331-Final-sol.docx

Umm Al-Qura Universtiy, Makkah Department of Electrical Engineering

Control (802331) Term 2; 2016/2017

Solution Final Exam

Dr. Waheed Ahmad Younis Student name:

May 22, 2017 Student ID:

Max Marks: 80 Section 1 or 2

Q1. [4, 6] Consider the system shown. Find:

a. The damping ratio and undamped natural frequency (for above figure).

b. To improve the damping, the system is changed into (figure below). Find the value of K such that the new damping ratio is 1/2.

Solution:

a.

C(s)

R(s)= G(s) 1+G(s)H(s)=

10 s(s+1) 1+ 10

s(s+1)

= 10

s(s+1)+10= 10

s²+s+10= (

√

¹⁰⁾²

s²+2

(

2

√

¹10

) √

^10s+(

√

¹⁰⁾²

Hence ω_n=

√

^{10=3.16rad/s∧ξ= 1}

2

√

^10=0.158

b. Consider the inner loop first: G(s) 1+G(s)H(s)=

10 s+1 1+10K

s+1

= 10

s+1+10K

Now consider the outer loop:

C(s)

R(s)= G(s) 1+G(s)H(s)=

10 s(s+1+10K)

1+ 10

s(s+1+10K)

= 10

s(s+1+10K)+10= 10

s²+(1+10K)s+10= (

√

¹⁰⁾²

s²+2

(

¹⁺¹⁰2

√

^10K

) √

^10s+(

√

¹⁰⁾²

Hence ξ=1+10K

2

√

^{10 =0.5⇒K=}

√

¹⁰⁻¹

10 =0.216

Q2. [1, 3, 6] Consider an unstable system G(s)= 1

s²−1 . Your goal is to design a controller such that the undamped natural frequency be ω_n=

√

^2rad/s and damping ration be ξ=1/

√

^{2 .}

a. Find the required pole location.

b. Can this goal be achieved by a simple controller G_c(s)=K ? c. Design a PD controller G_c(s)=K(s−z) to achieve this goal.

Solution:

a. Required pole location: −ξ ω_n± jω_n

√

^1−ξ2=−1

√

^{2 (}

√

2)± j

√

2

√

¹⁻

( √

¹²

)

^{2=−1± j}

b. Assume

G_c(s)=K . The root locus does not pass through the required poles. So, this goal cannot be achieved by simple adjustment of gain K.

c. Let the controller be G_c(s)=K(s−z) .

180=∠G_c(s)G(s)H(s)=∠

[

^Ks^(s−2−1^z)

]

^{=∠K+∠(s−z)−∠(s−1)−∠(s+1)=∠(−1+j−z)−∠(−1+j−1)−∠(−1+j+1)=tan−1−1−1 z−153.43−90}

-5 -4 -3 -2 -1 0 1 2

-3 -2 -1 0 1 2 3

x

x Root Locus

Real Axis (seconds^-1) Imaginary Axis (seconds-1 )

tan⁻¹ 1

−1−z=180+90+153.43=423.43≡63.43

z= −1

tan 63.43−1=−1.5 Hence G_c(s)=K(s+1.5)

Now we will find the value of K which will achieve the goal.

|

^Gc(s)G(s)H(s)

|

^=1⇒

|

^{(s+K1)(s+1.5s(s−1))}

|

⁼¹

⇒K=

|

^{(s+1s+1.5)(s−1)}

|

⁼

|

^{(−1+−1+j+1) (−1+j+1.5j−1)}

|

⁼

|

^{(j)(−20.5++jj)}

|

⁼

√ √

_0.5²²⁺²₊₁¹²²⁼²

Hence G_c(s)=2(s+1.5)

The root locus for this controller

Root Locus

Real Axis (seconds^-1) Imaginary Axis (seconds-1)

-5 -4 -3 -2 -1 0 1 2

-3 -2 -1 0 1 2 3

System: sys Gain: 2.01 Pole: -1.01 + 1i Damping: 0.709 Overshoot (%): 4.25 Frequency (rad/s): 1.42

Q3. [6, 2, 1]

Consider the liquid flow system shown. The transfer function from in-

flow-rate q_i(t) to out-flow-rate q_o(t) is found to be Q_o(s)

Q_i(s)= 1

R₁C₁R₂C₂s²+

(

^R1C₁+R₂C₂+R₂C₁

)

^s+1

Assume R₁=R₂=1s/m²,C₁=1m², C₂=0.25m² . The in-flow-rate of 1m³/s started at t=0. Find a. Find the out-flow-rate as a function of time.

b. Find the out-flow-rate after 2 sec.

c. Find the steady-state value of the out-flow-rate.

Solution:

a. Q_o(s)

Q_i(s)= 1

R₁C₁R₂C₂s²+

(

^R1C₁+R₂C₂+R₂C₁

)

^s+1=

1

0.25s²+2.25s+1= 4

s²+9s+4 . Hence ξ=2.25>1

Q_o(s)= 4

s²+9s+4Q_i(s)= 4

s(s+8.531)(s+0.469) Using the table to find inverse Laplace transform:

1

ab

[

^1+a−b1

(

^{b e−at−a e−bt}

) ]

^s(s+a1)(s+b)

q_o(t)= 4

(8.531) (0.469)

[

¹⁺8.531−0.469¹

(

0.469e^−8.531t−8.531e^−0.469t

) ]

¿1+0.0582e^−8.531t−1.0582e^−0.469t

b. q_o(10)=1+0.0582e^−8.531(2)−1.0582e^−0.469(2)=0.586m³/s c. qo(∞)=1m³/s

Q4. [3, 3, 4] Consider the system shown. Plot the root locus. Find all important parameters.

Solution:

G(s)H(s)= s+1

s

(

s²+2s+6

) (

s+¹1

)

⁼_s

(

_s²₊¹_2s+6

)

⁼_s(s+1+j

√

₅¹_{) (s+1−j}

√

₅₎

We have 3 poles and no zeros. So 3 asymptotes.

Center of asymptotes: σ_c=

∑

^p−

∑

^z

n−m =(0−1+j

√

^5−1+j

√

⁵)−(0)

3−0 =−2

3

Angles of asymptotes: β=180,±60

Angle of departure: θD=180+∠GH=180+∠

[

^s(s+1+j

√

^{51) (s+1−j}

√

⁵⁾

]

¿180−∠s−∠(s+1+j

√

⁵⁾−∠(s+1−j

√

⁵⁾=180−∠(−1+j

√

⁵⁾−∠(−1+j

√

^5+1+j

√

⁵⁾−∠(−1+j

√

^5+1−j

√

⁵⁾=180−114.09−90=−24.09

-5 -4 -3 -2 -1 0 1 2 -5

-4 -3 -2 -1 0 1 2 3 4 5

x

x

x

Q5. [10] Again consider the system of Q4. Using Routh’s stability test, find the range of K for the system to be stable.

Solution:

The close-loop transfer function will be C(s)

R(s)= G(s) 1+G(s)H(s)=

K(s+1) s

(

s²+2s+6

)

1+ K(s+1)

s

(

s²+2s+6

) (

s+¹1

)

=

K(s+1) s

(

s²+2s+6

)

1+ K

s

(

s²+2s+6

)

= K(s+1)

s

(

s²+2s+6

)

+K= K(s+1) s³+2s²+6s+K

s³ 1 6

s² 2 K

s¹ 6-K/2

s⁰ K

Hence K>0 and 6-K/2>0 or 12>K 0<K<12

Q6. [6, 4] Consider a PID controller with ^kp=2,k_I=1∧k_D=3 . The error signal is a combination of step and ramp: e(t)=2u(t)+t u(t) . Find the control signal generated by the PID controller.

Solution:

Transfer function of the PID controller: W(s)

E(s)=k_P+k_I

s +k_Ds=2+1 s+3s E(s)=2

s+1

s² . ^W(s)=

(

²⁺¹s+3s

) (

^2s+s¹²

)

^=6+7s+s⁴²+1 s³ . Control signal: w(t)=6δ(t)+

(

7+4t+0.5t²

)

u(t)

Q7. [4, 2, 4] Consider

a. What will be the output of the following MATLAB commands?

n=[1 0 1];

d=[1 2 3 0];

k=0:10;

y=step(n, d, k);

y

b. What is the difference between following two MATLAB commands?

1. n=[1 0 1];

2. n=[1 0 1]

c. Consider the following system:

G(s)= s+2

s(s+1)³, H(s)=1

Write MATLAB command to plot the root locus of this system.

Solution:

a. First 2 lines will define the numerator and denominator of a transfer function: G(s)= s²+1 s³+2s²+3s Third line will generate a vector k=[0 1 2 3 4 5 6 7 8 9 10].

Forth line will compute the step response of the system G(s) at times t=0, 1, … , 10 and store the output in ‘y’.

Last line will print the calculated values of the step response on the screen. No plot will be generated.

b. Command in the first line will make a vector n=[1 0 1] but do not print it on the screen.

Command in the second line will make a vector n=[1 0 1] and echo it on the screen.

c. Here

G(s)H(s)= s+2

s(s+1)³= s+2

s

(

s³+3s²+3s+1

)

⁼

s+2 s⁴+3s³+3s²+s Hence the following command will do the job:

rlocus([1 2], [1 3 3 1 0]);

Q8a. [5] Compare open-loop system with close-loop system. Give at least 3 points.

Ans.

Open loop control system Close loop control system

Cheaper Simple Less accurate

Easier/cheaper to maintain No issue of stability

Expensive Complicated More accurate

Difficult/expensive to maintain Stability is an important issue

Q8b. [5] Give at least 3 examples of open-loop systems and close-loop systems.

Ans.

Open loop control system Close loop control system

Speed control of DC motor without velocity feedback.

Open loop system for position control.

Toaster

Washing machine

Speed control of DC motor using velocity feedback.

Position control of DC motor using position feedback.

Position control of DC motor using position and velocity feedback.

Temperature control using thermostat.

Laplace transform

f (t) F(s)

Unit impulse: δ(t) 1

Unit step: 1 1

s

Unit ramp: t 1

s² tⁿ⁻¹

(n−1)!(n=1,2, 3,⋯) 1

sⁿ

tⁿ(n=1, 2, 3,⋯) n!

sⁿ⁺¹

e^−at 1

s+a 1

a

(

1−e^−at

)

¹

s(s+a)

t e^−at 1 (s+a)²

1

a²

(

1−e^−at−at e^−at

)

¹

s(s+a)²

sinωt ω

s²+ω²

cosωt s

s²+ω²

1−cosωt ω_n²

s

(

^s2+ωn 2

)

1− e^{−ξ ωnt}

√

^1−ξ2sin

(

^ωn

√

^{1−ξ2t+cos−1ξ}

)

^{(0<ξ<1) ωn}

2

s

(

^{s2+2ξ ω}ns+ω_n²

)

1

a²

(

1−e^−at−at e^−at

)

¹

s(s+a)² 1

ab

[

^1+a−b1

(

^{b e−at−a e−bt}

) ]

^{s(s+a)(s+1 b)}

sinω_nt ^ωn

s²+ω_n²

ω_n

√

^1−ξ2e

−ξω_nt

sin

(

^ωn

√

^1−ξ2t

)

^{(0<ξ<1) ωn}

2

s²+2ξ ω_ns+ω_n²

t e^{−at 1}

(s+a)² 1

b−a

(

e^−at−e^−bt

)

¹

(s+a)(s+b)

 Close loop system with feedback: C(s) R(s)= G

1+GH

 For root locus:

o Center of asymptotes:

σ_c=

∑

i=1 n

p_i−

∑

i=1 m

z_i n−m

Angles of asymptotes:

β=180+360k n−m

o Break-away/break-in point:

∑

i=1

n 1

σ_b−p_i=

∑

i=1

m 1

σ_b−z_i

o Departure/arrival angles: θ_D=180+∠GH ; θ_A=180−∠GH