Fourier transform of a scaled and shifted triangular function

[LancsPhys14]

We have a graph as shown in the picture. We want to find the Fourier transform of this function.

This function is given by: 2Tri((x-2)/3)+3Tri((x-5)/3) where Tri(x) is a traingle function

We know:

F[Tri(x/p)]=psinc^2(ps)

F[g(x-a)]=e^(-i2pias)G(s)

Does this mean F[Tri((x/p)-a)]=e^(-i2pias)psinc^2(ps)?

If anyone could provide a method or expression for how to find the Fourier transform of a shifted and scaled triangular function, I would appreciate it. Alternatively, if the function found for the graph is incorrect I would appreciate knowing.

 

[Romsek]

\(\displaystyle
f(x-a) \Leftrightarrow e^{-i a \omega}F(x)\\
f\left(\dfrac x p\right) \Leftrightarrow \dfrac{1}{|p|} F\left(\dfrac \omega p\right)\\

f\left(\dfrac{x-a}{p}\right) \Leftrightarrow \dfrac{1}{|p|} e^{-i a \frac \omega p} F\left(\dfrac \omega p\right)
\)

You can do the plug and chug for \(\displaystyle \Lambda(x)\)

 

[LancsPhys14]

So would I be right in saying that for [MATH]2\Lambda(\frac{x-2}{3})[/MATH] the Fourier transform would be given:

[MATH]2F[\Lambda(\frac{x-2}{3})](s)=2[\frac{1}{3}e^{\frac{-2si}{3}}sinc^{2}(\frac{s}{3})][/MATH]

 

[Romsek]

You really should be using \(\displaystyle \omega\) for Fourier transforms. \(\displaystyle s\) is used for Laplace transforms.

I know they are just variables but it does help keep things clearer.

Yes, the expression looks correct, assuming you have the correct Fourier transform of the Tri function. I didn't check that.