/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Consider the first-order differe... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the first-order difference equation $$y[n]+2 y[n-1]=x[n]$$ Assuming the condition of initial rest (i.e., if \(x[n]=0\) for \(n < n_{0},\) then \(y[n]=0\) for \(n < n_{0}\) ), find the impulse response of a system whose input and output are related by this difference equation. You may solve the problem by rearranging the difference equation so as to express \(y[n]\) in terms of \(y[n-1]\) and \(x[n]\) and generating the values of \(y[0], y[+1], y[+2], \ldots\) in that order.

Short Answer

Expert verified
The impulse response is given by \(h[n] = (-2)^n \delta[n]\).

Step by step solution

01

Understand the Problem

We have a first-order difference equation given by \(y[n] + 2y[n-1] = x[n]\). We need to find the impulse response of this system assuming initial rest conditions. The impulse response, \(h[n]\), is the output when the input \(x[n] = \delta[n]\), where \(\delta[n]\) is the Kronecker delta function.
02

Rewrite the Difference Equation

Rewrite the equation \(y[n] + 2y[n-1] = x[n]\) to express \(y[n]\) in terms of \(y[n-1]\) and \(x[n]\). The equation becomes \(y[n] = x[n] - 2y[n-1]\).
03

Find h[n] for n=0

To find the impulse response, we set \(x[n] = \delta[n]\), which means \(x[0] = 1\) and \(x[n] = 0\) for \(n eq 0\). Applying this to the equation \(y[0] = x[0] - 2y[-1]\); given initial rest, \(y[-1] = 0\). Therefore, \(y[0] = 1\).
04

Find h[n] for n=1

Use \(x[1] = 0\) because \(\delta[n]\) is 0 when \(n eq 0\). Substitute into the equation: \(y[1] = x[1] - 2y[0] = 0 - 2\cdot 1 = -2\).
05

Find h[n] for n=2

Now substitute for \(n=2\): \(y[2] = x[2] - 2y[1] = 0 - 2\cdot(-2) = 4\).
06

Find h[n] for n=3

For \(n=3\): \(y[3] = x[3] - 2y[2] = 0 - 2\cdot 4 = -8\).
07

Pattern Recognition

Observe the pattern of the impulse response: \(h[0] = 1\), \(h[1] = -2\), \(h[2] = 4\), \(h[3] = -8\). This is a geometric sequence where each term \(h[n] = (-2)^n\).

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Key Concepts

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

Difference Equation
A difference equation is a tool used to express relationships between successive terms in a sequence or set of discrete values. In our problem, the difference equation is \( y[n] + 2y[n-1] = x[n] \). It is termed a first-order difference equation because the highest delayed term involved is \( y[n-1] \), which is just one time step in the past. Difference equations play a similar role in discrete time systems as differential equations do in continuous time systems. They're essential for analyzing systems in which the input and output values occur at distinct intervals or points rather than continuously over time. This makes them particularly useful for digital signal processing and control systems.To solve a difference equation like \( y[n] = x[n] - 2y[n-1] \), we express the current output \( y[n] \) in terms of the previous output \( y[n-1] \) and the current input \( x[n] \). This allows us to iteratively calculate values and understand the behavior of the system over time.
Initial Rest Conditions
Initial rest conditions refer to the state of a system where all previous outputs are zero until an input stimulus is applied. In our context, this means if \( x[n] = 0 \) for \( n < n_0 \), then \( y[n] = 0 \) for \( n < n_0 \).These conditions are crucial in simplifying the analysis of a system's response because they provide a baseline from which changes occur. When a system is assumed to be at rest initially, the impact of an input can be directly attributed to the system's properties rather than any pre-existing conditions. In the exercise, the initial rest condition is used to determine that \( y[-1] = 0 \) when finding \( y[0] \) given the impulse input. Without this foundation, determining the system's true response to an input would be more challenging.
Kronecker Delta Function
The Kronecker delta function, denoted \( \delta[n] \), is fundamental in signal processing and systems analysis. It is a discrete function defined as \( \delta[n] = 1 \) when \( n = 0 \) and \( \delta[n] = 0 \) for all other values of \( n \). This creates an impulse at \( n = 0 \), used to test system responses.When the input \( x[n] \) is the Kronecker delta function, finding the output provides the impulse response, \( h[n] \). The impulse response fully characterizes the system because any input signal can be broken down into a series of these impulses.In our exercise, the Kronecker delta function \( \delta[n] \) is applied, triggering the system starting from rest, allowing us to calculate sequential outputs and determine the system's behavior over time.
Geometric Sequence
A geometric sequence is a pattern of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In our exercise, the impulse response \( h[n] \) forms a geometric sequence.The sequence displayed is \( h[0] = 1 \), \( h[1] = -2 \), \( h[2] = 4 \), and \( h[3] = -8 \). The common ratio in this sequence is \(-2\), as each term is obtained by multiplying the previous term by \(-2\). Therefore, the general form of the sequence can be represented as \( h[n] = (-2)^n \).Understanding this geometric sequence is vital because it encapsulates the system's response to an impulse, ultimately providing insights into its behavior without needing to compute each consecutive value manually.

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Most popular questions from this chapter

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 $$y[4]=5, \quad y[14]=0$$

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) \(h(t)=\left(2 e^{-t}-e^{(t-100) t 100}\right) u(t)\)

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) \(h_{2}[n]=3^{n} u[-n+10]\)

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 \(n<0\)

Draw block diagram representations for causal LTI systems described by the following difference equations: (a) \(y(n)=\frac{1}{3} y[n-1]+\frac{1}{2} x[n]\) (b) \(y[n]=\frac{1}{3} y[n-1]+x[n-1]\)

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