Homework 7  Solutions


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1 Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the gain, K, for the uncompensated system to operate with a 30% overshoot. Find the peak time and K v for the uncompensated system. Design a laglead compenator to decrease the peak time by a factor of 2, decrease the percent overshoot by a factor of 2, and improve the steadystate error by a factor of 30. Specify all poles, zeros and gains. Solution: Searching the 30% overshoot line (ζ = 0.358, arcsin (0.358) = ) for angle sum of 180 yields pole at, j3.818 and gain, K = T p = π = π = seconds. ω d K v = lim s 0 sg(s) = = (c) Required specifications: %OS = 15% T p = seconds K v = = Lead Design: Using the required specifications for %OS and T p, we can calculate the damping ratio and the natural frequency: ln(%os/100) ζ = π 2 + ln 2 (%OS/100) = ω d = π T p = ω n 1 ζ 2 = Hence ω n = Thus, the desired pole is located at ζω n + jω n 1 ζ 2 = j7.634 Assume a lead compensator zero at 5. Summing the angles of the uncompensated system s poles as well as the compensator zero at 5 yields Therefore, the compensator pole Rev. 1.0, 04/14/ of 13
2 must contribute ( ) = 8.8. Using the geometry below, tan(8.8 ) = p c Hence, p c = The compensated openloop transfer function is now: K s(s + 11)(s ) Evaluating the gain for this function at the desired pole, we get K = Lag Design: The lead compensated K v is given by, K v = lim s 0 sg(s) = = Thus, we need an improvement over the lead compensated system of /7.469 = Choosing p c = 0.001, we get z c = Thus, the compensator is given by G lag (s) = s s The final openloop transfer function is 2. Compensators (9 points) The unity feedback system shown in Problem 1, with G(s) = to meet the following specifications: 4430(s ) s(s + 11)(s )(s ) K (s 2 is to be designed + 4s + 8)(s + 10) Overshoot: Less than 25% Settling Time: Less than 1 second K p = 10 Do the following: (c) Evaluate the performance of the uncompensated system operating at 10% overshoot. Design a passive compensator to meet the desired specifications. Use MATLAB to simulate the compensated system. Compare the response with the desired specifications. Solution: Searching along the 10% overshoot line (ζ = 0.591) the operating point is found to be j2.53 with K = A third pole is at Thus, the estimated performance before Rev. 1.0, 04/14/ of 13
3 EE C128 / ME C134 Spring 2014 HW7  Solutions compensation is: 10% overshoot, Ts = 4/1.85 = 2.16 seconds, and Kp = 21.27/(8 10) = Lead Design: Since the overshoot of the uncompensated system is already lower than the desired specification, we can keep the same damping ratio (ζ = 0.591). Place compensator zero arbitrarily at 3. Now, with the required specifications, we can find the natural frequency: 4 4 = = ωn = ζts p Thus the desired pole is located at ζωn + jωn 1 ζ 2 = 4 + j5.46. The angular contribution of the system poles and compensator zero at the design point is Thus, the compensator pole must contribute = Using the geometry below, tan(13.04 ) = Hence, pc = pc 4 K(s + 3) + 4s + 8)(s + 10)(s ) Evaluating this function at 4+j5.46 yields K = 1092 with higher order poles at and Lag Design: For the lead compensated system, Kp = Thus, we need an improvement of a factor of 10/1, 485 = Choosing the compensator pole at 0.01, we get Glag (s) = s Finally, the equivalent forward path transfer function is: s Thus, the compensated openloop transfer function is Gc (s) = Rev. 1.0, 04/14/2014 (s2 (s2 1092(s + 3)(s ) + 4s + 8)(s + 10)(s )(s ) 3 of 13
4 EE C128 / ME C134 Spring 2014 HW7  Solutions (c) System Response: Uncompensated system Compensated system Rev. 1.0, 04/14/ of 13
5 3. Bode Form (3 points) Express the following transfer functions in their Bode forms. 1 G(s) = s(s + 2)(s + 4) G(s) = (c) G(s) = Solution: Thus, (s + 5) (s + 2)(s + 4) (s + 3)(s + 5) s(s + 2)(s + 4) G(s) = G(jω) = M(ω) = 1 s(s + 2)(s + 4) 1 6ω 2 + j(8ω ω 3 ) 1 (8ω ω 3 ) 2 + (6ω 2 ) 2 Thus, φ(ω) = π arctan 8 ω2 6ω G(s) = G(jω) = (s + 5) (s + 2)(s + 4) jω + 5 ω 2 + 6jω ω 2 M(ω) = (8 ω 2 ) ω 2 (c) Thus, φ(ω) = arctan ω 5 G(s) = G(jω) = (s + 3)(s + 5) s(s + 2)(s + 4) (15 ω2 ) + 8jω 6ω6 2 + j(8ω ω 3 (15 ω M(ω) = 2 ) ω 2 36ω 4 + (8ω ω 3 ) 2 φ(ω) = arctan 8ω 8 ω2 arctan 15 ω2 6ω Rev. 1.0, 04/14/ of 13
6 EE C128 / ME C134 Spring 2014 HW7  Solutions 4. Sketching Bode Plots (9 points) For each of the functions given in Problem 3, sketch the Bode asymptotic magnitude and asymptotic phase plots. Solution: Rev. 1.0, 04/14/ of 13
7 EE C128 / ME C134 Spring 2014 HW7  Solutions 5. Nyquist Plots (6 points) Sketch the Nyquist diagram for each of the systems given below, and using the Nyquist Criterion, determine whether or not each system is stable. Solution: P = 0, N = 0 Using Nyquist Criterion, Z = P N = 0. Thus, the system is stable. Rev. 1.0, 04/14/ of 13
8 EE C128 / ME C134 Spring 2014 HW7  Solutions P = 0, N = 2 Using Nyquist Criterion, Z = P N = 2. Thus, the system is unstable. 6. Bode Plot Analysis (6 points) For the Bode plots shown below, determine the transfer functions by hand, and confirm your answers via MATLAB. Rev. 1.0, 04/14/ of 13
9 Solution: Step 1: Initial slope (ω 0) is 20 db/decade. Thus, we can say that there exists a pure integration, 1/s. G(s) = K s Step 2: Final slope (ω ) is 40 db/decade. Looking for the intersection of the asymptotes K yields that the pole is at 5.3 rad/s. G(s) = s(s + 5.3) Step 3: 1/s at ω = 0.01 rad/s should contribute +40 db/decade. From the magnitude plot, we see that G(j0.01) = 66.5 db/decade. Thus, writing the function in bode form: G(s) = K 1 we see that K/5.3 contributes 26.5 db/decade ), s( s 5.3 Therefore, the transfer function is given by: 20 log K 5.3 = 26.5 K = G(s) = s(s + 5.3) Rev. 1.0, 04/14/ of 13
10 Step 4: MATLAB Simulation Step 1: By inspection, ω n = 10 rad/s and there exists a second order term of the form Kω 2 n s 2 + 2ζω n + ωn 2. And there is a zero at 1 rad/s. Therefore, there is a (s + 1) term in the numerator. 100K(s + 1) Step 2: Let ζ = 0.3, ω n = 10rad/s, we get: s 2 + 6s To compensate for the slope near the natural frequency, add a pole at 8.5 db. Therefore, a (s + 8.5) term exists in the denominator. Step 3: To compensate for the steep slope after the natural frequency, we add a zero at 23. Thus, a (s + 23) term exists in the numerator. Since these terms do not affect the gain of the bode plot, we need to make the DC gain of these terms equal to 1. Thus, we need a 8.5/23 850(s + 1)(s + 23) term. Therefore, G(s) = K 23(s 2 + 6s + 100)(s + 8.5) Rev. 1.0, 04/14/ of 13
11 Step 4: Check the Bode Plot to add appropriate gain: At ω 0, G(jω) = 10. Therefore, 20 log K = 10 or K = Therefore, 850(s + 1)(s + 23) G(s) = (s 2 + 6s + 100)(s + 8.5) = (s + 1)(s + 23) (s 2 + 6s + 100)(s + 8.5) Step 5: MATLAB Simulation Rev. 1.0, 04/14/ of 13
12 7. Bode Plot Analysis (6 points) The Bode plots for a plant, G(s), used in a unity feedback system are shown below. Do the following: Find the gain margin, phase margin, zero db frequency, 180 frequency, and the closedloop bandwidth. Use your results in Part to estimate the damping ratio, percent overshoot, settling time, and peak time. Solutions: From the Bode Plots: Gain Margin = db; phase margin = ; zero db frequency = rad/s; 180 frequency = rad/s; bandwidth (@ 7 db point) = 3.8 rad/s. Using the following relation (10.73 in Nise): φ M = tan 1 2ζ 2ζ ζ 4 we get, ζ = Using the damping ratio, we find the %OS = 17.93%. Using the following relation (10.55 in Nise): ω BW = 4 (1 2ζ T s ζ 2 ) + 4ζ 4 4ζ Rev. 1.0, 04/14/ of 13
13 we get, T s = 2.84 s. Using the following relation (10.56 in Nise): we get T p = 1.22s. π ω BW = (1 2ζ 2 ) + 4ζ 4 4ζ T p 1 ζ 2 Rev. 1.0, 04/14/ of 13
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