ECE-320 Linear Control SystemsHomework 8Due: Tuesday November 2, 2004If matrix P is given asP ="a bc d#ThenP−1=1ad − bc"d −b−c a#and the determinant of P is given by ad − bc.The general for for writing a continuous time state variable system is˙x(t) = Ax(t) + Bu(t)y(t) = Cx(t) + Du(t)Now assume we are using state variable feedback, so that u(t) = kpfr(t) − kx(t). Here r (t) isour new reference input, kpfis a scaling factor, and k = [k1k2] is the feedback gain matrix.With this state variable feedback, we have the system˙x(t) = Ax(t) + B(kpfr(t) − kx(t))or˙x(t) =˜Ax(t) +˜Br(t)where r(t) is the new input. For D = 0, the transfer matrix is given byG(s) = Ch(sI −˜A)−1i˜BFor each of the systems below,• determine the transfer function when there is state variable feedback• determine if k1and k2exist to allow us to place the poles arbitrarily. That is, can wemake the denominator look like s2+ a1s + a2for any a1and any a2.You can use Maple to check your answers, but I expect you to be able to do this withoutMaple.1 LetA ="1 01 1#, B ="01#, C = [0 1] , D = 0Ans. G(s) =(s−1)kpf(s−1)(s−1+k2)2 LetA ="0 10 1#, B ="01#, C = [0 1] , D = 0Ans. G(s) =skpfs2+(k2−1)s+k13 LetA ="0 11 1#, B ="01#, C = [1 0] , D = 0Ans. G(s) =kpfs2+(k2−1)s+(k1−1)4 LetA ="0 11 0#, B ="11#, C = [0 1] , D = 0Ans. G(s) =(s+1)kpf(s+k1)(s+k2)−(k1−1)(k2−1)5 Preparation for Lab:Consider the following model of the two DOF system we will be using.c cm mkk k1231212F(t)x (t)1x (t)2a) Draw freebody diagrams of the forces acting on the two masses.b) The equations of motion for the two masses can be writtenm1¨x1+ c1˙x1+ (k1+ k2)x1= F + k2x2(1)m2¨x2+ c2˙x2+ (k2+ k3)x2= k2x1(2)If we define q1= x1, q2= ˙x1, q3= x2, and q4= ˙x2, show that we get the following stateequations˙q1˙q2˙q3˙q4=0 1 0 0−³k1+k2m1´−³c1m1´ ³k2m1´00 0 0 1³k2m2´0 −³k2+k3m2´−³c2m2´q1q2q3q4+01m100FIn order to get the A and B matrices, we need to determine all of the quantities in the abovematrices.c) Now we want to rewrite equations 1 and 2 in terms of ζ1, ω1, ζ2, and ω2as¨x1+ 2ζ1ω1˙x1+ ω21x1=k2m1x2+1m1F (3)¨x2+ 2ζ2ω2˙x2+ ω22x2=k2m2x1(4)We will get our initial estimates of the parameters ζ1, ω1, ζ2and ω2using the log-decrementmethod. Assuming we measure these parameters, show how A2,1, A2,2, A4,3and A4,4can bedetermined.d) By taking the Laplace transforms of equations 3 and 4, show that we get the followingtransfer functionX2(s)F (s)=³k2m1m2´(s2+ 2ζ1ω1s + ω21)(s2+ 2ζ2ω2s + ω22) −k22m1m2(5)e) It is more convenient to write this transfer function asX2(s)F (s)=³k2m1m2´(s2+ 2ζaωas + ω2a)(s2+ 2ζbωbs + ω2b)(6)By equating coefficients of powers of s in the denominators in these two transfer functions(equations 5 and 6), you should be able to write down four equations. The equations cor-responding to the coefficients of s3, s2, and s do not seem to give us any new information,but they will be used later to get consistent estimates of ζ1and ω1. The equation for thecoefficients of s0will give us a new relationship fork22m1m2in terms of parameters we will bemeasuring.f) We will actually be fitting the frequency resp onse data to the following form of the transferfunction:X2(s)F (s)=K2(1ω2as2+2ζaωas + 1)(1ω2bs2+2ζbωbs + 1)(7)What is K2in terms of the parameters given in equation 6?g) Using equation 6 and the Laplace transform of equation 4, show that we can writeX1(s)F (s)=1m1(s2+ 2ζ2ω2s + ω22)(s2+ 2ζaωas + ω2a)(s2+ 2ζbωbs + ω2b)(8)h) This equation is more convenient to write in the formX1(s)F (s)=K1(1ω22s2+2ζ2ω2s + 1)(1ω2as2+2ζaωas + 1)(1ω2bs2+2ζbωbs + 1)(9)What is K1in terms of the quantities given in equation 8?i) Show thatA4,1=k2m2=K2K1ω22(10)j) Show thatA2,3=k2m1=ω21ω22− ω2aω2bA4,1(11)k) All that’s left is to find1m1, which is b2. It’s important to understand that this parameteralso includes “scaling” on F (s). Now assume we look at the closed loop response to a simpleproportional type controller. Hence we have the system shown below:-½¼¾»-kp-X2F-6+-For a step response of amplitude Amp , show that the steady state value of x2(t), x2,ssisx2,ss=K2kpAmp1 + K2kp(12)(Hint: It’s easiest to use equation 7 for X2/F )and that by rearranging this equation we getb2=1m1=x2,sskp(Amp − x2,ss)ω2aω2bA4,1(13)Summary1) Fit frequency response data toX2(s)F (s)=K2(1ω2as2+2ζaωas + 1)(1ω2bs2+2ζbωbs + 1)(14)this will give us estimates for K2, ζa, ωa, ζb, and ωb.2) Using the above parameters, fit frequency response data toX1(s)F (s)=K1(1ω22s2+2ζ2ω2s + 1)(1ω2as2+2ζaωas + 1)(1ω2bs2+2ζbωbs + 1)(15)This will give us estimates for K1, ζ2, ω2.3) Using the relationships derived in part 8, find the values of ζ1and ω1consistent with therest of the parameters. Note that if the estimate of ζ1is less than one fourth the estimatedetermined by the log-decrement method (initial estimate), the value of ζ1is set to onefourth the initial estimate.4) Estimate all of the parameters in A.5) Look at the step response to estimate
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