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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….
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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
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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.
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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
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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