Linearity, Periodicity and Symmetry Properties of Discrete-Time Fourier Transform

---

---

Discrete-Time Fourier Transform (DTFT)

The Fourier transform of a discrete-time sequence is known as the discrete-time Fourier transform (DTFT). Mathematically, the discrete-time Fourier transform of a discrete-time sequence x(n) is defined as:

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

Linearity Property of Discrete-Time Fourier Transform

Statement

The linearity property of discrete-time Fourier transform states that, the DTFT of a weighted sum of two discrete-time sequences is equal to the weighted sum of individual discrete-time Fourier transforms. Therefore, if:

$$\mathrm{F[x_1(n)] \:\overset{FT}\longleftrightarrow\: X_1(\omega) \quad \text{and} \quad F[x_2(n)] \:=\: X_2(\omega)}$$

Then:

$$\mathrm{F[a x_1(n) \:+\: b x_2(n)] \:=\: a X_1(\omega) \:+\: b X_2(\omega)}$$

Proof

From the definition of the discrete-time Fourier transform, we have:

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

$$\mathrm{\therefore\:F[a x_1(n) \:+\: b x_2(n)] \:=\: \sum_{n=-\infty}^{\infty}\: \left[ a x_1(n) \:+\: b x_2(n) \right] e^{-j \omega n}}$$

$$\mathrm{F[a x_1(n) \:+\: b x_2(n)] \:=\: \sum_{n=-\infty}^{\infty}\:a \:x_1(n) \:e^{-j \omega n} \:+\: \sum_{n=-\infty}^{\infty}\:b\: x_2(n)\: e^{-j \omega n}}$$

$$\mathrm{F[a x_1(n) \:+\: b x_2(n)] \:=\: a X_1(\omega) \:+\: b X_2(\omega)}$$

Periodicity Property of Discrete-Time Fourier Transform

The periodicity property of discrete-time Fourier transform states that the DTFT $\mathrm{X(\omega)}$ is periodic in $\mathrm{\omega}$ with period $\mathrm{2\pi}$, that is:

$$\mathrm{X(\omega) \:=\: X(\omega \:+\: 2n\pi)}$$

Therefore, using the periodicity property of DTFT, we need only one period of $\mathrm{X(\omega)}$ for the analysis and not the whole range $\mathrm{-\infty\: \lt\: \omega\: \lt\: \infty}$.

Symmetry Property of Discrete-Time Fourier Transform

The discrete-time Fourier transform (DTFT) $\mathrm{X(\omega)}$ is a complex function of $\mathrm{\omega}$ and hence can be expressed as:

$$\mathrm{X(\omega) \:=\: X_r(\omega) \:+\: j X_i(\omega)}$$

Where:

• $\mathrm{X_r(\omega)}$ is the real part of $\mathrm{X(\omega)}$, and
• $\mathrm{X_i(\omega)}$ is the imaginary part of $\mathrm{X(\omega)}$.

Now, from the definition of the DTFT, we have:

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

$$\mathrm{X(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: \cos(\omega n) \:-\: j \sum_{n=-\infty}^{\infty}\: x(n)\: \sin(\omega n)}$$

$$\mathrm{\Rightarrow\:X_r(\omega) \:+\: j X_i(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: \cos(\omega n) \:-\: j \sum_{n=-\infty}^{\infty}\: x(n)\: \sin(\omega n)}$$

On comparing LHS and RHS, we get:

$$\mathrm{X_r(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: \cos(\omega n)}$$

And

$$\mathrm{X_i(\omega) \:=\: - \sum_{n=-\infty}^{\infty}\: x(n) \:\sin(\omega n)}$$

Since $\mathrm{\cos(-\omega n) \:=\: \cos(\omega n)}$ and $\mathrm{\sin(-\omega n) \:=\: -\sin(\omega n)}$, we have:

$$\mathrm{X_r(-\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: \cos(-\omega n) \:= \:\sum_{n=-\infty}^{\infty}\: x(n)\: \cos(\omega n)}$$

$$\mathrm{\Rightarrow\:X_r(-\omega) \:=\: X_r(\omega)}$$

i.e, the real part of DTFT $\mathrm{X_r(\omega)}$ is an even function of $\mathrm{\omega}$, i.e., it has even symmetry property.

Also

$$\mathrm{X_i(-\omega) \:=\: - \sum_{n=-\infty}^{\infty}\: x(n)\: \sin(-\omega n) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: \sin(\omega n)}$$

$$\mathrm{\therefore\:X_i(-\omega) \:=\: - X_i(\omega)}$$

Therefore, the imaginary part of DTFT $\mathrm{X_i(\omega)}$ is an odd function of $\mathrm{\omega}$, i.e., it has odd symmetry property.