pi}\int_{-\infty}^{\infty} 2\pi x(-\omega) e^{j\omega t} \mathrm{d}\omega$$

Which is almost the correct result except for the additional minus sign. Where did I go wrong? Thank you in advance.

I would also like to add that I feel uncomfortable about simply changing $t$ and $\omega$ in the functions' argument because they have different dimensions.


2 Answers 2


You've answered your first question yourself: if you use the notation $X(j\omega)$ then you'll need to use $X(jt)$, because you simply replace $\omega$ by $t$.

The sign error in your derivation comes from the fact that you forgot to change the sign of the limits of the integral. After substituting $\omega=-u$ you should get

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

  • $\begingroup$ And using that notation and the duality property, how could we prove the duality of the differentiation property for instance: the Fourier Transform of $-jt x(t)$ is $\frac{dX(j\omega)}{d\omega}$ $\endgroup$
    – user33568
    Feb 17, 2018 at 1:58
  • $\begingroup$ Or would this require another post? $\endgroup$
    – user33568
    Feb 17, 2018 at 2:01
  • 1
    $\begingroup$ @0MW: I don't think that the fact $X(t)\Longleftrightarrow 2\pi x(-\omega)$ helps a lot in deriving the differentiation property. I think it's easier to derive it directly. $\endgroup$
    – Matt L.
    Feb 18, 2018 at 16:07

Two notes about notation and some philosophy of mathematics.

  1. Notation: Mostly for the utility and elegance regarding the property of Duality, I prefer this consistent definition of the continuous Fourier Transform: $$ X(f) \triangleq \mathscr{F} \Big\{ x(t) \Big\} \triangleq \int\limits_{-\infty}^{+\infty} x(t) \, e^{-j 2 \pi f t} \ \mathrm{d}t $$ and inverse: $$ x(t) \triangleq \mathscr{F}^{-1} \Big\{ X(f) \Big\} = \int\limits_{-\infty}^{+\infty} X(f) \, e^{+j 2 \pi f t} \ \mathrm{d}f $$ Even with different signs on $j$, the elegant symmetry between the forward transform and inverse should be clear. $$ $$

  2. While there is a quantitative difference between $-j$ and $+j$ (they are non-zero and they are negatives of each other), there is no qualitative difference between the two. They are both totally "imaginary", they have the same magnitude and, in reflection, they have the same angle. And most saliently, they are imaginary numbers that have equal claim to squaring to be $-1$. With any non-zero number, there are two square roots to $-1$ and both $-j$ and $+j$ have equal claim to being that $\sqrt{-1}$. $$ $$It's like saying that clockwise and counterclockwise have equal claim to being "positive" rotation in mathematics. They are equivalent but opposite. Left and Right are similar. Whether the number line runs from left to right or opposite is just a matter of convention. Now this does not mean that the real numbers $+1$ and $-1$ are equivalent. One is the multiplicative identity and the other is not. But there is absolutely nothing essential you can say about $+j$ that is not also true for $-j$. You could go to every mathematics book and swap $+j$ and $-j$ everywhere and every theorem would be just as valid as it was before.

That said, it should be clear that there is no qualitative difference between the Fourier Transform and the Inverse Fourier Transform as expressed above. But that change in sign will mean a change in sign for either $f$ or $t$ (but not both, take your pick). Now if either $x(t)$ or $X(f)$ have even symmetry, then you need not worry about the change in sign with $t$ or $f$.

Like it's pretty easy to show that

$$ \mathscr{F} \Big\{ \operatorname{rect}(t) \Big\} = \operatorname{sinc}(f) $$


$$ \operatorname{rect}(u) \triangleq \begin{cases} 1 \qquad & \mathrm{for} \ |u| < \tfrac12 \\ \tfrac12 \qquad & \mathrm{for} \ |u| = \tfrac12 \\ 0 \qquad & \mathrm{for} \ |u| > \tfrac12 \\ \end{cases} $$


$$ \operatorname{sinc}(u) \triangleq \begin{cases} 1 \qquad & \mathrm{for} \ u = 0 \\ \frac{\sin(\pi u)}{\pi u} \qquad & \mathrm{for} \ u \ne 0 \\ \end{cases} $$

So, accordingly,

$$ \mathscr{F}^{-1} \Big\{ \operatorname{sinc}(f) \Big\} = \operatorname{rect}(t) $$

Gee, I wonder what $ \mathscr{F} \Big\{ \operatorname{sinc}(t) \Big\} $ is?

As Matt pointed out, you did miss swapping the integral limits, but I would recommend changing your definitions to the above and your life with Duality will get a whole lot easier:

If $X(f) = \mathscr{F} \Big\{ x(t) \Big\}$, then $x(-f) = \mathscr{F} \Big\{ X(t) \Big\}$.

Your life with $X(0)$ and $x(0)$ and with Parseval's Theorem will get easier, too:

$$ X(0) = \int\limits_{-\infty}^{+\infty} x(t) \ \mathrm{d}t $$

$$ x(0) = \int\limits_{-\infty}^{+\infty} X(f) \ \mathrm{d}f $$

$$ \int\limits_{-\infty}^{+\infty} \Big| x(t) \Big|^2 \ \mathrm{d}t = \int\limits_{-\infty}^{+\infty} \Big| X(f) \Big|^2 \ \mathrm{d}f$$

NO nasty asymmetrical scaling factors to worry about!! (Just remember the $2\pi$ in the exponent.)

And you will be able to answer the question of how many functions of $t$ are exactly the same as their Fourier transform (with $f$ substituted for $t$). It's not just the Gaussian:

$$ x(t) = e^{-j \pi t^2} $$

which happens to have an identical Fourier Transform:

$$ X(f) = e^{-j \pi f^2} $$

Just everything in your continuous F.T. life will be easier if you use this notation.


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