# What happens to Fourier Transform of function when the function's time scale is changed?

When a function $f(t)=exp(-|t|)$ for example undergoes Fourier Transformation, it gives $F(w)=\frac{-2}{1+w^2}$

But what happens to the result if the time scale is scaled and shifted, so that $t \rightarrow\ t^* =at+b$ ?

How will the Fourier Transformation of the function change?

Edit: Following is the approach I took but is unsure about it's correctness

$Since \ t \rightarrow\ t^* =at+b \\ f(t) \rightarrow\ f(at+b) = e^{-|at+b|}) \\ therefore \\ F(w) = \frac{e^{-iwb}}{|-a|} \ * \frac{2}{1+(\frac{w}{-a})^2}$

The part I'm most uncertain about is $e^{-|at+b|}$ where there is the absolute value of at+b. I'm only treating it as a bracket at the moment, I'm not sure if that would change anything.

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Homework? Did you try to work it out? –  leonbloy Jun 20 '12 at 1:51
@leonbloy Just self revision, but anyway, I've added more information to what I have tried, please take a look. Thanks –  Synia Jun 20 '12 at 2:11
Why don't you write out the formula for the Fourier transform of $g(t) = f(at+b)$ as $$\int_{-\infty}^{\infty}f(at+b)\exp(-i2\pi ft)\,\mathrm dt$$ and then make a change of variable $\tau = at+b$? –  Dilip Sarwate Jun 20 '12 at 3:00

There is a property of the Fourier Transform which explains this. It's the time scale property $:$ $$\mathscr{F}\{f(at)\} = \frac{1}{|α|} F\Big(j\frac{ω}{α}\Big)$$
Where $α$ is the scaling constant. For example considering the standard function$: f(t)=e^{-kt}$ and it's fourier transform which is $: F(jω)=\frac{1}{jω + k}$ if you scale that function, for example with a scaling constant of $-1$, then it's fourier transform would become$:$ $$\mathscr{F}\{f(-t)\} = \frac{1}{|-1|} \frac{1}{1+j\frac{ω}{-1}} = \frac{1}{1-jω}$$ In this case the absolute value in your function makes no difference.
Now i'm not sure whether scaling with a factor of the form$: αt+β$ (as you said) is considered proper scaling. Because this isn't just scaling, it's also a time shift and there's another Fourier property about this. If you're time-shifting a function, then you multiply the fourier transform of that function by $e^{-jωt_o}$, such that $$\mathscr{F}\{f(t-t_o)\} = F(jω)e^{-jωt}$$
$(ω=2πf)$ So you would have to apply both properties etc. to get the Fourier transform properly. I'm not certain about this last one, but i believe this is the case.