Compensator Assignment (Without introduction and conclusion)

Topics: Control theory, Transfer function, Electronic amplifier Pages: 9 (985 words) Published: February 9, 2014
Abstract
In this assignment, lead compensator is implemented to reduce the percentage of peak overshot to less than 5% and the settling time to 2 second with 2% accuracy. First of all, to implement lead compensators into a circuit, a pole–zero pair is introduced into the open loop transfer function. Lead compensator is a component in a control system that improves an undesirable frequency response in a feedback and control system. Besides that, it also provides phase lead in its frequency response. Lead compensator also helps to improve transient response and a small change in steady-state accuracy.Therefore, modification of the circuit are to be taken place to meet the desired performance specification. Introduction

Product Design Description

Figure 1: The circuit of the assignment before modification

Bill Of Material ( BOM )
Item
Designation
Component Description
Value
Maker/Part Number
Supplier
Quantity
Per Unit Cost (RM)
Total Unit Cost (RM)

Original Part
1
R1
Resistor, Lead Type
1kΩ
N/A
I.E
4
0.05
0.20
2
R2
Resistor, Lead Type
1MΩ
N/A
I.E
1
0.05
0.05
3
R3
Resistor, Lead Type
500kΩ
N/A
I.E
1
0.05
0.05
4
R4
Resistor, Lead Type
400kΩ
N/A
I.E
2
0.05
0.10
5
R3
Resistor, Lead Type
80kΩ
N/A
I.E
1
0.20
0.05
6
C1
Ceramic Capacitor
1μF
N/A
I.E
3
0.20
0.60
7
TL082
Op-Amp
N/A
TL082
I.E
2
0.80
1.60
Gain Part
8
Ra
Resistor, Lead Type
500Ω
N/A
I.E
1
0.05
0.05
9
Rb
Resistor, Lead Type
620kΩ
N/A
I.E
1
0.05
0.05
Compensator Part
10
R1
Resistor, Lead Type
21kΩ
N/A
I.E
1
0.05
0.05
11
R2
Resistor, Lead Type
2.4kΩ
N/A
I.E
1
0.05
0.05
12
C1
Ceramic Capacitor
22μF
N/A
I.E
1
0.20
0.20
 
Total Amount:
RM 3.05

Part by Part Derivation
 
 

 

 
 The Figure below shows the block diagram of the system before modification.

Based on the derivation part by part, we know that:
Open Loop Gain =
Characteristic equation:
Number of poles = 3
Number of finite zeros = 0
Pole location: 0,-2,-5
The original circuit was tested using stimulation software (LTSpice and MATLAB). By using LTSpice, the step input applied to the circuit has result as shown in Figure 3 below.

Figure 3 shows the step response of the plant.
By using MATLAB, the step input applied to the transfer function has a result shown in figure 2 below.

Figure 4 Figure 5.
Figure 4 shows MATLAB result when a step input was applied to the transfer function Figure 5 shows MATLAB source code

Test Plan
Based on the assignment, the circuit is expected to achieve the performance specifications of less than 5% overshoot and settling time with 2% accuracy of less than 2 seconds. But after the stimulation has done, the output doesn't seems to meet both of the specification. From the specification above, we try to find the damping ratio (ζ) and frequency (ωn) first to know where the poles should be to achieve the specification. Finding ζ ,

 Plot the root locus using MATLAB,
We choose ζ = 0.7 and ωn=2.8 to plot inside the root locus.

Figure above shows the MATLAB source code to find the root locus

Figure above shows the root locus of the original circuit

The ζ must greater than 0.69 and the ωn must be greater than 2.79s. From the root locus above, our root locus must touch any point in the X region to achieve the performance specification of overshoot less than 5% and settling time less than 2 seconds of accuracy 2%.

 
Finding ωn,

So, a compensator need to be cascaded into the original circuit to bring the breakaway point to the left side so that it meet the specification. So, we need to calculate the gain for the compensator as the equation shown below.

First, a new characteristic equation is form by using the ζ and ωn found above.

After the poles was found, then they are use to draw the new root locus.
Figure 6 shows how to find the a and b...
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