The properties aren't entirely clear to me, sorry for the basic question.

I know the Fourier Transform of one function. Say, $\text{rect}(x,y) \Leftrightarrow \frac{\sin \pi u}{\pi u} \frac{\sin \pi v}{\pi v}$. Now, I have some related function and I want to easily get the FT based on simple properties.

For example, now I have something like $\frac{1}{3}\text{rect}(8x - 4, 4y - 2)$. How does this shift and scale inside the function affect its (continuous) Fourier Transform?

  • 2
    $\begingroup$ If this is homework, please add the homewark tag. $\endgroup$ Commented Feb 9, 2012 at 19:57

2 Answers 2


For one-dimensional Fourier transforms, $$\begin{align*} x(t) &\leftrightarrow X(f)\\ x(t-a) &\leftrightarrow X(f)\exp(-j2\pi fa)\\ x(bt) &\leftrightarrow \frac{1}{|b|}X\left(\frac{f}{b}\right) \end{align*}$$ all of which are proved by change of variables in the definition of the Fourier transform $$X(f) = \int_{-\infty}^{\infty} x(t)\exp(-j2\pi ft) \mathrm dt.$$ It is good exercise to work out the details of each of these results at least once in one's life, and once one has done so, the two-dimensional case is as easy as rolling off a log. If you always look up the answers in a table of Fourier transform properties, the right result will never come to mind just when you need it most.

More generally, it is a good idea to remember these results in graphic terms. If $x(t)$ is a signal with Fourier transform $X(f)$, and $b > 1$, say $b = 2$, then $y(t) = x(2t)$ is a time-compressed form of $x(t)$: while $y(0) = x(0)$, $y(3) = x(6)$, that is, $6$ seconds worth of $x(t)$ have been squeezed down into $3$ seconds worth of $y(t)$. Thus, the frequencies have doubled, and so $Y(f)$ should be something like $X(f/2)$ (but see below). More generally,

Compressing the time axis by a factor of $b$ expands the frequency axis by a factor of $b$.

But that is not the complete answer. One of the more boring results in Fourier transform that everyone tends to ignore is $$y(0) = \int_{-\infty}^{\infty} Y(f) \mathrm df.$$ But, $$y(0) = x(0) = \int_{-\infty}^{\infty} X(f) \mathrm df$$ and since we have stretched the horizontal axis by a factor of $b$ in going from $X(f)$ to $X(f/b)$, we better reduce the magnitude of each frequency component by a factor of $b$ so that the two integrals (= area under the curve) work out to have the same value, that is, $$Y(f) = \frac{1}{b}X\left(\frac{f}{b}\right), ~~ b > 0.$$ The result for $b < 0$ can be worked out similarly.


More generally, we have:

$$y(t) = x(at + b)$$

You may want to combine both properties listed by Dilip Sarwate into one equation. Using the definition of the Fourier transform, we can insert our input into the Fourier integral:

$$Y(jw) = \int^∞_{-∞}x(at+b)e^{jwt}dt$$

Using a change of variables such that T = at + b, we get:

$$Y(jw) = \int^∞_{-∞}\frac{1}{a}x(T)e^{\frac{jw(T-b)}{a}}dT$$

And pulling out constants we get:

$$Y(jw) = e^{-jw\frac{b}{a}} \int^∞_{-∞}\frac{1}{a}x(T)e^{\frac{jwT}{a}}dT$$

Which gives us the time-scaling property on the right, and the time-shifting property on the left.

$$Y(jw) = e^{-jw\frac{b}{a}} \frac{1}{|a|}X(\frac{jw}{a})$$

EDIT: minus sign when pulling out exponential constant w.r.t. T


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