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Which of the following impulse responses correspond(s) to stable \(\&\) T stems? (a) \(\left.h_{1} | n\right]=n \cos \left(\frac{\pi}{4} n\right) u[n]\) (b) \(h_{2}[n]=3^{n} u[-n+10]\)

Short Answer

Expert verified
Both impulse responses \( h_1[n] \) and \( h_2[n] \) are not stable.

Step by step solution

01

Definition of Stability

A discrete-time system is considered stable if its impulse response is absolutely summable. This means that \( \sum_{n=-\infty}^{\infty} |h[n]| < \infty \). We shall check the stability condition for each impulse response given in the exercise.
02

Evaluate Stability of \( h_1[n] \)

The impulse response is given by \( h_1[n] = n \cos \left(\frac{\pi}{4} n\right) u[n] \), where \( u[n] \) is the unit step function. This means \( h_1[n] \) is nonzero only for \( n \geq 0 \). For stability, we need to check if \( \sum_{n=0}^{\infty} |n \cos \left(\frac{\pi}{4} n\right)| \) converges. Given \( |\cos \left(\frac{\pi}{4} n\right)| \leq 1 \), the term \( n \cos \left(\frac{\pi}{4} n\right) \) increases without bound, implying this series diverges. Thus, \( h_1[n] \) is not stable.
03

Evaluate Stability of \( h_2[n] \)

The impulse response is given by \( h_2[n] = 3^n u[-n+10] \), where \( u[-n+10] \) is the reversed step function, nonzero for \( n \leq 10 \). In this case, the terms evaluated are from \( n = -\infty \) to \( n = 10 \). Since \( 3^n \) grows as \( n \) becomes more negative, this series does not have a finite sum. It diverges, indicating \( h_2[n] \) is also not stable.

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Key Concepts

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

Stability of Discrete-Time Systems
In discrete-time systems, stability is an essential property defined primarily through the concept of impulse responses. Simply put, a discrete-time system is stable if its output remains bounded for every bounded input. To determine this practically, we rely on the system's impulse response, represented by the function \( h[n] \). To ascertain stability, we check if the impulse response is **absolutely summable**. This implies that the sum of the absolute values of all elements in \( h[n] \) must be finite. Mathematically, this is expressed as:\[\sum_{n=-\infty}^{\infty} |h[n]| < \infty\]If the above summation holds true, then the discrete-time system can cherish a stable behavior. Stability ensures that the system responds predictably and does not result in uncontrollable outputs. This is crucial in ensuring the reliable functioning of systems ranging from basic signaling to complex engineering applications.
Impulse Response Evaluation
Evaluating the impulse response of a discrete-time system involves examining its behavior when subjected to the simplest input - a single impulse signal (often denoted as \( \delta[n] \)). The impulse response, \( h[n] \), characterizes the system's reaction and is fundamental in predicting how the system handles more complex inputs. For example, in the given exercises, two impulse responses are evaluated:
  • \( h_1[n] = n \cos \left(\frac{\pi}{4} n\right) u[n] \)
  • \( h_2[n] = 3^{n} u[-n+10] \)
With \( h_1[n] \), it is assessed for stability by analyzing if the series \( \sum_{n=0}^{\infty} |n \cos \left(\frac{\pi}{4} n\right)| \) converges. Because \( n \cos \left(\frac{\pi}{4} n\right) \) increases indefinitely, it diverges, confirming instability.Similarly, for \( h_2[n] \), the series from \( n = -\infty \) to \( n = 10 \) is checked. Here, \( 3^n \) increases sharply as \( n \) comes more negative, leading to divergence, also indicating instability.This process of evaluation enables us to verify if a system is capable of stability when faced with an impulse input, guiding engineers in designing systems that are reliable and predictable.
Absolute Summability
Absolute summability is a concept central to determining the stability of discrete-time systems. In practical terms, it involves taking the absolute values of all components in the system's impulse response and summing them up. If this total is finite, the system is said to possess absolute summability, which directly relates to stability.When expressed mathematically, the condition is:\[\sum_{n=-\infty}^{\infty} |h[n]| < \infty\]This criterion ensures that every output the system generates remains within limits, regardless of how complex or prolonged the input may be. This characteristic is vital in preventing the system from producing erratic or infinite outputs that could lead to system failures or inefficiencies.In the exercise example, neither of the given impulse responses, \( h_1[n] \) and \( h_2[n] \), meet the absolute summability condition as both series diverge. Therefore, they are not stable, highlighting the importance of this criterion in assessing the long-term reliability of discrete-time systems.

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Most popular questions from this chapter

Let $$x[n]=\delta[n]+2 \delta[n-1]-\delta[n-3] \text { and } h[n]=2 \delta[n+1]-2 \delta[n-1]$$ Compute and plot each of the following convolutions: (a) \(y_{1}[n]=x[n] * h[n]\) (b) \(y_{2}[n]=x[n+2] * h[n]\) (c) \(y_{3}[n]=x[n] * h[n+2]\)

One important use of inverse systems is in situations in which one wishes to remove distortions of some type. A good example of this is the problem of removing echoes from acoustic signals. For example. if an auditorium has a perceptible echo, then an initial acoustic impulse will be followed by attenuated versions of the sound at regularly spaced intervals. Consequently, an often-used model for this phenomenon is an LTI system with an impulse response consisting of a train of impulses, is... $$h(t)=\sum_{k=0}^{x} h_{k} \delta(t-k T)$$ Here the echoes occur \(T\) seconds apart, and \(h_{k}\) represents the gain factor on the \(k\) th echo resulting from an initial acoustic impulse. (a) Suppose that \(x(t)\) represents the original acoustic signal (the music produced by an orchestra, for example) and that \(y(t)=x(t) * h(t)\) is the actual signal that is heard if no processing is done to remove the echoes. In order to remove the distortion introduced by the echoes, assume that a microphone is used to sense \(y(t)\) and that the resulting signal is transduced into an electrical signal. We will also use \(y(t)\) to denote this signal, as it represents the electrical equivalent of the acoustic signal, and we can go from one to the other via acoustic-electrical conversion systems. The important point to note is that the system with impulse response given by eq. (P2.64-1) is invertible. Therefore, we can find an LTI system with impulse response \(g(r)\) such that $$y(t) * g(t)=x(t)$$ and thus, by processing the electrical signal \(y(t)\) in this fashicn and then converting back to an acoustic signal, we can remove the troublesome echoes. The required impulse response \(g(t)\) is also an impulse train: $$g(t)=\sum_{k=0}^{x} g_{k} \delta(t-k T)$$ Determine the algebraic equations that the successive \(g_{k}\) must satisfy, and solve these equations for \(g_{0}, g_{1},\) and \(g_{2}\) in terms of \(h_{k}\) (b) Suppose that \(h_{0}=1, h_{1}=1 / 2,\) and \(h_{t}=0\) for all \(i \geq 2\) What is \(g(t)\) in this case? (c) A good model for the generation of echocs is illustrated in Figure P2.64. Hence, each successive echo represents a fed-back version of \(y(t),\) delayed by \(T\) seconds and scaled by \(\alpha .\) Typically, \(0<\alpha<1,\) as successive echoes are attendated. (i) What is the impulse response of this systern's (Assume initial rest, i.e., \(y(t)=0\) for \(t<0\) if \(x(t)=0\) for \(t<0 .\) (ii) Show that the system is stable if \(0<\alpha<1\) and unstable if \(\alpha>1\) (iii) What is \(g(t)\) in this case? Construct a realization of the inverse system using adders, coefficient multipliers, and \(T\) -second delay elements. (d) Although we have phrased the preceding discussion in terms of continuous- time systems because of the application we have been considering, the same general ideas hold in diserete tome. That is, the CTI systern with impulse response $$h[n]=\sum_{k=0}^{\infty} \delta | n-k N$$ is invertible and has as its inverse an LTI sysiern with impulse response $$g[n]=\sum_{k=0}^{\infty} g_{k} \delta | n-k N$$ It is not difficult to check that the \(g_{r}\) satisfy the same algebrace equations as in part (a) Consider bow the discrete-time LTI system with impulse response $$h[n]=\sum_{k=-\infty}^{\infty} \delta[n-k N]$$ This system is nor invertible. Find two inputs that produce the same output.

Consider a causal LTI system \(S\) whose input \(x[n]\) and output \(y[n]\) are related by the difference equation $$2 y[n]-y[n-1]+y[n-3]=x[n]-5 x[n-4]$$ (a) Verify that \(S\) may be considered a cascade connection of two causal LII systems \(S_{1}\) and \(S_{2}\) with the following input-output relationship: $$\begin{aligned} &S_{1}: 2 y_{1}[n]=x_{1}[n]-5 x_{1}\\{n-4\\}\\\ &S_{2}: y_{2}[n]=\frac{1}{2} y_{2}[n-1]-\frac{1}{2} y_{2}[n-3]+x_{2}[n] \end{aligned}$$ (b) Draw a block diagram representation of \(S_{1}\) (c) Draw a block diagram representation of \(S_{2}\) (d) Draw a block diagram representation of \(S\) as a cascade connection of the block diagram representation of \(S_{1}\) followed by the block diagram representation of \(S_{2}\) (e) Draw a block diagram representation of \(S\) as a cascade connection of the block diagram representation of \(S_{2}\) followed by the block diagram representation of \(S_{1}\) (f) Show that the four delay elements in the block diagram representation of \(S\) obtained in part (e) may be collapsed to three. The resulting block diagram is referred to as a Direct Form \(I I\) realization of \(S,\) while the block diagrams obtained in parts \((\mathrm{d})\) and (e) are referred to as Direct Form \(I\) realizations of \(S\).

Determine whether each of the following statements concerning LTI systems is me or false. Justify your answers. (a) If \(h(r)\) is the impulse response of an LTI system and \(h(t)\) is periodic and nonzero. the system is unstable. (b) The inverse of a causal LTI system is always causal. (c) \(\mathbf{f}|h[n]| \leq K\) for each \(n\), where \(K\) is a given number, then the \(L T\) is system with \(h[n]\) as its impulse response is stable. (d) If a discrete-time LTI system has an impulse response \(h[n]\) of finite duration. the system is stable (e) If an LTI system is causal, it is stable. (f) The cascade of a noncausal LTI system with a causal one is necessarily noncausal. (g) A continuous-time LTI system is stable if and only if its step response \(s(t)\) is absolutely integrable - that is, if and only if $$\int_{-\infty}^{+\infty}|s(t)| d t<\infty$$ (h) A discrete-time LTI systew is causal if and only if its slep response \(s[n]\) is zero for \(n<0\)

Suppose that $$x(r)=\left\\{\begin{array}{ll} 1, & 0 \leq t \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ and \(h(t)=x(t / \alpha),\) where \(0<\alpha \leq 1\) (a) Determine and sketch \(y(t)=x(t) * h(t)\) (b) If \(d y(t) / d t\) contains only three discontinuities, what is the value of \(\alpha ?\)

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