Problem 7 - Complex Analysis for Mathematics and Engineering

Question

The single-pole low-pass filter is \(y[n]=K x[n]+(1-K) y[n-1]\), where constant \(K\) is between 0 and \(1 .\) (a) Use \(K=\frac{1}{4}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\). (b) Use \(K=\frac{1}{10}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\). (c) Use \(K=\frac{1}{16}\) to find \(A(\theta), A(0), A\left(\frac{\pi}{4}\right), A\left(\frac{\pi}{2}\right)\), and \(A(\pi)\).

Step-by-Step Solution

Answer

For each K value: A(0)=1, A(π/4)=0.461, 0.186, 0.113; A(π/2)=1/3, 1/9, 1/15; A(π)=1/7, 1/19, 1/31.

Step 1: Understanding the system function

The form of a low-pass filter is given by the recurrence relation \(y[n] = K x[n] + (1-K) y[n-1]\). The transfer function in the Z-domain is \[H(z) = \frac{Y(z)}{X(z)} = \frac{K}{1 - (1-K)z^{-1}}\]. By replacing \(z\) with \(e^{j\theta}\), we can find the frequency response \(H(e^{j\theta}) = \frac{K}{1 - (1-K)e^{-j\theta}}\). The magnitude \(A(\theta) = |H(e^{j\theta})|\).

Step 2: Compute for K = 1/4

For \(K = \frac{1}{4}\), we have \(H(e^{j\theta}) = \frac{1/4}{1 - 3/4e^{-j\theta}}\). The magnitude is \(A(\theta) = \left|\frac{1/4}{1 - 3/4e^{-j\theta}}\right| = \frac{1/4}{\sqrt{(1 - 3/4 \cos(\theta))^2 + (3/4 \sin(\theta))^2}}\).

Step 3: Calculate specific amplitudes for K = 1/4

Using the magnitude expression, calculate:- \(A(0) = \frac{1/4}{|1 - 3/4|} = 1\).- \(A(\frac{\pi}{4}) = \frac{1/4}{0.541} \approx 0.461\).- \(A(\frac{\pi}{2}) = \frac{1/4}{3/4} = \frac{1}{3}\).- \(A(\pi) = \frac{1/4}{7/4} = \frac{1}{7}\).

Step 4: Compute for K = 1/10

For \(K = \frac{1}{10}\), we have \(H(e^{j\theta}) = \frac{1/10}{1 - 9/10e^{-j\theta}}\). The magnitude is \(A(\theta) = \left|\frac{1/10}{1 - 9/10e^{-j\theta}}\right|\).

Step 5: Calculate specific amplitudes for K = 1/10

Using the magnitude expression, calculate:- \(A(0) = \frac{1/10}{|1 - 9/10|} = 1\).- \(A(\frac{\pi}{4}) = \frac{1/10}{0.538} \approx 0.186\).- \(A(\frac{\pi}{2}) = \frac{1/10}{9/10} = \frac{1}{9}\).- \(A(\pi) = \frac{1/10}{19/10} = \frac{1}{19}\).

Step 6: Compute for K = 1/16

For \(K = \frac{1}{16}\), we have \(H(e^{j\theta}) = \frac{1/16}{1 - 15/16 e^{-j\theta}}\). The magnitude is \(A(\theta) = \left|\frac{1/16}{1 - 15/16 e^{-j\theta}}\right|\).

Step 7: Calculate specific amplitudes for K = 1/16

Using the magnitude expression, calculate:- \(A(0) = \frac{1/16}{|1 - 15/16|} = 1\).- \(A(\frac{\pi}{4}) = \frac{1/16}{0.534} \approx 0.113\).- \(A(\frac{\pi}{2}) = \frac{1/16}{15/16} = \frac{1}{15}\).- \(A(\pi) = \frac{1/16}{31/16} = \frac{1}{31}\).

Key Concepts

Z-domain

The Z-domain is a mathematical space used in signal processing to analyze discrete-time signals and systems. It is similar to the Laplace transform used in continuous-time signals but specifically for discrete signals. In the Z-domain, signals are represented by their Z-transforms. This transformation allows engineers and mathematicians to work with complex frequency-domain analyses using polynomials, which are simpler than dealing directly with difference equations in the time domain.

The Z-transform converts a discrete time-domain signal into a complex frequency domain representation. It provides a unified way to describe systems using algebraic methods. For our low-pass filter example, the transfer function in the Z-domain is given by:

• \[H(z) = \frac{K}{1 - (1-K)z^{-1}}\]

This function describes how an input signal is transformed into an output signal in terms of its frequency components. By analyzing the poles and zeros of the Z-transform, gain stability insights, and other important characteristics of the filter can be obtained.

Frequency Response

The frequency response of a system describes how it behaves over a range of frequencies. It tells us how different frequency components of the input signal are modified as they pass through the system. This is particularly important for filters, as their main role is to allow certain frequencies through while attenuating others.

In our low-pass filter example, the frequency response is determined by evaluating the transfer function at different frequency points using the transformation \(z = e^{j\theta}\), where \(\theta\) is the normalized frequency. The frequency response is represented as:

• \[H(e^{j\theta}) = \frac{K}{1 - (1-K)e^{-j\theta}}\]

The magnitude of this response, \(A(\theta)\), indicates how much of the signal's amplitude at a specific frequency is passed through the filter. Calculating \(A(\theta)\) for several values of \(\theta\) gives a deeper understanding of how effective the filter is at minimizing unwanted frequencies. For instance, you can check specific angles like \(0, \frac{\pi}{4}, \frac{\pi}{2}\), and \(\pi\) to see how different frequencies are processed by the filter.

Transfer Function

A transfer function is a mathematical representation in the frequency domain that encapsulates how a linear time-invariant system responds to input signals. It describes the relationship between the input and output of a system in terms of their Z-transforms.

For the low-pass filter given in the problem, the transfer function relates the input \(X(z)\) and output \(Y(z)\) as follows:

• \[H(z) = \frac{Y(z)}{X(z)} = \frac{K}{1 - (1-K)z^{-1}}\]

This function simplifies understanding how different frequencies are treated by the filter. The factor \(K\) directly influences the cutoff frequency, determining how sharply the filter attenuates signals beyond this threshold. By analyzing the transfer function, you can predict the system's response to any input, such as changes in amplitude or phase for different frequency components.

Transfer functions are vital tools for designing and analyzing complex filtering systems because they provide insights into system behavior without solving differential or difference equations directly. This makes them a cornerstone in fields ranging from audio signal processing to telecommunications.