Fourier Transform (basic table lookup)

  • Thread starter FrogPad
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  • #1
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I have a practice exam I'm going through, and I am stumped on one of the basic problems.

How is this a transform pair?

[tex] 10 X(jt) [/tex] <-----> [tex] 20 \pi x (-\omega) [/tex]

I don't see how one can make this relation. What is the [tex] 10 X (jt) [/tex].

thanks in advance
 

Answers and Replies

  • #2
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Divide both sides by 10 to simplify!

X(jt)<-----> 2pi x(-w)

Still seems wrong to me. The 2pi is normally included in the defn. of the FT.
What's j ?
 
  • #3
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I'll post the exact question. I don't understand the X(jt) as I posted. Note: [tex] j = \sqrt{-1} [/tex]

437273077_752ab8a123_o.jpg
 
  • #4
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OK, I got it- but I don't like the j notation. Everyone else writes

x(t) <-----> X(omega)

Define FT[x(t)]= Int[e^-iwt x(t)] dt
FT-1[X(w)]=(1/2pi)Int[e^iwt X(w)] dw

(you may use a different convention, but the prefactors of FT and FT-1 must multiply to (1/2pi) no matter what convention you use)


Start with FT[x(t)]=X(w)

Int [ e^-iwt x(t)] dt = X(w)

substitute w=t', t=w'

Int [ e^-iw't' x(w') ] dw' =X (t')

which is the same as writing

Int [ e^-iwt x(w) ] dw =X (t)

Substitute w -> -w

-(1/2pi) Int [ e^iwt x(-w) ] dw =1/(2pi) X (t)

The LHS is the inverse FT, which I'll call FT-1

-FT-1 [x(-w)]=1/2pi X(t)

Take 2pi to other side
X(t)=-2pi FT-1 [x(-w)]

FT both sides

FT[ X(t) ] =-2pi x(-w)

X(t) <----> - 2pi x(-w)

I seem to have an extra minus sign- but who cares? You can have fun checking if it's me or the teacher who messed it up.

Edit- the minus sign comes from changing the limits from +ininity,-infinity to -infinity,+infinity in the integral, when w-> -w

You can then turn the integral 'the right way up' with the introduction of a minus sign.
That always trips me up.
 
Last edited:
  • #5
chroot
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The conjugation rule is :

[itex]
g\left( t \right) = \overline {f\left( t \right)} \Rightarrow G\left( \omega \right) = \overline {F\left( { - \omega } \right)}
[/itex]


- Warren
 
  • #6
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The conjugation rule is :

[itex]
g\left( t \right) = \overline {f\left( t \right)} \Rightarrow G\left( \omega \right) = \overline {F\left( { - \omega } \right)}
[/itex]


- Warren
Is that a proof?
 
  • #7
chroot
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Oops, actually, I read this too quickly. The example given is a form of the inversion rule:

[itex]
g\left( t \right) = F\left( t \right) \Rightarrow G\left( \omega \right) = f\left( { - \omega } \right)
[/itex]

And no, it's not a "proof," it's an rule of the so-called Fourier transform calculus.

- Warren
 

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