© 2021 GMU CYSE-220 CYSE-220 Systems Modeling Chapter 08: System Analysis in the Time Domain George Mason University © 2021 GMU CYSE-220 • Let us to pull together our modeling knowledge, our analytical, and computer solution methods. • The purpose of this chapter is to integrate this knowledge with emphasis on understanding system behavior in the time domain. System Analysis in the Time Domain © 2021 GMU CYSE-220 • The forcing functions commonly used to model real inputs or to test a system’s response in the time domain are: • The impulse, • the step, and • the ramp functions. • The impulse models a suddenly applied and suddenly removed input. • The step function models a suddenly applied input that remains constant. • The ramp models an input that is changing at a constant rate. • In this chapter we will show how to analyze systems subjected to these inputs. System Analysis in the Time Domain © 2021 GMU CYSE-220 • The transfer function and the characteristic polynomial are the principal tools for analyzing linear system response, and so it is now appropriate to review these concepts. System Analysis in the Time Domain © 2021 GMU CYSE-220 • Consider the following general form of a single, constant-coefficient, linear ordinary differential equation: System Analysis in the Time Domain where an = 0 and m ≤n. The response (the output) is y(t), and f (t) is the forcing function (the input). From the initial value properties of the Laplace transform, assuming that y(t) = 0 for t < 0, we have In our applications, the initial values (at t =0) of the input f (t) and its derivatives are zero; thus Applying the Laplace transform to the differential equation gives © 2021 GMU CYSE-220 System Analysis in the Time Domain © 2021 GMU CYSE-220 • Let us examine the response of first-order systems to step and ramp inputs. • The free response of a model is its solution in the absence of an input. We have seen that the free response of the model can be obtained with the substitution method or by using the Laplace transform. With f (t) = 0, the latter method gives where v(0) is the initial value of the response v(t), and m and c are constants. • When c/m <0 the solution grows exponentially; this is the unstable case. • If c/m = 0, the model is neutrally stable, and v(t) = v(0). • If c/m is positive, the model is stable, and the solution decays exponentially. 8.1 RESPONSE OF FIRST-ORDER SYSTEMS © 2021 GMU CYSE-220 • For the stable case, we introduced a new parameter τ for • The free response of can be written as: 8.1.1 THE TIME CONSTANT The free response © 2021 GMU CYSE-220 • For the stable case, we introduced a new parameter τ for • The free response of can be written as: 8.1.1 THE TIME CONSTANT The free response © 2021 GMU CYSE-220 • The new parameter τ has units of time and is the model’s time constant. • It gives a convenient measure of the exponential decay curve. • After a time equal to one time constant has elapsed, v has decayed to 37% of its initial value. We can also say that v has decayed by 63%. • At t = 4τ , v(t) has decayed to 2% of its initial value. At t = 5τ , v(t) has decayed to 1% of its initial value. Parameter τ © 2021 GMU CYSE-220 First Order Systems Having the Model Form • In this form, the time constant may be identified as τ = a/b. • This form can be used to determine the time constant expressions for each of the systems shown in Figure 8.1.2. © 2021 GMU CYSE-220 First Order Systems Having the Model Form Figure 8.1.2. • In this form, the time constant may be identified as τ = a/b. • This form can be used to determine the time constant expressions for each of the systems shown in Figure 8.1.2 © 2021 GMU CYSE-220 • The step response of the model ሶ + = , which has no input derivative, can be obtained with the substitution method or the Laplace transform. • Using the transform we obtain: where F is the magnitude of the step input. Solve for V(s) to obtain This gives: Note that the response is the sum of the free and the forced responses. 8.1.2 STEP RESPONSE OF A FIRST-ORDER MODEL … © 2021 GMU CYSE-220 • The time constant is useful also for analyzing the response when the forcing function is a step. • We can express the solution in terms of the time constant τ by substituting c/m = 1/τ to obtain • where vss = F/c. • The solution approaches the constant value F/c as t →∞. This is called the steady-state response, denoted vss . STEP RESPONSE Figure 8.1.3 © 2021 GMU CYSE-220 • The response is plotted in Figure 8.1.3 for v(0) = 0. • At t = τ , the response is 63% of the steady-state value. • At t = 4τ , the response is 98% of the steady-state value, and • at t = 5τ , it is 99% of steady state. • Thus, because the difference between 98% and 99% is so small, for most engineering purposes we can say that v(t) reaches steady state at t =4τ , although mathematically, steady state is not reached until t=∞. STEP RESPONSE Figure 8.1.3 © 2021 GMU CYSE-220 • If v(0) ≠ 0, the response is shifted by v(0)e−t/τ. • At t =τ , 37% of the difference between the initial value v(0) and the steady-state value remains. • At t = 4τ , only 2% of this difference remains. • Figure 8.1.4 shows the complete response. Response Figure 8.1.4 © 2021 GMU CYSE-220 • For the stable case (c/m > 0), we can separate the step response into the sum of a term that eventually disappears (the transient response) and a term that remains (the steady-state response). • Referring to • • we see that 8.1.3 TRANSIENT AND STEADY-STATE RESPONSES © 2021 GMU CYSE-220 Table 8.1.1 Free, step, and ramp response of © 2021 GMU CYSE-220 Step response for two values of c © 2021 GMU CYSE-220 • When using the time constant to compare the response times of two systems, the comparison should be done only for systems whose steady- state responses are identical. • Note that for the model: because τ =m/c, a small value of the damping constant c produces a large time constant, which indicates a sluggish system. • The result is counter to our intuition, which tells us that a small amount of damping should correspond to a system with a fast response. • This counterintuitive result is explained by noting that a larger steady-state response F/c results from a smaller damping constant. • Thus, for a large value of c, even though the time constant is smaller, the steady-state response is smaller, and thus it takes less time to reach a specific steady-state value. 8.1.4 RESPONSE TIME AND THE TIME CONSTANT © 2021 GMU CYSE-220 Response time Figure 8.1.5 for v(0) = 0 with two values of c. © 2021 GMU CYSE-220 • The step function is an approximate description of an input that can be switched on in a time interval that is very short compared to the time constant of the system. • A good example of a step input is the voltage applied to a circuit due to the sudden closure of a switch. 8.1.5 THE STEP FUNCTION APPROXIMATION © 2021 GMU CYSE-220 • The impulse response of is found as follows, where A is the impulse strength, or the area under the impulse versus time curve. • We can see that the effect of the impulse is to increase the effective initial condition by A/m. 8.1.6 IMPULSE RESPONSE © 2021 GMU CYSE-220 • Figure 8.1.7 shows a circuit representation of a telegraph line. • The resistance R is the line resistance and L is the inductance of the solenoid that activates the receiver’s clicker. • The switch represents the operator’s key. Assume that when sending a “dot,” the key is closed for 0.1 s. • Using the values R = 20 and L = 4 H, obtain the expression for the current i (t) passing through the solenoid. EXAMPLE 8.1.2: Analysis of a Telegraph Line Figure 8.1.7 © 2021 GMU CYSE-220 EXAMPLE 8.1.2: Analysis of a Telegraph Line From the voltage law we have where vi (t) represents the input voltage due to the switch and the 12-V supply. • We could model vi (t) as a rectangular pulse of height 12 V and duration 0.1 s, but the differential equation (1) is easier to solve if we model vi (t) as an impulsive input of strength 12(0.1) = 1.2 V・ s. • This model can be justified by the fact that the circuit time constant, L/R = 4/20 = 0.2, is greater than the duration of vi (t). • Thus we model vi (t) as vi (t) = 1.2δ(t). • The Laplace transform of equation (1) with i (0) = 0 gives © 2021 GMU CYSE-220 • The ramp function models an input that is changing at a constant rate. Thus a ramp input function has the form f (t) = mt, where m is the slope of the ramp. • We can obtain the response to the ramp using the Laplace transform. 8.1.7 RAMP RESPONSE © 2021 GMU CYSE-220 • You can obtain a list of all the performance parameters with the stepinfo function. • After you have created the model in either transfer function form or state space form, you type stepinfo(sys), where sys is the model name. • For example, the model • has the transfer function: MATLAB Applications © 2021 GMU CYSE-220 MATLAB Note that stepinfo gives the overshoot as a percent, the settling time using the 2% criterion, and the rise time as the 10%–90% rise time. You can change the definition of the rise time to the 0–100% value by typing © 2021 GMU CYSE-220 MATLAB You can change the definition of the rise time to the 0–100% value by typing: You can also change the definition of the settling time from the default value of 2% to 5% by entering stepinfo(sys,’SettlingTimeThreshold’,0.05) The stepinfo function assumes zero initial conditions. The Overshoot is calculated using the steady state response as the reference value, just as explained in Section 8.3. Here it is calculated as © 2021 GMU CYSE-220 Problem 8.3 © 2021 GMU CYSE-220 Problem 8.3a Solution x(0) = 6 m= 16, c= 14, f=0 => No force response Time constant τ = m/c = 16/14 = 8/7 c/m > 0, Stable © 2021 GMU CYSE-220 Problem 8.3b Solution x(0) = 3 m= 12, c= 5, f=15 => force response F/c = 15/5 = 3 τ = m/c = 12/5 Free Response c/m > 0, Stable © 2021 GMU CYSE-220 Problem 8.3d Solution x(0) = 9 m= 7, c= -5, f=0 => no force response F/c = 0 τ = 7/-5 = - 7/5 < 0 No time constant is defined because the model is unstable Free Response c/m < 0, unStable © 2021 GMU CYSE-220 Problem 8.5 © 2021 GMU CYSE-220 Problem 8.5a Obtain the steady-state response of each of the following model, and estimate how long it will take the response to reach steady-state. m=5, c=5, τ = m/c=6/5 1/τ=5/6 F=20, Vss = F/c = 20/5=4, Complete response Vss =4, the initial condition does not affect the steady-state response t = time it will take the response to reach steady-state = 4τ = 4(6/5)=24/5 x(t)= 4(1-e-5t/6)
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