FS coefficients; time shifting and conjugation

Information provided by Fourier series coefficients

The coefficient $a_k$ gives the amount (magnitude $|a_k|$) and the phase ($\angle a_k$) of the $k$th harmonic $e^{jk\omega_0 t}$ in the signal. This makes the coefficients the frequency-domain picture of the signal. $a_0$ is the DC (average) value, $|a_k|^2$ is the power at frequency $k\omega_0$ (by Parseval, $P = \sum|a_k|^2$), and the spread of the $a_k$ shows the bandwidth of the signal.

Time shifting property

Statement: if $x(t) \xleftrightarrow{FS} a_k$ with period $T$ and $\omega_0 = 2\pi/T$, then

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

Proof: the coefficients of $y(t) = x(t - t_0)$ are

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

Put $\tau = t - t_0$ (so $dt = d\tau$). The range of $\tau$ is still one full period:

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

Interpretation: $|b_k| = |a_k|$, so the magnitude spectrum does not change. The phase changes by $-k\omega_0 t_0$, which is proportional to the harmonic number.

Conjugation property

Statement:

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

Proof: the coefficients of $x^*(t)$ are

$$\begin{aligned} b_k &= \frac{1}{T}\int_T x^*(t)e^{-jk\omega_0 t}dt = \left[\frac{1}{T}\int_T x(t)e^{+jk\omega_0 t}dt\right]^* \\ &= \left[\frac{1}{T}\int_T x(t)e^{-j(-k)\omega_0 t}dt\right]^* = a_{-k}^* \end{aligned}$$

Consequence: if $x(t)$ is real, then $x(t) = x^*(t)$, which gives $a_{-k} = a_k^*$ (conjugate symmetry). The magnitude spectrum is then even and the phase spectrum is odd.