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EEGR 431: Linear Control Systems Matlab Assignment 2 (Due Monday, November 5, 2014) Review the computer-aided learning examples in the text: (p.57) C1.7-1.9, (p.125) C2.1, C2.3, C2.5 A: For problems D2.8 (a) and (c) from the text, in each case 1. Plot the step response, using matlab. 2. Calculate the damping ratio. 3. On your plots, indicate and determine the values of Mp, Tp, Tr and Ts (as specified by problem D2.8). 4. Using the above variables, discuss any tradeoffs by comparing the speed of response and overshoots, etc. B : Effect of varying gain in a closed-loop system (Proportional control) Consider the feedback system shown, consisting of a plant G(s) in cascade with a gain K (proportional controller). + K G(s) R(s) - Y(s) Controller Plant Let the Plant be G(s) = 1/ (s(s+2)) 1) Is this plant stable? Calculate and plot the step response of this plant (i.e. Open-loop step response, with K=0 in closed-loop characteristic polynomial). 2) Find the transfer function of this closed-loop system. 3) Use matlab to plot the closed loop step responses for the following gains: K= -1, ¼ , 1, 4 , 16. 4) Find the damping ratio, , for each K-value and use this to explain the type of response in each case. 5) For the K-value, which gives an underdamped response, determine the values of Mp, Tp, Tr and Ts. R(S) = 52 + 16 Then =S SY($) = $T(s)R(S) = 10(s + 1) 6s (s + 2)(s + 3) (s2 + 16) To employ the step command in vector form, the "conv” (meaning convolve) command may be used to multiply polynomial vectors snum=60*conv([1 0 0], [1 1]) den1=conv([1 2], [1 3]) den=conv(deni, [1 0 16]) sy=tf(snum, den) step(sy) Using the “zpk” form may sometimes be easier: sy=zpk ( [0 0 -1], [-2 -3 j*4 -j*4], 60) step(sy) C2.1 Use the step command to get the step response of the following: 5 (a) T(s) = (s + 4)2 5(s + 6) (b) T(s) = (s +1+34)(s +1 -04) Ans. (a) ta=zpk ( [], [-4 -4], 5) step(ta) (b) num=[5 30] den=[1 2 17] tb=tf(num, den) step(tb) C2.2 Use the step command to get the time response of the following: 32 (a) Y(S) = (s + 2)2 (s2 + 16) (b) Y(s) = T(s)R(s), where 48 ESC 1 a 2 Q HIGHER-ORDER SYSTEM RESPONSE D2.8 Using the curves given in the text, find approximately the percent overshoot, peak time, rise time, and settling time of the following systems when driven by a step input. = (a) 100 T(s) = S2 + 4s + 100 Ans. 54%, 0.32, 0.12, 1.1 (b) 49 T(S)= S2 + 4s + 49 Ans. 40%, 0.47, 0.18, 1.5 (c) 60 T(S) 2s2 + 8s + 30 Ans. 18%, 0.95, 0.44, 1.29 (d) 75 T(S) S2 + 3s + 20 Ans. 32%, 0.75, 0.34, 1.7 It is important to realize the amount of information that can be obtained from the T(S) function. In Drill Problem D2.8, the percent overshoot, peak time, rise time, and settling time are found from the Laplace-transformed form of T(s). In most control system designs, it is not necessary to convert the Laplace-transformed T(s) unction into the tim Ti be analyzed and designed by stay- CONTINUOUS-TIME SYSTEM RESPONSE Ans. (a) sya=zpk ( [0], [-2 -2 -j*4 j*4] , 32) 126 step(sya) (b) snum=[48 0]*3 deni-conv([1 3), [1 1 16]) den=conv(deni, [1 3]) stb=tf (snum, den) step(stb) independer d²y dt² As can be decays to must all zero, and way and a m transforms. These symbolic capabilities of MATLAB provide another In Chapter 1, we used symbolic techniques to get Laplace and inverse Laplace using the "ezplot" command to get the time response. If we want to solve T(S and the Y(s) = T(s)R(S) 10(s + 1) Y(s) = 52 +58 +6 r(t) = 6 cos 4t Y we can use symbolic commands syms st The ch T=10*(s+1)/(s^2+5*s+6) R=laplace (6*cos (4*t)) Y=R*T ysilaplace (Y) ezplot (y, [O 10]) s1 with r We need to specify the time range (here from 0 to 10 s). That range may require some trial-and-error work. S 2.3. C2.3 Find the time response of C2.2(a) and C2.2(b) using symbolic methods and the "ezplot" command. Ans. (a) syms st Ya=32/(s+2) 2/((s^2+16) ya=ilaplace(Ya) ezplot(ya, [0 15]) (b) syms s t T=48/(s+3)/(s^2+s+16) r=3*exp(-3*t) R=laplace (r) If th unit is Yb=T*R yb=ilaplace (Yb) ezplot (yb, [O 2]) 2.3 Response of Second-Order Systems wh w W 2.3.1 Time Respon 0.4 7 8 0.2 9 6 5 4 3 0 0 1 2 with constant transfer Figure 2.8 Normalized step response of a second-order system function numerator. (b) (b) numb=[10 8] denb=[1 4 4] tb=tf (numb, denb) step(tb) ( (c) numc=[-3 17] denc=[1 2 17] tc=tf(numc, denc) step(tc) C2.5 Use MATLAB Commands to obtain Figure 2.8. For convenience, le Wn = 1, so the time axis becomes t = @nt. Ans. num=1 deno=[1 0 1] den1=[1 .2 1] den3=[1 .6 1] den5=[1 1 1] den7=(1 1.4 1] den9=(1 1.8 1] den 10=f1.2 1] to=tf(num, deno) t1=tf(num, deni) t3=tf(num, den3) t5=tf(num, den5)
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