# Fourier series - time shift and scaling

What will be the new Fourier series coefficients when we shift and scale a periodic signal? Scaling alone will only affect fundamental frequency. But how to calculate new coefficients of shifted and scaled version. I tried searching, but couldn't find an answer where both properties are used. Please help.

Consider a continuous-time periodic signal $$x_1(t)$$ whose fundamental period is $$T_1$$, fundamental radian frequency is $$\omega_1 = \frac{2\pi}{T_1}$$ and CTFS (continuous-time Fourier series) coefficients are: $$x_1 \longleftrightarrow c_k = \frac{1}{T_1} \int_{} x_1(t) e^{-j\frac{2\pi k}{T_1}t} dt ~~ , ~~\text{for}~~ k=0,\pm 1, \pm 2...$$

We want to find the new CTFS coefficients, denoted as $$d_k$$, associated with the new signal $$x_2(t)$$ , with period $$T_2$$ and fundamental radian frequency $$\omega_2 = \frac{2\pi}{T_2}$$, and related to $$x_1(t)$$ as follows: $$x_2(t) = x_1( a(t-b) )$$ where $$a$$ and $$b$$ are constants, assuming $$a > 0$$. Then we say ; $$x_2 \longleftrightarrow d_k$$

We shall proceed from the above definition to show the relation between the coefficients $$d_k$$ and $$c_k$$ as follows: $$d_k = \frac{1}{T_2} \int_{-T_2/2}^{T_2/2} x_2(t) e^{-j\frac{2\pi}{T_2} k t} dt$$

First note that $$T_2 = T_1/a$$. And replacing $$x_2$$ with $$x_1$$ yields the following:

$$d_k = \frac{a}{T_1} \int_{t=-T_1/{2a}}^{t=T_1/{2a}} x_1(a(t-b)) e^{-j\frac{2\pi a}{T_1} k t} dt$$ Now apply a substitution $$t-b = t'$$, and then replace $$t'$$ with $$t$$ again (note that $$dt'=dt$$)

$$d_k = \frac{a}{T_1} \int_{t=-\frac{T_1}{2a}-b}^{t=\frac{T_1}{2a}-b} x_1(at) e^{-j\frac{2\pi a}{T_1} k (t+b)} dt$$ We know that for a periodic signal, as long as the integration involves a single period, we can redefine its limits:

$$d_k = e^{-j \frac{2\pi a}{T_1} k b} \left( \frac{a}{T_1} \int_{t=-T_1/{2a}}^{t=T_1/{2a}} x_1(at) e^{-j\frac{2\pi a}{T_1} k t} dt \right)$$ Now apply substitution $$at = t'$$ (note that $$dt' = a dt$$) and then replace $$t'$$ with $$t$$ again which yields:

$$d_k = e^{-j \frac{2\pi a}{T_1} k b} \left( \frac{1}{T_1} \int_{t=-T_1/{2}}^{t=T_1/{2}} x_1(t) e^{-j\frac{2\pi}{T_1} k t} dt \right)$$ Finally recognise that the integral equals $$T_1 c_k$$ and the relationship between $$d_k$$ and $$c_k$$ is:

$$d_k = e^{-j \frac{2\pi}{T_1} a b k} c_k$$

Note that the coefficients $$d_k$$ are for a periodic signal with fundamental frequency $$\omega_2 = \frac{2\pi}{T_2}$$, whereas $$c_k$$ are for the frequency $$\omega_1 = \frac{2\pi}{T_1}$$. So this means that eventhough, for example, $$|d_3|=|c_3|$$, i.e., both coefficients have the same magnitude for $$k=3$$, those coefficients are placed at different frequencies; $$d_3$$ is associated with $$\omega = \frac{2\pi}{T_1}6$$ whereas $$c_3$$ is associated with the frequency $$\omega=\frac{2\pi}{T_1}3$$... It can be seen that the new coefficients $$d_k$$ have the same magnitudes, i.e., $$|d_k| = |c_k|$$, but different phases.