Fourier transform of $ h(x)=f(ax-b)??$

Question

For any real number $b$, if $h(x) = f(x−b )$ ,  then $ {\displaystyle {\hat {h}}(\xi )=e^{-i\,2\pi \,b\,\xi }{\hat {f}}(\xi ).} $

For a non-zero real number $a$, if $h(x)=f(ax)$ , then $ {\displaystyle {\hat {h}}(\xi )={\frac {1}{|a|}}{\hat {f}}\left({\frac {\xi }{a}}\right).}$

now we want to obtain Fourier transform of $ h(x)=f(ax-b)??$

Is it correct to write :

$ {\displaystyle {\hat {h}}(\xi )=\frac{1}{a}e^{-i\,2\pi \,\frac{b}{a}\,\xi }{\hat {f}}(\frac{\xi}{a}).} $

Answer 1

Write $f(ax-b)$ as

$$f(ax-b)=f(a(x-b/a))$$

Now, apply the scaling theorem to the Fourier Transform of $f(x)$ shows

$$\mathscr{F}(f(ax))=\frac{1}{|a|}F(\xi/a)$$

Next, applying the shift theorem to $f(a(x-b/a))$, we find

$$\mathscr{F}(f(a(x-b/a)))=\frac{1}{|a|} e^{-i2\pi \xi b/a}F(\xi/a)$$

Answer 2

Note that $f(ax-b) = f \left(a\left( x-\frac{b}{a} \right) \right)$. You can then use the appropriate theorems to obtain the correct result: scale first, then shift.