CTFS shifting, scaling and conjugation

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

(a) Time shifting

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

Proof: put $\tau = t - t_0$:

$$b_k = \frac{1}{T}\int_T x(t-t_0)e^{-jk\omega_0 t}dt = e^{-jk\omega_0 t_0}\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau = e^{-jk\omega_0 t_0}a_k$$

The magnitudes $|b_k| = |a_k|$ stay the same. Only a linear phase $-k\omega_0 t_0$ is added.

(b) Time scaling

If $\alpha > 0$, then $x(\alpha t)$ is periodic with period $T/\alpha$ and fundamental frequency $\alpha\omega_0$. Its Fourier series is

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

Proof: with period $T' = T/\alpha$, put $\tau = \alpha t$:

$$b_k = \frac{\alpha}{T}\int_{T/\alpha} x(\alpha t)e^{-jk(\alpha\omega_0)t}dt = \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0\tau}d\tau = a_k$$

The coefficients do not change. What changes is the spacing of the harmonics, which becomes $\alpha\omega_0$ instead of $\omega_0$. Compressing a signal in time spreads its spectral lines further apart.

(c) Conjugation

$$x^*(t) \xleftrightarrow{FS} a_{-k}^*$$

Proof: take the conjugate of the synthesis equation:

$$x^*(t) = \sum_{k} a_k^* e^{-jk\omega_0 t} = \sum_{k} a_{-k}^* e^{jk\omega_0 t} \quad (k \to -k)$$

So the coefficients of $x^*(t)$ are $a_{-k}^*$.

Consequence (conjugate symmetry): if $x(t)$ is real, then $x(t) = x^*(t)$, so $a_k = a_{-k}^*$. This gives $|a_k| = |a_{-k}|$ (even magnitude) and $\angle a_{-k} = -\angle a_k$ (odd phase). If $x(t)$ is also even, the $a_k$ are real and even.