Let $x (n)$ be a Discrete Time periodic sequence with period $N.$ i.e $x(n+N)=x(n)$

$\mathrm{x}$ (n) can be represented in Fourier series as

$x[n]=\sum_{k=0}^{N-1} C_{k} \cdot e^{j 2 \pi k n / N} \cdots(1),$

which is called the Discrete Time Fourier series (DTFS).

Here $C_{k}$ are the DTFS coefficients and are given by

$C_{k}=\frac{1}{N} \sum_{n=0}^{N-1} x[n] \cdot e^{-j 2 \pi k n / N} \cdots(2),$

where $k=0,1,2\ldots N-1$

The frequency range of DT signals is unique over the interval $(-\pi, \pi)$ or $(0,2 \pi)$ . A DT signal of fundamental period $\mathrm{N}$ can consist of frequency components separated by $\frac{2 \pi}{N}$ radians or $f=\frac{1}{N}$ cycles. Thus, the DTFS representation of a DT periodic signal contains at the most N frequency components.

Given:

$X(n)=\{0,1,2,3\}$

We assume $x(n)$ is periodic with period $N=4$

Consider, $e^{-j \pi}=\cos \pi-j \sin \pi=-1-0=-1 \cdots(3)$

$e^{-j \pi / 2}=\cos \frac{\pi}{2}-j \sin \frac{\pi}{2}=0-j \times 1=-j \cdots(4)$

Similarly, $e^{-j 2 \pi}=1 ; e^{-j 3 \pi / 2}=j ; e^{-j 3 \pi / 2}=j ; e^{-j 3 \pi}=-1; e^{-j 9 \pi / 2}=-j ; \cdots(5) $

$\therefore$ From (2),

$C_{k}=\frac{1}{4} \sum_{n=0}^{3} x[n] \cdot e^{-j 2 \pi k n / 4}$

$=\frac{1}{4}\left\{x[0] \cdot e^{0}+x[1] \cdot e^{-j 2 \pi k \times 1 / 4}+x[2] \cdot e^{-j 2 \pi k \times 2 / 4}+.x[3] \cdot e^{-j 2 \pi k \times 3 / 4}\right\}$

$\therefore C_{k}=\frac{1}{4}\left[0+1 e^{-j \pi k / 2}+2 e^{-j \pi k}+3 e^{-j 3 \pi k / 2}\right]$

For $\mathrm{k}=0, C_{0}=\frac{1}{4}\left[e^{0}+2 e^{0}+3 e^{0}\right]$

$=\frac{1}{4}[1+2+3]=\frac{3}{2}+0 j$

For $\mathrm{k}=1, C_{1}=\frac{1}{4}\left[e^{-j \pi / 2}+2 e^{-j \pi}+3 e^{-j 3 \pi / 2}\right]$

$=\frac{1}{4}[-j-2+3 j] (\text { From } 3,4 \& 5)$

$=\frac{-1}{2}+\frac{1}{2} j$

For $\mathrm{k}=2, C_{2}=\frac{1}{4}\left[e^{-j \pi}+2 e^{-j 2 \pi}+3 e^{-j 3 \pi}\right]$

$\left.=\frac{1}{4}[-1+2-3] \text { (From } 3 \& 5\right)$

$=\frac{-1}{2}+0 j$

For $\mathrm{k}=3, C_{3}=\frac{1}{4}\left[e^{-j 3 \pi / 2}+2 e^{-j 3 \pi}+3 e^{-j 9 \pi / 2}\right]$

$\left.=\frac{1}{4}[j-2-3 j] \text { (From } 4 \& 5\right)$

$=\frac{-1}{2}-\frac{1}{2} j$

$\therefore$ From (1),

$x[n]=\sum_{k=0}^{3} C_{k} \cdot e^{j 2 \pi k n / 4}$

$=C_{0} e^{0}+C_{1} e^{j 2 \pi 1 n / 4}+C_{2} e^{j 2 \pi 2 n / 4}+C_{3} e^{j 2 \pi 3 n / 4}$

$=\frac{3}{2}+\left(\frac{-1}{2}+\frac{1}{2} j\right) e^{j \pi n / 2}-\frac{1}{2} e^{j \pi n}+\left(\frac{-1}{2}-\frac{1}{2} j\right) e^{j 3 \pi n / 2}$

$\therefore$ DTFS is $x(n)=\frac{3}{2}+\frac{1}{2}(-1+j) e^{j \pi n / 2}-\frac{1}{2} e^{j \pi n}-\frac{1}{2}(1+j) e^{j 3 \pi n / 2}$