# Signals and Systems – Relation between Discrete-Time Fourier Transform and Z-Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

## Discrete-Time Fourier Transform

The Fourier transform of the discrete-time signals is known as the discrete-time Fourier transform (DTFT). The DTFT converts a time domain sequence into frequency domain signal. The DTFT of a discrete time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is given by,

$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-j\omega n}}\:\:\:\:\:\:...(1)}$$

## Z-Transform

The Z-transform is a mathematical which is used to convert the difference equations in time domain into the algebraic equations in z-domain. Mathematically, the Z-transform of a discrete time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is given by,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}\:\:\:\:\:\:...(2)}$$

## Relation between DTFT and Z-Transform

Since the DTFT of a discrete time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is given by,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}\:\:\:\:\:\:...(3)}$$

For the existence of the DTFT, the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ must be absolutely summable, thus the summation in the above equation should converge.

Also, the Z-transform of the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is given by,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}\:\:\:\:\:\:...(4)}$$

Where, z is a complex variable and it is given by,

$$\mathrm{\mathit{z}\:\mathrm{=}\:\mathit{r\:e^{j\omega}}}$$

Where, r is the radius of a circle. Therefore, by substituting the value of z in equation (4), we get,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{r\:e^{j\omega }}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathrm{\left(\mathit{r\:e^{j\omega}}\right)}^{-\mathit{n}}}$$

$$\mathrm{\Rightarrow \mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{r}^{-\mathit{n}}\right]}\mathit{e^{-j\omega n}}\:\:\:\:\:\:...(5)}$$

For the existence of the Z-transform,

$$\mathrm{\:\sum_{\mathit{n=-\infty}}^{\infty}\left|\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{r^{\mathit{-n}}} \right|<\infty}$$

That is, the summation should converge or $\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{r}^{-\mathit{n}}\right]}$ must be absolutely integrable. The equation (5) represents the discrete time Fourier transform of a signal $\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{r}^{-\mathit{n}}$.

Therefore, it can be said that the Z-transform of a discrete time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is same as the discrete-time Fourier transform (DTFT) of $\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{r}^{-\mathit{n}}$, i.e.,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{r}^{-\mathit{n}} \right ]}}$$

Again, for the existence of the discrete-time Fourier transform (DTFT), the discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ must be absolutely integrable, i.e.,

$$\mathrm{\:\sum_{\mathit{n=-\infty}}^{\infty}\left|\mathit{x}\mathrm{\left(\mathit{n}\right)} \right|<\infty}$$

Thus, the DTFT for many sequences may not exist but the Z-transform may exist.

Also, if r = 1, then the discrete time Fourier transform (DTFT) is same as the Z-transform. In other words, the DTFT is nothing but the Z-transform evaluated along the unit circle centred at the origin of the z-plane.

Updated on 07-Jan-2022 06:51:41