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Consider a causal and stable LTI system \(S\) whose input \(x[n]\) and output \(y[n]\) are related through the second-order difference equation $$y | n]-\frac{1}{6} y[n-1]-\frac{1}{6} y[n-2]=x[n]$$ (a) Deteranine the trequency response \(H\left(e^{j \omega}\right)\) for the system \(S\) (h) Determine the impulse response \(h[n]\) for the system \(S\).

Short Answer

Expert verified
The frequency response is \(H(e^{j\omega}) = \frac{1}{1 - \frac{1}{6}e^{-j\omega} - \frac{1}{6}e^{-j2\omega}}\). The impulse response \(h[n]\) is found by taking the inverse Z-transform of \(H(z)\), resulting in an expression with terms \(A r_1^n u[n] + B r_2^n u[n]\) where \(r_1\) and \(r_2\) are the roots of the characteristic equation.

Step by step solution

01

Finding the Characteristic Equation

Consider the given difference equation: \(y[n] - \frac{1}{6} y[n-1] - \frac{1}{6} y[n-2] = x[n]\). To find the frequency response \(H(e^{j\omega})\), rewrite the equation in terms of the system function, \(H(z)\). Convert the difference equation into its equivalent Z-transform form:\[Y(z) - \frac{1}{6}z^{-1}Y(z) - \frac{1}{6}z^{-2}Y(z) = X(z)\]This simplifies to:\[Y(z)(1 - \frac{1}{6}z^{-1} - \frac{1}{6}z^{-2}) = X(z)\]
02

Expressing the Transfer Function

Divide both sides of the equation by the polynomial in \(z\) to express \(H(z)\):\[H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 - \frac{1}{6}z^{-1} - \frac{1}{6}z^{-2}}\]This is the transfer function in terms of \(z\).
03

Finding the Frequency Response

To determine the frequency response \(H(e^{j\omega})\), substitute \(z = e^{j\omega}\) into \(H(z)\):\[H(e^{j\omega}) = \frac{1}{1 - \frac{1}{6}e^{-j\omega} - \frac{1}{6}e^{-j2\omega}}\]This expression represents how the system processes each frequency component in the input signal.
04

Calculate the Impulse Response using Inverse Z-transform

The impulse response \(h[n]\) of a system is the inverse Z-transform of \(H(z)\). Consider the roots of the denominator polynomial \(1 - \frac{1}{6}z^{-1} - \frac{1}{6}z^{-2}\):Factor the characteristic equation: \(z^2 - \frac{1}{6}z - \frac{1}{6} = 0\). Use the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1, b=-\frac{1}{6}, c=-\frac{1}{6}\).Calculate the roots to get a general solution for \(h[n]\).
05

Applying Partial Fraction Expansion and Finding Time-Domain Solution

Express \(H(z)\) using partial fraction expansion and then take the inverse Z-transform to find \(h[n]\). The impulse response will be a sum of exponentials:If roots are \(r_1\) and \(r_2\), then:\[h[n] = A r_1^n u[n] + B r_2^n u[n]\]where \(u[n]\) is the unit step function. \(A\) and \(B\) are constants determined from initial conditions or additional information given about the system.

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Key Concepts

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

Frequency Response
In linear time-invariant (LTI) systems, understanding how the system reacts to different frequencies of input is crucial. The **frequency response** is a tool that helps illustrate this behavior. It describes how each component of a frequency present in an input signal is scaled and shifted in phase when passed through the system.
In mathematical terms, the frequency response, denoted as \(H(e^{j\omega})\), is derived from the system's transfer function \(H(z)\). For our system, we find the frequency response by substituting \(z = e^{j\omega}\) into the transfer function:\[H(e^{j\omega}) = \frac{1}{1 - \frac{1}{6}e^{-j\omega} - \frac{1}{6}e^{-j2\omega}}\]
This formula reveals how input signals at different frequencies \(\omega\) are altered after passing through the system. The magnitude of \(H(e^{j\omega})\) tells us how much each frequency component is amplified or attenuated, while the argument (or angle) indicates the phase shift introduced. This is essential when designing and analyzing filters, as it helps us determine which frequencies are suppressed or passed through unaltered.
Impulse Response
An impulse response is a fundamental characteristic of an LTI system. It represents the output when an impulse input is applied. This is critical for describing the system entirely, as any input can be expressed as a combination of impulse inputs through superposition.
For the system at hand, finding the impulse response involves taking the inverse Z-transform of the transfer function \(H(z)\). We start by establishing the transfer function, then look at solving the characteristic equation obtained from the denominator:\[z^2 - \frac{1}{6}z - \frac{1}{6} = 0\]
Using the quadratic formula to find the roots, we simplify the process of performing partial fraction expansion. With this, the impulse response \(h[n]\) takes the form of a sum of exponentials, usually expressed as:
  • \(h[n] = A r_1^n u[n] + B r_2^n u[n]\)
The variables \(A\) and \(B\) are constants determined by initial conditions, and \(u[n]\) is the unit step function. This expression tells us how the system reacts over time to an initial 'kick' provided by the impulse.
Difference Equation
The **difference equation** is the mathematical backbone of describing relationships in discrete-time LTI systems. It is akin to a differential equation in continuous systems but is applicable to sequences rather than functions.
In our context, the difference equation is given by:
\[y[n] - \frac{1}{6} y[n-1] - \frac{1}{6} y[n-2] = x[n]\]
This equation asserts the relationship between the input signal \(x[n]\) and the output signal \(y[n]\) with past outputs. The coefficients in the equation determine how much the current and previous inputs and outputs influence the system's current output. The order of the system, which is the highest number of previous outputs in the equation, dictates the memory length of the system according to this equation.
  • This particular equation is second-order due to the terms \(y[n-1]\) and \(y[n-2]\).
  • The solution to this equation gives us insights into the stability and frequency behavior of the system.
Understanding and solving such equations are essential to predict a system's response to varying inputs over time.
Z-transform
The **Z-transform** is a powerful tool used in the analysis of discrete-time LTI systems. It transforms signals into the frequency domain, allowing easier manipulation of the system's behavior. Think of it as the discrete-time equivalent of the Laplace Transform used in continuous-time analysis.
When dealing with difference equations, employing the Z-transform allows us to convert these time-domain equations into algebraic equations, which are easier to solve. From our earlier steps, we used the Z-transform on the given difference equation to translate it into:
  • \[Y(z) - \frac{1}{6}z^{-1}Y(z) - \frac{1}{6}z^{-2}Y(z) = X(z)\]
This equation can be reordered to find the transfer function \(H(z)\), which helps in determining the system's frequency response and impulse response.
The Z-transform is beneficial for stabilizing analysis, simplifying convolution via multiplication, and realizing system design and filtering applications. It compresses complex sequences into a simple polynomial form, making it a crucial tool in digital signal processing.

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Most popular questions from this chapter

(a) Suppose we want to design a discrete-time LTI system which has the property that if the input is $$\left.x\\{n]=\left(\frac{1}{2}\right\\}^{n} u f n\right\\}-\frac{1}{4}\left(\frac{1}{2}\right)^{n-1} u\\{n-1\\}$$ then the output is \\[ y[n]=\left(\frac{1}{3}\right)^{n} u[n] \\] (i) Find the impulse response and frequency response of a discrete-time LTI system that has the foregoing property. (ii) Find a difference equation relating \(x[n\\}\) and \(y[n]\) that characterizes the system. (b) Suppese that a systent has the response \((1 / 4)^{n} u[n]\) to the input \((n+2)(1 / 2)^{n} u[n]\) If the output of this system is \(\delta[n]-(-1 / 2)^{n} x[n] .\) what is the input?

Let \(x|n|\) and \(h[n]\) be two signals, anct let \(y[n]=x[n] * h[n] .\) Write two expressions for \(y[0],\) ane (using the convolution sum directly) in terms of \(x[n]\) and \(h[n],\) and one (using the convolution property of Fourier transforms) in terms of \(X\left(\varphi^{\prime \omega}\right)\) and \(H\left(e^{I \omega}\right) .\) Then. by a judicious choice of \(h[n],\) use these two expressions to derive Parseval's relation - that is, $$\sum_{n=-\infty}^{r \times 2}|x[n]|^{2}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}\left|X\left(e^{j \omega}\right)\right|^{2} d \omega$$ In a similar fashon, derive the following generalization of Parseval's relation: $$\sum_{n=1}^{+\infty} x[n] z^{*}[n]=\frac{1}{2 \pi} \int_{-\neg}^{\pi} X\left(e^{j \omega}\right) Z^{*}\left(e^{j \omega}\right) d \omega$$

Consider a system consisting of the cascade of two LTI systems with frequency responses $$H_{1}\left\langle e^{j \omega}\right\rangle=\frac{2-e^{-j \omega}}{1+\frac{1}{2} e^{-j \omega}}$$ and $$H_{2}\left(e^{j \alpha}\right)=\frac{1}{1-\frac{1}{2} e^{-j \omega}+\frac{1}{4} e^{-j 2 \omega}}$$ (a) Find the difference equation describing the overall system. (b) Determine the impulse response of the overall system.

Compute the Fourier transform of each of the following signals: $$\text { (a) } x[n]=u[n-2]-u[n-6]$$ $$\text { (b) } x[n]=\left(\frac{1}{2}\right)^{-n} u[-n-1]$$ $$\text { (c) } x[n]=\left\\{\frac{1}{3}\right)^{|n|} u[-n-2]$$ $$\text { (d) } x[n]=2^{n} \sin \left(\frac{\pi}{4} n\right) u[-n]$$ $$\text { (e) } x[n]=\left(\frac{1}{2}\right)^{|n|} \cos \left(\frac{\pi}{8}(n-1)\right)$$ $$\text { (P) } x[n]=\left\\{\begin{array}{ll} n, & -3 \leq n \leq 3 \\ 0, & \text { otherwise } \end{array}\right.$$ $$\text { (g) } x[n]=\sin \left(\frac{\pi}{2} n\right)+\cos (n)$$ $$\text { (h) } x[n]=\sin \left(\frac{5 \pi}{3} n\right)+\cos \left(\frac{7 \pi}{3} n\right)$$ (i) \(x[n]=x[n-6],\) and \(x[n]=u[n]-u[n-5]\) for \(0 \leq n \leq 5\) $$\text { (j) } x[n]=(n-1)\left(\frac{1}{3}\right)^{|n|}$$ $$\text { (k) } x[n]=\left(\frac{(\ln (\pi r / 5)}{\pi n}\right) \cos \left(\frac{7 \pi}{2} n\right)$$

(a) Let $$x_{1}[n]=\cos \left(\frac{\pi n}{3}\right)+\sin \left(\frac{\pi n}{2}\right)$$ be a signal, and let \(X_{1}\left(e^{J \omega}\right)\) denote the Fourier uansform of \(x_{1}[n] .\) Sketch \(x_{1}[n]\) together with the signais with the following Founer transfortns: (i) \(\quad X_{2}\left\\{e^{j \omega}\right)=X_{1}\left(e^{j \omega}\right) e^{j \omega}, \quad|\omega|<\pi\) (ii) \(\quad X_{3}\left(e^{j \omega}\right)=X_{1}\left(e^{j \omega}\right) e^{-j 3 \omega / 2}, \quad|\omega|<\pi\) (b) Let $$w(t)=\cos \left(\frac{\pi \mathrm{r}}{3 T}\right)+\sin \left(\frac{\pi t}{2 T}\right)$$ be a continuous- time signal. Note that \(x_{1}[n]\) can be regarded as a sequence of evenly spaced samples of \(w(t) ;\) that is \\[ x_{1}[n]=w(n T) \\] Show that \\[ x_{2}[n]=w(n T-\alpha) \\] and \\[ x_{3}(n]=w(n T-\beta) \\] and specify the values of \(\alpha\) and \(\beta .\) From this result we can conclude that \(x_{2}[n]\) and \(x_{3}[n]\) are also evenly spaced samples of \(w(t)\)

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