control system labs all

8/7/2019 control system labs all

1/87

Object: To become familiar with the basics of Mat lab.

Tools:

1) IBM PC

2) Mat lab 7 Version

Description:

What is MATLAB

The name MATLAB is short for Matrix Laboratory. It is a commercial software

package whose main function is to perform calculations on matrices, row vectors and

column vectors. It is widely used in both industry and academic institutions. It

possesses many of the features of a high level, numerically oriented programming

language, but in addition has a large collection of built-in functions for performing

matrix and vector operations in a way that is very simple for the user. For example, tofind the determinant of a matrix A one need only enter:

det(A)

MATLAB commands can be entered one line at a time, or the user can write programs

of MATLAB code and define his or her own functions. In this session, we shall only

use the one-line-at-a-time interpretive mode for simplicity, but the regular user will

find that the power of the package is considerably extended by writing his or her own

programs and functions.

Entering and quitting MATLAB

To enter MATLAB, simply double click on the MATLAB icon. To leave MATLAB

and return to the PCs operating system

Simply type: quit

Lab Practical # 01

An Introduction to MATLAB


2/87

Aisha Muslim:

BS (TE)-044-07F

v = [2 4 7 5];w = [1 3 8 9];

Exercise#1:Investigate the effect of the following commands:

(a) v(2) (b) sum = v + w (c) diff = w - v (d) vw = [v w]

(e) vw(2:6) (f) v'

(a) v(2)

ans =

4

(b) sum = v + w

sum =

3 7 15 14

(c) diff = w - v

diff =

-1 -1 1 4

(d) vw = [v w]

vw =

2 4 7 5 1 3 8 9

(e) vw(2:6)

ans =

4 7 5 1 3

(f) v'

ans =

2

4

7

5

Control systems lab manual


3/87

Aisha Muslim:

BS (TE)-044-07F

Now the matlab window shows all these commands excuted in command window

=

0

0

1

1

z

v = [2 4 7 5];

Exercise#2: Investigate the effect of the following commands

(i) z' (b) z*v (c) [v; w] (d) v*z (e) [z; v] (f) z + v

(i) z'

ans =

1 1 0 0

(b) z*v

ans =



4/87

Aisha Muslim:

BS (TE)-044-07F

2 4 7 5

2 4 7 5

0 0 0 00 0 0 0

(c) [v; w]

ans =

2 4 7 5

1 3 8 9

(d) v*z

ans =

6




5/87

Aisha Muslim:

BS (TE)-044-07F

(e) [z; v']

ans =

1

10

0

2

4

7

5

(f) z + v'

ans =

3

5

7

5

Now the matlab window shows all these commands window executed in command

window;



6/87

Aisha Muslim:

BS (TE)-044-07F

=

43

21M

Exercise#3:Investigate the effect of the following commands:

(a) N = inv(M) (b) M*N (c) I = eye(2) (d) M + I (e) M*z(1:2) (f) v(3:4)*M

(g) M(1,1) (h) M(1:2,1:2) (i) M(:,1) (j) M(2,:)

(a) N = inv(M)

N =

-2.0000 1.00001.5000 -0.5000

(b) M*N

ans =



7/87

Aisha Muslim:

BS (TE)-044-07F

1.0000 0

0.0000 1.0000

(c) I = eye(2)

I =

1 0

0 1

(d) M + I

ans =

2 2

3 5




8/87

Aisha Muslim:

BS (TE)-044-07F

(e) M*z(1:2)

ans =

3

7

(f) v(3:4)*M

ans =

22 34

(g) M(1,1)

ans =

1



9/87

Aisha Muslim:

BS (TE)-044-07F

(h) M(1:2,1:2)

ans =

1 2

3 4

(i) M(:,1)

ans =

1

3

(j) M(2,:)

ans =

3 4




10/87

Aisha Muslim:

BS (TE)-044-07F

Exercise#4: Use the help command to find out about the following built-in functions

and make up your own simple examples:1. ones 2. zeros 3. det 4. Linspace 5. Logspace

1. ones

ONES(N) is an N-by-N matrix of ones.

ONES(M,N) or ONES([M,N]) is an M-by-N matrix of ones.

2. zeros

ZEROS Zeros array.ZEROS(N) is an N-by-N matrix of zeros.

ZEROS(M,N) or ZEROS([M,N]) is an M-by-N matrix of zeros.

3. det

DET Determinant.

DET(X) is the determinant of the square matrix X.

4. Linspace

LINSPACE Linearly spaced vector.

LINSPACE(X1, X2) generates a row vector of 100 linearly

equally spaced points between X1 and X2.

5. Logspace

LOGSPACE Logarithmically spaced vector

LOGSPACE(X1, X2) generates a row vector of 50 logarithmically

equally spaced points between decades 10^X1 and 10^X2. If X2is pi, then the points are between 10^X1 and pi.

t=linspace(0,20,100);

Exercise#5:

1. Generate a vector yc containing the values of cos(t) for the time range givenabove.

The program code is



11/87

Aisha Muslim:

BS (TE)-044-07F

ys = sin(t);

To plot the set of points type:Plot(t,ys)The basic graph can be prettied up using the following self explanatory

sequence of commands:

xlabel('Time in seconds')ylabel('sin(t)')title(' Roll No=44')

grid



12/87

Aisha Muslim:

BS (TE)-044-07F

2. From the help entry for plot find out how to get both ys and yc plotted against t

on the same axis.

PLOT Linear plot.

PLOT(X,Y) plots vector Y versus vector X. If X or Y is a matrix,

then the vector is plotted versus the rows or columns of the matrix,

whichever line up. If X is a scalar and Y is a vector, length(Y)

disconnected points are plotted.

3. From the help entry for subplot find out how to plot ys and yc plotted on two

separate graphs, one above the other.

SUBPLOT Create axes in tiled positions.

H = SUBPLOT(m,n,p), or SUBPLOT(mnp), breaks the Figure window

into an m-by-n matrix of small axes, selects the p-th axes for

for the current plot, and returns the axis handle. The axes

are counted along the top row of the Figure window, then thesecond row, etc. For example,

SUBPLOT(2,1,1), PLOT(income)

SUBPLOT(2,1,2), PLOT(outgo)



13/87

Aisha Muslim:

BS (TE)-044-07F

Exercise#1:

(a) Find zeros of transfer function (if any)

Solution:

This function can be stored in the MATLAB workspace by typing

num = 100;den = [1 14 100];

zeros=roots (num)

zeros =

Empty matrix: 0-by-1

(b) Is the above system stable? Why?Answer:

yes, the system is stable because the real parts of complex poles lie in the left half

plane of the s plane.

(c) Use the plot function to plot the poles of the above transfer function marked by

the symbol x. Record this plot in your lab book with your Roll. No on the top

of the plot.


G(S) ==100

S2+14S+100

Lab Practical #02

Basic Control Functions


14/87

Aisha Muslim:

BS (TE)-044-07F

Exercise#2:

(a) What is the difference between the built-in function conv and series?

Answer:

Conv: this function first computes the numerator and denominator of theresultant transfer function separately by multiplying the respective entity.

Series: this is the built in function in mat lab to compute the transfer function oftwo functions attached in series by accepting the four parameters of the num and

denominator of series functions.

(b) Reduce each of the following block diagrams into a single block form. In each

case find poles and zeros and sketch pole zero diagram. In which block diagrams

you cannot use the function series? Why? Also determine whether the system is

stable or not?

(i)

In these diagrams we can use the series function because there are only two blocks connected

in series. So the code for single block is

num1 = 99;

den1 = [9 27];

num2 = 5*[3 27];

den2 = 7* [1 12 9 27];

[num12, den12] = series (num1, den1, num2, den2);printsys (num12, den12)

num/den =

1485 s + 13365

-------------------------------------------

63 s^4 + 945 s^3 + 2835 s^2 + 3402 s + 5103


999S + 27

5(3S+27)

7(S3+12S

2+9S+27


15/87

Aisha Muslim:

BS (TE)-044-07F

poles=roots(den12)

poles =

-11.4189

-3.0000

-0.2905 + 1.5100i

-0.2905 - 1.5100i

zeros=roots(num12)

zeros =

-9

pzmap(poles,zeros)



16/87

Aisha Muslim:

BS (TE)-044-07F

The system is stable because all poles lie on the left half plane.

(ii)

In these diagrams we can not use the series function because there are three blocks connected inseries. So the code for single block is

num1 = [2 4];

den1 = [1 0 0 0];

num2 = 4/5;

den2 = [1 3];

num3=[1 0 0];

den3=[1 -6];

num12 =conv(num1,num2);

den12 = conv(den1,den2);


4/5S + 3

2S + 4

S3

S2

S - 6


17/87

Aisha Muslim:

BS (TE)-044-07F



printsys (num123, den123)

num/den =

1.6 s^3 + 3.2 s^2

--------------------

S^5 - 3 s^4 - 18 s^3

poles=roots(den123)

poles =

0

0

0

6.0000

-3.0000

zeros=roots(num123)

zeros =

0

0

-2

The mat lab command window for above question is following;



18/87

Aisha Muslim:

BS (TE)-044-07F

pzmap(poles,zeros)



19/87

Aisha Muslim:

BS (TE)-044-07F

The system is unstable because pole of the function lie on the right half plane at s=6as well as on the right half plane.

(iii)

In these diagrams we can not use the series function because there are three

blocks connected in series. So the code for single block is

num1 = [1 0 0];

den1 = [1 3];

num2 = [1 6];

den2 = [1 1 1];

num3=[9 0];

den3=[1 8];

num4=22;

den4=[1 5.6];


22S + 5.6

S + 6

S2+S+1

S2

S + 3

9SS + 8


20/87

Aisha Muslim:

BS (TE)-044-07F








num/den =

198 s^4 + 1188 s^3

-----------------------------------------------

s^7 + 10.6 s^6 - 14 s^5 - 379.2 s^4 - 806.4 s^3

poles=roots(den1234)

poles =

0

0

0

6.0000

-8.0000

-5.6000

-3.0000

zeros=roots(num1234)

zeros =

0

0

0

-6



21/87

Aisha Muslim:

BS (TE)-044-07F

The mat lab window of the above question is following;

pzmap(poles,zeros)



22/87

Aisha Muslim:

BS (TE)-044-07F

Exercise#3(a) Consider the following block diagram. Find the closed loop transfer function.

Solution:num1 = 9;

den1 = [1 5];

num2 = 1;

den2 = 1;

[n,d]=feedback(num1,den1,num2,den2);

printsys(n,d)

num/den =9


C(S)R(S)-

9S + 5


23/87

Aisha Muslim:

BS (TE)-044-07F

------

s + 14

(b) Find the closed loop transfer function of the following block diagram.

Solution:num1 = [1 2];

den1 = [1 3 12];

num2 = 1;

den2 = 1;

[n,d]=feedback (num1, den1, num2,den2,+1);

printsys(n,d)

num/den =

s + 2

--------------

s^2 + 2 s + 10

(c) Find the closed loop transfer function of the following block diagram.

Solution:


C(S)R(S)+

S + 2S2 +3S +12

C(S)R(S)-

1S +1

2S


24/87

Aisha Muslim:

BS (TE)-044-07F

num1 = 1;

den1 = [1 1];

num2 = 2;

den2 = [1 0];


printsys(n,d)

num/den =

s

------------

s^2 + s + 2

(d) Find the closed loop transfer function of the following block diagram.

num1 = [1 0];

den1 = [1 6];

num2 = [2 1];

den2 = [1 2];


printsys(n,d)

num/den =

s^2 + 2 s

----------------

3 s^2 + 9 s + 12

The mat lab command window executing the above commands is shown asfollowing;


C(S)R(S)-

SS +6

2S+1S+2


25/87

Aisha Muslim:

BS (TE)-044-07F

Introduction

Simulink is a software package for modeling, simulating, and analyzing dynamic

systems. It supports linear and nonlinear systems, modeled in continuous time,

sampled time, or a hybrid of the two. One can easily build, analyze and visualize

models.

Thousands of engineers around the world using it to model and solve real problems.

Simulink provides a graphical user interface (GUI) for building models as blockdiagrams, using click-and-drag mouse operations. With this interface, you can draw


Lab Practical # 03

Introduction to Simulink


26/87

Aisha Muslim:

BS (TE)-044-07F

the models just as you would with pencil and paper. Knowledge of this tool will serve

you well throughout your professional career.

Exercise#1: - Perform the indicated tasks using SIMULINK.

(a) Find the step response of the following transfer function.

Solution:

First in the untitled page of mat lab simulinks we created this model by copying the

blocks into the model from the respective Simulink block libraries:

Sources library (the step response)

Sinks library (the Scope block)

Continuous library (the transfer function block)


S+1S+1

G(S)G(S) == 11


27/87

Aisha Muslim:

BS (TE)-044-07F

The result of above after choosing the Start from the Simulation menuand watch the simulation output on the Scope.



28/87

Aisha Muslim:

BS (TE)-044-07F

(c) Differentiate the ramp function and check the result on the scope.

Solution:

First in the untitled page of mat lab simulinks, we created this model by copying the


Sources library (the step response)


Continuous library (the transfer function block)



29/87

Aisha Muslim:

BS (TE)-044-07F

This is shown as below



30/87

Aisha Muslim:

BS (TE)-044-07F

The result of above after choosing the Start from the Simulation menuand watch the simulation output on the Scope.

(d) Amplify sine wave using gain block.Solution:

First in the untitled page of mat lab simulinks, we created this model by copying the


Sources library (the sine wave)




31/87

Aisha Muslim:

BS (TE)-044-07F

This is shown as below

After double clicking the gain block we get its properties block and setting the

gain from 1 to 2 first we get the amplified sine wave with the amplitude 2 even the

amplitude of the sine wave in the properties block is still 1 .



32/87

Aisha Muslim:

BS (TE)-044-07F

Now the result on the scope is shown below after clicking the start from the

simulation menu.



33/87

Aisha Muslim:

BS (TE)-044-07F

(e) Simulate the JK flip-flop in toggle mode and copy resultant waveform in youranswer book.SOLUTION:



34/87

Aisha Muslim:

BS (TE)-044-07F

(e) Explain briefly the function of each block you have used.

(f) Also explain in which library or sub-library this block is present?

1. Step block:Generate a step function

LibrarySources

Description:The Step block provides a step between two definable levels at a specified time. If the

simulation time is less than the Step time parameter value, the block's output is the

Initial value parameter value. For simulation time greater than or equal to the Step

time, the output is the Final value parameter value.

2. Transfer Fcn:It Implements a linear transfer function

Library:Continuous

Description:The Transfer Fcn block implements a transfer function where the input (u) and output

(y) can be expressed in transfer function form.A Transfer Fcn block takes a scalar input. If the numerator of the block's transfer

function is a vector, the block's output is also scalar. However, if the numerator is a

matrix, the transfer function expands the input into an output vector equal in width to

the number of rows in the numerator. For example, a two-row numerator results in a

block with scalar input and vector output. The width of the output vector is two.

3. Scope:Display signals generated during a simulation

Library

SinksDescription:The Scope block displays its input with respect to simulation time. The Scope

block can have multiple axes (one per port); all axes have a common time range

with independent y-axes. The Scope allows to adjust the amount of time and the

range of input values displayed. We can move and resize the Scope window and

you can modify the Scope's parameter values during the simulation. When we

start a simulation, Simulink does not open Scope windows, although it does

write data to connected Scopes. As a result, if we open a Scope after a

simulation, the Scope's input signal or signals will be displayed. If the signal is



35/87

Aisha Muslim:

BS (TE)-044-07F

continuous, the Scope produces a point-to-point plot. If the signal is discrete,

the Scope produces a stair-step plot.

4. Gain :Multiply the input by a constant

LibraryMath Operations

Description :The Gain block multiplies the input by a constant value (gain). The input and

the gain can each be a scalar, vector, or matrix. You specify the value of the

gain in the Gain parameter. The Multiplication parameter lets you specifyelement-wise or matrix multiplication. For matrix multiplication, this parameter

also lets you indicate the order of the multiplicands. The gain is converted from

doubles to the data specified in the block mask offline using round-to-nearest

and saturation. The input and gain are then multiplied, and the result is

converted to the output data type using the specified rounding and overflow

modes.

5. Sine Wave:Generate a sine wave

Library:Sources

Description:The Sine Wave block provides a sinusoid. The block can operate in either time-

based or sample-based mode.

Time-Based Mode:

The output of the Sine Wave block is determined by Time-based mode has two

submodes: continuous mode or discrete mode. The value of the Sample time

parameter determines whether the block operates in continuous mode or discrete

mode: 0 (the default) causes the block to operate in continuous mode. >0 causesthe block to operate in discrete mode.

Sample-Based Mode:

Sample-based mode uses the following formula to compute the output of the

Sine Wave block.

Y = a sin(2*pi*(k+0)/p+b)

where

A is the amplitude of the sine wave.

p is the number of time samples per sine wave period.

k is a repeating integer value that ranges from 0 to p-1.o is the offset (phase shift) of the signal.



36/87

Aisha Muslim:

BS (TE)-044-07F

b is the signal bias.

In this mode, Simulink sets k equal to 0 at the first time step and computes the

block's output, using the preceding formula. At the next time step, Simulinkincrements k and recomputes the output of the block. When k reaches p,

Simulink resets k to 0 before computing the block's output. This process

continues until the end of the simulation.



37/87

Aisha Muslim:

BS (TE)-044-07F

1. Introduction:A Control system can be represented as:

d/dx = Ax + BU

d/dx = Cx + DUWhere A, B, C and D are matrices, x is the state vector, y is the output vector and u is

the input vector. The above representation can be obtained from the differential

equation or transfer function of the system under considerations.

MATLAB has built-in functions tf2ss (transfer function to state space) andss2tf(state

space to transfer function) to convert transfer function model into state spacerepresentation and vice versa. Following examples demonstrate the use of both of

these functions.

Exercise#1: -Find the state space representation of the following transfer functions.

1).

Solution:

num= [1 2];den= [1 7 12];

[A, B, C, D]=tf2ss (num, den);

printsys (A, B, C, D)


G(S) =

S+2

S 2+ 7S + 12

Lab Practical # 04State Space Representation


38/87

Aisha Muslim:

BS (TE)-044-07F

After executing the above code in command window of mat lab we get the

following result

a =x1 x2

x1 -7.00000 -12.00000

x2 1.00000 0

b =

u1

x1 1.00000

x2 0

c =

x1 x2

y1 1.00000 2.00000

d =

u1

y1 0

2).

Solution:

num= [1 3 2];

den= [1 3 1 0];

[A, B, C, D]=tf2ss (num, den);printsys (A, B, C, D)


following result

a =

x1 x2 x3

x1 -3.00000 -1.00000 0

x2 1.00000 0 0

x3 0 1.00000 0


G(S) =

=

(S+1)(S +2)

S(S 2+3S+1)


39/87

Aisha Muslim:

BS (TE)-044-07F

b =

u1x1 1.00000

x2 0

x3 0

c =

x1 x2 x3

y1 1.00000 3.00000 2.00000

d =

u1

y1 0

Now the mat lab window shows the results of above questions:



40/87

Aisha Muslim:

BS (TE)-044-07F

Exercise#2:

Find the transfer functions of the following systems

(a).

Solution:

A= [-1 0 0; 0 -2 -3; 0 0 3];

B= [3;-6; 3];

C= [1 1 1];

D=0;

[num, den] =ss2tf (A, B, C, D);printsys (num, den)


following result

num/den =

4.2188e-015 s^2 + 9 s - 3

-------------------------S^3 - 7 s - 6



41/87

Aisha Muslim:

BS (TE)-044-07F

2).Solution:

A= [0 1 0; 0 0 1;-6 -11 -6];

B= [0; 0; 0];

C= [1 0 0];

D=0;

[num, den]=ss2tf (A, B, C, D);

printsys (num, den)

After executing the above code in command window of mat lab we get thefollowing result

num/den =

4.2188e-015 s^2 + 9 s - 3

-------------------------

S^3 - 7 s - 6

Eigen values

Exercise#3:

Find the Eigen values of the systems with

A=

A= [0 2 0 0 0;0.1 0.35 0.1 0.1 0.75;0 0 0 2 0;0.4 0.4 -0.4 -0.4 0;0 -0.03 0 0 -1];

B=eig (A)

After executing the above commands in mat lab command window we get;

B =

0.8061


0 2 0 0 0

0.1 0.35 0.1 0.1 0.75

0 0 0 2 0

0.4 0.4 -0.4 -0.4 0

0 -0.03 0 0 -1


42/87

Aisha Muslim:

BS (TE)-044-07F

-0.2223 + 0.9426i

-0.2223 - 0.9426i

-0.4321-0.9794

Now the mat lab window showing the above commands is given below;

b)... compute the eign values of the system given in exercise 2(a) and 2(b).1).

Solution:

A= [-1 0 0; 0 -2 -3; 0 0 3];

B=eig (A)

After execution we get

B =



43/87


44/87

Aisha Muslim:

BS (TE)-044-07F

(a): System given in (a) at t=4;

Solution:

t=4;

A= [-1 0 0; 0 -2 -3; 0 0 3];

B=expm (A*t)

After execution of the above code we have

B =

1.0e+005 *

0.0000 0 0

0 0.0000 -0.9765

0 0 1.6275

b) System given in (b) at t=0;

Solution:

t=0;A= [0 1 0; 0 0 1;-6 -11 -6];

B=expm (A*t)

After execution of the above code we have

B =

1 0 0

0 1 0

0 0 1



45/87

Aisha Muslim:

BS (TE)-044-07F

Now the mat lab window shows the above all programs;



46/87

Aisha Muslim:

BS (TE)-044-07F

1. Introduction

When a time domain signal is applied to a control system it produces a response of

that signal in time domain. This response of the system is called time response of a

control system. At the first instant the system exhibits some transients and after some

time it comes to steady state. While designing a control system the information of

transient response characteristics and the steady state characteristics is very important.

In this lab we will apply different test signals (Step, Impulse and Ramp) to a particular

control system and from the response of that signal we will acquire the information

about the peak response, steady state, settling time, rise time and steady state error.

Exercise#1: - Find transient and steady state characteristics of the following transfer

functions.

a).

num = [1 2];

den = [1 7 12];

step(num,den)grid on

After executing in command window we get the following result


Lab Practical # 05Time Response of a Control System


47/87

Aisha Muslim:

BS (TE)-044-07F

b).

num = [1 3 2];den = [1 3 1 0];

step(num,den)grid on



48/87

Aisha Muslim:

BS (TE)-044-07F

Exercise#2: - Reduce the following open loop systems into a single block and find

time response of the system.

(i)

Solution:

num1= [1 0];den1= [9 17];num2= [9 27];den2= [2 9 27];[num12, den12]=series (num1, den1, num2, den2);printsys(num12, den12)

After executing the above code we get the single transfer function as


S9S + 17

9(S+3)

2S2

+ 9s +27


49/87

Aisha Muslim:

BS (TE)-044-07F

num/den =

9 s^2 + 27 s------------------------------18 s^3 + 115 s^2 + 396 s + 459

>>

Now for the transient and steady state response in terms of step input is

step (num12, den12)

The result in mat lab figure is following;

(ii)

num1 = [2 4];


4/5S + 3

2S + 4

S3

S2

S - 6


50/87

Aisha Muslim:

BS (TE)-044-07F

den1 = [1 0 0 0];

num2 = 4/5;

den2 = [1 3];

num3=[1 0 0];

den3=[1 -6];






After executing the above code we get the single transfer function as

num/den =

1.6 s^3 + 3.2 s^2

--------------------

S^5 - 3 s^4 - 18 s^3Now for the transient and steady state response in terms of step input isstep(num123,den123)




51/87

Aisha Muslim:

BS (TE)-044-07F

Exercise#3: - Find time response of following closed loop systems.

(i)

Solution:

num1 = 9;

den1 = [1 5];

num2 = 1;

den2 = 1;[n,d]=feedback (num1, den1, num2, den2);


C(S)R(S)-

9S + 5


52/87

Aisha Muslim:

BS (TE)-044-07F

printsys(n,d)

For the equilent transfer function of the feedback system we get the following result

from mat lab command window;

num/den =

9

------

s + 14


step(n,d)




53/87

Aisha Muslim:

BS (TE)-044-07F

(ii)

Solution:num1 = 1;

den1 = [1 1];

num2 = 2;

den2 = [1 0];


printsys(n,d)

For the equilent transfer function of the feedback system we get the following result

from mat lab command window;

num/den =

s

------------

s^2 + s + 2


step(n,d)


C(S)R(S)-

1S +1

2S


54/87

Aisha Muslim:

BS (TE)-044-07F

The result in mat lab figure is following



55/87

Aisha Muslim:

BS (TE)-044-07F

Introduction

Generalized transfer function of second order system is represented by the following

equation

Where n is the natural un-damped frequency and is damping ratio. Thus the

response of the second order system depends upon both factors. In this lab we will

analyze the effect of variation of damping ratio and natural un-damped frequency on

the response of the system.

Exercise#1: -

(a) Find Step response of the system when

n = 0.1 rad/sec

= 0.2, 0.5, 0.81, 2.5

After putting = 0.2 and n = 0.1 rad/sec we get the transfer function


G(S) =S2 + 2nS +n

2

n2

G(S) =S2 + 2nS +n

2

n2

Lab Practical # 06Time domain analysis of Second Order System


56/87

Aisha Muslim:

BS (TE)-044-07F

By entering the following MATLAB code we can find the step response of the system

num = 0.01den = [1 0.04 0.01];

step (num, den)grid on

The command window shows the value of the numerator as

num =

0.0100

The step response figure is given below


G(S) =0.01

S 2+0.04S+0.01


57/87

Aisha Muslim:

BS (TE)-044-07F



num = 0.01den = [1 0.01 0.01];

step (num, den)

grid onThe command window shows the value of the numerator as

num =

0.0100



G(S) =0.01

S 2+0.01S+0.01


58/87

Aisha Muslim:

BS (TE)-044-07F



num = 0.01

den = [1 0.162 0.01];step (num, den)grid on


num =

0.0100



G(S) =0.01

S 2+0.162S+0.01


59/87

Aisha Muslim:

BS (TE)-044-07F



num = 0.01den = [1 0.5 0.01];

step (num, den)

grid onThe command window shows the value of the numerator as

num =

0.0100


G(S) =0.01

S 2+0.5S+0.01


60/87

Aisha Muslim:

BS (TE)-044-07F


(c).What is the effect of increasing the value of .

Answer:

The effect of increasing the value of damping ratio . in part (1) of the question is

that the systems transient response is more stable and more time the system willtake to reach the steady state.

Exercise#2: -

(a) Find impulse response of the system when

n = 0.1 rad/sec

= 0.2, 0.5, 0.81, 2.5



G(S) =S2 + 2nS +n

2

n2


61/87

Aisha Muslim:

BS (TE)-044-07F

By entering the following MATLAB code we can find the impulse response of the

system

num = 0.01den = [1 0.04 0.01];

impulse(num,den)grid on


num =

0.0100

The impulse response figure is given below



G(S) =0.01

S 2+0.04S+0.01


62/87

Aisha Muslim:

BS (TE)-044-07F


system

num = 0.01den = [1 0.01 0.01];

impulse(num,den)

grid on


num =

0.0100



G(S) =0.01

S 2+0.01S+0.01


63/87

Aisha Muslim:

BS (TE)-044-07F



system

num = 0.01den = [1 0.162 0.01];impulse(num,den)

grid on


num =

0.0100



G(S) =0.01

S 2+0.162S+0.01


64/87

Aisha Muslim:

BS (TE)-044-07F



system

num = 0.01den = [1 0.5 0.01];

impulse(num,den)grid on


num =

0.0100


G(S) =0.01

S 2+0.5S+0.01


65/87

Aisha Muslim:

BS (TE)-044-07F


(b) What is the effect of increasing the value of .

Answer:

The effect of increasing the value of damping ratio . in part (1) of the question is

that the systems transient response is more stable and more time the system will

take to reach the steady state.

Exercise#3: -

(a) Find ramp response of the system when

n = 0.1 rad/sec

= 0.2, 0.5, 0.81, 2.5


G(S) =S2 + 2nS +n

2

n2


66/87

Aisha Muslim:

BS (TE)-044-07F


By using the following MATLAB code we can find the ramp response of the system

t=0:0.01:10;r=t;num = 0.01

den = [1 0.04 0.01];plot(r,t)

hold onlsim(num,den,r,t)hold off

grid on


num =

0.0100

The ramp response figure is given below


G(S) =0.01

S 2+0.04S+0.01


67/87

Aisha Muslim:

BS (TE)-044-07F



t=0:0.01:10;r=t;num = 0.01

den = [1 0.01 0.01];plot(r,t)

hold onlsim(num,den,r,t)

hold offgrid on


G(S) =0.01

S 2+0.01S+0.01


68/87


69/87

Aisha Muslim:

BS (TE)-044-07F


t=0:0.01:10;r=t;

num = 0.01den = [1 0.162 0.01];

plot(r,t)hold onlsim(num,den,r,t)hold off

grid on


num =

0.0100

The ramp response figure is given below



70/87


71/87

Aisha Muslim:

BS (TE)-044-07F

(f) What is the effect on steady state error with the increasing the value of .

Answer:

The slope of the transient response of the system is more steepest after increasing

the value of .

Exercise#4: -

(a) Find step, ramp and impulse response of the system whenn = 0.1, 0.2, 0.5 and 1 rad/sec = 0.2

Step responses:

After putting n = 0.1 and = 0.2 rad/sec we get the transfer function


G(S) =S2 + 2nS +n

2

n2

G(S) =0.01

S 2+0.04S+0.01


72/87

Aisha Muslim:

BS (TE)-044-07F


num = 0.01



num =

0.0100




73/87

Aisha Muslim:

BS (TE)-044-07F



num = 0.04



num =


G(S) =0.04

S 2+0.08S+0.04


74/87

Aisha Muslim:

BS (TE)-044-07F

0.0400




num = 0.25

den = [1 0.2 0.25];step (num, den)

grid on


G(S) =0.25

S 2+0.2S+0.25


75/87

Aisha Muslim:

BS (TE)-044-07F


num =

0.2500


After putting n = 1 and = 0.2 rad/sec we get the transfer function


G(S) = 1S 2+0.4S+1


76/87


77/87

Aisha Muslim:

BS (TE)-044-07F

Impulse responses:After putting n = 0.1 and = 0.2 rad/sec we get the transfer function


system

num = 0.01den = [1 0.04 0.01];

impulse (num, den)grid on


num =

0.0100



G(S) =0.01

S 2+0.04S+0.01


78/87

Aisha Muslim:

BS (TE)-044-07F



system

num = 0.04den = [1 0.08 0.04];impulse (num, den)

grid on


num =


G(S) =0.04

S 2+0.08S+0.04


79/87

Aisha Muslim:

BS (TE)-044-07F

0.0400




system

num = 0.25den = [1 0.2 0.25];

impulse (num, den)grid on



G(S) =0.25

S 2+0.2S+0.25


80/87

Aisha Muslim:

BS (TE)-044-07F

num =

0.2500



system

num = 1;

den = [1 0.4 1];impulse (num, den)

grid on


G(S) = 1S 2+0.4S+1


81/87

Aisha Muslim:

BS (TE)-044-07F


num =

1.0000


Ramp responses:


By entering the following MATLAB code we can find the ramp response of the

system

t=0:0.01:10;r=t;


G(S) =0.01

S 2+0.04S+0.01


82/87

Aisha Muslim:

BS (TE)-044-07F

num = 0.01

den = [1 0.04 0.01];

plot(r,t)hold onlsim(num,den,r,t)

hold offgrid on


num =

0.0100

The ramp response figure is given below:



G(S) =0.04

S 2+0.08S+0.04


83/87

Aisha Muslim:

BS (TE)-044-07F


system

t=0:0.01:10;r=t;

num = 0.04den = [1 0.08 0.04];


grid on


num =

0.0400




84/87

Aisha Muslim:

BS (TE)-044-07F



system

t=0:0.01:10;r=t;

num = 0.25;den = [1 0.02 0.25];


grid on


G(S) =0.25

S 2+0.2S+0.25


85/87

Aisha Muslim:

BS (TE)-044-07F


num =

0.2500


After putting n = 1 and = 0.2 rad/sec we get the transfer function


system

t=0:0.01:10;

r=t;num = 1;den = [1 0.4 1];

plot(r,t)hold on


G(S) =1

S 2+0.4S+1


86/87

Aisha Muslim:

BS (TE)-044-07F

lsim(num,den,r,t)

hold off

grid on


num =

1.0000


(b) What is the effect of increasing the value of .



87/87

Aisha Muslim:

BS (TE)-044-07F

control system labs all

Documents