Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that the Fourier transform is as follows:$$\hat{f}(\xi)= \int_{-\infty}^{\infty}\exp(-\mathrm ix\xi)f(x)\mathrm{d}x$$ but I couldent understand why should use complex number $i$ in the integration. Is that means I have a real number function and after fourier transformation I get a complex function? I know that $\hat{f}(\xi)$ stand for the amplitude of each frequency. But how to understand when the amplitude is a complex number?

share|improve this question
    
I'm not sure if there's any good geometric intuition behind the Fourier transform of a complex function, but if you write $f(x) = g(x) + i h(x)$, with $g,h$ real, then the $\hat{f} = \hat{g} + i \hat{h}$, and I guess the Fourier transform of a real-valued function is easier to visualize. –  Christopher A. Wong Jan 10 '13 at 12:31
1  
@ChristopherA.Wong. I think maple is asking about the Fourier transform of a real function, but wonders why the operation itself involves the complex number $i$. –  Thomas E. Jan 10 '13 at 12:35
    
maybe $\exp(-iz)=\cos(z) - i\sin(z)$ helps... –  draks ... Jan 10 '13 at 12:43
    
They don't have to use complex numbers. See the Hartley transform: en.wikipedia.org/wiki/Hartley_transform . It has the same information in the sense that H[g] = Re[F[g]] - Im[F[g]], where g is a function and H and F are the Hartley and Fourier transform. But it never uses complex anything. –  Jess Riedel Mar 25 at 21:56

1 Answer 1

up vote 6 down vote accepted

You need to ask yourself why we use Fourier transforms. We want to transfer the signal from the space or time domain to another domain - the frequency domain. In this domain, the signal has two "properties" - magnitude and phase. If we want to get only the signal's "power" in a specific frequency bin, we indeed need only to take the absolute value of the Fourier transform, which is real. But, the Fourier transform gives as the phase of each frequency as well. While the first's (magnitude) importance is immediate, the phase is sometimes just as important. For example, for images, most of the information is contained in the phase and NOT in the amplitude. Also, frequency responses (fourier transform) are used in digital and analog filters, and the phase plays a major role here as well, especially for audio filters where a linear phase is required: This is what enables an audio filter to process all frequencies and output them without a different delay for each frequency (which will distort the sound - imagine a filter that makes your bass sound come a little before your treble...).

So I hope I convinced you the phase is important as well as the magnitude. And in order to get these two properties, we need something other than just real numbers, we need something with magnitude and phase. Something like a complex number.

share|improve this answer
    
thank you. It's really help –  maple Jan 10 '13 at 13:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.