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.


1 Answer 1


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.

  • $\begingroup$ can you please confirm your answer? I tried applying your Dk's formula to find FS coefficient of cos(3t+1) from cos(t) but answers are not in agreement. The Oppenheim's book has similar problem and solution manual has different answer then suggested by this, I'm really stuck it'd be nice if you can confirm your results. $\endgroup$
    – Deep
    Sep 11, 2019 at 17:54
  • $\begingroup$ nevermind there's nothing wrong with your solution, I mistakenly put k=3 (thinking that new coefficient would be at thrice frequency), I got my answers matched now, thanks a lot. $\endgroup$
    – Deep
    Sep 11, 2019 at 18:25

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