Dtyjlui

The periodic pulse function can be represented as a Fourier series as,

$$f_f(t) = a_0 + sum_{i=1}^inf (a_n cos(nomega_0t))$$

where

$$a_0 = Afrac{T_p}{T}$$

$$a_n = 2frac{A}{npi}sin(npifrac{T_p}{T})$$

with period $T$, amplitude $A$ and pulse width $T_p$.

Two periodic pulse functions with different pulse widths/duty cycles represented as Fourier series can be summed as shown graphically here. In these functions, $omega_2 = omega_1N$ where $N$ is an arbitrary number. The duty cycles are given as $frac{T_{p1}}{T_{1}} = frac{1}{N}$, and $frac{T_{p2}}{T_{2}} = 0.5$. Function 1 is negative.

However, I would like the second Fourier series representation to be time shifted, so that the summed function is as shown graphically here.

Is this possible to do with a Fourier series representation, if so, how?

We can get a general answer for a time-shifted trigonometric Fourier series as follows

$$ f(t - tau) = a_v + sum_{n = 1}^{infty}left[a_n cos(nomega_c (t-tau)) + b_n sin(nomega_c (t - tau))right]$$

Let $alpha = cos(nomega_c (t-tau))$ and let $beta= sin(nomega_c (t-tau))$. Then

begin{align*}
alpha &= cos(nomega_c t - nomega_c tau) = cos(nomega_c t)cos(nomega_c tau) + sin(nomega_c t)sin(nomega_c tau) \
beta &= sin(nomega_c t - nomega_c tau) = sin(nomega_c t)cos(nomega_c tau) - cos(nomega_c t)sin(nomega_c tau)end{align*}

We can then combine these results as follows

$$a_nalpha + b_nbeta = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }cos(nomega_c t) + { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }sin(nomega_c t). $$

We can define the following to make things simpler

$$Av_f = { a_v }_f = { frac{a_0}{2} }_f\
A_f(n,tau) = { a_ncos(nomega_c tau) - b_nsin(nomega_c tau) }_f \
B_f(n,tau) = { a_nsin(nomega_c tau) + b_ncos(nomega_c tau) }_f.$$

Using the above definitions, we can define a time-shifted trigonometric Fourier series as follows

$$ f(t - tau) = Av_f + sum_{n = 1}^{infty}left[A_f(n,tau) cos(nomega_c t) + B_f(n,tau) sin(nomega_c t)right].$$

This will allow us to add two trigonometric Fourier series, each with their own distinct time-shifts as follows

$$ g(t - tau) = Av_g + sum_{n = 1}^{infty}left[A_g(n,tau) cos(nomega_c t) + B_g(n,tau) sin(nomega_c t)right] \
h(t - lambda) = Av_h + sum_{n = 1}^{infty}left[A_h(n,lambda) cos(nomega_c t) + B_h(n,lambda) sin(nomega_c t)right] $$

$$ f(t) = g(t - tau) + h(t - lambda) $$

We can setup some more definition to make things simpler as follows

begin{align*}
Av_f &= Av_g + Av_h \
A_f(n, tau, lambda) &= { A_g(n, tau) + A_h(n, lambda)} \
B_f(n, tau, lambda) &= { B_g(n, tau) + B_h(n, lambda)}. end{align*}

Then

$$
f(t) = Av_f + sum_{n = 1}^{infty}left[A_f(n, tau, lambda) cos(nomega_c t) + B_f(n, tau, lambda) sin(nomega_c t)right].
$$