ECE-320: Linear Control Systems Homework 5 Due: Wednesday April 19 at 3:30 PM 1) For the following problem, assume we are using the following control system + -()cGs()pGs where the plant is given by 211()429(25)(25pGsss s js j==++ +− ++) For the following controllers, sketch the root locus with arrows showing the direction of travel as k increases. If there are any poles going to zeros at infinity, you need to compute the centroid of the asymptotes (cσ) and the angles of the asymptotes. You may (and should) check your answers with Matlab (use the rlocus command), but you need to do this by hand. a) Gs ( proportional (P) controller) ( )ck= b) ()ckGs (an integral (I) controller) s= c) (()cks z)s+=Gs (a proportional + integral (PI) controller) Write the centroid cσas a function of z. For what values of z will the two asymptotes be in the right half plane? (For plotting purposes, assume z is equal to 2.) d) Gs (a proportional+derivative (PD) controller) ( ) ( )cks z=+e) 1()(()cks z s z2)s++=Gs (a proportional+integral+derivative (PID) controller) Sketch this for the case where both zeros are real and then when both zeros are complex conjugates.f) (()()cks zGssp+=+) ( a lead controller, ) Write an expression for pz>cσas a function of the distance between the pole and the zero, lpz=−. What happens to the asymptotes as l gets larger? (For plotting purposes, assume p is 5 and z is 1.) 2) For the following problem, assume we are using the following control system + -()cGs()pGs where the plant is given by 1()3pGss=+ Want to determine if it is possible to meet the following constraints • 1secsT ≤• Steady state error for a unit step input < 0.1 with P, I, PD, and PID (real zeros and complex conjugate zeros) controllers. For each of these controllers, you need to • Sketch the root locus with arrows showing the direction of travel as increases. If there are any poles going to zeros at infinity, you need to compute the centroid of the asymptotes (kcσ) and the angles of the asymptotes. • State whether it is possible, based only on the closed loop pole locations, to meet the constraints. If it is necessary to put constraints on , , or to meet the constraints, you must specify them. (At this point you can only put conditions on k to meet the steady state error constraints.) kpz • Determine if it is possible for the output to oscillate (nonzero dω) You may (and should) check your answers with Matlab (use the rlocus command), but you need to do this by hand.3) A different approach to the root locus... Consider the following closed loop control system with a lead controller ( ), where k, p, and z are values to be determined. We would like the pz>dominant poles of the closed loop system to be at and the62j−± steady state error (for a unit step) to be less than or equal to 0.08. We will assume k, p, and z are positive values. + -()ks zsp++5(1)(2)ss++ a) Show that the closed loop poles are roots of the equation 510(1)(2)szksp s s⎛⎞⎛ ⎞++=⎜⎟⎜ ⎟+++⎝⎠⎝ ⎠ b) Since s is a complex variable, we can write the results for part (a) as a magnitude condition and a phase condition. Show that these conditions are given as magnitude condition: || 51|||1||2|szksps s+=+++ phase condition: ( ) 5 ( ) ( 1) ( 2) 180osz sp s s∠+ +∠−∠+ −∠+−∠+ =± c) Since we want the point to be on the root locus, show that when we evaluate the above expressions using we get the equations 62j−+62s=− + j magnitude condition: 22(6)44.817(6)4pkz−+=−+ phase condition: 1122tan tan 131.666ozp−−⎛⎞⎛⎞−=⎜⎟⎜⎟−−⎝⎠⎝⎠ Note that once we choose p and z that meet the phase condition, we can always find a k to meet the magnitude condition. Hence the phase condition is the most difficult to meet.d) Now use the "identity" () ()11 1121212tan tan tan1rrrrrr−− −⎛⎞−−=⎜⎟+⎝⎠ to show we can write the phase condition as 7.775 404.224zpz−=− Note that this identity does not take into account the two argument nature of the arctangent function. Hence the results can sometimes be off by 180 degrees. However, for most cases it is quite helpful. e) The numerator of the above function is zero when z = 5.144, and the denominator is zero when p = 4.224. Usually places where the numerator or denominator is zero indicate where there is some type of change. To answer the following questions, if may prove useful to put in a value of z and find the corresponding p value to determine what is wrong. Remember we want a lead controller and the dominant poles at 62j−± i) If z > 5.144 we will not get an acceptable lead controller. Why? ii) If 4.224 < z < 5.144 we will not get an acceptable controller. Why? iii) If z < 2, we will not get an acceptable lead compensator. Why? (Hint: Sketch a root locus plot) f) Show that the condition to meet the steady state error requirement can be written as 4.6kzp> g) We must determine k, p, and z to satisfy the requirements 2 4.224z<<7.775 404.224zpz−=− 22(6)44.817(6)4pkz−+=−+ 4.6kzp> Make a table showing your guess for z, the corresponding p and k, and then kz/p. If you think about your results this should not take very long. h) Determine the location of all of the closed loop poles and verify that the dominant poles are at (the Matlab command rlocus may help) 62j−±4) One of the things that will be coming up in lab more and more is the limitation of the amplitude of the control signal, or the control effort. This is also a problem for most practical systems. In this problem we will do some simple analysis to better understand why Matlab's sisotool won't give us a good estimate of the control effort for some types of systems, and why dynamic prefilters can often really help us out here. a) For the system below, ()pfGs+ -()cGs()pGs()Us()Rs show that U(s) and R(s) are related by () ()() ()1()()cpfpcGsG sUs RsGsGs=+ b) For many types of controllers, the maximum value of the control signal is just after the step is applied, at . Although most of the time we are concerned with steady state values and use the final value Theorem in the s-plane, in this case we want to use the initial value Theorem, which can be written as 0t+= 0lim ( ) lim ( )stut sUs+→∞→= If the system input is a step of amplitude A, show that () ()(0 ) lim1()(cpfspc)AGsG suGsGs+→∞=+ This result shows very clearly that the