PI&PD Controller Sample Problem

7/27/2019 PI&PD Controller Sample Problem

1/28

Problem 1Solution

I

s=tf('s');G=1/((s+2.5)*(s+4.5)*(s+7));rlocus(G);grid on

II

-20 -15 -10 -5 0 5-15

-10

-5

0

5

10

150.140.280.420.560.70.82

0.91

0.975

0.140.280.420.560.70.82

0.91

0.975

2.557.51012.51517.5

Root Locus

Real Axis (seconds-1)

ImaginaryAx

is(seconds-1)

-2.19 -2.18 -2.17 -2.16 -2.15 -2.14 -2.13 -2.12 -2.11

3.66

3.67

3.68

3.69

3.7

3.71

3.72

3.73

3.74

3.75

3.76

0.490.4930.4970.50.503

0.506

0.51

0.512

4.22

4.24

4.26

4.28

4.3

System: G

Gain: 101

Pole: -2.15 + 3.72i

Damping: 0.5

Overshoot (%): 16.3

Frequency (rad/s): 4.3

Root Locus

Real Axis ( seconds-1)

Imag

inaryAxis(seconds-1)


2/28

K = 101

Check the third pole for K=101, closed loop TF:

M(s) =101

(:2.)(:4.)(:7)

s=tf('s');M = 101/((s+2.5)*(s+4.5)*(s+7));F = feedback(M,1);pole(F)

ans =

-9.6988

-2.1506 + 3.7294i

-2.1506 - 3.7294i

Since the third pole s = -9.6988 is NOTgreater than five times the real part of the dominant pole

(-2.1506), therefore the approximation that the system is a second-order is not valid. In this case the

actual settling time/peak time of original equation which can be found using the plot of the step

response must be considered.


3/28

s=tf('s');M = 101/((s+2.5)*(s+4.5)*(s+7));F = feedback(M,1);step(F)

From the graph above it was found that the actual settling time Ts= 1.95 (sec).

III.

Considering the desired parameter that Ts(new) 5 x Ts ;

Taking the maximum possible value

Ts(new) = 5 x 1.95 = 9.75 (sec)

IV.

Desired location for the compensated poles:

Real part dnew 4

Ts(new) . (sec)

Imaginary part dnew .n .

Step Response

Time (seconds)

Amplitude

0 0.5 1 1.5 2 2.5 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

System: F

Settling Time (seconds): 1.95


4/28

V.System with PI controller looks like

Graph the locus of the compesated sytem with varying values of zero to find a suitable value of K,

then plot the step response of each system until we come up with the our desired value of settling

time (Ts ).

Test point 1

M(s) =(:0.1)

(:2.)(:4.)(:7)

K = 100

Taking the step response using the obtained value of K

s=tf('s');M = (100*(s+0.1))/(s*(s+2.5)*(s+4.5)*(s+7));F=feedback(M,1);step(F)

Ts = 54.1 (sec)

-5 -4 -3 -2 -1 0

-1

0

1

2

3

4

5

60.140.280.420.56

0.7

0.82

0.91

0.975

0.975

1

2

3

4

5

1

Root Locus


ImaginaryAxis(seconds-1)


5/28

Test point 2

M(s) =(:0.)

(:2.)(:4.)(:7)

K = 99.1


s=tf('s');M = (99.1*(s+0.5))/(s*(s+2.5)*(s+4.5)*(s+7));F=feedback(M,1);step(F)

Ts = 9.91 (sec)

Test point 3

M(s) =(:0.6)

(:2.)(:4.)(:7)

K = 98.7


s=tf('s');M = (98.7*(s+0.6))/(s*(s+2.5)*(s+4.5)*(s+7));F=feedback(M,1);step(F)

Ts = 8.06 (sec)

This value of settling time satisfies our the condition T s(new) 5 x Ts. Therefore it is safe to say that the

location of the zero for this system should be at -0.6.

VI.

Gc(s) =

9.7(:0.6)

Gc(s) = K1 +2

Therefore

K1 = 98.7

K2= 59.22


6/28

VII.Validation of second order system approximation

s=tf('s');M = (98.71*(s+0.6))/(s*(s+2.5)*(s+4.5)*(s+7) );F=feedback(M,1);pole(F)

ans =

-9.5760

-2.0229 + 3.5020i

-2.0229 - 3.5020i

-0.3781

Pole and zero dont cancel each other, second order approximation is not valid. We have to

simulate compensated system to make sure all requirements have been met.

VIII.Plot step responses for uncompensated and compensated system.

s=tf('s');G=101/((s+2.5)*(s+4.5)*(s+7));F1=feedback(G,1);H=(98.7*(s+0.6))/(s*(s+2.5)*(s+4.5)*(s+7));

F2=feedback(H,1);step(F1,F2)


7/28

IX.

Uncompensated Simulation Pi-Compensated Simulation

TF, G(s)

H(s)

()

( . 5)( .5)( )

()

( .)

( .5)( . 5)(

Dom. Poles -2.15 j3.72 -2.02 j3.5

Gain, K 101 98.7

0.5 0.5

%OS 16.3 14.4 16.3

Ts 1.95 1.95 9.75 8.06

Tp 0.962 0.962 4.42 >16

Kp 1.28

e 0.44 0

Other Poles -9.6988 -0.3781

Zeroes None -0.6

Comments Second-order approx.

Step Response

Time (seconds)

Amplitude

0 2 4 6 8 10 12 14 160

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

System: F2


System: F1


System: F1

Peak amplitude: 0.643

Overshoot (%): 14.4

At time (seconds): 0.962


8/28

not valid

*some values (poles and zeroes etc.) were obtained using MatLab, codes and obtained values can be

found from the preceding process.

Kp

(.5)(.5)() .8

e

K p

.8 .

Ts = 1.95

Kp 9.7(0.6)

0

e

K p

Ts= 5 x 1.95 = 9.75 (sec)

dnew 4

Ts(new) . (sec)

dnew .n .

Tp

d

. .


9/28

Problem 2

Solution

I.

s=tf('s');G=1/(s*(s+15)*(s+6));rlocus(G);grid on

II

-60 -50 -40 -30 -20 -10 0 10 20-40

-30

-20

-10

0

10

20

30

400.160.340.50.640.76

0.86

0.94

0.985

0.160.340.50.640.760.86

0.94

0.985

1020304050

Root Locus

Real Axis (seconds -1)


Root Locus



-2.06 -2.05 -2.04 -2.03 -2.02 -2.01 -2 -1.99

3.93

3.94

3.95

3.96

3.97

3.98

3.99

4

System: G

Gain: 338

Pole: -2.03 + 3.96i

Damping: 0.456

Overshoot (%): 20


0.4460.4490.4510.4530.45658

0.46

0.463

4.41

4.42

4.43

4.44

4.45

4.46

.


10/28

K = 388


M(s) =3

(:1)(:6)

s=tf('s');G=388/(s*(s+15)*(s+6));F=feedback(G,1);pole(F)

ans =

-16.8746

-2.0627 + 4.0163i

-2.0627 - 4.0163i

The third pole s = -16.8746 is twelve times farther from the j-axis than the dominant poles, so the

approximation is valid and we can use the peak time.

III. Peak time of uncompensated system is

Tp =

d

3.96 . (sec)

We need to reduce settling time by1

2so;

Tp (new) .

.5 (sec)

IV. Desired location for the compensated poles:

Imaginary part dnew

()7.95

Real part dnew dnew

n.8

.

Desired compensated poles:


11/28

V. System with PD compensator looks like:

We need to find gain K and zero z. To find location of zero se use an angle criteria:

-16 -14 -12 -10 -8 -6 -4 -2 0 2

-2

0

2

4

6

8

10

12

14

16 0.160.340.50.64

0.76

0.86

0.94

0.985

0.985

246810121416

Root Locus


ImaginaryAxis(second

s-1)

116.74deg


12/28

-16 -14 -12 -10 -8 -6 -4 -2 0 2

-2

0

2

4

6

8

10

12

14

16 0.160.340.50.64

0.76

0.86

0.94

0.985

0.985

246810121416

Root Locus


ImaginaryAxis(seco

nds-1)

116.74deg


13/28

VI. Value of PD Compensator

s=tf('s');G=(388*(s+10.86))/(s*(s+15)*(s+6));rlocus(G);grid on

K=92.6

VII Designed PD Compensator

Gc(s) = 92.6(s+10.86)

K1 = 1000.56

K2= 92.6

-4.06 -4.05 -4.04 -4.03 -4.02 -4.01 -4

7.75

7.8

7.85

7.9

7.95

8

System: G

Gain: 92.6

Pole: -4.03 + 7.85i

Damping: 0.456

Overshoot (%): 20


0.4470.4490.452

0.454

0.456

0.458

0.461

0.463

8.7

8.75

8.8

8.85

8.9

8.95

Root Locus




14/28

VIII Validation of secnd order system approximation

s=tf('s');G=(92.6*(s+10.86))/(s*(s+15)*(s+6));F=feedback(G,1);pole(F)

ans =

-4.7845 +18.5937i

-4.7845 -18.5937i

-11.4311

Second order system approximation is not valid.

IX Step responses for uncompensated and compensated system

s=tf('s');G=388/(s*(s+15)*(s+6));F2=feedback(G,1);H=(92.6*(s+10.86))/(s*(s+15)*(s+6));F1=feedback(H,1);step(F1,F2)

Step Response

Time (seconds)

A

mplitude

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

System: F2


System: F1


System: F2


Overshoot (%): 23


System: F1


Overshoot (%): 22



15/28

*F1 Compensated

*F2 Uncompensated

X

Uncompensated Simulation PD-Compensated Simulation

TF, G(s) H(s) G(s)=

(:1)(:6) G(s)=

(:10.6)

(:1)(:6)

Dominant Poles -2.03 j3.96 -4.07 j7.95

Gain, K 338 92.7

0.45 0.45

%OS 20% 23% 20% 22

Ts 1.97 1.82 0.983 0.936

Tp 0.79 0.79 0.395 0.388

Kv 3.76 11.19

e 0.27 0.09

Other Poles -16.8491 -12.8844

Zeroes none -10.86

Comments 2ndis valid 2ndis not valid

Kv 8

5() .

e

Kv

.

.

Tp

d

. .

Ts

d

. .


16/28

Kv .(.8)

5() .

e

Kv

. .

Tp Tp

.

.5

Tp

d

d

Tp

.5 .5

d d

n .8 .

Ts

d

. .8


17/28

Problem 3

Solution

PD Controller

I

s=tf('s');G=1/(s*(s+4)*(s+6)*(s+10));rlocus(G)

grid on

II

-25 -20 -15 -10 -5 0 5 10 15-20

-15

-10

-5

0

5

10

15

200.160.340.50.640.76

0.86

0.94

0.985

0.160.340.50.640.76

0.86

0.94

0.985

5101520

Root Locus




18/28

K=413


M(s) =413

(:4)(:6)(:10)

s=tf('s');M = 413/(s*(s+4)*(s+6)*(s+10));F = feedback(M,1);pole(F)

ans =

-9.1188 + 1.9910i

-9.1188 - 1.9910i

-0.8812 + 1.9910i

-0.8812 - 1.9910i

Root Locus


ImaginaryAxis(seco

nds-1)

-0.895 -0.89 -0.885 -0.88 -0.875 -0.87 -0.865 -0.861.975

1.98

1.985

1.99

1.995

2

2.005

2.01

System: G

Gain: 413

Pole: -0.878 + 1.99iDamping: 0.404

Overshoot (%): 25


0.3940.3960.3990.4010.4040.406

0.408

0.411

2.15

2.16

2.17

2.17

2.17

2.18

2.19


19/28

The approximation of the system to be a second order is valid.

III Settling time of uncompensated system

Ts 4

4

0.7 .5 (sec)

Ts 2 (sec)

IV. Desired location for the compensated poles:

Real part dnew 4

Ts

Imaginary part dnew n. .5

V. System with PD compensator looks like:

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

-2

-1

0

1

2

3

4

5

6

7 0.160.340.50.64

0.76

0.86

0.94

0.985

0.985

1

2

3

4

5

6

7

1

2

Root Locus




20/28

Location of zero

VI Gain(K) of th PD Compensator

s=tf('s');M = (s+2.96)/(s*(s+4)*(s+6)*(s+10));rlocus(M)grid on

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

-2

-1

0

1

2

3

4

5

6

7 0.160.340.50.64

0.76

0.86

0.94

0.985

0.985

1

2

3

4

5

6

7

1

2

Root Locus




21/28

K=294

VII Designed PD Compensator

Gc(s) = 294(s+2.96)

K1 = 870.24

K2= 294

VIII Validation of the second order assumption

ans =

s=tf('s');M = (294*(s+2.96))/(s*(s+4)*(s+6)*(s+10));F=feedback(M,1);pole(F)

-2.06 -2.04 -2.02 -2 -1.98 -1.96 -1.94 -1.92

4.46

4.48

4.5

4.52

4.54

4.56

4.58

4.6

4.62

4.64

System: M

Gain: 294

Pole: -2 + 4.53i

Damping: 0.404Overshoot (%): 25

Frequency ( rad/s): 4.95

0.3840.3890.3940.3990.4040.409

0.413

0.418

4.85

4.88

4.9

4.92

4.95

4.98

5

Root Locus

Real Axis (seconds -1)



22/28

-13.3382

-1.9991 + 4.5275i

-1.9991 - 4.5275i

-2.6637

Second order system approximation is valid.

IX Step responses for uncompensated and compensated system

Upon graphing the step responses, Ts=1.78 (sec) which satisfies our desired conditions (T s )

and (overshoot response 5%).

Step Response

Time (seconds)

Amplitude

0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

1.4System: F


Overshoot (%): 18.7


System: H


Overshoot (%): 23.6


System: F


System: HSettling Time (seconds): 4.07


23/28

PI Controller

Test point 1

s=tf('s');G=((s+0.1)*(s+2.96))/(s^2*(s+4)*(s+6)*(s+10));rlocus(G)grid on

K = 292


s=tf('s');G=(292*(s+2.96)*(s+0.1))/(s^2*(s+4)*(s+6)*(s+10));F=feedback(G,1);step(F)

Ts > 3.5 (sec)

Test point 2


K=286


24/28



Ts = 5.05 (sec)

Based on the first to test points as we increse the value of zero the value of Ts also increases, which

we should avoid in order to achieve the desired parameter. So this time we try to adjust the decimal

point backwards.

Test point 3


K=294



Ts = 1.78 (sec)

Ts , Therefore at this location of zero is valid for our PI design.

Designed PID Compensator

K1 + K2s+3

=

294(:2.96)(:0.003)

= 871.122+294s++

2.61


25/28

Validation of second order system approximation

s=tf('s');G=(294*(s+0.003)*(s+2.96))/(s^2*(s+4)*(s+6)*(s+10));F=feedback(G,1);pole(F)

ans =

-13.3378

-1.9980 + 4.5262i

-1.9980 - 4.5262i

-2.6632

-0.0030

Approximation is valid.

Plot of responses (Uncompensated, PD, PID)

s=tf('s');G=(294*(s+0.003)*(s+2.96))/(s^2*(s+4)*(s+6)*(s+10));PID=feedback(G,1);H= (413)/(s*(s+4)*(s+6)*(s+10));U=feedback(H,1);J=(294*(s+2.96))/(s*(s+4)*(s+6)*(s+10));PD=feedback(J,1);step(U,PD, PID)


26/28

Computation of some parameters:

Kp =

es = 0

Kv = 413/(4)(6)(10) = 1.72

er =1/1.72 = 0.58

Kp=

es = 0

Kv= 294(2.96)/(4)(6)(10) = 3.63

er = 1/3.63 = 0.275

Mark Jayson Mercado()

20128048

Step Response

Time (seconds)

Amplitude

0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

1.4

System: U


System: PID


System: U


Overshoot (%): 23.6


System: PID


Overshoot (%): 18.8


System: PD


Overshoot (%): 18.7


System: PID

Time (seconds): 1.88

Amplitude: 0.995


27/28

Control Engineering()

Homework 8


28/28

PI&PD Controller Sample Problem

Documents

Transcript of PI&PD Controller Sample Problem