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Verification of pronumerals in Fourier Transform

  • Thread starter 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 ωp???

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
 

Answers and Replies

  • #2
uart
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[itex]\omega[/itex] is the transform (frequency) variable of your FT while [itex]\omega_p[/itex] is just a particular value of [itex]\omega[/itex], that is, a constant.

Kind of like the distinction between t and [itex]t_0[/itex] in the equation [itex]f(t) = sin(t - t_0)[/itex],

or the distinction between [itex]x[/itex] and [itex]x_1[/itex] in the equation [itex](y - y_1) = m(x - x_1)[/itex]
 
  • #3
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Thanks. So I can factor [tex]\omega_p[/tex] out of the integrals when it's in the denominator. I shall play around and see what I get
 
  • #4
vela
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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
You can write it as
$$\frac{\sin \omega T}{\omega} = T\,\text{sinc }\omega T$$ where $$\text{sinc }x = \frac{\sin x}{x}.$$ Read about sinc here:

http://en.wikipedia.org/wiki/Sinc_function

In particular, look at the relationship of the function to the Dirac delta function.
 
  • #5
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okay, looked that up.
the sinc function relates the the dirac delta function as a limit bla bla all good.

now i just need to get my integration from the fourier transform to the sinc function.

taking the first integral again

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

integrate and sub in bounds bla bla becomes

[tex]\frac{e^{iωT}-e^{-iωT}}{2iω}[/tex]

factor out ω in denominator and use eulers formula

[tex]\frac{1}{ω}sin(ωT)[/tex]

then apply sinc equality to get

[tex]Tsinc(ωT)[/tex]

now looking at the relationship between sinc and the dirac delta function

as T → 0 [itex]lim \frac{1}{T}sinc(\frac{ω}{T})[/itex] = [itex]\delta(ω)[/itex]

am i on the right track? the limit i need to sketch is as T → ±∞

and do i attack the other two terms the same way?

thanks
 
  • #6
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i now have completed (i think) the entire integration of the transform F(ω) and then reduced them to sinc functions and got

[tex]F(ω) = Tsinc(ωT) + \frac{1}{2}Tsinc(T(2ω_p-ω)) + \frac{1}{2}Tsinc(T(ω+2ω_p))[/tex]

now i need to sketch that and then sketch the limiting form as T → ±∞

the relationship to the dirac delta function seems inverse to what i have.

[tex]lim as T→0 \frac{1}{T}sinc(\frac{x}{T}) = delta function[/tex]

but i have Tsinc(xT) and T→±∞.

its practically opposite.
what does that mean?

and hints on sketching any of this?

thanks
 
  • #7
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the Sinc function approaches zero as x goes towards infinity, with the envelope of sinc(x) tapering off as 1/x.
tapering off? what does that imply? that when x approaches infinity then sinc(x) = 1/x?
 
  • #8
vela
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It might help you to not use T in both cases. The limit of your Fourier transform will have terms of the form
$$\lim_{T \to \infty} T\,\text{sinc }\omega T$$
The representation of the delta function is
$$\lim_{a \to 0} \frac{1}{a} \text{sinc }{\frac{\omega}{a}}$$

As far as sketching goes, remember that ω is your independent variable here and T is some parameter. You'll want to recall what you learned way back in algebra about scaling and translating functions. What happens to the graph of T sinc(ωT) as you vary T?
 
  • #9
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okay so the hint implied by not using T in both cases if for me to notice that if i let T=1/a then have another look.

T = 1/a and what happens as 1/a approaches infinity. as a approaches 0 then 1/a approaches infinity. so its really the same.

as for shifting, varying T results in horizontal dilation from the T in the argument, and vertical dialation from the T out front. so as T gets larger the function squashes up and becomes infinitely vertical. this is the delta function and is at the three locations given by the arguments in the sinc functions.

so as the sinc functions approach the limiting form there are 3 spikes.

one is at ω
another is at 2ωp
and the last is at ω+2ωp

so like a partial comb function yeah?
 
  • #10
uart
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so as the sinc functions approach the limiting form there are 3 spikes.

one is at ω
another is at 2ωp
and the last is at ω+2ωp

so like a partial comb function yeah?
Yes. at T goes to infinity it just becomes the continuous function [itex]\cos^2(w_p t)[/itex] (as opposed to just a rectangular windowed piece of this). And since,

[tex]\cos^2(x) = \frac{1}{2} + \frac{1}{2} \cos(2x)[/tex]

then you should just end up with the Fourier transform of the DC component (a delta centered at [itex]w=0[/itex]) plus the Fourier transform of the "cos(2x)" term, which are the "deltas" centered at plus and minus [itex]2 w_p[/itex].
 
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