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

That is to say, I understand if $x(t) = X(\omega)$, $x(at) = \frac{1}{|a|}X(\frac{\omega}{a})$ 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: $$ x(at-t_0) = x(a(t-\frac{t_0}{a})) \\ \text{let } v(at) = x(a(t-\frac{t_0}{a})) \\ \text{and if } v(at) \leftrightarrow \frac{1}{|a|}V(\frac{\omega}{a}) \\ \text{then } x(a(t-\frac{t_0}{a}) = \frac{1}{|a|}V(\omega)e^{-jw\frac{t_0}{a}} \\ \therefore x(at-t_0) = \frac{1}{|a|}X(\frac{\omega}{a})e^{-jw\frac{t_0}{a}} $$

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.

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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.

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  • $\begingroup$ Okay, thanks! I couldn't find anything wrong with it but I wanted to make sure my thinking was right. $\endgroup$ – tamul Nov 12 '15 at 8:42

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