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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.

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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: $$ x_1 \longleftrightarrow c_k = \frac{1}{T_1} \int_{<T_1>} x_1(t) e^{-j\frac{2\pi k}{T_1}t} dt ~~ , ~~\text{for}~~ k=0,\pm 1, \pm 2...$$

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.

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