CTFS time shifting and time scaling

Let $x(t)$ be periodic with period $T$, $\omega_0 = 2\pi/T$, and $x(t) \overset{FS}{\longleftrightarrow} a_k$, where $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$.

Time shifting property

Statement:

$$x(t - t_0) \overset{FS}{\longleftrightarrow} a_k\, e^{-jk\omega_0 t_0}$$

Proof: $y(t) = x(t - t_0)$ has the same period $T$. Its coefficients are

$$\begin{aligned} b_k &= \frac{1}{T}\int_T x(t - t_0)e^{-jk\omega_0 t}dt \qquad (\tau = t - t_0) \\ &= \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0(\tau + t_0)}d\tau \\ &= e^{-jk\omega_0 t_0}\,\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0\tau}d\tau = e^{-jk\omega_0 t_0}\,a_k \end{aligned}$$

So $|b_k| = |a_k|$ (magnitude unchanged) and the phase changes by $-k\omega_0 t_0$.

Time scaling property

Statement: for $\alpha > 0$, $x(\alpha t)$ is periodic with period $T/\alpha$ and fundamental frequency $\alpha\omega_0$, and

$$x(\alpha t) = \sum_{k=-\infty}^{\infty}a_k e^{jk(\alpha\omega_0)t}$$

The coefficients stay $a_k$; only the fundamental frequency changes.

Proof: let $y(t) = x(\alpha t)$, $T' = T/\alpha$:

$$\begin{aligned} b_k &= \frac{1}{T'}\int_0^{T'}x(\alpha t)e^{-jk(\alpha\omega_0)t}dt \qquad (\tau = \alpha t,\ dt = d\tau/\alpha) \\ &= \frac{\alpha}{T}\int_0^{T}x(\tau)e^{-jk\omega_0\tau}\frac{d\tau}{\alpha} = a_k \end{aligned}$$

Summary

Example: $x(t) = \cos\omega_0 t$ ($a_{\pm1} = 1/2$); $x(t - t_0)$ has $a_{\pm1} = \frac{1}{2}e^{\mp j\omega_0 t_0}$; $x(2t) = \cos 2\omega_0 t$ still has $a_{\pm1} = 1/2$ but at fundamental $2\omega_0$.