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Chapter 10: Problem 54

By means of a specific filter design procedure, a nonideal continuous-time lowpass filter with frequency response \(H_{0}(j \omega),\) impulse response \(h_{0}(t),\) and step response \(s_{0}(t)\) has been designed. The cutoff frequency of the filter is at \(\omega=2 \pi \times 10^{2}\) rad/sec, and the step response rise time, defined as the time required for the step response to go from \(10 \%\) of its final value to \(90 \%\) of its final value, is \(\tau_{r}=10^{-2}\) second. From this design, we can obtain a new filter with an arbatrary cutoff frequency \(\omega_{c}\) by the use of frequency scaling. The frequency response of the resulting filter is then of the form \\[ H_{\mathrm{I} \rho}(j \omega)=H_{0}(j a \omega) \\] where \(a\) is an appropriate scale factor. (a) Determine the scale factor a such that \(H_{1 p}(j \omega)\) has a cutoff frequency of \(\omega_{c}\) (b) Determine the impulse response \(h_{| p}(t)\) of the new filter in terms of \(\omega_{c}\) and \(h_{0}(t)\) (c) Determine the step response \(s_{1 p}(2)\) of the new fiter in terms of \(\omega_{c}\) and \(s_{0}(f)\) (d) Determine and sketch the rise time of the new filler as a function of its cutoff frequency \(\omega_{c}\) This is one illustration of the trade-off between fine-domain and frequency- domain characteristics. In particular, as the cutoff frequency decreases, the rise time tends to increase

Short Answer

Expert verified
The scale factor is \( a = \frac{2\pi \times 10^2}{\omega_c} \). The new impulse response and step response are scaled by this factor, and the rise time increases as the cutoff frequency decreases.

Step by step solution

01

Understanding Initial Conditions

We start with a basic understanding of the given parameters. The original filter has a cutoff frequency of \( \omega = 2\pi \times 10^2 \) rad/sec and a rise time \( \tau_r = 10^{-2} \) seconds for the step response. This sets the initial conditions for the original filter design.
02

Frequency Scaling Transformation

To obtain a new filter with a different cutoff frequency \( \omega_c \), we use frequency scaling. The frequency response transformation is given by \( H_{I_\rho}(j \omega) = H_0(j a \omega) \). The parameter \( a \) scales the frequency and is determined by setting \( a \omega_c = 2\pi \times 10^2 \). Therefore, \( a = \frac{2\pi \times 10^2}{\omega_c} \).
03

Calculating Impulse Response

The impulse response of the new filter, \( h_{|p}(t) \), is related to the original impulse response \( h_0(t) \) through frequency scaling. The relationship can be defined as \( h_{|p}(t) = \frac{1}{a} h_0\left(\frac{t}{a}\right) \), substituting the value of \( a \) from Step 2.
04

Calculating Step Response

The step response \( s_{1p}(t) \) for the new filter also scales similarly. Use the expression \( s_{1p}(t) = s_0\left(\frac{t}{a}\right) \). This uses the same scaling factor \( a \) determined with \( a = \frac{2\pi \times 10^2}{\omega_c} \).
05

Determining and Sketching New Rise Time

The rise time for the new filter is affected by the scale factor \( a \). Since the original rise time is \( 10^{-2} \, ext{s} \) for \( \omega = 2\pi \times 10^2 \), the new rise time \( \tau_{r}' \) can be expressed as \( \tau_{r}' = a \times 10^{-2} \). Substitute \( a = \frac{2\pi \times 10^2}{\omega_c} \) to get \( \tau_{r}' = \frac{2\pi \times 10^2}{\omega_c} \times 10^{-2} \). Sketching this relationship will show that as \( \omega_c \) decreases, \( \tau_{r}' \) increases.

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Key Concepts

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

Frequency Scaling
In the world of continuous-time filters, frequency scaling is a crucial concept. It allows for the transformation of a filter's characteristics by adjusting the frequencies at which it operates. When we have a filter with a specific frequency response, such as the original filter in our exercise with a cutoff frequency of \( \omega = 2\pi \times 10^2 \) rad/sec, frequency scaling lets us modify this cutoff frequency to a new desired value \( \omega_c \).

Frequency scaling is performed by introducing a scale factor, denoted in our solution by \( a \). This factor is calculated by the equation \( a = \frac{2\pi \times 10^2}{\omega_c} \). Here, \( \omega_c \) represents the new desired cutoff frequency. Once the scale factor \( a \) is determined, it adjusts the filter's operation such that the frequency response \( H_{I_\rho}(j \omega) = H_0(j a \omega) \).
  • Enables flexible design of filters.
  • Adjusts filter's cutoff frequency to meet system requirements.
  • Ensures that the frequency characteristics of a filter are easily adaptable.
Cutoff Frequency
The cutoff frequency is a key parameter in filter design as it defines the threshold frequency beyond which the filter significantly attenuates the signal. In a lowpass filter, it marks the transition between the passband and the stopband. For the original filter in this exercise, the cutoff frequency is \( \omega = 2\pi \times 10^2 \) rad/sec.

When designing a new filter through frequency scaling, the cutoff frequency can be set to an arbitrary value \( \omega_c \). This is accomplished by recalibrating the filter鈥檚 frequency response using the scale factor \( a = \frac{2\pi \times 10^2}{\omega_c} \). Setting a new cutoff frequency allows the filter to be tuned specifically to handle different signal processing tasks, which can be critical in applications where specific bandwidth limits are necessary.
  • Determines the filtering performance and its suitability for different applications.
  • Critical for controlling which frequencies are allowed to pass and which are attenuated.
  • A pivotal factor in designing signal processing systems.
Impulse Response
In signal processing, the impulse response of a system, denoted as \(h(t)\), is its output when presented with a brief input signal known as an impulse. For the original filter, the impulse response is \( h_0(t) \).

When we alter the cutoff frequency of the filter using frequency scaling, the impulse response adapts accordingly. The relationship for the new impulse response \( h_{|p}(t) \) with respect to the original is defined by \( h_{|p}(t) = \frac{1}{a} h_0\left(\frac{t}{a}\right) \). This transformation accounts for the adjusted timing of signal passages through the filter due to frequency scaling, and it ensures that the altered filter retains its functionality tailored to the new cutoff frequency.
  • Indicates how the filter reacts to signals passing through it.
  • Important for understanding the timing and phase shifts introduced by the filter.
  • Highlights the temporal response characteristics of the filter.
Step Response
The step response of a filter shows how it responds to a sudden change in input, like a voltage or signal step. For our original filter design, this response is represented by \( s_0(t) \). The step response is a marker of the speed and stability with which a filter reaches a steady-state output when subjected to a sudden input change.

With frequency scaling and the introduction of a new cutoff frequency, the step response adjusts to \( s_{1p}(t) = s_0\left(\frac{t}{a}\right) \), where \( a \) is the scale factor \( a = \frac{2\pi \times 10^2}{\omega_c} \). When calculating this for the new filter, it's important that the new step response maintains proportional characteristics to the original response but aligned to the altered frequency operation.
  • Illustrates how evenly and quickly the output stabilizes after an input step.
  • Essential for evaluating the transient performance of filters.
  • Helps in determining the rise time, which indicates the system's responsiveness.

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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) \(\frac{z^{-2}\left(1-z^{-1}\right)}{\left(1-\frac{1}{4} z^{-1}\right)\left(1+\frac{1}{4} z^{-1}\right)}\)

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.

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.

Determine the unilateral \(z\) -transform of each of the following signals, and specify the corresponding regions of convergence: (a) \(x_{1}[n]=\left(\frac{1}{4}\right)^{n} u[n+5]\) (b) \(x_{2}[n]=\delta[n+3]+\delta[n]+2^{n} u[-n]\) (c) \(x_{3}[n]=\left(\frac{1}{2}\right)^{|n|}\)

Let \(x[n]\) be a signal whose rational \(z\) -transform \(X(z)\) contains a pole at \(z=1 / 2\) Given that \\[ x_{1}[n]=\left(\frac{1}{4}\right)^{n} x[n] \\] is absolutely summable and \\[ x_{2}[n]=\left(\frac{1}{8}\right)^{n} x[n] \\] is not absolutely summable, determine whether \(x[n]\) is left sided, right sided, or two sided.
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