- #1

- 809

- 0

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

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- Thread starter FrogPad
- Start date

- #1

- 809

- 0

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

- #2

- 529

- 1

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

- 809

- 0

- #4

- 529

- 1

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.

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

Staff Emeritus

Science Advisor

Gold Member

- 10,237

- 40

[itex]

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

[/itex]

- Warren

- #6

- 529

- 1

[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

Staff Emeritus

Science Advisor

Gold Member

- 10,237

- 40

[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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