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Chapter 10: Problem 41

A particular causal LTI system is described by the difference equation \\[ y[n]-\frac{\sqrt{2}}{2} y[n-1]+\frac{1}{4} y[n-2]=x[n]-x[n-1] \\] (a) Find the impulse response of this system. (b) Sketch the log magnitude and the phase of the frequency response of the system.

Short Answer

Expert verified
Impulse response is step-response difference equation solved for \( x[n]=\delta[n] \); frequency response from \( H(e^{j\omega}) \).

Step by step solution

01

Understand the System Equation

The given difference equation is \( y[n]-\frac{\sqrt{2}}{2} y[n-1]+\frac{1}{4} y[n-2]=x[n]-x[n-1] \). This relates the output \( y[n] \) to the input \( x[n] \) for a linear time-invariant (LTI) system. To find the system's impulse response, set \( x[n] = \delta[n] \), where \( \delta[n] \) is the Kronecker delta function (impulse).
02

Solve with Impulse Input

By substituting \( x[n] = \delta[n] \), the equation becomes \( y[n]-\frac{\sqrt{2}}{2} y[n-1]+\frac{1}{4} y[n-2]=\delta[n]-\delta[n-1] \). This is a homogeneous difference equation with particular solution and initial conditions based on the impulse input.
03

Find Homogeneous Solution and Particular Solution

The associated homogeneous equation is \( y[n]-\frac{\sqrt{2}}{2} y[n-1]+\frac{1}{4} y[n-2]=0 \). To solve it, assume \( y[n] = r^n \) which gives the characteristic equation \( r^2 - \frac{\sqrt{2}}{2}r + \frac{1}{4} = 0 \). The roots \( r_1 \) and \( r_2 \) are found using the quadratic formula.
04

Apply Quadratic Formula

The roots are computed as \( r = \frac{\frac{\sqrt{2}}{2} \pm \sqrt{(\frac{\sqrt{2}}{2})^2 - 4 \cdot \frac{1}{4}}}{2} \). Simplifying gives the roots \( r_1 = \frac{1}{2} \) and \( r_2 = \frac{1}{2} \). Since both roots are equal, the homogeneous solution is \( y_h[n] = (C_1 + C_2 n)\left(\frac{1}{2}\right)^n \).
05

Determine Particular Solution

For a particular solution \( y_p[n] \) to \( \delta[n] - \delta[n-1] \), use trial functions and initial conditions. The simplest particular solution starting from the impulse gives \( y_p[0] = 1 \) and \( y_p[1] \) adjusts such that it satisfies the difference equation on these terms.
06

Determine Impulse Response

Consider \( y_p[n] = 0 \) for \( n \geq 2 \) and solve for initial conditions. Calculate \( y[1] \) based on system equation with \( y[0] = 1 \) and \( y[-1] = 0 \). Utilize superposition of homogeneous and particular solutions to construct \( y[n] \).
07

Frequency Response Analysis

Express \( y[n] \) in terms of the impulse response \( h[n] \). Use the Z-transform to find \( H(z) \). Then, substitute \( z = e^{j\omega} \) to find the frequency response \( H(e^{j\omega}) \).
08

Sketch Frequency Response

Calculate \( |H(e^{j\omega})| \) and \( \arg(H(e^{j\omega})) \) over \( \omega \). Sketch the log magnitude \( 20 \log_{10} |H(e^{j\omega})| \) and phase response \( \arg(H(e^{j\omega}) \) for \( \omega \) ranging over typical frequency domains (e.g., \( [0, \pi] \)).

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Key Concepts

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

Impulse Response
In the study of Signals and Systems, the impulse response is a fundamental concept that reveals how a system responds to a very simple input, an impulse. An impulse, denoted as \( \delta[n] \), is a sudden spike at a single point in time, with all other values being zero. For Linear Time-Invariant (LTI) systems, the impulse response, \( h[n] \), fully characterizes the system.

By setting the input \( x[n] = \delta[n] \), we can analyze how an LTI system responds over time. For the given difference equation of our system:
  • \( y[n] - \frac{\sqrt{2}}{2} y[n-1] + \frac{1}{4} y[n-2] = x[n] - x[n-1] \).
we substitute the delta function to find \( y[n] \) at different values of \( n \). The impulse response is essentially finding the output \( y[n] \) when the input is this impulse signal. By solving this equation with \( x[n] = \delta[n] \), we determine the impulse response of our system, capturing its inherent properties.
Frequency Response
Frequency response illustrates how an LTI system reacts to different frequency components of an input signal. It involves analyzing the system's response in the frequency domain, providing insights into its filtering characteristics. For a given LTI system, its frequency response is derived from its impulse response through the use of complex exponentials.

We find the frequency response by performing the Z-transform on the impulse response, \( h[n] \), to obtain \( H(z) \). By substituting \( z = e^{j\omega} \), where \( \omega \) represents angular frequency, the frequency response \( H(e^{j\omega}) \) is determined. This response tells us how the system attenuates or amplifies each frequency component.

To fully understand the frequency response, it is common to evaluate:
  • The **magnitude response**, which shows the gain for each frequency and is typically plotted as \( 20 \log_{10} |H(e^{j\omega})| \).
  • The **phase response**, which illustrates the phase shift as a function of frequency and is plotted as \( \arg(H(e^{j\omega})) \).
By interpreting these plots, one can deduce whether the system acts more like a low-pass, high-pass, band-pass, or band-stop filter.
Difference Equation
A difference equation is a mathematical expression used to describe the behavior of discrete-time systems. It relates the output of the system to its past outputs and current or past inputs, much like a formula that governs behavior via time-recursive relations.

In our scenario, the given difference equation is:
  • \( y[n] - \frac{\sqrt{2}}{2} y[n-1] + \frac{1}{4} y[n-2] = x[n] - x[n-1] \)
This equation tells us how the current output \( y[n] \) is computed based on its previous outputs \( y[n-1] \) and \( y[n-2] \), as well as the current and past inputs \( x[n] \) and \( x[n-1] \). Solving difference equations often involves finding characteristic equations using trial functions, and then determining homogeneous and particular solutions for different input scenarios, such as impulse responses.

The difference equation is central in digital signal processing as it enables the modeling of systems like filters, predictors, or reverberators efficiently with discrete data.
Linear Time-Invariant Systems
Linear Time-Invariant (LTI) systems are a crucial class of systems in both continuous and discrete domains. They are distinctive since they adhere to properties of linearity and time-invariance, simplifying many analyses and system responses.

**Linearity** ensures that the principle of superposition applies. This means responses to individual inputs can be summed to find the response to a combined input. Mathematically, if two inputs \( x_1[n] \) and \( x_2[n] \) lead to outputs \( y_1[n] \) and \( y_2[n] \), then a combination \( ax_1[n] + bx_2[n] \) gives \( ay_1[n] + by_2[n] \).

**Time-Invariance** guarantees that the system's properties do not change over time, meaning shifting an input in time results in an equally shifted output. If an input \( x[n] \) produces output \( y[n] \), then \( x[n-k] \) results in \( y[n-k] \).

These properties make LTI systems particularly appealing, as solutions using convolution and frequency response analysis become straightforward. Understanding LTI systems helps design stable and predictable systems in engineering tasks such as communications, signal filtering, and even audio processing.

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Most popular questions from this chapter

The time constant provides a measure of how fast a first-order system responds to inputs. The idea of measuring the speed of response of a system is also important for higher order systems, and in this problem we investigate the extension of the time constant to such systems. (a) Recall that the time constant of a first-order system with impulse response \\[ h(t)=a e^{-a t} u(t), \quad a>0 \\] is \(1 / a,\) which is the amount of time from \(f=0\) that it takes the system step response \(\left.s(t) \text { to settle within } 1 / e \text { of is final value [i.e., } s(\infty)=\lim _{t \rightarrow x} s(t)\right]\) Using this same quantitative definition, find the equation that must be solved in order to determine the time constant of the causal LTI system described by the differential equation \\[ \frac{d^{2} y(t)}{d t^{2}}+11 \frac{d y(t)}{d t}+10 y(t)=9 x(t) \\] (b) As can be seen from part (a), if we use the precise definition of the time constant set forth there, we obtain a simple expression for the time constant of a firstorder system, but the calculations are decidedly more complex for the system of eq. \((P 649-1) .\) However, show that this system can be viewed as the parallel interconnection of two first-order systems. Thus, we usually think of the system of eq. \((P 6.49-1)\) as having mo time constants, corresponding to the two firstorder factors. What are the two time constants for this system? (c) The discussion given in part (b) can be directly generalized to all systems with impulse responses that are linear combinations of decaying exponentials. In any system of this type, one can identify the dominant time constants of the system, which are simply the largest of the time constants. These represent the slowest parts of the system response, and consequently, they have the dominant effect on how fast the system as a whole can respond. What is the dominant time constant of the system of eq. \((P 6.49-1) ?\) Substitute this time constant into the equation determined in part (a). Although the number will not satisfy the equation exactly, you should see that it nearly does, which is an indication that it is very close to the time constant defined in part (a). Thus, the approach we have outlined in part \((b)\) and here is of value in providing insight into the speed of response of LTI systems without requiring excessive calculation. (d) One important use of the concept of dominant time constants is in the reduction of the order of LTI systems. This is of great practical significance in problems involving the analysis of complex systems having a few dominant time constants and other very small time constants. In order to reduce the complexity of the model of the system to be analyzed, one often can simplify the fast parts of the system. That is, suppose we regard a complex system as a parallel interconnection of first- and second-order systems. Suppose also that one of these subsystems, with impulse response \(h(t)\) and step response \(s(t),\) is fast - -that is, that \(s(t)\) settles to its final value \(s(\infty)\) very quickly. Then we can approximate this subsystem by the subsystem that settles to the same final value instantaneously. That is, if \(f(t)\) is the step response to our approximation, then \\[ s(t)=s(\infty) w(t) \\] This is illustrated in Figure \(P 6.49\). Note that the impulse response of the approximate system is then \\[ \hat{h}(t)=s(\infty) \delta(t) \\] which jadicates that the approximate system is memory-less. Consider again the causal LTI system described by eq. (P6.49-1) and, in particular, the representation of it as a parallel interconnection of two first-order systems, as described in part (b). Use the method just outlined to replace the faster of the two subsystems by a memory-less system. What is the differential equation that then describes the resulting overall system? What is the frequency response of this system? Sketch \(|H(j \omega)|\) (not \(\log |H(j \omega)|\) ) and \(\angle H(j \omega)\) for both the original and approximate systems. Over what range of frequencies are these frequency responses nearly equal? Sketch the step responses for both systems. Over what range of time are the step responses nearly equal? From your plots. you will see some of the similarities and differences between the original system and its approximation. The utility of an approximation such as this depends upon the specific application. In particular, one must take into account both how widely separated the different time constants are and also the nature of the inputs to be considered. As you will see from your answers in this part of the problem, the frequency response of the approximate system is essentially the same as the frequency response of the original system at low frequencies. That is, when the fast parts of the system are sufficiently fast compared to the rate of fluctuation of the input, the approximation becomes useful.

The square of the magnitude of the frequency response of a class of continuous-time lowpass filters, known as Butterworth filters, is \\[ |B(j \omega)|^{2}=\frac{1}{1-\left(\omega / \omega_{t}\right)^{2 N}} \\] Let us define the passband edge frequency \(\omega_{p}\) as the frequency below which \(B(j \omega)\\}^{2}\) is greater than one-half of its value at \(\omega=0 ;\) that is, \\[ |B(j \omega)|^{2} \geq \frac{1}{2}|B(j 0)|^{2},|\omega|<\omega_{p} \\] Now let us define the stopband edge frequency \(w_{s}\) as the frequency above which \(|B(j \omega)|^{2}\) is less than \(10^{-2}\) of its value at \(\omega=0 ;\) that is. \\[ |B(j \omega)|^{2} \leq 10^{-2}|B\langle j 0)|^{2}, \quad|\omega|>\omega_{s} \\] The transition band is then the frequency range between \(\omega_{p}\) and \(\omega_{s}\). The ratio \(\omega_{s} / \omega_{p}\) is referred to as the transition ratio. For fixed \(\omega_{p},\) and making reasonable approximations, determined and sketch the transition ratio as a function of \(N\) for the class of Butterworth filters.

Consider a right-sided sequence \(x[n]\) with \(z\) -transform \\[ X(z)=\frac{1}{\left(1-\frac{1}{2} z^{-1}\right)\left(1-z^{-1}\right)} \\] (a) Carry out a partial-fraction expansion of eq. (P10.25-1) expressed as a ratio of polynomials in \(z^{\prime},\) and from this expansion, determine \(x\lfloor n\rfloor\) (b) Rewrite eq. \((\mathrm{P} 10.25-1)\) as a ratio of polynomials in \(z,\) and carry out a partial-fraction expansion of \(X\\{z \text { ) expressed in terms of polynomials in } z\). From this expansion, determine \(x[n],\) and demonstrate that the sequence obtained is identical to that obtained in part (a).

Consider the following algebraic expression for the \(z\) -transform \(X(z)\) of a signal \(x[n]\) \\[ X(z)=\frac{1+z^{-1}}{1+\frac{1}{3} z^{-1}} \\](a) Assuming the ROC to be \(|z|>1 / 3\), use long division to determine the values of \(x\\{0\\}, x[1],\) and \(x\\{2\\}\) (b) Assuming the ROC to be \(|z|<1 / 3,\) use long division to determine the values of \(x[0], x[-1],\) and \(x[-2]\)

Suppose that the algebraic expression for the \(z\) -transform of \(x[n]\) is \\[ X(z)=\frac{1-\frac{1}{4} z^{-2}}{\left(1+\frac{1}{4} z^{-2}\right)\left(1+\frac{5}{4} z^{-1}-\frac{3}{8} z^{-2}\right)} \\] How many different regions of convergence could correspond to \(X(z) ?\)
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