DTFS frequency shifting and convolution

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

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

Frequency shifting property

Statement: multiplying a signal by a harmonic exponential shifts its coefficients:

$$e^{jM\omega_0 n}x[n] \xleftrightarrow{FS} a_{k-M}$$

Proof: let $z[n] = e^{jM(2\pi/N)n}x[n]$. Its coefficients are

$$\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle} e^{jM\omega_0 n}x[n]e^{-jk\omega_0 n} \\ &= \frac{1}{N}\sum_{n=\langle N\rangle} x[n]e^{-j(k-M)\omega_0 n} = a_{k-M} \end{aligned}$$

So multiplying by $e^{jM\omega_0 n}$ moves the whole spectrum $M$ places to the right. This is the DT form of modulation.

Convolution property (periodic convolution)

Statement: periodic convolution in time corresponds to multiplication of the coefficients, scaled by $N$:

$$\sum_{r=\langle N\rangle} x[r]\,y[n-r] \xleftrightarrow{FS} N a_k b_k$$

Proof: let $z[n] = \sum_{r=\langle N\rangle} x[r]y[n-r]$, which is also periodic with period $N$. Then

$$\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle}\sum_{r=\langle N\rangle} x[r]\,y[n-r]\,e^{-jk\omega_0 n} \\ &= \frac{1}{N}\sum_{r=\langle N\rangle} x[r]e^{-jk\omega_0 r}\sum_{n=\langle N\rangle} y[n-r]e^{-jk\omega_0 (n-r)} \end{aligned}$$

Put $m = n - r$. Because $y[m]e^{-jk\omega_0 m}$ is periodic with period $N$, the inner sum over any $N$ consecutive $m$ is

$$\sum_{m=\langle N\rangle} y[m]e^{-jk\omega_0 m} = N b_k$$

Therefore

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

Dual (multiplication) property: $x[n]y[n] \xleftrightarrow{FS} \sum_{l=\langle N\rangle} a_l b_{k-l}$, which is periodic convolution of the coefficients.