# Fourier Transform (basic table lookup)

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?

$$10 X(jt)$$ <-----> $$20 \pi x (-\omega)$$

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

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 ?

I'll post the exact question. I don't understand the X(jt) as I posted. Note: $$j = \sqrt{-1}$$

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)

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:
chroot
Staff Emeritus
Gold Member
The conjugation rule is :

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

- Warren

The conjugation rule is :

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

- Warren

Is that a proof?

chroot
Staff Emeritus
$g\left( t \right) = F\left( t \right) \Rightarrow G\left( \omega \right) = f\left( { - \omega } \right)$