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Consider a causal LTI system whose frequency response is given as: \\[ H\left(e^{j \omega}\right)=e^{-j \omega} \frac{1-\frac{1}{2} e^{j \omega}}{1-\frac{1}{2} e^{-j \omega}} \\] (a) Show that \(\left|H\left(e^{j \omega)} |\text { is unity at all frequencies. }\right.\right.\) (b) Show that \\[ \Varangle H\left(e^{j \omega}\right)=-\omega-2 \tan ^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1-\frac{1}{2} \cos \omega}\right) \\] (c) Show that the group delay for this filter is given by \\[ \tau(\omega)=\frac{\frac{3}{4}}{\frac{5}{4}-\cos \omega} \\] Sketch \(\tau(\omega)\) (d) What is the output of this filter when the input is \(\cos \left(\frac{\pi}{3} n\right) ?\)

Short Answer

Expert verified
(a) Magnitude is unity for all frequencies. (b) Phase angle verified. (c) Group delay is \( \frac{3/4}{5/4 - \cos \omega} \). (d) Output is a phase-shifted \( \cos(\frac{\pi}{3}n) \).

Step by step solution

01

Show Magnitude is Unity

Given the frequency response \( H\left(e^{j \omega}\right) = e^{-j \omega} \frac{1 - \frac{1}{2} e^{j\omega}}{1 - \frac{1}{2} e^{-j\omega}} \), the magnitude is \(|H\left(e^{j \omega}\right)| = |e^{-j \omega}| \frac{|1 - \frac{1}{2} e^{j\omega}|}{|1 - \frac{1}{2} e^{-j\omega}|} \). Since \(|e^{-j\omega}| = 1\), the expression simplifies to \(|H\left(e^{j \omega}\right)| = \frac{|1 - \frac{1}{2}e^{j\omega}|}{|1 - \frac{1}{2}e^{-j\omega}|}\), where both numerator and denominator are complex conjugates, showing \(|H\left(e^{j \omega}\right)| = 1\).
02

Find the Phase Angle

Start from \( H\left(e^{j \omega}\right) = e^{-j \omega} \frac{1 - \frac{1}{2} e^{j\omega}}{1 - \frac{1}{2} e^{-j\omega}} \). The phase angle \( \Varangle H\left(e^{j \omega}\right) \) is given by subtracting the phase of the denominator from the phase of the numerator and the phase shift \(-\omega\).Calculate each:- \(-\omega\)- Phase of numerator: \( \Varangle (1 - \frac{1}{2}e^{j\omega}) = -\tan^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1 - \frac{1}{2}\cos \omega}\right)\)- Phase of denominator: \( -(-\tan^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1 - \frac{1}{2} \cos \omega}\right)) = \tan^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1 - \frac{1}{2} \cos \omega}\right)\)Thus, the total phase is \(-\omega - 2 \tan^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1 - \frac{1}{2} \cos \omega}\right)\).
03

Derive the Group Delay

The group delay \( \tau(\omega) \) is the derivative of the phase angle with respect to \( \omega \):\[ \tau(\omega) = -\frac{d}{d\omega}\left(\omega + 2 \tan^{-1}\left(\frac{\frac{1}{2}\sin \omega}{1 - \frac{1}{2}\cos \omega}\right)\right) \]First, differentiate the arctangent:- The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1+u^2} \cdot \frac{du}{d\omega} \), where- \( u = \frac{\frac{1}{2}\sin \omega}{1 - \frac{1}{2}\cos \omega} \)- Calculate \( \frac{du}{d\omega} \)Simplifying, you find:\[ \tau(\omega) = \frac{3/4}{5/4 - \cos \omega} \]
04

Sketch Group Delay

Sketch the graph of \( \tau(\omega) = \frac{3/4}{5/4 - \cos \omega} \). Note that as \( \cos \omega \to \frac{5}{4} \), the group delay approaches infinity, suggesting spikes at multiples of \(2\pi\). The behavior of the graph is governed by the function reaching higher values near these transitions, with the denominator never reaching zero due to the range of \( \cos \omega \).
05

Output of the Filter for Input \( \cos(\frac{\pi}{3}n) \)

Express \( x[n] = \cos(\frac{\pi}{3}n) \) as a complex exponential: \( x[n] = \frac{1}{2}(e^{j\frac{\pi}{3}n} + e^{-j\frac{\pi}{3}n}) \).Using the frequency response formula, the output is:- For \( e^{j\frac{\pi}{3}n} \): \( H(e^{j\frac{\pi}{3}})e^{j\frac{\pi}{3}n} \)- For \( e^{-j\frac{\pi}{3}n} \): \( H(e^{-j\frac{\pi}{3}})e^{-j\frac{\pi}{3}n} \)Combine both parts and simplify using the results from previous steps, where phase affects the result:\( y[n] = \frac{1}{2}(e^{j(\frac{\pi}{3}n + \Varangle H(e^{j\frac{\pi}{3}}))} + e^{-j(\frac{\pi}{3}n + \Varangle H(e^{-j\frac{\pi}{3}}))}) \). Using the calculated phase, simplify further. This represents a phase-shifted cosine wave with the same amplitude.

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Key Concepts

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

Frequency response
Frequency response is a critical concept in understanding how a causal linear time-invariant (LTI) system reacts to different frequencies present in an input signal. It's essentially a function describing how each frequency within the input is amplified or attenuated by the system. For any LTI system defined in the frequency domain by \( H(e^{j\omega}) \), the magnitude of \( H(e^{j\omega}) \) represents how the amplitude of that frequency is affected. In our specific system, this function has been designed to maintain a unity magnitude—meaning that for all frequencies \( \omega \), the amplitude does not change. This property is key for filters that are intended to manipulate phase or delay but should not impact the amplitude of the signals passing through them.
Phase angle
The phase angle in the context of frequency response is a measure of how much a sinusoidal input is shifted in time when passed through the system. It's represented mathematically as the angle of \( H(e^{j\omega}) \). In our scenario, the phase angle given by the system's response is calculated to be \( -\omega - 2\tan^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1 - \frac{1}{2} \cos \omega}\right) \). This function describes a frequency-dependent phase shift, with the primary component \( -\omega \) ensuring a basic linear delay consistent at all frequencies. The arctangent term provides additional frequency-specific shifts, adjusting how different frequency components realign as they move through the system.
Group delay
Group delay, \( \tau(\omega) \), is an important characteristic that describes how the phase of the system varies with frequency. It's the derivative of the phase angle with respect to frequency, often visualized as the time delay experienced by the envelope of a modulated signal. For our LTI system, the group delay is determined by the expression \( \tau(\omega) = \frac{\frac{3}{4}}{\frac{5}{4} - \cos \omega} \). Plotting this, we see the behavior of the group delay changes with \( \omega \), with significant "spikes" occurring in the vicinity of points where the denominator approaches zero but does not reach it due to the bounded range of \( \cos \omega \). These spikes signify areas where the system introduces considerable delay variation as a function of time.
Cosine input response
When a cosine input signal like \( \cos(\frac{\pi}{3}n) \) is fed into a causal LTI system, the output can be determined by evaluating the frequency response at the input's frequency components. Cosine wave inputs are split into complex exponentials to get \( x[n] = \frac{1}{2}(e^{j\frac{\pi}{3}n} + e^{-j\frac{\pi}{3}n}) \). Each term responds to the filter according to the system's frequency response: \( H(e^{j\frac{\pi}{3}}) \) and \( H(e^{-j\frac{\pi}{3}}) \). By applying these, the output is a recombination of these components, modified primarily by phase shifts, as the magnitude remains unaffected. The result is a cosine output with the same frequency and amplitude as the input, but potentially phase-shifted due to the behavior captured in the calculated \( \Varangle H(e^{j\omega}) \). This maintains the structure of the wave while adjusting its position in time.

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Most popular questions from this chapter

A three-point symmetric moving average, referred to as a weighted moving average, is of the form \\[ y[n]=b\\{a x[n-1]+x[n]+a x[n+1]\\} \\] (a) Determine, as a function of \(a\) and \(b\), the frequency response \(H\left(e^{j \omega}\right)\) of the three point moving average in eq. \((\mathrm{P} 6.47-1)\) (b) Determine the scaling factor \(b\) such that \(H\left(e^{j \omega}\right)\) has unity gain at zero frequency. (c) In many time-series analysis problems, a common choice for the coefficient \(a\) in the weighted moving average in eq. \((P 6.47-1)\) is \(a=1 / 2\) Determine and sketch the frequency response of the resulting filter.

Show that if \(h[n]\), the impulse response of a discrete-time LTI filter, is always greater than or equal to zero, the step response of the filter is a monotonically nondecreasing function and therefore will not have overshoot.

Suppose we are given the following five facts about a particular LTI system \(S\) with impulse response \(h[n]\) and \(z\) -transform \(H(z)\). I. \(h|n|\) is real. 2\. \(h[n]\) is right sided. 3\. \(\lim _{=-x} H(z)=1\) 4\. \(H\) i \(z\) ) has two zeros. 5\. \(H(z)\) has one of its poles at a nonreal location on the circle defined by \(|z|=3 / 4\) Answer the following two questions: (a) Is \(S\) causal? (b) Is \(S\) stable?

Let $$ y[n]=\left(\frac{1}{9}\right)^{n} u[n] $$ Determine two distinct signals such that each has a z-transform \(X\) iz ) which satisfies both of the following conditions: 1\. \([X(z)+X(-z)] / 2=Y\left(z^{2}\right)\) 2\. \(X(z)\) has only one pole and only one zero in the z-plane.

Using the method indicated, determine the sequence that goes with each of the following \(z\) -transforms: (a) Partial fractions: \(X(z)=\frac{1-2 z^{-1}}{1-\frac{5}{2} z^{-1}+z^{-2}},\) and \(x[n]\) is absolutely summable (b) Long division: \(X t z)=\frac{1-\frac{1}{2} z^{-1}}{1+\frac{1}{2} z^{-1}},\) and \(x[n]\) is right sided (c) Partial fractions: \(X(z)=\frac{3}{z-\frac{1}{4}-\frac{1}{8} z^{-1}},\) and \(x[n]\) is absolutely summable

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