lab assignment

Assignment

Submitted to sir Umer RanaSubmitted by Mariyam Zafar

Roll no 916

B.E 6

Lahore College for Women University Lahore

Experiment :

Objective:To evalute the effect of pole and zero location upon the time response of first and second order system.

Pre lab:1.Given the transfer function G(s)= a/ (s+a). Evalute settling time and rise time for the following values of a=1,2,3,4.plot the poles

Solution:

For a=1

In the command window

>>a=1

a =

1

>> numf=[a]

numf =

1

>> denf=[1 a]

denf =

1 1

>> t=tf([numf],[denf]) Transfer function: 1-----s + 1

Then draw the diagram in the simulink. Write simulink in the command window. then draw the block diagram of the transfer function. It will be like this

By clicking on the scope we get

In the command window, write>>p=pole(t)p =

-1>>z=zero(t)z =

Empty matrix: 0-by-1

>>y=pzmap(t)y =

-1

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, rise time and settling time.

For a =2


>>a=2

a =

2

>> num=[a]

num =

2

>> den=[1 a]

den =

1 2

>> t=tf([num],[den]) Transfer function: 2-----s + 2

Then draw the diagram in the simulink. Write simulink in the command window. then draw the block diagram of the transfer function. It will be like this.

By clicking on scope we get.

In the command window, write

>> p=pole(t)

p =

-2

>> z=zero(t)

z =


>> y=pzmap(t)

y =

-2

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, rise time and settling time

For a=3

In the command window ,write

>>a=3

a =

3

>> num=[a]

numf =

3

>> den=[1 a]

den =

1 3

>> t=tf([num],[den]) Transfer function: 3-----s + 3 Then draw the diagram in the simulink. Write simulink in the command window. then draw the block diagram of the transfer function. It will be like this.

By clicking on scope we get

In the command window write

p =pole(t)p=

-3

>> z=zero(t)

z =


>> y=pzmap(t)

y =

-3


For a=4


>>a=4

a =

4

>> num=[a]

num=

4

>> den=[1 a]

den =

1 4

>> t=tf([num],[den]) Transfer function: 4-----s + 4 Then draw the diagram in the simulink. Write simulink in the command window. then draw the block diagram of the transfer function. It will be like this



>> p=pole(t)

p =

-4

>> z=zero(t)

z =


>> y=pzmap(t)

y =

-4

For 2nd order system

2. Given the transfer function G(s)=b/(s*2+as+b)a) evaluate percent overshoot, settling time, peak time, and rise time for the following values, a=4, b= 25. Also plot the poles.

Solution:


>>a=4

a =

4

>> b=25

b =

25

>> num=[b]

num=

25

>> den=[1 a b]

den =

1 4 25

>> t1=tf([num],[den]) Transfer function:

25--------------

s^2 + 4 s + 25

then draw the diagram in the simulink. Write simulink in the command window. then draw the block diagram of the transfer function. It will be like this


In the command window, write

>> p=pole(t1)

p =

-2.0000 + 4.5826i -2.0000 - 4.5826i

>> z=zero(t1)

z =


>> y=pzmap (t1)

y =

-2.0000 + 4.5826i -2.0000 - 4.5826i

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, peak time rise time and settling time,peak time and overshoot

b) Calculate the values of a and b so that imaginary part of the poles remains the same but the real part is increased 2 times over that of (a) and repeat prelab(2a). Now put the value of a=8, b=37

Solution:


>>a=8

a =

8

>> b=37b =

37

>> num=[b]

numf=

37>> den=[1 a b]

den =

1 8 37


37--------------

s^2 + 8s + 37

Then draw the diagram in the simulink. Write simulink in the command window. Then draw the block diagram of the transfer function. It will be like this


>> p=pole(t1)p =

-4.0000 + 4.5826i -4.0000 - 4.5826i>> z=zero(t1)

z =


>> y=pzmap(t1)

y =

-4.0000 + 4.5826i -4.0000 - 4.5826i

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole,peak time,overshoot, rise time and settling time

c) Calculate the values of a and b so that the imaginary part of the poles are remain the same but the real part is decreased1/2 time over of (a) and repeat prelab(2a).

Solution: now the values of a and b are a=2, b=22


>>a=2a =

2

>> b=22b =

22

>> num=[b]

num ==

22>> den=[1 a b]

den =

1 2


22--------------

s^2 + 2s + 22



>> p=pole(t1)

p =

-1.0000 + 4.5826i -1.0000 - 4.5826i

>> z=zero(t1)

z =


>> y=pzmap(t1)

y =

-1.0000 + 4.5826i -1.0000 - 4.5826i

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, peak time,overshoot, rise time and settling time

3 .a) For the system of prelab 2(a) calculate the values of a and b so that the real

part of the poles remains the same but the imaginary part is increased 2 times ove that of prelab 2(a) and repeat prelab 2(a)

b)Solution :

Now the values of a and b are. A=4, b=88

In the command window w

>>a=4a =

4

>> b=88

b =

88

>> num=[b]

num =

88

>> den=[1 a b]

den =

1 4 88


88--------------

s^2 + 4s + 88


In the command window >> p=pole(t1)

p =

-2.0000 + 9.1652i -2.0000 - 9.1652i

z=zero(t1)z =

Empty matrix: 0-by-1>> y=pzmap(t1)

y =

-2.0000 + 9.1652i -2.0000 - 9.1652i

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, peak time,overshoot ,rise time and settling time

b) For the system of prelab 2(a) calculate the values of a and b so that the real part of the poles remains the same but the imaginary part is increased 4 times over that of prelab 2(a) and repeat prelab 2(a).

Solution :

Now the values of a and b are A= 4, b=340


>>a=4a =

4

>> b=340

b = 340

>> num=[b]

num=

340

>> den=[1 a b]

den =

1 4 340


340--------------

s^2 + 4s + 340



>> p=pole(t1)

p =

-2.0000 +18.3303i -2.0000 -18.3303i>> z=zero(t1)

z =


>> y=pzmap(t1)

y =

-2.0000 +18.3303i -2.0000 -18.3303i

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, rise time peak time,overshoot and settling time

4a) For the system of prelab 2(A) calculate the values of a and b so that tha

damping ratio remains the same but the natural frequency is increased 2 times over of prelab 2(a) and repeat prelab 2(a).

Solution :

Now the values of a and b are A= 4,b=25

In the command window >> a=4

a =

4

>> b=25

b =

25

>> omega=sqrt(b)

omega =

5

>> eeta=(a/(2*omega))

eeta =

0.4000

>> omega=10

omega =

10

>> eeta=0.4

eeta =

0.4000

>> b=omega*omega

b =

100

>> a=2*eeta*omega

a =

8

>> num=[b]

num =

100

>> den=[1 a b]

denf=

1 8 100

>> t=tf([num],[den]) Transfer function: 100---------------s^2 + 8 s + 100


In the command window write

p=pole(t)

p =

-4.0000 + 9.1652i -4.0000 - 9.1652i

>> z=zero(t)

z =


>> y=pzmap(t)

y =

-4.0000 + 9.1652i -4.0000 - 9.1652i

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, rise time, peak time,overshoot and settling time

b) For the system pf prelab 2 (a) calculate the values of a and b so that the damping ratio remains the same but the natural frequency is increases 4 times over that of prelab 2(a) and repeat prelab 2(a).

Solution :


>> eeta=0.4

eeta =

0.4000

>> omega=20

omega=

20

>> b=omega*omega

b =

400

>> a=2*eeta*omega

a =

16

>> num=[b]

num=

400

>> den=[ 1 a b]

den =

1 16 400

>> t=tf([num],[den]) Transfer function: 400----------------s^2 + 16 s + 400




>> p=pole(t)

p =

-8.0000 +18.3303i -8.0000 -18.3303i

>> z=zero(t)

z =

Empty matrix: 0-by-1>> y=pzmap(t)

y =

-8.0000 +18.3303i -8.0000 -18.3303i

Write ltiview in command window then import the above transfer function (G) the graph will be like this with its pole, rise time and settling time,peak time and overshoot.

LAB1) Using simulink set up the system of prelab 1 and plot the step response of

each of four transfer functions on a single graph by using the simulink LTI viwer. Also record the values of settling time and rise time for each step response.

Solution

Use the following command in command window>> a1=tf([1],[1 1]) Transfer function: 1-----s + 1

>> a2=tf([2],[1 2])

Transfer function: 2-----s + 2

>> a3=tf([3],[1 3]) Transfer function: 3-----s + 3 >> a4=tf([4],[1 4]) Transfer function: 4-----

s + 4

>> step(a1,a2,a3,a4)

Now for LTI VIEWER (for rise time and settling time)

>> ltiview Import the transfer functions one by one on tha single graph.

Settling time for a1=3.91 and rise time for a1=2.2.

Settling time for a2=1.96 and rise time for a2=1.1

Settling time for a3=1.3and rise time for a3=0.732

Settling time for a4=0.978and rise time for a4=0.549

1) Using simulink set up the system of prlab 2using the simulink LTI VIEWER plot the step response of each of the 3 transfer functions on a single graph . tlso record the values of percentage overshoot,setlling time,peak time and rise time of each step response

Use the following command in command window

>> b1=tf([25],[1 4 25]) Transfer function: 25--------------s^2 + 4 s + 25 >> b2=tf([37],[1 8 37]) Transfer function: 37--------------s^2 + 8 s + 37

>> b3=tf([22],[1 2 22]) Transfer function: 22--------------s^2 + 2 s + 22

>> step(g,g1,g2)

>> ltiview

B1Percent overshoot =25.4Settling time =1.3Rise time =0.732

B2Percent overshoot=6.44Settling time=0.988Rise time=o.329

B3Percent overshoot=50.4Settling time=3.64Rise time=0.261

1) Using simulink , set up the system of prelab 2 a)and prelab 3. Using the simulink LTI viewer plot the step response of each of the 3 transfer functions on a single graph . also record the values of percent overshoot , settling time,peak time and rise time for each step response.

Solution: Use the following commands in command window

>> c1=tf([25],[1 4 25]) Transfer function: 25--------------s^2 + 4 s + 25 >> c2=tf([88],[1 4 88]) Transfer function: 88--------------s^2 + 4 s + 88 >> c3=tf([340],[1 4 340]) Transfer function: 340---------------s^2 + 4 s + 340 >> step(f,f1,f2)

>> ltiview

for c1:Percent overshoot=25.4Settling time=1.3Rise time=0.732

For c2:Percent overshoot=50.4Settling time=1.82Rise time=0.131

For c3:Percent overshoot=70.9Settling time=1.91Rise time=0.0607

4) simulink set up the system of prelab 2 a) and prelab 4 using the simulink LTI viewer , plot the step response of ech of the three transfer functions on a sigle graph. Also record the values of percent overshoot, settling time, peaktime and a rise timr for each response

Solution:

Used the following commands in command window

>> d1=tf([25],[1 4 25]) Transfer function: 25--------------s^2 + 4 s + 25 >> d2=tf([100],[1 8 100]) Transfer function: 100---------------s^2 + 8 s + 100

>> d3=tf([400],[1 16 400]) Transfer function: 400----------------s^2 + 16 s + 400

>> step(r,r1,r2) >> ltiview

>> r=tf([25],[1 4 25]) Transfer function: 25--------------s^2 + 4 s + 25 >> r1=tf([100],[1 8 100]) Transfer function: 100---------------s^2 + 8 s + 100

>> r2=tf([400],[1 16 400]) Transfer function: 400----------------s^2 + 16 s + 400

In order to draw the three step responses on a single graph

>> step(r,r1,r2)

Now for LTI VIEWER (for rise time and settling time,over shoot and peak time)

>> ltiview

Import the transfer functions one by one on tha single graph

for d1:Percent overshoot=25.4Settling time=1.3Rise time=0.732

For d2:Percent overshoot=25.4Settling time=0.147Rise time=0.841

For d3:Percent overshoot=25.4Settling time=0.42Rise time=0.0734

lab assignment

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