Linear System Theory (Fall 2011): Homework 1. Solutions

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1 Linear System Theory (Fall 20): Homework Soltions De Sep. 29, 20 Exercise (C.T. Chen: Ex.3-8). Consider a linear system with inpt and otpt y. Three experiments are performed on this system sing the inpts (t), 2 (t) and 3 (t) for t In each case, the initial state at t = 0, x(0) is the same. The corresponding observed otpts are y (t), y 2 (t) and y 3 (t). Which of the following three predictions are tre if x(0) 0? a. If 3 = 2, then y 3 = y y 2. b. If 3 = 2 ( 2 ), then y 3 = 2 (y y 2 ). c. If 3 = 2, then y 3 = y y 2. Which of the above are tre if x(0) = 0? For each answer no, give a simple conterexample (nmerical or on paper). Soltion. The best soltion was given by Tsai-lin. If H is a linear system, it has state eqations sch as: ẋ = Ax B y = Cx The Laplace transform of the above by taking into accont the initial condition is (check yo basic control system corse notes): y(s) = C(sI A) B(s) C(sI A) x(0) Then, exploiting linearity, the following can be dedced:

2 Case. x(0) 0: a. Qestion: H( 2 ) =? H H 2? y = C(sI A) B C(sI A) x(0) y 2 = C(sI A) B 2 C(sI A) x(0) y 3 = C(sI A) B( 2 ) C(sI A) x(0) Hence, y 3 y y 3 if x(0) Answer: no. b. Qestion: H ( 2 2 ) =? H 2 H 2 2? y 3 = C(sI A) B 2 ( 2 ) C(sI A) x(0) = 2 C(sI A) B( 2 ) C(sI A) x(0) 2 y 2 y 2 = 2 [C(sI A) B C(sI A) x(0) Hence, y 3 = 2 (y y 2 ). Answer: yes! 2 [C(sI A) B 2 C(sI A) x(0) = 2 C(sI A) B[ C(sI A) x(0) c. Qestion: H( 2 )? = H H 2? The answer is the same as a. simply by replacing 2 with 2, i.e. no. Case 2. x(0) = 0: from the above, it is easy to see that the answer is yes for all a., b. and c. This is why the transfer-fnction representation mst assme that the system is initially relaxed.

3 Exercise 2 (C.T. Chen: Ex.3-4). The inpt and the otpt y of a system is described by: ÿ 2ẏ 3y = 2. What is the transfer fnction of the system? Find a state-space representation for the system. Soltion 2. Transfer fnction: y(s) (s) = 2s s 2s 3 Canonical controllable form: Canonical observable form: [ [ [ẋ 0 x = ẋ y = [ 2 [ x [ẋ = ẋ 2 [ y = [ 0 [ x [ x [ 0 [ 2 and there are other state-space representations.

4 Exercise 3 (C.T. Chen: Ex.3-23). Find the transfer fnction and a state-space representation of the network in the following figre. Simlate in MATLAB/Octave/Scilab the state-space representation for a constant A inpt. What happens to otpt and the internal state? Do yo think the transfer fnction is a good description of the system? V C2 i C2 F Z i L spply crrent i R Ω Ω F y = V C _ H V L _ Soltion 3. The circit eqations are (notice there are 3 differential eqations, so we expect a 3rd-order system): Choose the state variables Then i C i C2 =, i C2 i R =, i C = dv C dt, i C 2 = dv C 2 dt, V L i L = V C, V C2 = i R. x = V C = y, = V C2, x 3 = i L. ẋ = x 3, ẋ 2 =, ẋ 3 = x x 3. Ths, a state-space representation for the circit is: 0 0 ẋ = 0 0 x 0 0 y = [ 0 0 x The transfer fnction can be obtained from the state-space representation sing the formla H(s) = C(sI A) B, or by noticing that the relation between the inpt and

5 the otpt y is only dependent on the impedance Z of the parallel element across the otpt: Z = s //( s) = s s 2 s A simlation of the circit s response to a A step inpt at t =, with zero initial conditions at t = 0, is shown below. It can be seen that while the otpt voltage y = V C and the branch crrent i L are well-behaved, the crrent i C2 diverges. = V C2 x = V C = y x 3 = i L This means that by observing solely the otpt voltage, it is not possible to detect an exploding crrent (and voltage) across the inpt-side parallel element. Hence, the inptside element is not observable from the otpt. Electrically, this can be easily nderstood by noticing that the circit s transfer fnction is exactly the same as the one below, no matter what the inpt-side element is: spply crrent _ Ω H F y _

6 Exercise 4 (C.T. Chen: Ex.3-3 (modified)). Consider the below feedback system with sbsystems S and S 2 described by: (Sys ) [ẋ = ẋ 2 [ 2 0 y = [ 0 [ x [ [ [ x 4, 2 2 [ [ ; 2 and (Sys 2 ) [ [ẋ3 2 = 3, ẋ 4 [ [ [ y2 2 0 x3 =. y 3 x 4 [ r r 2 [ 2 _ Sys y [ y2 Sys 2 3 y 3 a. Withot sing any compter programme, write down the state-space representation of the feedback system, with the states (x,, x 3, x 4 ), the inpt (r, r 2 ) and the otpt y. b. Next, se MATLAB/Octave/Scilab to obtain the transfer fnctions of the sbsystems Sys and Sys 2, respectively. Then, calclate the transfer fnction of the feedback system (type help feedback at the command prompt). Print (or copy and paste) all the steps as yor answer. c. Explain why the state-space representation in qestion a. is of dimension 4, bt the transfer fnction in qestion b. is of dimension 2. What is missing? Soltion 4. a. A straight-forward way to obtain a state-space representation for the feedback system is to recycle all the state variables (x,, x 3, x 4 ) of the sbsystems, and se the formla given in class (or work it ot manally again) [ [ [ [Ẋ A B = C 2 X B R, () Ẋ 2 B 2 C A 2 B 2 D C 2 X 2 B 2 D Y = [ [ X C D C 2 D R, (2) X 2

7 specializing it to This yields:: ẋ ẋ 2 ẋ 3 ẋ 4 A 2 = [ = y = [ 0 x x 3 x 4 x x 3 x [ r r 2 b. The following are Scilab command history and reslts; similar reslts can be obtained in MATLAB: -->A=[-2 ; 0 -; B=[4 ; - 2; C=[0 ; D=[ -; -->A2=[0 0; 0 0; B2=[2;; C2=[2 0; -; D2=[0;0; ->sys=syslin( c,a,b,c,d) sys = sys() (state-space system:)!lss A B C D X0 dt! sys(2) = A matrix = sys(3) = B matrix = sys(4) = C matrix =. sys(5) = D matrix =. -. sys(6) = X0 (initial state) = sys(7) = Time domain =

8 c -->sys2=syslin( c,a2,b2,c2,d2) sys2 = sys2() (state-space system:)!lss A B C D X0 dt! sys2(2) = A matrix = 2.. sys2(3) = B matrix = sys2(4) = C matrix = sys2(5) = D matrix = sys2(6) = X0 (initial state) = c sys2(7) = Time domain = -->tf=ss2tf(sys) tf = s - s s s -->tf2=ss2tf(sys2) tf2 = 4 - s - s

9 It can be observed that we have transfer-fnctions matrices instead of scalar transfer fnctions, since Sys is 2-inpt--otpt, and Sys 2 is -inpt-2-otpt. The feedback system is therefor 2-inpt--otpt: ->tf/.tf2 ans = 2 2 s s - s s s 4s s The fact that the reslting transfer matrices of the feedback system are only second order, whereas the total nmber of states is 4, is de to the fact that both sbsystems have transfer matrices that are first order. This is becase both Sys and Sys 2 have redndant states. First, notice that y only depends on and (, 2 ), bt not on x, so the same inpt-otpt relation can be represented by the following first-order state-space eqations: ẋ 2 = [ 2 [, 2 (Sys a ) y = [ [. 2 It can be easily verified that Sys a has the same transfer matrix as Sys : -->sysa=syslin( c,[-,[- 2,[,[ -); -->tfa=ss2tf(sysa) tfa = s - s s s Next, Sys 2 can be redced to 2nd order by noticing that x 3 and x 4 are linearly dependent if they both have zero initial conditions (remember, we want to find the transfer matrix, which has to assme a relaxed system). In fact, if x 3 (0) = x 4 (0) = 0,

10 then x 3 (t) = x 4 (t) = t 0 t o 2(t)dt (t)dt x 3 (t) = 2x 4 (t), y 2 = 2x 3 = 4x 4, y 3 = x 3 x 4 = x 4. Ths, Sys 2 can be redced to the following first-order system (provided we have zero initial conditions!) ẋ 4 = 3, [ [ (Sys 2a ) y2 4 = x 4. It can be easily verified that Sys 2a has the same transfer matrix as Sys 2 : y 3 -->sys2a=syslin( c,[0,[,[4;,[0;0); -->tf2a=ss2tf(sys2a) tf2a = 4 - s - s In later lectres, we will see that Sys is not observable, and Sys 2 is not controllable.