1
$\begingroup$

I understand how shifting and scaling work separately, but I'm unsure how to put them together.

That is to say, I understand if $$x(t) = X(\omega)$$ $$x(at) = \frac{1}{|a|}X\left(\frac{\omega}{a}\right)$$ and $$x(t-t_0) = e^{-j\omega{t_0}}X(\omega)$$ But if $x(t)$ is both scaled and shifted, as in $x(at-t_0)$, how would the Fourier Transform be determined using properties of the Fourier Transform?

What I tried was along the lines of: $$ x(at-t_0) = x\left(a\left(t-\frac{t_0}{a}\right)\right) \\ \text{let } v(at) = x\left(a\left(t-\frac{t_0}{a}\right)\right) \\ \text{and if } v(at) \leftrightarrow \frac{1}{|a|}V\left(\frac{\omega}{a}\right) \\ \text{then } x\left(a\left(t-\frac{t_0}{a}\right)\right) = \frac{1}{|a|}V(\omega)e^{-jw\frac{t_0}{a}} \\ \therefore x(at-t_0) = \frac{1}{|a|}X\left(\frac{\omega}{a}\right)e^{-jw\frac{t_0}{a}} $$

But I feel as though my logic rewriting $x(t)$ in terms of $v(t)$ isn't quite right and would appreciate if someone could either validate my logic or explain where my mistake is.

$\endgroup$

1 Answer 1

1
$\begingroup$

Your logic is correct. There are two ways to determine the answer. Yours is one of them:

$$\begin{align}v(t)=x(at)&\Longleftrightarrow \frac{1}{|a|}X(\omega/a)\\v(t-t_0/a)=x(at-t_0)&\Longleftrightarrow\frac{1}{|a|}X(\omega/a)e^{-j\omega t_0/a}\end{align}$$

The other option is

$$\begin{align}w(t)=x(t-t_0)&\Longleftrightarrow X(\omega)e^{-j\omega t_0}\\w(at)=x(at-t_0)&\Longleftrightarrow \frac{1}{|a|}X(\omega/a)e^{-j(\omega/a) t_0}\end{align}$$

Of course, both options lead to the same result.

$\endgroup$
1
  • $\begingroup$ Okay, thanks! I couldn't find anything wrong with it but I wanted to make sure my thinking was right. $\endgroup$
    – tamul
    Nov 12, 2015 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.