[Math] Interpreting the sign of fourier coefficients

fourier analysisfourier series

I am studying the Fourier series right now. Hopefully it's going okay. Now I have been playing a little bit with taking the product of a wave function (a sine or cosine with some phase) with a sine and integrating the result (using a convolution sum) over the period of the function ($2 \pi$ in this case).

All is good and I believe I can make a good sense of the result. Generally the technique tells how much the input signal (a sine in my case) "contains" of the sine function of frequency $f$ and phase $\phi$. And we test the input signal with the cosine function, it indicates the same thing but for the cosine function.

Now what I haven't found in the references that I have on Fourier series, is an interpretation of the sign of the Fourier coefficients $a_n$ and $b_n$. I read that one possible interpretation of the coefficients is in terms of amplitude and phase. By my understanding is that this is true when you express the coefficients in terms of complex numbers, where the modulus $|z|$ of the number relates to the amplitude and the argument $arg(z)$ relates to the phase. Is that correct?

But what I am after really is to know if there's a possible interpretation of the sign of the coefficients $a_n$ and $b_n$. Can you tell something about the signal when for instance $b_n$ is negative instead of positive (doesn't it indicate something about the phase of our input signal in relation to the cosine for some given frequency, etc.)?

Thank you.

Best Answer

Yes, you're correct that taking $z_n=a_n+ib_n$ will allow you to express your signal in terms of amplitude and phase using $|z|$ and $\operatorname{arg}(z)$ respectively.

You can always switch from $a_n$ and $b_n$ to $z_n$ and see how changing the former parameter set affects the latter. Namely, if you change sign of $a_n$but not $b_n$, or vice-versa, you effectively conjugate $z_n$ (and also multiply by $-1$ if it's $a_n$which you change). Then it's easy to see how $\operatorname{arg}(z)$ will change.

Note that if you use complex parameter $z_n$ instead of real parameters $a_n$ and $b_n$, then should still have a pair of parameters for each $|f|$ where $f$ is frequency (i.e. you have to work with signed frequencies): $a_n+ib_n$ and $a_n-ib_n$, which after doing inverse Fourier transform would cancel out the imaginary part of each other and give you real signal.

And, answering your comment, yes, $\operatorname{arg}(z_n)$ is the phase shift of the harmonic $n$ from its purely cosine version.