Control Home Work-1

7/29/2019 Control Home Work-1

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1

0 1 2 3 4 5 6 7 8 9

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6(step response)

t (seconds)

y(t)

Step response of second order systems

clc

clear

clf

numa = 9;

dena = [1 1.2 9];

Ta = tf(numa, dena);

%percenta = exp(-zetaa*pi/ sqrt(1-zetaa^2))*100

step(Ta)

title('(step response)')xlabel(' t')

ylabel('y(t)')

sys = tf([9],[1 1.2 9]);

S = stepinfo(sys,'RiseTimeLimits',[0.01,1])

RiseTime : 0.5768 SettlingTime : 6.5273

Overshoot : 52.5701 SettlingMax : 1.5257

Undershoot : 0 SettlingMin : 0.7246Peak : 1.5257 PeakTime : 1.0492


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0 2 4 6 8 10 12-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

impulse response

Time (seconds)

Amplitude

Impulse response of second order systems

clf

clc

clear

numa = 9;

dena = [1 1.2 9];

Ta = tf(numa, dena);

%percenta = exp(-zetaa*pi/ sqrt(1-zetaa^2))*100

impulse(Ta)

title('impulse response')

sys = tf([9],[1 1.2 9]);


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0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Second Order Step Response

Normalized Time n

t

y(t)

= 0

= 0.1

= 0.2

= 0.4

= 0.5

= 0.6

= 0.8

= 1

Step Response of Second-Order Systems Versus clf

clc

cleart = [0:.01:12];

y = zeros(max(size(t)),8);

zetas = [0,0.1,0.2,0.4,0.5,0.6,0.8,1]

for index = 1:8

y(:,index) = step(tf([1],[1 2*zetas(index) 1]),t);

end;

plot(t,y(:,1),t,y(:,2),t,y(:,3),t,y(:,4),t,y(:,5),t,y(:,6),t,y(:,7),t,y(:,8))

title('Second Order Step Response')xlabel('Normalized Time \omega_n t')

ylabel('y(t)')

legend('\zeta = 0 ','\zeta = 0.1','\zeta = 0.2','\zeta = 0.4','\zeta = 0.5',...

'\zeta = 0.6 ','\zeta = 0.8 ','\zeta = 1 ')

grid


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4

Discussion step response of the second order system:

1. The damping frequency (actual frequency) inverse proportional to the

damping ratio where damping frequency decreases slightly as the damping

ratio increases.

2. The response oscillate for very low damping while for large damping

shows no oscillation

3. The system becomes more stable with increasing of damping ratio, and vice

versa.

4.


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5

Plot of the peak overshoot Mp versus the damping ratio

for the second order system

Matlab command

clf

clc

clear

>> fplot('100*exp(-pi*h/sqrt(1-h^2))', [0, 1])


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6

-7 -6 -5 -4 -3 -2 -1 0 1 2-5

-4

-3

-2

-1

0

1

2

3

4

50.160.30.460.60.72

0.84

0.92

0.98

0.160.30.460.60.720.84

0.92

0.98

123456

System: sy s

Gain: 1.43

Pole: -0.29 + 0.711i

Damping: 0.378Overshoot (%): 27.8

Frequency (rad/s): 0.767

Root Locus

Real Axis (seconds-1

)

ImaginaryAxis(seconds

-1)

ROOT LOCUS PLOT

>> %MATLAB Program

>> clf

>> clc>> clear

>> num = [01 ];>> den = [1 3 2 0];

>> rlocus(num, den);


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-1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2-1.5

-1

-0.5

0

0.5

1

1.5

0.97

0.120.240.360.480.620.76

0.88

0.97

0.20.40.60.811.21.41.6

0.120.240.360.480.620.76

0.88

Root Locus

Real Axis (seconds-1

)

ImaginaryAxis

(seconds-1)

>> %MATLAB Program

clf;

num = [1 0 ];

den = [1 0 1 ];

rlocus(num, den);


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-80

-60

-40

-20

0

20

40

Magnitude(dB)

10-1 100 101 102

-180

-135

-90

-45

0

Phase(deg)

Bode Diagram

Frequency (rad/s)

Magnitude and phase of a second-order system

clf

clcclear

H1=tf([0 9],[1 0.3 9])

H2=tf([0 9],[1 0.6 9])

H3=tf([0 9],[1 1.2 9])

H4=tf([0 9],[1 1.8 9])

H5=tf([0 9],[1 3 9])

H6=tf([0 9],[1 4.2 9])

H7=tf([0 9],[1 5.4 9])

bode(H1,H2,H3,H4,H5,H6,H7)


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-150

-100

-50

0

50

Magnitude(dB)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Phase(deg)

Bode DiagramGm = 6.02 dB (at 1 rad/s) , Pm = 21.4 deg (at 0.682 rad/s)

Frequency (rad/s)

-100

-50

0

50

100

Magnitude(dB)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Phase(deg)

Bode Diagram

Gm = -7.96 dB (at 1 rad/s) , Pm = -23.2 deg (at 1.52 rad/s)

Frequency (rad/s)

GM and PM for the following transfer function for (K=1 ,K=5)

()

()


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10

Calculate gain constant (K) for the following transfer function

()

(), when a PM of (i) 45

o, and (ii) 70

o

Solution steps:

1- Draw the magnitude and phase margin for (k=1) by MATLAB program as

follows

>>hd = tf([0 1],[1 2 1 0])

>> [Gm,Pm,Wg,Wp] = margin(hd);

>>margin(hd)

2- By click on the curve of PM and adjust the angle to (-110, -135) we can

find the conjugate frequency.

3- By click on the curve of GM and adjust the frequency which appears on

the curve to the same value that we find in the second step.

4- Read the value of GM appears on the curve after adjust the frequency as

mentioned in previous step and apply the following equation to calculate

gain constant

GM dB=20 log10 GM

K= 1/GM


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Bode Diagram

Gm = 6.02 dB (at 1 rad/s) , Pm = 21.4 deg (at 0.682 rad/s)

Frequency (rad/s)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

System: hd


Phase (deg): -110

Phase(deg)

-150

-100

-50

0

50

System: hd


Magnitude (dB): 14.7

Magnitude(dB)

Bode Diagram

Gm = 6.02 dB (at 1 rad/s) , Pm = 21.4 deg (at 0.682 rad/s)

Fre uenc rad/s

-150

-100

-50

0

50System: hd


Magnitude (dB): 6.17

Magn

itude(dB)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

System: hd


Phase (deg): -135

P

hase(deg)


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12

Effect of closed-loop pole position in the S-plane on system transient

response

The systems response to any input will also include these features.

The following figures show a selection of pole locations, with their corresponding

contribution to the total response.

Note:

1- The real part of the pole, determines both stability and the time

constant, .

2- The imaginary part of the pole, !, determines the damped natural

frequency (actual frequency of oscillation) in rad/sec.

3- The magnitude of the pole determines the natural frequency.

4- The argument of the pole determines the damping ratio

1- Zeta (1)2- System is stable3- Root is complex conjugate

1- Zeta ( =1)2- System is stable3- Root is real


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13

1- Zeta ( = 0)2- System is stable3- Root is imaginary

1- Zeta ( =1)2- System is stable3-

Root is real

1- Zeta (1)2- System is unstable3- Root is complex conjugate

1- Zeta (1)2- System is unstable3- Root is complex conjugate

1- Zeta (1)2- System is unstable3- Root is real


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1- Zeta ( =0)2- System is stable3- Root is real

1- Zeta (1)2-

System is unstable3- Root is real


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15

Stability Theorem

It should be clear from these examples that if any of the poles of G(s) have a

positive real part then the impulse response will have a term that blows up

exponentially (consider the partial fraction expansion of G(s)). Also, if G(s) has a

repeated imaginary axis pole then the impulse response will have a term that still

blows up, although more slowly. In either of these cases, the system is unstable.

Isolated poles on the imaginary axis, on the other hand, give rise

to terms in the impulse response which remain bounded (e.g. steps or sinusoids).

In this case the system is not asymptotically stable but is nevertheless marginallystable (provided it has no RHP or repeated imaginary axis poles). In fact, (for

systems with proper rational transfer functions) it can

be shown that Stability Theorem:

1- A system is asymptotically stable if all its poles have negative real parts.

2- A system is unstable if any pole has a positive real part, or if there are any

repeated poles on the imaginary axis.

3- A system is marginally stable if it has one or more distinct poles

on the imaginary axis, and any remaining poles have negative real

parts. Repeated poles on the imaginary axis.

Control Home Work-1

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