ME 375 System Modeling and Analysis. Homework 11 Solution. Out: 18 November 2011 Due: 30 November 2011 = + +

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1 Out: 8 November Due: 3 November Problem : You are given the following system: Gs () =. s + s+ a) Using Lalace and Inverse Lalace, calculate the unit ste resonse of this system (assume zero initial conditions). A B C Y() s = = + + ( s+ )( s+ ) s s s+ s+ 5/4 /4 = + + s s+ s+ 5 t t yt () = e + e 4 4 b) Show the oles of the system on a comlex lane. Have Matlab do this for you: zma. Pole-Zero Ma Imaginary Axis (seconds - ) Real Axis (seconds - ) c) Based on the oles of this system, estimate the time constant and the % settling time? Calculate these. Don t read them from a time resonse lot (that s later). Hint: Focus on the slow decaying ole. Since the slow decaying ole is at -, τ = / sec. = 4τ = sec ts

2 Out: 8 November Due: 3 November d) Plot the unit ste resonse using Matlab. (Hint: You can first use the command ste. You can then also lot your solution in art a and comare with what ste gives you. That way you can check your answer in art a. But turn in one lot.) Ste Resonse Amlitude Problem (continues from roblem ): a) Now you are interested in achieving a settling time that is less than sec, and you realize that you have to use feedback control to do this. Draw the block diagram of the system with a roortional feedback controller (K ) and calculate the closed-loo-transfer function (CLTF). + R(s) - K s + s+ Y(s) Y() s KGs ( ) K K = = = R s K G s s s K s s K ( ) + ( ) ( + )

3 Out: 8 November Due: 3 November b) Determine the oles of the CLTF with K =,.,.4,.6,.8,,, 4, and lot them on a comlex lane. Use zma in Matlab (hint: use the command hold on. You can tye hel hold and read about what the command does). Pole-Zero Ma 8 6 Imaginary Axis (seconds - ) Real Axis (seconds - ) c) Now determine the K value that will give you just about sec settling time (Make your slow decaying ole settle at sec). 4 ts = = / τ = 4 ole = 4 / τ K =.6 gives the oles -4 and -8. d) Now lot the unit ste resonse of the resulting system with this K. Comare the settling time with art d of roblem. Is the settling time better than that in Problem? But now there is a roblem that didn t exist in the original oen loo system. What is it?

4 Out: 8 November Due: 3 November.4 Ste Resonse Amlitude Yes. The % settling time is better ~ sec: (it used to be sec). But now, the final value is far from the desired reference which is (there is a steady state error.) e) Now suose you want to hel mitigate the steady-state-error roblem. Look at your closed loo transfer function and calculate the steady state resonse to a unit ste using the final value theorem. What do you need to do to K to hel with the steady state error roblem? K = = yss lims s s + s + + K + K K You want K as high as you can to reduce steady state error. Problem 3 (continues from roblem ). a) Crank your K way u: make it. Now lot the ste resonse again. What haened to the steady state error? But what s wrong now? So far, what is your conclusion on the effect of K on the steady state error and %OS?

5 Out: 8 November Due: 3 November Ste Resonse Amlitude Steady state error has reduced significantly: our final value is about. But now, we have a lot of overshoot (we now have oscillations). K hels reduce steady state error but introduces oscillations and increases overshoot. b) Now make the K ridiculously high:,,. Let s see the ste resonse again. Does increasing the K from to,, shorten the settling time even more (comare this lot with the revious lot)? But back in roblem, increasing it from to just around.6 made an imortant difference. Why is K not shortening the settling time anymore? Answer this question by first looking back at the ole-zero lot you made in Problem art b and then calculating the theoretical % settling times in this roblem s art b and a.

6 Out: 8 November Due: 3 November Ste Resonse Amlitude Increasing K only imroves (shortens) the settling time when the oles are still real. Once they become imaginary, increasing K simly ushes the oles u and down the imaginary axis (doesn t change their real comonent), and hence doesn t affect their decaying rate. Theoretical t s in both cases is 4/6 =.67 s (-6 is the oint achieved with K =.8. Poles break u from the real axis with K being any higher.)