Solution to Homework Assignment 1

ECE382/ME482 Spring 2012 Homework 1 Solution February 24, 2012 1

6. For the system shown in Figure 2.79, page 131 of the text, with s + 50 Gc(s) = 10,H(s) = 1, G(s)= , (1) s2 + 60s + 500 we are asked to ﬁnd the steady state value of the output assuming the input noise and disturbance are zero.

Solution: From the reference input R(s), the system has forward path Gc(s)G(s) and negative feedback path H(s). Thus

s+50 Y (s) Gc(s)G(s) 10 s2+60s+500 10s + 500 T (s)= = = = . (2) R(s) 1+ H(s)Gc(s)G(s) s+50 s2 + 70s + 1000 1 + 10 s2+60s+500 Using the ﬁnal value theorem we conclude that 1 10s + 500 1 yss = lim y(t) = lim sT (s)R(s) = lim s = . (3) t→∞ s→0 s→0 s s2 + 70s + 1000 2

7. We are given the same system topology and asked to ﬁnd the impulse response to the reference input, assuming the initial conditions, the disturbance and the noise are zero. This time s + 4 Gc(s) = 20,H(s) = 1, G(s)= . (4) s2 − 12s − 65 Solution We ﬁnd that

Y (s) Gc(s)G(s) 20s + 80 10 10 T (s)= = = = + , (5) R(s) 1+ H(s)Gc(s)G(s) s2 + 8s + 15 s + 5 s + 3

where we have taken the partial fraction expansion of T(s) as a prelude to obtaining the inverse Laplace transform. (The Laplace transform of an impulse of unit area is one.) Inverting the Laplace transform we obtain

− t − t y(t) = 10e 5 + 10e 3 , t ≥ 0. (6)

8. We are given a new block diagram, see Figure 2.80, page 132 and asked to ﬁnd the closedloop transfer function. Solution: This problem can be solved by analyzing the block diagram or by drawing and analyzing the equivalent signal ﬂow graph. We choose one and then use the second to check our work. To analyze the block diagram, we redraw the diagram as having an inner and an outer loop. We then apply the feedback formula twice. For the inner loop, the forward path transfer function is 10 s + 5 ECE382/ME482 Spring 2012 Homework 1 Solution February 24, 2012 2

and the negative feedback path transfer function is 5. The resulting closed loop gain is 10 . s + 55 For the outer loop we then have forward path 10 1 s + 55 s and negative unity feedback. Thus the closed-loop transfer function is 10 T (s)= . (7) s2 + 55s + 50

Alternatively, we can use Mason’s signal ﬂow formula as follows. Multiplying the gains of the segments of the path from R(s) to Y (s) we ﬁnd that 10 P = . 1 s2 + 5s The two loop transfer functions are −50 1 −10 L = and L = . (8) 1 s + 5 2 s s + 5 Note that this is the only possible way to choose the loops. The loops must be traversed in the direction of the arrows on the segments. The graph determinant is

∆ = 1 − L1 − L2 (9)

since there are no non-touching loops. After some algebra, we ﬁnd

s2 + 55s + 10 ∆ = 1 − L − L = . (10) 1 2 s2 + 5s

The co-factor of the path P1 is just ∆1 = 1, the graph determinant after the loops that contain P1 are broken. Thus P ∆ 10 T (s)= 1 1 = . (11) ∆ s2 + 55s + 10

9. For the system shown in Figure 2.79, page 131 of the text, with s + 5 Gc(s) = 4,H(s) = 1, G(s)= , (12) s2 + 10s + 5 we are asked to ﬁnd the closed loop transfer function. Solution: We reason as in Problem 6 and obtain 20 T (s)= . (13) s2 + 10s + 25 ECE382/ME482 Spring 2012 Homework 1 Solution February 24, 2012 3

10. For the system of the previous problem, we are asked to find the unit step response of the closed-loop system. Solution To find the unit step response we must find the partial fraction expansion and inverse transform of 20 20 20/25 −20/25 −4 T (s)R(s)= = = + + . (14) s(s2 + 10s + 25) s(s + 5)2 s s + 5 (s + 5)2 The inverse transform is − t − t y(t) = 20/25 − 20e 5 /25 − 4te 5 , t ≥ 0. (15)

11. For the previous problem we are asked to ﬁnd the steady state value of the output y(t). Solution: The exponential goes to zero faster than t goes to inﬁnity, thus we have

ss y =t lim→∞ y(t) = 20/25. (16)

12. We are given the differential equation y¨ + 2y ˙ + y = u, y(0) =y ˙(0) = 0, (17) and asked to find the poles of the system. We are told that the input is a unit step. Solution: The value of the input is irrelevant. The Laplace transform of the differential equation is s2Y (s) − sy(0) − y˙(0) + 2(sY (s) − y(0)) + Y (s)= U(s). (18) Thus Y (s) 1 1 = = , (19) U(s) s2 + 2s + 1 (s + 1)2 i.e. the system has a repeated pole at −1. Note that if the initial conditions had been nonzero we would have gotten a different transfer function. 13. A truck tows a cart of mass m and the coupling element is modelled as parallel connection of a spring with spring constant k and a damper with damping constant b. We are asked to find the transfer function between the speeds of the vehicles. We are given m = 1000 kg, k = 20, 000 N/m, and b = 200 Ns/m, and told that the truck moves with constant acceleration a = 0.7 m/s2. Solution Applying Newton’s Law, we find the differential equation describing the system to be mx¨c + b(x ˙ c − x˙ t)+ k(xc − xt) = 0, (20)

where xc is the position of the cart, and xt is the position of the truck. Taking the Laplace transform we obtain 2 ms Xc(s)+ bs(Xc(s) − Xt(s)) + k(Xc(s) − Xt(s)) = 0, (21) which yields the transfer function

Xc(s) bs + k = (22) Xt(s) ms2 + bs + k Since v =x ˙, V (s)= sX(s) − x(0), thus if we deﬁne our coordinate system to originate at the original location of the cart we have Vc(s)/Vc(s)= Xc(s)/Xt(s). ECE382/ME482 Spring 2012 Homework 1 Solution February 24, 2012 4

14. For the system shown in Figure 2.79, page 131 of the text, with 1000 Gc(s) = 15,H(s) = 1, G(s)= , (23) s3 + 50s2 + 4500s + 1000 we are asked to ﬁnd the closed-loop transfer function and its zeros and poles. Solution: Using the method of Problem 6, we ﬁnd that 15000 T (s)= (24) s3 + 50s2 + 4500s + 16000 which has no zeros and 3 poles. According to Matlab, the poles are −23.1519 ± 61.5861i and −3.6961. 15. For the system shown in Figure 2.79, page 131 of the text, with K(s + 0.3) 1 Gc(s)= ,H(s) = 2s, G(s)= , (25) s (s − 2)(s2 + 10s + 45

we are asked to ﬁnd the closed-loop transfer function from the disturbance Td(s) to the output Y (s). Solution: Using the method of Problem 6, the forward path transfer function is G(s) and the negative feedback path transfer function is G(s)H(s)Gc(s). We obtain after some algebra that Y (s) G(s) 1 = = . (26) Td(s) 1+ G(s)H(s)Gc(s) s3 + 8s2 + (2K + 25)s + (0.6K − 90)

P2.25 For the block diagram of Figure P2.25(b), we are asked to determine the transfer functions from the reference and disturbance inputs to the output. Solution: Working backward from Y (s), we ﬁnd that 1 Y (s)= G(s) R(s) − (G(s)R(s)+ Td(s)) + Td(s)+ G(s)R(s)= G(s)R(s), (27) G(s) so the transfer function from reference input to the output is G(s) and the transfer function from the disturbance input to the output is zero.

AP2.2 We are asked to find the transfer function from input r1 to output y2 in the coupled systems of Figure AP2.2., and then select G5(s) to decouple the two systems. Solution: The problem can be solved using the block diagram or the signal flow graph. However, it is easier to solve it using the signal flow graph. We have two paths from R1(s) to Y2(s),

P1 = G1G5G4 (28)

P2 = G1G2G6G3G4, (29) and two loops

L1 = −G1G2H1 (30)

L2 = −G3G4H2. (31) ECE382/ME482 Spring 2012 Homework 1 Solution February 24, 2012 5

Accordingly the graph determinant is

∆ = 1−L1 −L2 +L1L2 = 1−(−G1G2H1)−(−G3G4H2)+(−G1G2H1)(−G3G4H2). (32)

Since both loops touch both paths, the path co-factors are both one. Thus Y P ∆ + P ∆ G G G + G G G G G 2 = 1 1 2 2 = 1 5 4 1 2 6 3 4 (33) R1 ∆ ∆

and if G5(s)= −G2(s)G6(s)G3(s), the numerator will be zero. Since there is no path from r2 to r1, this will suﬃce to decouple the systems. DP2.3 We are given the input r(t)= t, t ≥ 0, and the output

−t − t y(t)= e − (1/4)e 2 − 3/4 + (1/2)t, t ≥ 0,

and asked to ﬁnd the transfer function G(s)= Y (s)/R(s). Solution: We simply take the ratio of the Laplace transforms, and after some algebra obtain 1 1 − 3 1 Y (s) s+1 + s 4s + s2 1 G(s)= = 4( +2) 2 = . (34) R(s) 1/s2 (s + 1)(s + 2)

CP2.5 We are asked to use Matlab to compute (a) the closed loop transfer function, (b) the step response to a 10 degree step input, (c) the step response with a diﬀerent moment of inertia, and to comment on the results of the closed step response with a prescribed controller. Solution A Matlab script that accomplishes the computations and plots the results is given below.

%Part (a) Create feedback system a=1; b=8; k=10.8e+08; J=10.8e+08; num=k*[1 a]; den=J*[1 b 0 0]; sys=tf(num,den); sys_closedloop=feedback(sys,[1]);

% Part (b) and (c) t=[0:0.1:100]; % Nominal case angle=10*pi/180; %Convert angle to radians sys_angle=sys_closedloop*angle; y=step(sys_angle,t);

% Case J reduced by 80% J=10.8e+08*0.8; den=J*[1 b 0 0]; sys=tf(num,den); sys_closedloop=feedback(sys,[1]); sys_angle=sys_closedloop*angle; y1=step(sys_angle,t);

% Case J reduced by 50% J=10.8e+08*0.5; den=J*[1 b 0 0]; ECE382/ME482 Spring 2012 Homework 1 Solution February 24, 2012 6

sys=tf(num,den); sys_closedloop=feedback(sys,[1]); sys_angle=sys_closedloop *angle; y2=step(sys_angle,t); % figure (1) clf; plot(t,y*180/pi,t,y1*180/pi,’--’,t,y2*180/pi,’:’),grid xlabel(’Time (sec)’) ylabel(’Spacecraft attitude (deg)’) title(’Response to step change of 10 deg. in desired attitude’) legend(’Nominal (solid)’,’J reduced by 80% (dashed)’,... ’J reduced by 50% (dotted)’)

The spacecraft simulations are shown in Figure below. We see that as J is decreased, the time to settle down decreases. Also, the overshoot from 10 degrees decreases as J decreases. Thus, the performance seems to get better (in some sense) as J decreases.

Response to step change of 10 deg. in desired attitude 18 Nominal (solid) 16 J reduced by 80% (dashed) J reduced by 50% (dotted) 14

6 Spacecraft attitude (deg)

0 0 20 40 60 80 100 Time (sec)

Figure 1: Comparison of Step Responses of Satellite Attitude