Although my question is of engineering purposes, I still wanted to grasp the mathematical principle and gain some intuition for the following problem.
For a given periodic pulse train composed of square waves, where each pulse width (or duty cycle) is the same, the Fourier analysis says, that the frequency spectrum consists of a constant value , the fundamental frequency of the pulse train and its odd harmonics, whose amplitudes depend on the pulse width or duty cycle. That is shown in the following picture.Dependancy of the harmonics amplitudes on the square wave pulse width. That part makes sense to me.
However, what's hard for me to grasp is the scenario where the pulse train consists of square waves, with each of them having a different pulse width, while the fundamental pulse train frequency stays the same. These would "simulate" a periodic, yet discreet sine wave, like it is shown in this picture. Pulse train with variable pulse width to simulate a sine wave.
Because the fundamental pulse train frequency and its harmonics change amplitude according to the current pulse width, they would do this at certain frequencies, which according to the first picture are multiples of the frequency of the discreet sine wave that is incorporated into the pulse train. For example, the fundamental frequency of the pulse train would change its amplitude at are rate which would be two times greater than of the sine wave frequency. The 1st harmonic would do it at a four times higher rate etc. Is my reasoning here correct?
But what about the frequency spectrum of the sine wave? Because now, it seems to me like only the pulse train frequency gets distorted with harmonics, which seem to be dependant solely on the frequency of the sine wave.
I would appreciate some help on how to transfer my thinking into a more logical mathematical form to obtain the actual frequency components of the spectrum.
This answer illustrates the case of $6$ (instead of $14$) rectangular pulses in the interval $0<x<2 \pi$ because the formulas are simpler and the Fourier series converges faster due to fewer transitions in this interval ($12$ versus $28$).
Consider the function
$$f(x)=\sum\limits_{n=1}^3 \left(\left(\theta \left(x-\left(\left(\frac{n \pi}{3}-\frac{\pi}{6}\right)-\frac{a_n}{2}\right)\right)-\theta \left(x-\left(\left(\frac{n \pi}{3}-\frac{\pi}{6}\right)+\frac{a_n}{2}\right)\right)\right)-\left(\theta\left(x-\pi-\left(\left(\frac{n \pi}{3}-\frac{\pi}{6}\right)-\frac{a_n}{2}\right)\right)-\theta\left(x-\pi-\left(\left(\frac{n \pi}{3}-\frac{\pi}{6}\right)+\frac{a_n}{2}\right)\right)\right)\right)\tag{1}$$
defined on the interval $0<x<2 \pi$ where $\theta(x)$ is the Heaviside step function and
$$a_n=\int\limits_{\frac{1}{3} \pi (n-1)}^{\frac{\pi n}{3}} \sin(x) \, dx\tag{2}$$
resulting in the values
$$\begin{array}{cc} n & a_n \\ 1 & \frac{1}{2} \\ 2 & 1 \\ 3 & \frac{1}{2} \\ \end{array}$$
for $a_n$.
Figure (1) below illustrates $f(x)$ defined in formula (1) above in blue and the function $\sin(x)$ in green.
Figure (1): Illustration of $f(x)$ (blue) and $\sin(x)$ (green)
The Fourier series for $f(x)$ defined in formula (1) above is as follows
$$f(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N b_n\, \sin(n x)\right)\tag{3}$$
where
$$b_n=\frac{2 \left(\cos\left(\left(\frac{1}{4}-\frac{5 \pi}{6}\right) n\right)-\cos\left(\frac{1}{2} (1+\pi) n\right)-\cos\left(\frac{1}{12} (3+2 \pi) n\right)-\cos\left(\frac{1}{12} (3+10 \pi) n\right)+\cos\left(\frac{1}{12} (3-2 \pi) n\right)+\cos\left(\frac{1}{2} (n-\pi n)\right)\right)}{\pi n}\tag{4}$$
resulting in $b_n=0$ for even $n$.
Figure (2) below illustrates the function $f(x)$ defined in formula (1) above in blue, the Fourier series for $f(x)$ defined in formula (3) above in orange, and the function $\sin(x)$ in green where the sum in formula (3) above is evaluated over $N=50$ terms.
Figure (2): Illustration of $f(x)$ (blue), Fourier series for $f(x)$ (orange), and $\sin(x)$ (green)
Figure (3) below illustrates the first $50$ values of $b_n$ defined in formula (4) above. Note $b_n=0$ for even $n$ as mentioned above.
Figure (3): Illustration of $b_n$ values