How does both shifting and scaling a function affect the Fourier Transform?

Question

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.

Answer 1

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.