0
$\begingroup$

I'm new here and I hope you can help me. I started studying the Fourier Transform now at University and I have a lot of doubts about this subject. My teacher sent to us one list of exercises and I don't know how I can solve the Fourier Transform of this function:

$$ x(t) = \delta \left(\frac{t-T}{a}\right) $$

I don't know how to use the properties of scaling and shifting in this case, it would be great if you could explain to me the steps to solve this question. Thanks for your help :)

$\endgroup$
3
$\begingroup$

Technically, the impulse $\delta(t)$ is called a distribution, and not a function, but for the purposes of your first course in Fourier transforms, what you need to know is that $\delta(t)$ has the sifting property

$$\displaystyle\int_{-\infty}^\infty x(t)\delta(t) \,\mathrm dt = x(0) ~\text{provided that }x(t) ~\text{is continuous at }t=0 \tag 1$$

which sifts the single value $x(0)$ out the function $x(t)$ and discards all other values. Also, any integral involving impulses can be manipulated via the change of variable technique into an integral of the form above to get the value of the integral. For example,

\begin{align} \int_{-\infty}^\infty x(t)\delta(t-T) \,\mathrm dt &= \int_{-\infty}^\infty x(\tau+T)\delta(\tau) \,\mathrm d\tau &\scriptstyle{\text{set }\tau = t-T, ~ dt = d\tau}\\ &= x(T) \end{align}

and so we can think of $\delta(t-T)$ as a time-shifted impulse that plucks the value $x(T)$ (instead of $x(0)$) out of the function $x(t)$. Similarly, for $a >0$, \begin{align} \int_{-\infty}^\infty x(t)\delta(at) \,\mathrm dt &= \int_{-\infty}^\infty x\left(\frac{\tau}{a}\right)\delta(\tau) \,\frac 1a \mathrm d\tau &\scriptstyle{\text{set }\tau = at, ~ dt = \frac 1a d\tau}\\ &= \frac 1a x(0) \end{align} Figure out why we can think of $\delta(at)$ as the magnitude-scaled impulse $\frac 1{|a|}\delta(t)$ and then apply both of the above ideas to evaluating the Fourier transform integral $$\int_{-\infty}^\infty e^{-j\omega t}\delta\left(\frac{t-T}{a}\right) \, \mathrm dt$$

$\endgroup$
2
$\begingroup$

The Fourier transform properties we need here are: $$\begin{align*} x(t-a) &\xrightarrow{\mathscr{F}} X(j\omega)e^{-j\omega a}\\ x(bt) &\xrightarrow{\mathscr{F}} \frac{1}{|b|}X\left(\frac{j\omega}{b}\right) \end{align*}$$

where $X(j\omega)$ is the Fourier transform of $x(t)$.

We can write the signal you have as $\delta\left(\frac{t}{a}-\frac{T}{a}\right)$ and analyze the scaling and the shifting separately.

Remember that the transform of a Dirac delta equals $1$. The Fourier transform of a shifted by $T/a$ delta would be

$$\delta(t-T/a) \xrightarrow{\mathscr{F}} e^{-j\omega \frac{T}{a}}$$

But the signal is also scaled by $1/a$, so

$$\delta\left(\frac{t}{a}-\frac{T}{a}\right) \xrightarrow{\mathscr{F}} |a|e^{-j\omega T}$$

$\endgroup$
  • $\begingroup$ Your statement about the scaled impulse is not right. Please correct it. $\endgroup$ – Fat32 Feb 13 '17 at 0:32
  • $\begingroup$ @Fat32 Sorry about that, I didn't know that property of the delta function. I've already changed completely my answer. Thanks for pointing that out! $\endgroup$ – Tendero Feb 13 '17 at 1:21

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.