DTFS linearity and time shift

Let $x[n]$ and $y[n]$ be periodic with period $N$ ($\omega_0 = 2\pi/N$), with $x[n] \xleftrightarrow{FS} a_k$ and $y[n] \xleftrightarrow{FS} b_k$, where

$$a_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n]e^{-jk\omega_0 n}$$

Linearity

Statement: for any constants $A$ and $B$,

$$z[n] = A x[n] + B y[n] \xleftrightarrow{FS} c_k = A a_k + B b_k$$

Proof: $z[n]$ is also periodic with period $N$. Then

$$\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle}\left(A x[n] + B y[n]\right)e^{-jk\omega_0 n} \\ &= A\cdot\frac{1}{N}\sum_{n=\langle N\rangle} x[n]e^{-jk\omega_0 n} + B\cdot\frac{1}{N}\sum_{n=\langle N\rangle} y[n]e^{-jk\omega_0 n} \\ &= A a_k + B b_k \end{aligned}$$

Example: if $x[n] = \cos(2\pi n/N)$ ($a_{\pm1} = 1/2$) and $y[n] = 1$ ($b_0 = 1$), then $3x[n] + 2y[n]$ has $c_0 = 2$ and $c_{\pm1} = 3/2$.

Time shifting

Statement: a delay of $n_0$ samples multiplies each coefficient by a linear-phase term:

$$x[n - n_0] \xleftrightarrow{FS} a_k e^{-jk\omega_0 n_0} = a_k e^{-jk(2\pi/N)n_0}$$

Proof: let $z[n] = x[n - n_0]$. Its coefficients are

$$c_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n-n_0]e^{-jk\omega_0 n}$$

Put $m = n - n_0$, so $n = m + n_0$. As $n$ runs over one period, so does $m$:

$$\begin{aligned} c_k &= \frac{1}{N}\sum_{m=\langle N\rangle} x[m]e^{-jk\omega_0 (m+n_0)} \\ &= e^{-jk\omega_0 n_0}\cdot\frac{1}{N}\sum_{m=\langle N\rangle} x[m]e^{-jk\omega_0 m} = e^{-jk\omega_0 n_0}a_k \end{aligned}$$

Meaning: $|c_k| = |a_k|$, so a time shift does not change the magnitude spectrum. It only adds a phase of $-k\omega_0 n_0$, which is linear in $k$.