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Let \(x[n]\) be an absolutely summable signal with rational \(z\) -transform \(X(z) .\) If \(X(z)\) is known to have a pole at \(z=1 / 2,\) could \(x[n]\) be (a) a finite-duration signal? (b) a left-sided signal? (c) a right-sided signal? (d) a two-sided signal?

Short Answer

Expert verified
(a) No, (b) No, (c) Yes, (d) No.

Step by step solution

01

Understand the question

First, you must understand that you need to determine the possibility of the signal being finite-duration, left-sided, right-sided, or two-sided given the location of the pole at \( z = \frac{1}{2} \). A pole at \( z = \frac{1}{2} \) influences stability and causality.
02

Recall properties of poles and signal type

A pole inside the unit circle (\( |z| < 1 \)) indicates a stable signal if it is right-sided. A finite duration signal has no poles. A left-sided signal is stable if all poles are outside the unit circle (\( |z| > 1 \)). A two-sided signal requires special consideration as the entire signal needs to be studied for possible convergence.
03

Evaluate finite-duration signal possibility

A finite-duration signal leads to a z-transform that has no poles, as the transform is a polynomial. Since there's a pole at \( z = \frac{1}{2} \), \( x[n] \) cannot be a finite-duration signal.
04

Evaluate left-sided signal possibility

For a left-sided signal, the poles should be outside the unit circle (\( |z| > 1 \)). Since \( z = \frac{1}{2} \) is inside the unit circle, \( x[n] \) cannot be a left-sided signal.
05

Evaluate right-sided signal possibility

A right-sided signal with poles inside the unit circle indicates stability. Since \( z = \frac{1}{2} \) is inside the unit circle, \( x[n] \) could possibly be a right-sided signal.
06

Evaluate two-sided signal possibility

A two-sided signal can have poles inside and outside the unit circle, but the absolute convergence of \( x[n] \) becomes difficult to satisfy. Given the pole at \( z = \frac{1}{2} \) inside the unit circle, it cannot be a two-sided signal due to convergence issues.

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Key Concepts

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

Poles and Zeros
In signal processing, poles and zeros are fundamental concepts used to understand and manipulate the behavior of systems, particularly in the context of z-transforms. A _pole_ is a value of \( z \) that makes the denominator of a transfer function zero. Conversely, a _zero_ is a value of \( z \) that makes the numerator zero.
  • Poles determine the system's stability and response characteristics.
  • Zeros can be thought of as frequencies that are attenuated or nullified in the system's output.
In this exercise, we consider a pole at \( z = \frac{1}{2} \). The location of the pole—in this case, inside the unit circle—tells us about the potential stability for right-sided signals. Poles outside the unit circle would indicate potential stability for left-sided signals. For finite-duration signals, the concept of poles does not exist, as their z-transform is simply a polynomial with no denominator.
Signal Stability
_Signal stability_ refers to whether the output of a system remains bounded for any bounded input. This stability is determined through the locations of poles.
  • For a right-sided signal, stability is assured if all poles are inside the unit circle \(|z| < 1\).
  • For left-sided signals, the poles must lie outside the unit circle \(|z| > 1\) to ensure stability.
  • A two-sided signal, which extends indefinitely in both directions, requires complex analysis to determine stability, as some poles can be inside and some outside the unit circle.
In our exercise, the given pole at \( z = \frac{1}{2} \) suggests that the signal could be potentially stable if right-sided since the pole is inside the unit circle. However, if the signal were two-sided or left-sided, it would not meet the stability criteria due to pole locations.
Right-Sided Signals
A _right-sided signal_ is one where the non-zero values exist, or are primarily significant, for non-negative time indices (\( n \geq 0 \)). This type of signal's characteristics include:
  • They tend to be stable if their poles are located inside the unit circle.
  • They are often associated with causal systems, which only respond after input.
  • Right-sided signals can converge and provide meaningful transforms easily when poles are within the unit circle.
Given the pole at \( z = \frac{1}{2} \), which is within the unit circle, the signal in question could be a right-sided signal. It meets the stability criterion because such a signal would not diverge, aligning well with usual expectations for discrete-time, right-sided signals.
Discrete-Time Signals
_Discrete-time signals_ are sequences of values defined at discrete intervals, which is to say time progresses in separate steps. These are fundamental to digital signal processing (DSP) since most practical systems are digital and thus naturally deal with discrete-time signals.
  • A valuable feature of these signals is their representation in the z-domain for analysis.
  • They allow us to study behavior across frequency components through z-transforms.
  • In evaluating such signals, the location of poles and zeros in the z-plane is crucial for understanding their duration and stability.
In considering a discrete-time signal's characteristics, it's essential to assess the pole location critically, as in this exercise. The pole at \( z = \frac{1}{2} \) offers insight into the signal's duration and whether it's likely stable as a right-sided signal. Such assessments help in designing filters and analyzing systems effectively.

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Most popular questions from this chapter

Consider a discrete-time lowpass filter whose impulse response \(h | n]\) is knownto be real and whose frequency response magnitude in the region \(-\pi \leq \omega \leq \pi\) is given as: \\[ \left|H\left(e^{j \alpha}\right)\right|=\left\\{\begin{array}{ll} 1, & |\omega| \leq \frac{\pi}{4} \\ 0, & \text { otherwise } \end{array}\right. \\] Determine and sketch the real-valued impulse response \(h[n]\) for this filter when the corresponding group delay function is specified as: (a) \(\tau(\omega)=5\) (b) \(\tau(\omega)=\frac{5}{2}\) (c) \(\tau(\omega)=-\frac{5}{2}\)

Find the inverse \(z\) -transform of \\[ X(z)=\frac{1}{1,024}\left\\{\frac{1,024-z^{-10}}{1-\frac{1}{2} z^{-1}}\right\\},|z|>0 \\].

For each of the following algebraic expressions for the z-transform of a signal. determine the number of zeros in the finite \(z\) -plane and the number of zeros at infinity. (a) \(\frac{z^{-1}\left(1-\frac{1}{2} z^{-1}\right)}{\left(1-\frac{1}{3} z^{-1}\right)\left(1-\frac{1}{4} z^{-1}\right)}\) (b) \(\frac{\left(1-z^{-1}\right)\left(1-2 z^{-1}\right)}{\left(1-3 z^{-1}\right)\left(1-4 z^{-1}\right)}\) (c) \(\frac{z^{-2}\left(1-z^{-1}\right)}{\left(1-\frac{1}{4} z^{-1}\right)\left(1+\frac{1}{4} z^{-1}\right)}\)

Consider the following system functions for stable LTI systems. Without utilizing the inverse \(z\) -transform, determine in each case whether or not the corresponding system is causal. (a) \(\frac{1-\frac{4}{3} z^{-1}+\frac{1}{2} z^{-2}}{z^{-1}\left(1-\frac{1}{2} z^{-1}\right)\left(1-\frac{1}{z} z^{-1}\right)}\) (b) \(\frac{z-\frac{1}{2}}{z^{2}+\frac{1}{2} z-\frac{3}{16}}\) (c) \(\frac{z+1}{z+\frac{4}{3}-\frac{1}{2} z^{-2}-\frac{2}{3} z^{-3}}\)

By means of a specific filter design procedure, a nonideal continuous-time lowpass filter with frequency response \(H_{0}(j \omega),\) impulse response \(h_{0}(t),\) and step response \(s_{0}(t)\) has been designed. The cutoff frequency of the filter is at \(\omega=2 \pi \times 10^{2}\) rad/sec, and the step response rise time, defined as the time required for the step response to go from \(10 \%\) of its final value to \(90 \%\) of its final value, is \(\tau_{r}=10^{-2}\) second. From this design, we can obtain a new filter with an arbatrary cutoff frequency \(\omega_{c}\) by the use of frequency scaling. The frequency response of the resulting filter is then of the form \\[ H_{\mathrm{I} \rho}(j \omega)=H_{0}(j a \omega) \\] where \(a\) is an appropriate scale factor. (a) Determine the scale factor a such that \(H_{1 p}(j \omega)\) has a cutoff frequency of \(\omega_{c}\) (b) Determine the impulse response \(h_{| p}(t)\) of the new filter in terms of \(\omega_{c}\) and \(h_{0}(t)\) (c) Determine the step response \(s_{1 p}(2)\) of the new fiter in terms of \(\omega_{c}\) and \(s_{0}(f)\) (d) Determine and sketch the rise time of the new filler as a function of its cutoff frequency \(\omega_{c}\) This is one illustration of the trade-off between fine-domain and frequency- domain characteristics. In particular, as the cutoff frequency decreases, the rise time tends to increase

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