Control Lab 6 PDF

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Shahzaib Qadir FA18-EPE-Osama Idress FA18-EPE-Shahid Akhtar FA18-EPE-Class Control Systems Lab.Section CTeacher’s Name Dr. Uzair khanLab Engineer’s Name Muhammad Asad Javaid KhanDEPARTMENT OF ELECTRICAL ENGINEERINGCOMSATS UNIVERSTY ISLAMABD (ABBOTTABAD CAMPUS)Lab:06 Transient analysis of first order...

Description

Shahzaib Qadir

FA18-EPE-087

Osama Idress

FA18-EPE-117

Shahid Akhtar

FA18-EPE-100

Class

Control Systems Lab.

Section

C

Teacher’s Name

Dr. Uzair khan

Lab Engineer’s Name

Muhammad Asad Javaid Khan

DEPARTMENT OF ELECTRICAL ENGINEERING COMSATS UNIVERSTY ISLAMABD (ABBOTTABAD CAMPUS)

Lab:06 Transient analysis of first order systems Objective: The following lab would help the students to get familiar with first order systems and also with specifications of the first order systems.

Task 1 Find the pole location, rise time and settling time of the following PART A A)

G ( s )=

1 S+ 1

CODE a=1; ta=1/a; t=0:0.01:5*ta; x=(1-exp(-t)).*unit(t); plot(t,x); grid on; xlabel('Time'); ylabel('C(t)'); title('Transiant Analysis'); n=[1]; dn=[1 1]; figure; zplane(n,dn) title ('Pole Zero') figure; sys=tf(n,dn); pole(sys) grid on pzmap(n,dn) figure; step(sys) grid on stepinfo(sys) grid on

FIGURES:

Figure 1. Transiant Analysis of G ( s )=

1 S+1

Figure 2. Pole Zero graph of G ( s )=

1 S+ 1

Figure 3. Step response of

G ( s )=

1 S+1

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PART B B)

G ( s )=

5 s+ 2

CODE a=2; ta=1/a; t=0:0.01:5*ta; x=((5/2)-(2/5)*exp(-2*t)).*unit(t); plot(t,x); grid on; xlabel('Time'); ylabel('C(t)'); title('Transiant Analysis'); n=[5]; dn=[1 2]; figure; zplane(n,dn) grid on title ('Pole Zero') figure; sys=tf(n,dn); pole(sys)

grid on pzmap(n,dn) figure step(sys) grid on stepinfo(sys) grid on

FIGURES:

Figure 4. Transiant Analysis of G ( s )=

5 S+2

Figure 5. Pole Zero graph of G ( s )=

Figure 6. Step response of G ( s )= RiseTime SettlingTime SettlingMin SettlingMax Overshoot Undershoot Peak PeakTime

: : : : : : : :

5 S+2

5 S+2

1.0985 1.9560 2.2613 2.4999 0 0 2.4999 5.2729

TABLE: Pole Location -1 -2

Rise time calculated 2.2 1.1

Rise time observed 2.190 1.098 Table No:1

CONCLUSION:

Setting time calculated 4 2

Setting time observed 3.912 1.956

When pole move towards the left side Rise Time and Setting time will decreases. When rise time decreases the speed of system increases which means system will be fast and it settling time is less which means it will required less time to become stable. -----------------------------------------------------------------------------------------------------------------------

Task 2 For the RC circuit given below, find equation for output voltage, pole location, rise time and settling time when the applied voltage is �� (�) = step voltage, For the circuit 50٠and C =5Mf.

SOLUTION: To Find the equation we have to apply the VDR

EQUATIONS:

CODE: n=[1]; dn=[0.25 1]; zplane(n,dn) figure title('Pole zero'); sys=tf(n,dn) pole(sys) pzmap(n,dn); grid on figure step(sys); grid on figure stepinfo(sys); grid on figure

FIGURES:

Figure 1. Pole zero graph of the RC circuit

Figure 1. Step Response graph of the RC circuit

TABLE: Pole location Rise time calculated Rise time observed Setting time calculated Setting time observed

-4 0.55 0.549 1 0.978 Table No:2

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Task 3

A motor is an electromechanical component that yields an angular displacement output, θm , for an input voltage, e a , that is, a mechanical output generated by an electrical input. The motor’s schematic is shown in Figure 6.4(a) and its transfer function for motor speed response is shown in Figure 6.4(b).

The transfer function will be;

EQUATION:

Case 1 When K=1 and alpha varies from 1-5

CODE: for alpha = 1:5 k=1; s=tf('s'); G=k/(s+alpha); step(G) grid on hold on end

FIGURE:

Figure 1. Step response of electrical system connected with motor for Case 1

Case 2: When K=5 and alpha varies from 1-5

CODE: for alpha = 1:5 k=5; s=tf('s'); G=k/(s+alpha); step(G) grid on hold on end

FIGURE:

Figure 1. Step response of electrical system connected with motor for Case 2

CONCLUSION:

When pole move towards the right side Rise Time and Setting time will increases. When rise time increases the speed of system decreases which means system will be slow and its settling time is more so it will requires more time to become stable. On the other hand when pole move towards the left side shown in Rise Time and Setting time will decrease. When rise time decreases the speed of system increases which means system will be fast and it settling time is less so it will require less time to become stable. K effect the stability value of the system.

THE END...