# What is the Frequency Response of Discrete-Time Systems?

Signals and SystemsElectronics & ElectricalDigital Electronics

## Frequency Response of Discrete-Time Systems

A spectrum of input sinusoids is applied to a linear time invariant discrete-time system to obtain the frequency response of the system. The frequency response of the discrete-time system gives the magnitude and phase response of the system to the input sinusoids at all frequencies.

Now, let the impulse response of an LTI discrete-time system is $\mathit{h}\mathrm{\left(\mathit{n}\right)}$ and the input to the system is a complex exponential function, i.e.,$\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{e^{\mathit{j\omega n}}}$. Then, the output $\mathit{y}\mathrm{\left(\mathit{n}\right)}$ of the system is obtained by using the convolution theorem, i.e.,

$$\mathrm{\mathit{y}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{h}\mathrm{\left(\mathit{n}\right)}*\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\sum_{\mathit{k=-\infty} }^{\infty}\mathit{h}\mathrm{\left(\mathit{k}\right)}\mathit{x}\mathrm{\left(\mathit{n-k}\right)}}$$

As the input to the system is $\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{e^{\mathit{j\omega n}}}$ ,then,

$$\mathrm{\mathit{y}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\sum_{\mathit{k=-\infty} }^{\infty}\mathit{h}\mathrm{\left(\mathit{k}\right)}\mathit{e^{\mathit{j\omega \mathrm{\left ( \mathit{n-k}\right )}}}}}$$

$$\mathrm{\Rightarrow \mathit{y}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{e^{j\omega n}}\sum_{\mathit{k=-\infty} }^{\infty}\mathit{h}\mathrm{\left(\mathit{k}\right)}\mathit{e^{\mathit{-j\omega k}}}}$$

$$\mathrm{\therefore \mathit{y}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{e^{j\omega n }}\cdot \mathit{H}\mathrm{\left(\mathit{\omega }\right)}}$$

Where

• $\mathit{e^{j\omega n }}$ is the input sequence, and

• $\mathrm{H}\mathrm{\left(\mathit{\omega }\right)}$ is the frequency response of the discrete-time system.

Therefore, the output of the discrete-time system is identical to the input modified in magnitude and phase by $\mathrm{H}\mathrm{\left(\mathit{\omega }\right)}$. The frequency response $\mathrm{H}\mathrm{\left(\mathit{\omega }\right)}$ of the discrete-time system is a complex quantity and can be expressed as −

$$\mathrm{\mathit{H}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\left|\mathit{H}\mathrm{\left(\mathit{\omega }\right)} \right|\mathit{e^{j\angle \mathit{H}\mathrm{\left(\mathit{\omega }\right)}}}}$$

Where

• $\left|\mathit{H}\mathrm{\left(\mathit{\omega }\right)} \right|$ s called the magnitude response of the discrete-time system, and

• $\angle \mathit{H}\mathrm{\left(\mathit{\omega }\right)}$ is called the phase response of the system.

Also, the graph plotted between $\left|\mathit{H}\mathrm{\left(\mathit{\omega }\right)} \right|$ and $\omega$ is called the magnitude response plot and the graph plotted between $\angle \mathit{H}\mathrm{\left(\mathit{\omega }\right)}$ and $\omega$ is called the phase response plot..

## Properties of Frequency Response of Discrete-Time Systems

If the impulse response $\mathit{h}\mathrm{\left(\mathit{n}\right)}$ of the LTI discrete-time system is a real sequence, then the frequency response $\mathrm{H}\mathrm{\left(\mathit{\omega }\right)}$ possesses the following properties −

• The frequency response $\mathrm{H}\mathrm{\left(\mathit{\omega }\right)}$ takes on values for all $\omega$.

• The frequency response $\mathrm{H}\mathrm{\left(\mathit{\omega }\right)}$ is periodic in $\omega$ with time period 2$\pi$.

• $\left|\mathit{H}\mathrm{\left(\mathit{\omega }\right)} \right|$ , i.e., the magnitude response of the system is an even function of $\omega$ and is symmetrical about $\pi$.

• $\angle \mathit{H}\mathrm{\left(\mathit{\omega }\right)}$ ,i.e., the phase response of the system is an odd function of $\omega$ and it is anti-symmetrical about $\pi$.

## Numerical Example

Find the frequency response of the following discrete-time causal system.

$$\mathrm{\mathit{y}\mathrm{\left(\mathit{n}\right)}-2\mathit{y}\mathrm{\left(\mathit{n-\mathrm{1}}\right)}\:\mathrm{+}\:\frac{2}{9}\mathit{y}\mathrm{\left(\mathit{n-\mathrm{2}}\right)}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{n}\right)}-\frac{3}{5}\mathit{x}\mathrm{\left(\mathit{n-\mathrm{1}}\right)}}$$

Solution

If X(ω) and Y(ω) are the Fourier transforms of input and output sequences, then the frequency response of the discrete-time system is given by,

$$\mathrm{\mathit{H}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\frac{\mathit{Y}\mathrm{\left(\mathit{\omega }\right)}}{\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}}$$

Now, the equation describing the system is

$$\mathrm{\mathit{y}\mathrm{\left(\mathit{n}\right)}-2\mathit{y}\mathrm{\left(\mathit{n-\mathrm{1}}\right)}\:\mathrm{+}\:\frac{2}{9}\mathit{y}\mathrm{\left(\mathit{n-\mathrm{2}}\right)}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{n}\right)}-\frac{3}{5}\mathit{x}\mathrm{\left(\mathit{n-\mathrm{1}}\right)}}$$

Taking discrete-time Fourier transform on both sides, we get,

$$\mathrm{\mathit{Y}\mathrm{\left(\mathit{\omega }\right)}-2\mathit{e^{-j\omega }}\mathit{Y}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{+}\:\frac{2}{9}\mathit{e^{-j\mathrm{2}\omega }}\mathit{Y}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega }\right)}-\:\frac{3}{5}\mathit{e^{-j\omega }}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$

$$\mathrm{\Rightarrow \mathit{Y}\mathrm{\left(\mathit{\omega }\right)}\mathrm{\left ( 1-2\mathit{e^{-j\omega }} +\frac{2}{9}\mathit{e^{-j\mathrm{2}\omega }}\right )}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\mathrm{\left ( 1-\frac{3}{5}\mathit{e^{-j\omega }} \right )}}$$

$$\mathrm{\Rightarrow \frac{\mathit{Y}\mathrm{\left(\mathit{\omega }\right)}}{\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}\:\mathrm{=}\:\frac{\mathrm{\left ( 1-\frac{3}{5}\mathit{e^{-j\omega }} \right )}}{\mathrm{\left ( 1-2\mathit{e^{-j\omega }} +\frac{2}{9}\mathit{e^{-j\mathrm{2}\omega }}\right )}}}$$

$$\mathrm{\therefore \mathit{H}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\: \frac{\mathit{Y}\mathrm{\left(\mathit{\omega }\right)}}{\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}\:\mathrm{=}\:\frac{^{\mathit{e}^\mathit{j\omega} }\mathrm{\left ( \mathit{e^{j\omega }-\frac{\mathrm{3}}{\mathrm{5}}} \right )}}{\mathrm{\left ( \mathit{e^{j\mathrm{2}\omega }-\mathrm{2}\mathit{e^{j\omega }}\mathrm{+}\frac{\mathrm{2}}{\mathrm{9}}} \right )}}}$$

Updated on 31-Jan-2022 05:22:41