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Chapter 10: Problem 16

Consider the following system functions for stable LTI systems. Without utilizing the inverse \(z\) -transform, determine in each case whether or not the corresponding system is causal. (a) \(\frac{1-\frac{4}{3} z^{-1}+\frac{1}{2} z^{-2}}{z^{-1}\left(1-\frac{1}{2} z^{-1}\right)\left(1-\frac{1}{z} z^{-1}\right)}\) (b) \(\frac{z-\frac{1}{2}}{z^{2}+\frac{1}{2} z-\frac{3}{16}}\) (c) \(\frac{z+1}{z+\frac{4}{3}-\frac{1}{2} z^{-2}-\frac{2}{3} z^{-3}}\)

Short Answer

Expert verified
(a) Causal, (b) Causal, (c) Complex analysis, requires further matching inspection.

Step by step solution

01

Understand the concept of causality for LTI systems

A causal LTI system's impulse response is zero for all negative time indices, meaning the system's behavior depends solely on present and past inputs and not on future inputs. In terms of system functions or transfer functions, a causal system's pole-zero cancellation, if any, should not involve zeros with a magnitude greater than that of the poles (inside the unit circle for causal systems).
02

Analyze the polynomial denominators for poles and zeros (Part a)

The given system function is \( \frac{1-\frac{4}{3} z^{-1}+\frac{1}{2} z^{-2}}{z^{-1}\left(1-\frac{1}{2} z^{-1}\right)\left(1-\frac{1}{z} z^{-1}\right)} \). Rewrite the denominator for easier analysis: it is \( z^{-1} \left(1-\frac{1}{2} z^{-1}\right)\left(z^{-1} - 1\right)\). The poles derived are at \( z = 0, \frac{1}{2}, 1 \). There are no zero factors outside these poles, and thus this system is causal since there are no zeros outside the unit circle to cancel poles inside.
03

Analyze the polynomial denominators for poles and zeros (Part b)

The given system function is \( \frac{z-\frac{1}{2}}{z^{2}+\frac{1}{2} z-\frac{3}{16}} \). We need to determine the poles from the denominator, which requires factoring the quadratic. The poles can be found via the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = \frac{1}{2}, c = -\frac{3}{16} \). Solving this gives poles inside the unit circle. The zero given by \( z = \frac{1}{2} \) confirms there are no zeros cancelling with larger magnitude poles, indicating causality.
04

Analyze the polynomial denominators for poles and zeros (Part c)

The given system function is \( \frac{z+1}{z+\frac{4}{3}-\frac{1}{2} z^{-2}-\frac{2}{3} z^{-3}} \). Observe that the poles from this more complex rational expression can be analyzed by analyzing the behavior near \( z \rightarrow \infty \), intending to check for zero crossing with significant poles. This specific form allows evaluating systems advantageously by unconventional analysis without overt reduction, focusing understanding on the pole-zero interplay. The system demonstrates complexity when poles exist without zero cancellations or enhancements, ensuring stability but needing thorough pole-zero versus zero-pole matching.

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Key Concepts

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

Causality in Systems
Causality refers to the sequence of cause and effect. For linear time-invariant (LTI) systems, causality is a key property. It implies that the system's output at any time depends only on its past and current inputs, not future ones. This is intuitive if you think of a system responding in real-time to input signals.
In mathematical terms, an LTI system is causal if its impulse response is zero for all discrete-time indices that are less than zero. Effectively, no future inputs can influence the present output. When analyzing a system's transfer function, causality dictates that poles, the values of \(z\) that make the denominator zero, should have no corresponding zeroes outside the unit circle in the complex plane. This ensures that the output depends solely on inputs received till that point, proving causality.
Linear Time-Invariant (LTI) Systems
LTI systems are a mathematical model used for analyzing and designing signal processing systems. They are characterized by two constant features: linearity and time-invariance. Linearity suggests that the system's response to a weighted sum of inputs is the weighted sum of the responses to those individual inputs. Time-invariance implies that the system's behavior and characteristics do not change over time.
  • Properties: LTI systems are predictable and stable, making them fundamental in control systems, communications, and electronic devices.
  • Convolution: The operation called convolution provides a way to calculate an output from the system's impulse response and input signal, emphasizing the utility of knowing a system's characteristics.
LTI systems are pivotal in simplifying the problem-solving approach in different domains, as they can be analyzed with the same mathematical tools efficiently.
Pole-Zero Analysis
Pole-zero analysis serves as a core analytical tool for understanding the behavior of LTI systems. It entails graphically inspecting the poles and zeroes of a system's transfer function in the complex plane. This technique is indispensable as each pole and zero gives insights into the system dynamics.
  • Poles: These are the points where the system's denominator equals zero, meaning the transfer function is undefined. They govern the stability and response of a system. For instance, poles within the unit circle imply a stable system.
  • Zeros: Conversely, zeroes are points that make the numerator zero, indicating frequencies where the system output is nullified. This means input signals at these frequencies produce no output.
The placement of poles and zeros can directly affect whether an LTI system is causal. Escape of any zero outside the unit circle without an involving pole often breaks the causality in the system.
Understanding the Z-Transform
The z-transform is a powerful mathematical tool used mostly for analyzing discrete-time LTI systems. It's similar to the Laplace transform used in continuous-time systems, applying to sequences rather than continuous functions.
  • Definition: The z-transform represents a sequence as an infinite series in terms of the complex frequency variable \(z\). Mathematically, a sequence \(x[n]\) transforms into \(X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}\).
  • Region of Convergence (ROC): This is pivotal as it defines the values of \(z\) for which the z-transform converges. The ROC is directly linked to the system's stability and causality.
The z-transform aids in converting complex differential equations into manageable algebraic ones, expediting the process of understanding signal behaviors in the spectral domain.

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Most popular questions from this chapter

For each of the following algebraic expressions for the z-transform of a signal. determine the number of zeros in the finite \(z\) -plane and the number of zeros at infinity. (a) \(\frac{z^{-1}\left(1-\frac{1}{2} z^{-1}\right)}{\left(1-\frac{1}{3} z^{-1}\right)\left(1-\frac{1}{4} z^{-1}\right)}\) (b) \(\frac{\left(1-z^{-1}\right)\left(1-2 z^{-1}\right)}{\left(1-3 z^{-1}\right)\left(1-4 z^{-1}\right)}\) (c) \(\frac{z^{-2}\left(1-z^{-1}\right)}{\left(1-\frac{1}{4} z^{-1}\right)\left(1+\frac{1}{4} z^{-1}\right)}\)

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).

In many filtering applications, it is often undesirable for the step response of a filter to overshoot its final value. In processing pictures, for example, the overshoot in the step response of a linear filter may produce flare-that is, an increase in intensity at sharp boundaries. It is possible, however. to eliminate overshoot by requiring that the impulse response of the filter be positive for all time. Show that if \(h(t),\) the impulse response of a continuous-time LTT filter, is always greater than or equal to zero, the step response of the filter is a monotonically nondecreasing function and therefore will not have overshoot.

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.
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