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ProPatto16

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## Homework Statement

Determine Fourier Transform of

[tex]f(t) = cos^2 ω_p t .... for |t|<T[/tex]

also, for |t|>T, f(x) = 0, although i dont think you need to do anything with that.

## The Attempt at a Solution

okay so:

[tex]f(t) = cos^2 ω_p t .... for |t|<T[/tex]

becomes

[tex]f(t) = \frac{1}{2} + \frac{e^{2iω_pt}}{4} + \frac{e^{-2iω_pt}}{4}[/tex]

and so now the Fourier Tranform F(ω) is this

[tex]F(ω)= \frac{1}{2}\int_{-T}^{T}e^{-iωt}dt + \frac{1}{4}\int_{-T}^{T}e^{2iω_pt}e^{-iωt}dt + \frac{1}{4}\int_{-T}^{T}e^{-2iω_pt}e^{-iωt}dt[/tex]

my question is,

What is the different between ω and ω

_{p}???

also another question. when performing the integrals, if i treat t and ω as seperate variables and not as say x=ωt then this happens, just taking the first integral

[tex]\frac{1}{2}\int_{-T}^{T}e^{-iωt}dt = \frac{e^{iωT}-e^{-iωT}}{2iω}[/tex]

now if that ω wasnt in the denominator then it would easy reduce to sin(ωT)... but with it there im not sure what to do with it? and since i have no values given and the last part of the question says "make a sketch of F(ω), then sketch its limiting form as T → ±∞." i believe something has to be done with it? or does it not matter? also noted that for sin functions, the limit to infinity doesnt exist.

thanks