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Chapter 2: Problem 5

Let $$x[n]=\left\\{\begin{array}{ll} 1, & 0 \leq n \leq 9 \\ 0, & \text { etsewhere } \end{array} \text { and } h[n]=\left\\{\begin{array}{ll} 1, & 0 \leq n \leq N \\ 0, & \text { elsewhere } \end{array}\right.\right.$$ where \(N \leq 9\) is an integer. Determine the value of \(N\), giventhat \(y[n]=x[n] * h[n]\) and $$y[4]=5, \quad y[14]=0$$

Short Answer

Expert verified
The value of \(N\) is 4.

Step by step solution

01

Understand the problem

We need to determine the value of \(N\) for which the convolution of \(x[n]\) and \(h[n]\), denoted as \(y[n]\), satisfies the conditions \(y[4]=5\) and \(y[14]=0\). The signals \(x[n]\) and \(h[n]\) are defined over finite ranges, and the convolution operation is given by \( y[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k] \).
02

Set up the convolution

For any \(n\), the convolution \(y[n] = x[n] * h[n]\) can be expressed as: \[y[n] = \sum_{k=0}^{9} x[k] h[n-k]\] This is because \(x[k]\) is non-zero only from 0 to 9, and \(h[n-k]\) is non-zero wherever \(0 \le n-k \le N\). Both \(x[k]\) and \(h[n-k]\) must be non-zero for each term to contribute to the sum.
03

Use given condition y[4]=5

For \(y[4]=5\), evaluate the convolution at \(n=4\): \[y[4] = \sum_{k=0}^{9} x[k] h[4-k] = \sum_{k=0}^{4} 1 = 5\]This implies that \(h[4-k] = 1\) for \(0 \leq k \leq 4\), meaning the range of \(h[n]\) is at least \(4\). Thus, \(N \geq 4\).
04

Use given condition y[14]=0

For \(y[14]=0\), we evaluate the convolution at \(n=14\): \[y[14] = \sum_{k=5}^{9} x[k] h[14-k] = 0\]This tells us that none of the terms satisfy both \(0 \leq k \leq 9\) and \(0 \leq 14-k \leq N\). The non-zero contributions would be from \(k=5\) to \(k=9\), which means either \(h[14-5]\) to \(h[14-9]\) are zero, or \(h[n]\) should not overlap with \(x[n]\) in this range. Since \(x[k]=1\) for \(5 \le k \le 9\), it implies \(N < 5\).
05

Determine the value of N

From Step 3, we determined \(N \geq 4\). From Step 4, we determined \(N < 5\). Combining these, we conclude that \(N = 4\).

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Key Concepts

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

Discrete-time signals
Discrete-time signals are sequences of numbers, representing samples of a continuous signal taken at distinct time intervals. They are crucial in digital systems, where continuous signals are processed by computers or other digital devices. These signals are indexed by time or position, such as the signal \(x[n]\) and \(h[n]\) given in the exercise above.
In this context, \(x[n]\) represents a finite pulse that is non-zero between 0 and 9, while \(h[n]\) is another pulse function that is non-zero between 0 and \(N\). Discrete-time signals allow for precise digital signal processing, as each sample in the sequence can be individually manipulated to achieve a desired outcome.
  • They provide the framework for sampling, which is essential for converting analog signals to digital.
  • Easy manipulation of each individual sample makes it efficient for computations in discrete-time systems.
Finite impulse response
Finite Impulse Response (FIR) filters are a class of filters in signal processing whose impulse response is of finite duration, eventually reaching zero. This is because they depend solely on the current and past input values.
In the exercise given, the signal \(h[n]\) can be considered an impulse response of a system. The fact that it is non-zero only for a finite number of points (from 0 to \(N\)) makes it a perfect example of a finite impulse response. FIR filters like this are crucial in digital signal processing for producing desired changes to a signal while maintaining stability, as they do not rely on past outputs.
  • FIR filters are inherently stable since they depend only on present and past input samples.
  • They offer a simple structure but can be computationally intensive if the width \(N\) is large.
  • Due to their finite duration, they do not introduce recursion into systems, avoiding aliasing effects.
Signal processing
Signal processing involves the manipulation, analysis, and transformation of signals to extract information or modify them according to specific needs. In this exercise, we're performing convolution of discrete-time signals, which is a fundamental operation in linear systems, characterizing how one signal modifies another.
Convolution combines two signals into one, determining how the shape of one is "smeared" onto another. The result, in this case \(y[n]\), represents how the system with impulse response \(h[n]\) transforms the input signal \(x[n]\). Signal processing applications are numerous and include filtering noise from data, compressing audio and video, and even in communications systems.
  • Convolution is essential for understanding how systems respond to inputs in both time and frequency domains.
  • Enables the design of filters that can enhance or inhibit specific features of the input signal.
  • Signals, like the examples given, often define the range and duration the processing will affect.
Discrete-time systems
Discrete-time systems process discrete-time signals. They consist of a network of operators like delays, multipliers, or adders that transform input sequences into output sequences. An example in this task is the system described by the convolution of \(x[n]\) and \(h[n]\) to produce \(y[n]\).
Such systems are manipulated mathematically using discrete equations, like the convolution equation given. These systems are fundamental in DSP (Digital Signal Processing), allowing for practical implementations in various fields such as telecommunications, audio processing, and control systems.
  • Discrete-time systems are tailored to handle signals at specific intervals, making them essential in digital communication.
  • They rely on algorithms like FIR filters discussed earlier, for prediction and analysis.
  • Performing operations on sequences ensures signals processed remain within the capabilities of digital systems.

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Most popular questions from this chapter

Consider a discrete-time \(L T I\) system with unit sample response $$h[n]=(n+1) \alpha^{n} u[n]$$ where \(|\alpha|<1 .\) Show that the step response of this system is $$s[n]=\left[\frac{1}{(\alpha-1)^{2}}-\frac{\alpha}{(\alpha-1)^{2}} \alpha^{n}+\frac{\alpha}{(\alpha-1)}(n+1) \alpha^{n}\right] u[n]$$ (Hint: Note that $$\sum_{k=0}^{N}(k+1) \alpha^{k}=\frac{d}{d \alpha} \sum_{k=0}^{N+1} \alpha^{k}$$)

The following are dee impulse responses of continuous-time LTI systems. Determine whether each system is causal and/or stable. Justify your answers. (a) \(h(t)=e^{-4 t} u(t-2)\) (b) \(h(t)=e^{-6 r} u(3-t)\) (c) \(h(t)=e^{-2 t} u(t+50)\) (d) \(h(t)=e^{2 t} u(-1-t)\) (e) \(h(t)=e^{||6 |}\) (f) \(h(t)=i e^{-i} u(t)\) (g) \(h(t)=\left(2 e^{-t}-e^{(t-100) t 100}\right) u(t)\)

Another application in which matched filters and correlation functions play an important role is radar systems. The underlying principle of radar is that an electromaguetic pulse transmitled at a target will be reflected by the targe and will subsequently return to the sender wath a delay proportional to the distance to the target Ideally, the received sigral wil? simply be a shifted and possibly scaled version of the original transmitted signal Let \(p(t)\) he the original pulse that is sent out. Show that $$\phi_{\mu p}(0)=\max \phi_{, p}(t)$$ if the waveform that comes back to the sender is $$x(t)=\alpha p\left(t-t_{0}\right)$$ where \(\alpha\) is a positive constant, then $$\phi_{r p}\left(t_{t}\right)=\max _{t} \phi_{r p}(t)$$ (Hint: Use Schwartz's inequality.) Thus, the way in which simple radar ranging systems work is hased on using a matched filter for the transmitted waveform \(p(t)\) and noting the tome at which the output of this system reaches its maximum value.

Consider the evaluation of $$y[n]=x_{1}[n] * x_{2}[n] * x_{3}[n]$$ where \(x_{1}[n]=(0.5)^{n} u[n], x_{2}[n]=u[n+3],\) and \(x_{3}[n]=\delta[n]-\delta[n-1]\) (a) Evaluate the convolution \(x_{1}[n] * x_{2}[n]\) (b) Convolve the result of part (a) with \(x_{3}[n]\) in order to evaluate \(y[n]\) (c) Evaluate the convolution \(x_{2}[n] * x_{3}[n]\) (d) Convolve the result of part (c) with \(x_{1}[n]\) in order to evaluate \(y[n]\)

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.
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