hw2_sol

UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN

Department of Electrical and Computer Engineering

ECE 486: Control Systems

HOMEWORK 2 - SOLUTION

Spring 2012

Problem 3.23

For the electric circuit shown in Fig. 3.56, find the following:

(a) The time-domain equation relating i(t) and v1(t);

(b) The time-domain equation relating i(t) and v2(t);

(c) Assuming all initial conditions are zero, the transfer function V2(s)/V1(s) and the dampingratio ζ and undamped natural frequency ωn of the system;

(d) The values of R that will result in v2(t) having an overshoot of no more than 25%,assuming v1(t) is a unit step, L = 10 mH, and C = 4 µF.

Solution:

(a) By the Kirchhoff’s voltage law (KVL),

v1(t) = Ldi

dt+Ri+

1

C

∫

u(t)dt.

(b) v2(t) is equal to the voltage around the capacitor.

v2(t) =1

C

∫

i(t)dt.

(c) Apply Laplace transform to the equations in (a) and (b),

V1(s) = sLI(s) +RI(s) +I(s)

sC

V2(s) =I(s)

sC⇒ I(s) = sCV2(s), plug-in to the above equation

V1(s) = s2LCV2(s) + sRCV2(s) + V2(s)

V2(s)

V1(s)=

1sC

sL+R+ 1sC

=1

s2LC + sRC + 1=

1/LC

s2 + (R/L)s + 1/LC=

ω2n

s2 + 2ζωns+ ω2n

.

ωn =1√LC

.

ζ =1

2ωn· RL

=R

2√

LC

.

(d) For 25% overshoot ζ ≈ 0.4 (see Page 118 Figure 3.23),

0.4 ≈ ζ =R

2√

LC

R = 2ζ

√

L

C= (2)(0.4)

√

10× 10−3

4× 10−6= 40 Ω.

Problem 3.25

For the unity feedback system shown in Fig. 3.58, specify the gain and pole location of thecompensator so that the overall closed-loop response to a unit-step input has an overshoot ofno more than 25%, and a 1% settling time of no more than 0.1 sec. Verify your design usingMATLAB.

Solution:

Y (s)

R(s)=

100K(s+a)(s+25)

1 + 100K(s+a)(s+25)

=100K

s2 + (25 + a)s + 25a+ 100K=

100K

s2 + 2ζωns+ ω2n

.

2

Using the given information:

R(s) =1

sunit step,

Mp ≤ 25%,

ts ≤ 0.1sec.

Solve for ζ:

Mp = e−πζ/√

1−ζ2 ,

ζ =

√

(lnMp)2

π2 + (lnMp)2

∣

∣

∣

∣

∣

Mp=0.25

≥ 0.4037.

Solve for ωn:e−ζωnts = 0.01, for a 1% settling time.

ts ≤4.605

ζωn= 0.1,

⇒ ωn ≈ 114.07.

Now find a and K:2ζωn = (25 + a)

a = 2ζωn − 25 = 92.10 − 25 = 67.10

ω2n = (25a + 100K),

K =ω2n − 25a

100≈ 113.34.

The step response of the system using MATLAB is shown below.

a = 67.10;

K = 113.34;

% Compensator

num1 = K;

den1 = [1 a];

sys1 = tf(num1,den1);

% Plant

num2 = 100;

den2 = [1 25];


sys3 = series(sys1,sys2);

3

% The feedback system

num4 = 1;

den4 = 1;


sys = feedback(sys3,sys4,-1);

hold on;

[y,t] = step(sys);

plot(t,y);

title(’Step response’);

xlabel(’Time (sec)’);

ylabel(’y(t)’);

grid on;

0 0.02 0.04 0.06 0.08 0.1 0.12 0.140

0.2

0.4

0.6

0.8

1

1.2

1.4Step response

Time (sec)

y(t)

4

Problem 3.27

A certain servomechanism system has dynamics dominated by a pair of complex poles andno finite zeros. The time-domain specifications on the rise time (tr), percent overshoot (Mp),and settling time (ts) are given by,

tr ≤ 0.6sec,

Mp ≤ 17%,

ts ≤ 9.2sec.

(a) Sketch the region in the s-plane where the poles could be placed so that the system willmeet all three specifications.

(b) Indicate on your sketch the specific locations (denoted by ×) that will have the smallestrise-time and also meet the settling time specification exactly.

Solution:

(a)-(b)

tr =1.8

ωn≤ 0.6 =⇒ ωn ≥ 3

Mp = e−

ζπ√1−ζ2 = 0.17 =⇒ ζπ

√

1− ζ2= − log 0.17 ≈ 1.772

=⇒ θ = sin−1 ζ = tan−1

(

ζ√

1− ζ2

)

= tan−1

(

1.772

π

)

≈ 29.42

ts =4.6

σ≤ 9.2 =⇒ σ ≥ 0.5

5

Problem 3.30

The equation of motion for the DC motor shown in Fig. 2.32 were given in Eqs. (2.52-53) as

Jmθm +

(

b+KtKe

Ra

)

θm =Kt

Rava.

Assume that

Jm = 0.01 kg ·m2,

b = 0.001 N ·m · sec,Ke = 0.02 V · sec,Kt = 0.02 N ·m/A,

Ra = 10 Ω.

(a) Find the transfer function between the applied voltage va and the motor speed θm.

(b) What is the steady-state speed of the motor after a voltage va = 10 V has been applied?

(c) Find the transfer function between the applied voltage va and the shaft angle θm.

(d) Suppose feedback is added to the system in part (c) so that it becomes a position servodevice such that the applied voltage is given by

va = K(θr − θm),

where K is the feedback gain. Find the transfer function between θr and θm.

(e) What is the maximum value of K that can be used if an overshoot Mp < 20% is desired?

(f) What values of K will provide a rise time of less than 4 sec? (Ignore the Mp constraint.)

(g) Use MATLAB to plot the step response of the position servo system for values of the gainK = 0.5, 1, and 2. Find the overshoot and rise time for each of the three step responsesby examining your plots. Are the plots consistent with your calculations in parts (e) and(f)?

Solution:

Jmθm +

(

b+KtKe

Ra

)

θm =Kt

Rava.

(a)

JmΘms2 +

(

b+KtKe

Ra

)

Θms =Kt

RaVa(s)

6

sΘm(s)

Va(s)=

Kt

RaJm

s+ bJm

+ KtKe

RaJm

.

Jm = 0.01 kg ·m2,

b = 0.001 N ·m · sec,Ke = 0.02 V · sec,Kt = 0.02 N ·m/A,

Ra = 10 Ω.

sΘm(s)

Va(s)=

0.2

s+ 0.104

(b) Final Value Theorem

˙θm(∞) = sΘm(s)Va(s)|s=0 =s(10)(0.2)

s(s+ 0.104)

∣

∣

∣

∣

s=0

=2

0.104= 19.23.

(c)Θm(s)

Va(s)=

0.2

s(s+ 0.104).

(d)

Θm(s) =0.2K(Θr −Θm)

s(s+ 0.104).

Θm(s)

Θr(s)=

0.2K

s2 + 0.104s + 0.2K.

(e)

Mp = e−πζ/√

1−ζ2 ≤ 0.2 (20%),

ζ ≥ 0.4559.

Θm(s)

Θr(s)=

ω2n

s2 + 2ζωns+ ω2n

.

2ζωn = 0.104,

ωn ≤ =0.104

2ζ=

0.104

2(0.4559)= 0.114 rad/sec,

ω2n = 0.2K = (0.114)2,

K < 6.50× 10−2.

7

(f)

ωn ≥ 1.8

tr=

1.8

4, ωn ≈ 0.45

ω2n = 0.2K = (0.45)2

K ≥ 1.01.

(g) MATLAB

% Problem 3.30 FPE6e

clear all

close all

K1=[0.5 1.0 2.0 6.5e-2];

t=0:0.01:150;

for i=1:1:length(K1)

K = K1(i);

titleText = sprintf(’K= %1.4f ’, K);

wn = sqrt(0.2*K);

num=wn^2;

den=[1 0.104 wn^2];

zeta=0.104/2/wn;

sys = tf(num, den);

y= step(sys, t);

% Finding maximum overshoot

if zeta < 1

Mp = (max(y) - 1)*100;

overshootText = sprintf(’Max overshoot = %3.2f %’, Mp);

else

overshootText = sprintf(’No overshoot’);

end

% Finding rise time

idx_01 = max(.nd(y<0.1));

idx_09 = min(.nd(y>0.9));

t_r = t(idx_09) - t(idx_01);

risetimeText = sprintf(’Rise time = %3.2f sec’, t_r);

% Plotting

subplot(3,2,i);

plot(t,y);

grid on;

title(titleText);

text( 0.5, 0.3, overshootText);

text( 0.5, 0.1, risetimeText);

end

8

%%%%%%%%%%%%%%%%

% Function for computing rise time

function tr = risetime(t,y)

% A. Emami 2006

% normalize y to 1:

y = y/y(length(y));

idx1 = min(find(y>=0.1))

idx2 = min(find(y>=0.9))

if ~isempty(idx1) & ~isempty(idx2)

tr = t(idx2) - t(idx1);

else

tr = 0

end

0 50 100 1500

0.5

1

1.5

2K= 0.5000

Max overshoot = 59.23 Rise time = 3.70 sec

0 50 100 1500

0.5

1

1.5

2K= 1.0000


0 50 100 1500

0.5

1

1.5

2K= 2.0000


0 50 100 1500

0.5

1

1.5K= 0.0650

Max overshoot = 19.99

Rise time = 13.66 sec

For part (e) we concluded that K < 6.50 × 10−2 in order for Mp < 20%: This is consistentwith the above plots. For part (f) we found that K ≥ 1.01 in order to have a rise time of lessthan 4 seconds. We actually see that our calculations is slightly off and that K can be K ≥ 0.5,but since K ≥ 1.01 is included in K ≥ 0.5, our answer in part (f) is consistent with the aboveplots.

9

Problem 3.33

In aircraft control systems, an ideal pitch response (qo) versus a pitch command (qc) is describedby the transfer function

Qo(s)

Qc(s)=

τω2n(s+ 1/τ)

s2 + 2ζωns+ ω2n

.

The actual aircraft response is more complicated than this ideal transfer function; nevertheless,the ideal model is used as a guide for autopilot design. Assume that tr is the desired rise timeand that

ωn =1.789

tr,

1

τ=

1.6

tr,

ζ = 0.89.

Show that this ideal response possesses a fast settling time and minimal overshoot by plottingthe step response for tr = 0.8, 1.0, 1.2, and 1.5 sec.

Solution:

The following program statements in MATLAB produce the following plots:

% Problem 3.33 FPE6e

tr = [0.8 1.0 1.2 1.5];

t=[1:240]/30;

tback=fliplr(t);

clf;

for I=1:4,

wn=(1.789)/tr(I); %Rads/second

tau=tr(I)/(1.6); %tau

zeta=0.89; %

b=tau*(wn^2)*[1 1/tau];

a=[1 2*zeta*wn (wn^2)];

y=step(b,a,t);

subplot(2,2,I);

plot(t,y);

titletext=sprintf(’tr=%3.1f seconds’,tr(I));

title(titletext);

xlabel(’t (seconds)’);

ylabel(’Qo/Qc’);

ymax=(max(y)-1)*100;

msg=sprintf(’Max overshoot=%3.1f%%’,ymax);

text(.50,.30,msg);

yback=flipud(y);

10

yind=find(abs(yback-1)>0.01);

ts=tback(min(yind));

msg=sprintf(’Settling time =%3.1f sec’,ts);

text(.50,.10,msg);

grid;

end

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4tr=0.8 seconds

t (seconds)

Qo/

Qc

Max overshoot=2.3%

Settling time =2.3 sec

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4tr=1.0 seconds

t (seconds)

Qo/

Qc

Max overshoot=2.3%


0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4tr=1.2 seconds

t (seconds)

Qo/

Qc

Max overshoot=2.3%


0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4tr=1.5 seconds

t (seconds)

Qo/

Qc

Max overshoot=2.3%


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