91Ó°ÊÓ

(a) Consider two LT systems with the following frequency responses: \\[ \begin{array}{l} H_{1}\left(e^{j \theta+}\right)=\frac{1+\frac{1}{2} e^{-j \omega}}{1+\frac{1}{4} e^{-j \omega}} \\ H_{2}\left(e^{j \omega}\right)=\frac{\frac{1}{2}+e^{-j \omega}}{1+\frac{1}{4} e^{-j \omega}} \end{array} \\] Show that both of these frequency responses have the same magnitude function \(\left.\text { [i.e., }\left.| H_{1} \text { ( } e^{f \omega}\right)|=| H_{2}\left(e^{J \omega}\right)\right]\), but the group delay of \(H_{2}\left(e^{j \omega}\right)\) is greater than the group delay of \(H_{1}\left(e^{j \omega}\right)\) for \(\omega>0\) (b) Determine and sketch the impulse and step responses of the two systems. (c) Show that \\[ H_{2}\left(e^{j \omega}\right)=G\left(e^{j \omega}\right) H_{1}\left(e^{j \omega}\right) \\] where \(\left.G\left(e^{j \omega}\right) \text { is an all-pass system [i.e., }\left|G\left(e^{j \omega}\right)\right|=1 \text { for all } \omega\right\\}\).

Step by step solution

01

General Structure of Magnitude Function

The magnitude of a frequency response, \( H(e^{j\omega}) \), is determined by the modulus of the numerator divided by the modulus of the denominator. Thus, for our systems, \(|H_1(e^{j\omega})| = \left| \frac{1+\frac{1}{2}e^{-j\omega}}{1+\frac{1}{4}e^{-j\omega}} \right|\) and \(|H_2(e^{j\omega})| = \left| \frac{\frac{1}{2}+e^{-j\omega}}{1+\frac{1}{4}e^{-j\omega}} \right|\). We will evaluate these magnitudes next.

02

Compute Magnitude of \(H_1(e^{j\omega})\)

Calculate \(|H_1(e^{j\omega})|\) by replacing \(e^{-j\omega}\) with \(e^{j\omega}\) conjugate terms: \( |1 + \frac{1}{2}e^{-j\omega}| = \sqrt{1 + \frac{1}{4} + \cos \omega} \) and \( |1 + \frac{1}{4}e^{-j\omega}| = \sqrt{1 + \frac{1}{16} + \frac{1}{2}\cos \omega} \). Thus, \(|H_1(e^{j\omega}| = \frac{\sqrt{1.25 + \cos \omega}}{\sqrt{1.0625 + 0.5 \cos \omega}}\).

03

Compute Magnitude of \(H_2(e^{j\omega})\)

For \(H_2(e^{j\omega})\), \(|\frac{1}{2}+e^{-j\omega}| = \sqrt{\left(\frac{1}{2}\right)^2 + 1 + 1\cos \omega}\) and \(|1 + \frac{1}{4}e^{-j\omega}|\) was already calculated in Step 2. Thus, the expression becomes \(|H_2(e^{j\omega})| = \frac{\sqrt{1.25 + \cos \omega}}{\sqrt{1.0625 + 0.5 \cos \omega}}\), which equals \(|H_1(e^{j\omega}|)\).

04

Analyze Group Delay

Group delay \( \tau_g(\omega) \) is derivative of the phase of a filter. Compute phase delays: \( \angle H_1(e^{j\omega}) \) and \( \angle H_2(e^{j\omega}) \) by differentiating corresponding phase terms. Generally, adding more poles/zeros increases delay. \(H_2\) has an additional zero at origin, causing a higher group delay.

05

Impulse and Step Response Sketching

Converting \( H_1(e^{j\omega}) \) and \( H_2(e^{j\omega}) \) to time domain using inverse Fourier transform gives the impulse response. For step response, convolve impulse response with Step function. Plot these.

06

Express \( H_2 \) as Scaled \( H_1 \) and Verify All-pass

Express \(G(e^{j\omega}) = \frac{H_2(e^{j\omega})}{H_1(e^{j\omega})}\). Simplifying gives: \( G(e^{j\omega}) = \frac{\frac{1}{2} + e^{-j\omega}}{1 + \frac{1}{2} e^{-j\omega}} \). Then confirm \(|G(e^{j\omega})| = 1\), proves it's an allpass filter.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency Response

In linear time-invariant systems, the frequency response is a crucial concept that describes how the system reacts to various frequencies of input signals. It is mathematically represented by the function \( H(e^{j\omega}) \), where \( \omega \) is the angular frequency. The frequency response tells us how much the system will amplify or attenuate each frequency and what phase shift, if any, is introduced.

An important aspect of the frequency response is analyzing its magnitude and phase. The magnitude shows the strength of the output signal compared to the input, while the phase provides insight on how the signal is shifted in time. For instance, in our exercise, we have two systems \( H_1(e^{j\omega}) \) and \( H_2(e^{j\omega}) \) with specific mathematical expressions for their frequency responses, which help us derive key insights into their dynamics.

Magnitude Function

The magnitude function is a part of the frequency response and describes how the amplitude of each frequency component is altered by the system. This is calculated as the absolute value of the frequency response, denoted as \(|H(e^{j\omega})|\).

In the given systems, both \( H_1 \) and \( H_2 \) share the same magnitude function. This is shown by simplifying each frequency response's expression:

• The magnitude of \( H_1(e^{j\omega}) \) was computed as \( \frac{\sqrt{1.25 + \cos \omega}}{\sqrt{1.0625 + 0.5 \cos \omega}} \).
• Similarly, \( H_2(e^{j\omega}) \) also results in the same expression, demonstrating equal influence on different frequency amplitudes across systems.

Thus, despite potential differences in other properties, both systems will output signals with the same amplitude scaling for every frequency.

Group Delay

Group delay is a measure of how different frequency components of a signal are delayed as they pass through a system. It indicates the delay a particular frequency will experience, which is crucial for preserving the shape of signal pulses.

Mathematically, group delay \( \tau_g(\omega) \) is the negative derivative of the phase response. Between \( H_1(e^{j\omega}) \) and \( H_2(e^{j\omega}) \), even though both have the same magnitude function, \( H_2 \) exhibits a larger group delay than \( H_1 \). This increased delay results from the additional zero present in \( H_2 \), implying more significant distortion in the time it takes for signals to propagate through the system, especially for frequencies greater than zero.

Impulse Response

Impulse response is a fundamental analysis tool that characterizes the output of a system when subjected to an impulse input. This short burst signal contains all frequencies, making it perfect for analyzing a system's behavior.

For systems \( H_1(e^{j\omega}) \) and \( H_2(e^{j\omega}) \), we derive the impulse response by performing an inverse Fourier transform on their respective frequency responses. The impulse response exposes how the system will react in real-time, giving insights into temporal behavior and stability characteristics.

The resulting sketch of an impulse response provides a visual interpretation of this characteristic, making it easier to grasp changes and transients that occur within the system's output.

Step Response

A step response describes how a system responds to a sudden, constant input, often visualized as a step function. The step response is essential in understanding how a system reaches a steady state after a change.

To find the step response, you convolve the impulse response with a step function. This process reveals how system outputs evolve from an initial reaction to settling into a new state over time.

• For \( H_1 \) and \( H_2 \), the step responses illustrate the time-dependent behavior post any input change.
• It helps deduce latency factors and anticipate system performance under persistent shifts in signals.

Graphs of these responses offer easy-to-follow depictions of how long it takes each system to stabilize and provides a clearer understanding of dynamic reaction tendencies within these systems.