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What is the Frequency Response of Discrete-Time Systems?
Frequency Response of Discrete-Time Systems
A spectrum of input sinusoids is applied to a linear time-invariant (LTI) 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.
Let the impulse response of an LTI discrete-time system be $\mathrm{h(n)}$ and the input to the system be a complex exponential function, i.e., $\mathrm{x(n) \:=\: e^{j \omega n}}$. Then, the output $\mathrm{y(n)}$ of the system is obtained by using the convolution theorem, i.e.,
$$\mathrm{y(n) \:=\: h(n)\:\cdot\: x(n) \:=\: \sum_{k=-\infty}^{\infty}\: h(k) x(n\:-\:k)}$$
As the input to the system is $\mathrm{x(n) \:=\: e^{j \omega n}}$, then:
$$\mathrm{y(n) \:=\: \sum_{k=-\infty}^{\infty}\: h(k)\: e^{j \omega (n-k)}}$$
$$\mathrm{y(n) \:=\: e^{j \omega n}\: \sum_{k=-\infty}^{\infty}\: h(k)\: e^{-j \omega k}}$$
$$\mathrm{\therefore\:y(n) \:=\: e^{j \omega n} \:\cdot\: H(\omega)}$$
Where:
- $\mathrm{e^{j \omega n}}$ is the input sequence, and
- $\mathrm{H(\omega)}$ 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 $\mathrm{H(\omega)}$. The frequency response $\mathrm{H(\omega)}$ of the discrete-time system is a complex quantity and can be expressed as:
$$\mathrm{H(\omega) \:=\: |H(\omega)| \:e^{j \angle H(\omega)}}$$
Where:
- $\mathrm{|H(\omega)|}$ is called the magnitude response of the discrete-time system, and
- $\mathrm{\angle\: H(\omega)}$ is called the phase response of the system.
Also, the graph plotted between $\mathrm{|H(\omega)|}$ and $\mathrm{\omega}$ is called the magnitude response plot, and the graph plotted between $\mathrm{\angle H(\omega)}$ and $\mathrm{\omega}$ is called the phase response plot.
Properties of Frequency Response of Discrete-Time Systems
If the impulse response $\mathrm{h(n)}$ of the LTI discrete-time system is a real sequence, then the frequency response $\mathrm{H(\omega)}$ possesses the following properties:
- The frequency response $\mathrm{H(\omega)}$ takes on values for all $\mathrm{\omega}$.
- The frequency response $\mathrm{H(\omega)}$ is periodic in $\mathrm{\omega}$ with a time period $\mathrm{2\pi}$.
- $\mathrm{|H(\omega)|}$\, i.e., the magnitude response of the system, is an even function of $\mathrm{\omega}$ and is symmetrical about $\mathrm{\pi}$.
- $\mathrm{\angle H(\omega)}$, i.e., the phase response of the system, is an odd function of $\mathrm{\omega}$ and it is anti-symmetrical about $\mathrm{\pi}$.
Numerical Example
Find the frequency response of the following discrete-time causal system:
$$\mathrm{y(n) \:-\: 2y(n\:-\:1) \:+\: \frac{2}{9}y(n\:-\:2) \:=\: x(n) \:-\: \frac{3}{5}x(n\:-\:1)}$$
Solution
If $\mathrm{X(\omega)}$ and $\mathrm{Y(\omega)}$ are the Fourier transforms of the input and output sequences, then the frequency response of the discrete-time system is given by:
$$\mathrm{H(\omega) \:=\: \frac{Y(\omega)}{X(\omega)}}$$
Now, the equation describing the system is:
$$\mathrm{y(n) \:-\: 2y(n\:-\:1) \:+\: \frac{2}{9}y(n\:-\:2) \:=\: x(n) \:-\: \frac{3}{5}x(n\:-\:1)}$$
Taking the discrete-time Fourier transform on both sides, we get:
$$\mathrm{Y(\omega) \:-\: 2e^{-j \omega} Y(\omega) \:+\: \frac{2}{9}e^{-j 2 \omega} Y(\omega) \:=\: X(\omega) \:-\: \frac{3}{5}e^{-j \omega}\: X(\omega)}$$
$$\mathrm{\Rightarrow\:Y(\omega) \left( 1 \:-\: 2e^{-j \omega} \:+\: \frac{2}{9}e^{-j 2 \omega} \right) \:=\: X(\omega) \left( 1 \:-\: \frac{3}{5}e^{-j \omega} \right)}$$
$$\mathrm{\Rightarrow\:\frac{Y(\omega)}{X(\omega)}\:=\:\frac{\left(1\:-\:\frac{3}{5}e^{-j \omega}\right)}{\left(1\:-\:2e^{-j\omega}\:+\:\frac{2}{9}e^{-j2\omega}\right)}}$$
$$\mathrm{\therefore\:H(\omega) \:=\: \frac{Y(\omega)}{X(\omega)} \:=\: \frac{e^{j \omega}}{\left(e^{j \omega} \:-\: \frac{3}{5}\right)\left(e^{j 2 \omega} \:-\: 2e^{j \omega} \:+\: \frac{2}{9}\right)}}$$