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Chapter 6: Problem 9

Consider a continuous-time causal and stable LTI system whose input \(x(t)\) and output \(y(t)\) are related by the differential equation \\[\frac{d y(t)}{d t}+5 y(t)=2 x(t)\\] What is the final value \(s(x)\) of the step response \(s(t)\) of this filter? Also, determine the value of \(t_{0}\) for which $$s\left(t_{0}\right)=s(s)\left[1-\frac{1}{e^{2}}\right]$$

Short Answer

Expert verified
The final step response value is \(\frac{2}{5}\). The time \(t_0\) is \(\frac{2}{5}\).

Step by step solution

01

Find the Transfer Function

To find the step response, first determine the transfer function, \(H(s)\), of the system. Start by taking the Laplace Transform of both sides of the given differential equation, assuming zero initial conditions: \(sY(s) + 5Y(s) = 2X(s)\). Rearrange to find \(H(s) = \frac{Y(s)}{X(s)} = \frac{2}{s + 5}\).
02

Find the Step Response

The step response, \(s(t)\), is the inverse Laplace Transform of \(H(s) \cdot \frac{1}{s}\), as a step input is given by \(\frac{1}{s}\). So: \(s(t) = \mathcal{L}^{-1}\left\{\frac{2}{s(s+5)}\right\}\). Using partial fraction decomposition: \(\frac{2}{s(s+5)} = \frac{A}{s} + \frac{B}{s+5}\), solve to get \(A = \frac{2}{5}\) and \(B = -\frac{2}{5}\). Thus, \(s(t) = \frac{2}{5}(1 - e^{-5t})\).
03

Calculate Final Value of Step Response

The final value theorem states \(\lim_{t \to \infty} s(t) = \lim_{s \to 0} s \cdot S(s)\). Apply it to find: \(\lim_{s \to 0} s \cdot \frac{2}{s(s+5)} = \frac{2}{5}\). Thus, the final value \(s(\infty) = \frac{2}{5}\).
04

Determine Specific Time \(t_0\)

Set the equation \(s(t_0) = s(\infty) \left[1 - \frac{1}{e^2}\right]\). Plug in the values \(s(t) = \frac{2}{5} (1 - e^{-5t})\) and \(s(\infty) = \frac{2}{5}\). Solve \(\frac{2}{5} (1 - e^{-5t_0}) = \frac{2}{5} \left[1 - \frac{1}{e^2}\right]\) to find \(t_0 = \frac{2}{5}\).

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Key Concepts

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

Laplace Transform
The Laplace Transform is a mathematical technique used to transform signals from the time domain to the s-domain. This is particularly useful for analyzing linear time-invariant (LTI) systems, like the one described in the exercise.
The transformation changes a differential equation involving functions of time into an algebraic equation. This conversion greatly simplifies solving these types of problems because algebraic equations are much easier to manipulate than differential equations.
  • LTI systems can be described by differential equations.
  • Transforming these equations to the s-domain helps to find solutions using algebraic methods.
  • The Laplace Transform is denoted by \( \mathcal{L}\{x(t)\} = X(s)\), where \(X(s)\) is the image of \(x(t)\) in the s-domain.
In this exercise, by applying the Laplace Transform to the differential equation, the problem becomes solvable using algebraic operations, finding the transfer function, and analyzing the step response efficiently.
Step Response
The step response of a continuous-time LTI system describes how the system outputs to a step input, like a switch that suddenly turns on.
It is essential in understanding system behavior in reaction to sudden changes.
A step input is typically represented by the function \(u(t)\), which equals 1 when \(t \geq 0\) and 0 otherwise.
  • The step response is obtained by applying the inverse Laplace Transform to the product of the transfer function and the Laplace Transform of the step input.
  • The exercise uses \(H(s)\) as the transfer function to find the corresponding step response.
  • The inverse Laplace Transform leads to a time domain function that describes how the output behaves over time.
In this problem, the step response \(s(t)\) derived as \(\frac{2}{5} (1 - e^{-5t})\), reveals how the system reaches its steady-state after responding to an initial step signal.
Transfer Function
The transfer function is a fundamental concept in control theory and signal processing. It is defined as the Laplace Transform of the system's output over the Laplace Transform of the system's input, with initial conditions set to zero.
It's represented as \(H(s) = \frac{Y(s)}{X(s)}\).
  • The transfer function of a system provides crucial insights into the system's stability and behavior.
  • It allows us to understand how systems respond over a range of frequencies.
  • In this exercise, after applying the Laplace Transform to the differential equation and rearranging, we found \(H(s) = \frac{2}{s+5}\).
This transfer function demonstrates that the system has a pole at \(-5,\) indicating exponential decay in its step response, which assists in understanding the dynamics of the original equation over time.
Final Value Theorem
The Final Value Theorem is a useful tool in the analysis of LTI systems. It relates the behavior of a system as time approaches infinity to its Laplace Transform. This theorem provides a quick way to determine the steady-state value of a system's time domain function without requiring its complete solution.
  • The theorem is expressed as \(\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s),\)
  • where \(f(t)\) is the time domain function and \(F(s)\) its Laplace Transform.
  • It simplifies calculations by using properties of the transformed domain.
In the exercise, this theorem allowed us to find the final value of the step response: \(\frac{2}{5}\). Using this information, we could also solve for a specific time point \(t_0\) where the system reaches a fraction of its steady state, emphasizing the theorem's practical utility in solving real-world problems.

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Most popular questions from this chapter

Consider the following frequency response for a causal and stable LTI system: \\[H(j \omega)=\frac{1-j \omega}{1+j \omega}\\] (a) Show that \(|H(j \omega)|=A,\) and determine the value of \(A\) (b) Determine which of the following statements is true about \(\tau(\omega),\) the group delay of the system. (Note: \(\not < H\left(j \omega_{0}\right)\) , where \(\not < H\left(j \omega_{0}\right)\) is expressed in a form that does rot contain any discontinuities.) 1\. \(\tau(\omega)=0\) for \(\omega>0\) 2\. \(\tau(\omega)>0\) for \(\omega>0\) 3\. \(\tau(\omega)<0\) for \(\omega>0\)

Consider a causal LTI system whose frequency response is given as: \\[H\left(e^{j \omega}\right)=e^{-j \omega} \frac{1-\frac{1}{2} e^{j \omega}}{1-\frac{1}{2} e^{-j \omega}}\\] (a) Show that \(\left|H\left(e^{j \omega}\right)\right|\) is unity at all frequencies. (b) Show that $$\not < H\left(ej \omega_{}\right)=-\omega-2 \tan ^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1-\frac{1}{2} \cos \omega}\right)$$ (c) Show that the group delay for this filter is given by \\[\tau(\omega)=\frac{\frac{3}{4}}{\frac{5}{4}-\cos \omega}\\] Sketch \(\tau(\omega)\) (d) What is the ouput of this filter when the input is \(\cos \left(\frac{\pi}{3} n\right) ?\)

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{l}_{\mathrm{p}}}(j \omega)=H_{0}(j a \omega)\\] where \(a\) is an appropriate scale factor. (a) Determine the scale factor \(a\) such that \(H_{\mathrm{lp}}(j \omega)\) 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.

Let \(h_{d}[n]\) denole the unit sarple response of a desired ideal system with frequency response \(H_{d}\left(e^{j \omega}\right),\) and let \(h[n]\) denote the unit sample response for an FIR system of length \(N\) and with frequency response \(H\left(e^{j \omega}\right) .\) In this problem, we show that a rectangular window of length \(N\) samples applied to \(h_{d}[n]\) will produce a unit sample response \(h[n]\) such that the mean square error \\[\boldsymbol{\epsilon}^{2}=\frac{1}{2 \pi} \int_{-\pi}^{\mathbb{V}}\left|H_{d}\left(e^{j \omega}\right)-H\left(e^{j \omega}\right)\right|^{2} d \omega\\] is minimized. (a) The error function \(E\left(e^{t a}\right)=H_{d}\left(e^{t \omega}\right)-H\left(e^{j \omega}\right)\) can be expressed as the power series \\[E\left(e^{j \omega}\right)=\sum_{n=-\infty}^{\infty} e|n| e^{-j \omega n}\\] Find the coefficients \(e[n]\) in terms of \(h_{d}[n]\) and \(h[n]\) (b) Using Parseval's relation, express the mean square extor \(\epsilon^{2}\) in terms of the coefficients \(e[n]\) (c) Show that for a unit sample response \(h[n]\) of length \(N\) samples, \(\epsilon^{2}\) is minimized when \\[h[n]=\left\\{\begin{array}{ll}h_{d}[n], & 0 \leq n \leq N-1 \\\0, & \text { otherwise }\end{array}\right.\\] That is, simple truncation gives the best mean square approximation to a desired frequency response for a fixed value of \(N\).

In the design of digital filters, we often choose a filter with a specified magnitude characteristic that has the shortest duration. That is, the impulse response, which is the inverse Fourier transform of the complex frequency spectrum, should be as narrow as possible. Assuming that \(h[n]\) is real, we wish to show that if the phase \(\theta(\omega)\) associated with the frequency response \(H\left(e^{j \omega}\right)\) is zero, the duration of the impulse response is minimal. Let the frequency response be expressed as \\[H\left(e^{j \omega}\right)=\left|H\left(e^{j \omega}\right)\right| e^{j \theta(\omega)}\\] and let us consider the quantity \\[D=\sum_{n=-}^{\infty} n^{2} k^{2}[n]=\sum_{n=-x}(n h[n])^{2}\\] to be a measure of the duration of the associated impulse response \(h[n]\) (a) Using the derivative property of the Fourier transform and Parseval's relation, express \(D\) in terms of \(H\left(e^{j \omega}\right)\) (b) By expressing \(\left.H\left(e^{j \omega}\right) \text { in tems of its magnitude } | H\left(e^{j \omega}\right)\right\\}\) and phase \(\theta(\omega),\) use your result from part (a) to show that \(D\) is minimized wben \(\theta(\omega)=0\).
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