Relation between Discrete-Time Fourier Transform and Z-Transform

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 x(n) is given by,

$$\mathrm{F[x(n)] \:=\: X(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n) e^{-j\: \omega n} \quad \dotso\: (1)}$$

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 x(n) is given by,

$$\mathrm{Z[x(n)] \:=\: X(z) \:=\: \sum_{n=-\infty}^{\infty}\: x(n) z^{-n} \quad \dotso\: (2)}$$

Relation between DTFT and Z-Transform

Since the DTFT of a discrete time sequence x(n) is given by,

$$\mathrm{X(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n) e^{-j\: \omega n} \quad \dotso\: (3)}$$

For the existence of the DTFT, the sequence x(n) must be absolutely summable, thus the summation in the above equation should converge.

Also, the Z-transform of the sequence x(n) is given by,

$$\mathrm{X(z) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: z^{-n} \quad \dotso\: (4)}$$

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

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

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

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

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

For the existence of the Z-transform,

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

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

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

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

Again, for the existence of the discrete-time Fourier transform (DTFT), the discrete-time sequence x(n) must be absolutely integrable, i.e.,

$$\mathrm{\sum_{n=-\infty}^{\infty}\: |x(n)|\: \lt \:\infty}$$

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.