# Compensator Assignment (Without introduction and conclusion)

**Topics:**Control theory, Transfer function, Electronic amplifier

**Pages:**9 (985 words)

**Published:**February 9, 2014

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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