control final document

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1. Introduction

2. Discussion

Plant Transfer Function

)1093277.310124.60(

1093277.3932

9

sstf

For this transfer function an overshoot of 18% and zero steady state error was obtain. And the later

specifications were to reduce the overshoot to 5% at zero steady state error. The performance

improvements of this 2nd

order plant are presented in this document. The characteristics of P,I and Dcontrollers is given below to help when choosing what type of controller to design for.

Figure 1:P,I and D controller characteristics

3. Controller Design Calculations

The Integral controller Ki decreases the rise time increases both the overshoot and settling time but it

eliminates the steady state error of type zero system to zero.

A proportional integral controller will be used to eliminate the steady state error of the system and

keep it at zero. The general transfer function of a PI controller is as follows:

To calculate Kp a value of 0.01 steady state error is assumed.Therefore….



99

1

101.0

1

1

p

p

p

ss

K

K

K e

And the transfer function is

)1093277.310124.60(

1093277.3)(932

9

sss

KiKpstf

To find Ki again a steady state error of 0.01 is assumed. Therefore…..

100

101.0

1

i

i

i

ss

k

k

k e

i

f

p R

RK

Let R1 = 1kΩ thus,

k

k R f

99

991

uF

k C

C RK

f

f i

i

10

1001

1

1



3.2 Derivative Controller Design

The derivative control reduces both the overshoot and the settling time of the system.The derivative

part of the controller will be used to decrease the overshoot of the system.

The closed loop transfer function is

nd

d

d

d

d

wK

sK s

sK

sK ss

sK TF

21093277.310124.60

1093277.3)1093277.310124.60(

1093277.3)(

10277.93.3)(10277.93.310124.60

10277.93.3)(

93

9932

9

9932

9

For a 5% oveshoot the damping ration will be calculated using the formular below.

22)]100 / [ln(%

)100 / ln(%

OS

OS

For an overshoot of 5% a damping ratio of 0.32 is required

but practically this was not not achieved,

therefore a damping ratio of 2 is assumed.

6

339

93

107,620

10124.60)103.624)(2(21093277.3

21093277.310124.60

kd

kd

kd n

To calculate the resistor and capacitor values Let iC = 1uF

7.620

1

107.6206

uF R

C RK

f

i f d

3.3 Inverter

Because the transfer function of both the integrator and the differentiator is negative and there is also

a negative output from the summing point, an inverter will be used to invert the signal into a positive

signal to the plant.



4. Final Schematic

5.Simulation Results

5.1

MultiSim Results

5.2 Matlab Simulation Results

Root locus Plot of the plant

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

x 104

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5x 10

5

0.32

0.32

Root Locus

Real Axis

I m a g i n a r y A x i s



Matlab Script File

num =3932769154;

den = [1 60124.1748264 num];

Gp = tf(num,den)

Gc =tf ( [620.7e-6 99 100], [0 1 0 ]);

g1 = Gp*Gc;

Go = series(g1,1);

T = feedback (Go,1)

ltiview('step',T);

Transfer function:

3.933e009

-----------------------------

s^2 + 6.012e004 s + 3.933e009

Closed Loop Transfer Function

Transfer function:

2.441e006 s^2 + 3.893e011 s + 3.933e011

---------------------------------------------

s^3 + 2.501e006 s^2 + 3.933e011 s + 3.933e011

Overall Output Response

Step Response

Time (sec)

A m p l i t u d e

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 10-5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

System: T

Peak amplitude: 1.03

Overshoot (%): 2.66

At time (sec): 2.73e-006

System: T

Final Value: 1

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

x 105

-1.5

-1

-0.5

0

0.5

1

1.5x 10

5 Root Locus

Real Axis

I m a g i n a r y A x i s

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