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 couldn't understand why we should use the complex number $i$ in the integration. Does that mean I have a real function and after Fourier transformation, I get a complex function? I know that $\hat{f}(\xi)$ stand for the amplitude of each frequency. But what does it mean when the amplitude is a complex number?
[Math] Why do Fourier transforms use complex numbers
fourier analysis
Related Solutions
An analogy with probability may be helpful.
- When we have a discrete random variable, its distribution is described by numbers that are actual probabilities of certain events, $P(X=a)$.
- For a continuous random variable, we use a probability density function. It relates to probabilities as follows: $P(a\le X\le b)=\int_{a}^b p(x)\,dx$
Same with Fourier series and integrals:
- Fourier series describes a periodic function by numbers (coefficients of Fourier series) that are actual amplitudes (and phases) associated with certain frequencies.
- Fourier transform is a "frequency density function". It relates to amplitudes as follows: if all frequencies in the range $a\le \xi\le b$ align perfectly at some moment, the resulting amplitude is $\int_{a}^b |\widehat f(\xi)|\,d\xi$
In the paper, a function with 2 real variables is defined $$ I(x,y)=\frac1{2\pi i}\int_C\frac{F(u,v)}{u+iv-(x+iy)}d(u+iv). $$ Of this function, the Fourier transform in both variables is considered, it is assumed that the integration over the (compact) curve $C$ can be moved out of the Fourier integration \begin{align} \hat I(ω_x,ω_y)&=A_{Fourier}\int_{\Bbb R^2}\frac1{2\pi i}\int_C\frac{F(u,v)}{u+iv-(x+iy)}\,d(u+iv)\,e^{-i(xω_x+yω_y)}\,d(x,y) \\ &=\frac{A_{Fourier}}{2\pi i}\int_CF(u,v)\int_{\Bbb R^2}\frac{e^{-i(xω_x+yω_y)}}{u+iv-(x+iy)}\,d(x,y)\,d(u+iv) \\ &=\frac{A_{Fourier}}{2\pi i}\int_CF(u,v)e^{-i(uω_x+vω_y)}\int_{\Bbb R^2}\frac{e^{i((u-x)ω_x+(v-y)ω_y)}}{(u-x)+i(v-y)}\,d(x,y)\,d(u+iv) \\ &=\frac1{2\pi i}\int_CF(u,v)e^{-i(uω_x+vω_y)}\,d(u+iv)\cdot{A_{Fourier}}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{x+iy}\,d(x,y) \end{align} For the last factor one now finds \begin{align} A_{Fourier}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{x+iy}\,d(x,y) &={A_{Fourier}}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}(ω_x-iω_y)}{(xω_x+yω_y)+i(yω_x-xω_y)}\,d(x,y) \\ &=A_{Fourier}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{(xω_x+yω_y)+i(yω_x-xω_y)}\,\frac{d(xω_x+yω_y,yω_x-xω_y)}{ω_x+iω_y} \\ &=\frac{A_{Fourier}}{ω_x+iω_y}\int_{\Bbb R^2}\frac{e^{ix}}{x+iy}\,d(x,y) =\frac{A_{Fourier}}{ω_x+iω_y}\int_{\Bbb R^2}\frac{ix\sin(x)-iy\cos(x)}{x^2+y^2}\,d(x,y) \end{align} and somehow the last singular integral can be assigned a finite value.
Best Answer
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 only need to take the absolute value of the Fourier transform, which is real. But, the Fourier transform gives the phase of each frequency as well.
While the first (magnitude's) 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 transforms) 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.