State the relationship between DFS, DFT and Z Transform
1 Answer

Let $x (n)$ be a discrete sequence.

Part 1: DTFT and $Z T$

By definition of $Z-$transform, $X(z)=Z\{x[n]\}=\sum_{n=-\infty}^{\infty} x[n] \cdot z^{-n}$ and

By definition of Discrete Time Fourier transform, $X(\omega)=D T F T\{x[n]\}=\sum_{n=-\infty}^{\infty} x[n] \cdot e^{-j \omega n}$

It is found that, $X(z)=D T F T\left\{x[n] r^{-n}\right\}$

If $X(z)$ is evaluated on a unit circle i.e. $z=r e^{j \omega}=e^{j \omega}$ then $X(z)=\operatorname{DTFT}\{x[n]\}$

Hence, Fourier Transform of a discrete signal is equal to $Z-$ Transform evaluated on a unit circle.

Part 2: DTFT and DFT

DFT can be seen as the sampled version (in Frequency Domain) of the DTFT output. i.e. $X(k)=\left.X(\omega)\right|_{\omega=2 \pi k / N}$

Part 3: DFT and ZT

From Part I and II, DFT of a discrete signal is equal to $Z-$Transform evaluated on a unit circle calculated at discrete instant of Frequency.

Part 4: DFT and DTFS

Let sequence $x_{p}(n)$ is a periodic repetition of sequence $x(n)$ . Let $\mathrm{N}$ be period of $x_{p}(n)$

By definition, Fourier Series coefficients $C_{k}=D T F S\left\{x_{p}(n)\right\}=\frac{1}{N} \sum_{N} x_{p}(n) \cdot e^{-j 2 \pi k n / N},$ where $k=0,1,2 \ldots \mathrm{N}-1$

By definition, $X(k)=D F T[x(n)]=\sum_{N} x(n) \cdot e^{-j 2 \pi k n/N}$

$\therefore C_{k}=\frac{1}{N} X(k)$ or $N C_{k}=X(k)$

Thus, the N-point DFT provides the exact line spectrum of a periodic scquence with fundamental period $N .$

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