CTFS shifting properties; coefficients of A cos ω₀t

Let $x(t)$ have period $T$, $\omega_0 = 2\pi/T$, and $x(t) \overset{FS}{\longleftrightarrow} a_k$.

Time shifting property

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

Proof sketch: with $\tau = t - t_0$,

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

A delay leaves $|a_k|$ unchanged and adds a phase $-k\omega_0 t_0$ that grows linearly with $k$.

Frequency shifting property

$$e^{jM\omega_0 t}\,x(t) \overset{FS}{\longleftrightarrow} a_{k-M}$$

Proof:

$$b_k = \frac{1}{T}\int_T e^{jM\omega_0 t}x(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T x(t)e^{-j(k-M)\omega_0 t}dt = a_{k-M}$$

Multiplying by a complex exponential at the $M$th harmonic shifts the whole set of coefficients by $M$ places (this is the basis of modulation).

Fourier series coefficients of $x(t) = A\cos(\omega_0 t)$

Using Euler's formula:

$$x(t) = A\cos\omega_0 t = \frac{A}{2}e^{j\omega_0 t} + \frac{A}{2}e^{-j\omega_0 t}$$

Comparing with $x(t) = \sum_k a_k e^{jk\omega_0 t}$:

$$a_1 = a_{-1} = \frac{A}{2}, \qquad a_k = 0 \text{ for } k \ne \pm1$$

Check with the analysis equation for $k = 1$:

$$a_1 = \frac{1}{T}\int_0^T A\cos\omega_0 t\,e^{-j\omega_0 t}dt = \frac{A}{2T}\int_0^T\left(1 + e^{-j2\omega_0 t}\right)dt = \frac{A}{2}$$

 a_k
 A/2      |       |
      ----+---+---+----> k
         -1   0   1

Answer: $a_{\pm1} = A/2$ (real, zero phase), all other coefficients are zero.