Intuition for imaginary part of Fourier Transformation?

[159753x]

Hi there,

I'm having trouble understanding the Fourier transform of a function where the result in the frequency domain has imaginary components.

For example, if you take the Fourier transform of Sin[t] , the result is

I Sqrt[\[Pi]/2] DiracDelta[-1 + \[Omega]] - 
 I Sqrt[\[Pi]/2] DiracDelta[1 + \[Omega]]

What does this mean? I can't really graph it, so I am having trouble understanding it.

I can grasp a regular Fourier transform; it simply tells you what components are making up your signal wave. But when an imaginary I is thrown in there, what happens? How can the superposition of all those waves give you the real signal wave?

Does anybody have an intuition they could share?

 

[olivermsun]

Suppose you have a certain (real) signal which you can represent as a sum of sines and cosines. Using Euler's formula, you can combine each pair of cosine and sine terms into an e^ix term. Instead of two real coefficients, now you will have a real and imaginary part. One reason to do it this way is that it actually makes the algebra easier!

To inspect the Fourier transform, you can just graph the real and imaginary parts separately (just as you would have graphed the sine and cosine coefficients separately) or you can graph the magnitude and phase separately.

The superposition of waves will give you back a real signal because there are both "positive" and "negative" frequencies, and they have a certain symmetry: if A(x) is the coefficient at frequency x, then A(x) = conj(A(-x)), i.e., if A(x) = a + ib then A(-x) = a - ib.

Check out what you get when you try to put the signal back together, remembering also that cos(x) = cos(-x) and -sin(x) = sin(-x):

(a + bi) * e^ix + (a - bi) * e^-ix = (a + bi) * (cos x + i sin x) + (a - bi) * (cos x - i sin x) = …

 

[159753x]

Wow, that was a terrific explanation. You should consider becoming a teacher!

I believe I have a better understanding now, thanks friend!

 

[olivermsun]

You're welcome!