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Wikipedia, fount of all knowledge (Ha! LOL), gives a formula for a pulse wave here:

The formula is:

$$f(t)=\frac{\tau}{T}+\sum_{n=1}^{\infty}\frac{2}{n\pi}\sin\left(\frac{\pi n \tau}{T}\right)\cos\left(\frac{2\pi n}{T}t\right)$$

Given that the first term adds a non-zero value $\frac{\tau}{T}$ to the $\sum_{n=1}^\infty$ term, how can this formula produce $f(x)=0$ for the intervals between pulses?

Does the sigma element produce $-\frac{\tau}{T}$ for those intervals? Or is the formula wrong?

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  • $\begingroup$ Have you tried plotting this somewhere? I used this in wxMaxima: t:0.3$ wxplot2d(t+sum(2/n/%pi*sin(%pi*n*t)*cos(2*%pi*n*x),n,1,10),[x,0,2])$ (considerint T=1), then with plot(0.3+sum(2/n/pi*sin(pi*n*0.3)*cos(2*pi*n*x),n,1,10),x,0,2) in Wolfram Alpha. Since you didn't even question the validity of what you found, but immediately jumped to (wrong) conclusions, I am sorry, but -1 from me. $\endgroup$ – a concerned citizen Feb 5 '18 at 6:50
  • $\begingroup$ Actually, since this is your first post, I reverted it, but I did it with the hope that it will serve against future laziness. $\endgroup$ – a concerned citizen Feb 5 '18 at 6:52
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The formula is a Fourier series of a rectangular $T$-periodic pulse with pulse width $\tau$. The first term $\tau/T$ is the mean value, and so the second term has zero mean, and it can become positive and negative.

Note that since that formula is a Fourier series of a discontinuous function, it does not converge pointwise to the pulse wave, but it converges in the mean (i.e., in the $L^2$ norm). This means that if $f(t)$ is the pulse wave and if $g_N(t)$ is the $N^{th}$ partial sum of its Fourier series (i.e., the upper limit in the sum is replaced by $N$), then you have

$$\lim_{N\to\infty}\int_0^{T}\left|f(t)-g_N(t)\right|^2dt=0\tag{1}$$

This implies that the Fourier series representation will not be exactly zero at each point where the pulse wave equals zero, but the mean squared error $(1)$ will tend to zero as the number of terms in the sum goes to infinity.

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  • $\begingroup$ Many thanks to both of you. I clearly still have a lot to learn! I think we can consider this thread answered. Best wishes, Richard $\endgroup$ – Richard Burke-Ward Feb 5 '18 at 8:48

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