# Fourier-series odd vs even square wave?

I really need to know the difference when doing a fourier seires between even and odd square waves. I've been trying to understand but I just get the same results and the same spectrums... Is the difference in the formula? If so, I can't find any formulas online that match with that my professor gave me. They're all a little different.

• If you show your calculations for both kinds of waves, it'll be easier to tell if/where you might be making a mistake. – MBaz Dec 6 '16 at 23:05
• Okay here, updated – Alex Leonardi Dec 6 '16 at 23:13
• when you just shift a signal in time (to make it even for example) the Fourier series coefficients magnitude spectrum will be the same but their phase spectrum will change. – Fat32 Dec 6 '16 at 23:18

For a trigonometric Fourier series:

$$\tilde{x}(t) = a_0 + \sum\limits_{n=1}^{\infty} a_n \cos(n \omega_0 t) + \sum\limits_{n=1}^{\infty} b_n \sin(n \omega_0 t)$$ with $$a_0 = \frac{1}{T} \int_{-T/2}^{T/2} x(t) \ dt$$

$$a_n = \frac{2}{T} \int_{-T/2}^{T/2} x(t) \cos (n \omega_0 t) \ dt, \quad n = 1, 2, \ldots$$

$$b_n = \frac{2}{T} \int_{-T/2}^{T/2} x(t) \sin (n \omega_0 t) \ dt, \quad n = 1, 2, \ldots$$ and $\omega_0 = \frac{2\pi}{T}$ .

the Fourier series of an odd square wave will be a sum of only sine terms, and the Fourier series of an even square wave will be a sum of only cosine terms.

This is because the product of an odd and an even function is an odd function, which when integrated over a symmetric interval (i.e. from $-T/2$ to $T/2$), will return zero. For an odd square wave, this means that all the $a_n$ will be zero, and for an even square wave, all the $b_n$ will be zero.

You can save yourself time and potential mistakes by exploiting this fact and only computing the coefficients that are nonzero.