Time-scaling property of CTFS

Statement

If $x(t)$ is periodic with period $T$ (fundamental frequency $\omega_0 = 2\pi/T$) and

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

then for $\alpha > 0$, the time-scaled signal $x(\alpha t)$ is periodic with period $T/\alpha$ and fundamental frequency $\alpha\omega_0$, and its Fourier series coefficients are the same $a_k$:

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

Time scaling does not change the coefficients; it only changes the spacing of the spectral lines from $\omega_0$ to $\alpha\omega_0$.

Proof

Synthesis route: replace $t$ by $\alpha t$ in the synthesis equation:

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

This is a Fourier series with fundamental frequency $\alpha\omega_0$ and coefficients $a_k$.

Analysis route (check): let $y(t) = x(\alpha t)$, period $T' = T/\alpha$, $\omega_0' = \alpha\omega_0$:

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

Hence $b_k = a_k$.

Interpretation

• Compressing in time ($\alpha > 1$) spreads the spectral lines further apart in frequency; expanding ($\alpha < 1$) brings them closer.
• The amplitude and phase of each harmonic are unchanged, only its frequency $k\omega_0$ becomes $k\alpha\omega_0$.

Example: $x(t) = \cos(2t)$ has $a_{\pm1} = 1/2$ at $\omega_0 = 2$. Then $x(3t) = \cos(6t)$ still has $a_{\pm1} = 1/2$, now at $\omega_0' = 6$ rad/s.