basic electrical simulation lab › sites › default › files › pics › eee › eee-e... ·...

G.N.I.T.S. – EEE DEPARTMENT

BASIC ELECTRICAL SIMULATION LAB

Experiment No: 1 Cass: III year B.Tech EEE Ist Semester

Exp - 1 1 of 3

BASIC OPERATIONS ON MATRICES

AIM: Generate a matrix and perform basic operations on matrices using MATLAB software.

Software Required: MATLAB software

Theory:

MATLAB is a MATrix laboratory, It is widely used to solve different types of scientific problems. The basic data structure is a complex double

precision matrix.

MATLAB treats all variables as matrices. Vectors are special forms of

matrices and contain only one row or one column. A matrix with one row is

called row vector and a matrix with single column is called column vector. A matrix with only one row and one column is a scalar.

A matrix can be generated and portion of a matrix can be extracted

and stored in a smaller matrix by specifying the the rows and columns to

extract.

Following are some of the matrix building functions.

zeros(M,N) MxN matrix of zeros ones(M,N) MxN matrix of ones

eye(M) identity matrix of size M.

rand(M,N) MxN matrix of uniformly distributed random. numbers on (0,1)

Following are some of the commands on matrix and operations . det(A) -determinant of a square matrix A.

inv(A) - inverse of a matrix A.

rank(A) - rank of a matrix A.

eig(A) - eigenvalues and eigenvectors of Vector A. size(a) – size of matrix A

PROCEDURE:-

Open MATLAB

Open new M-file Type the program

Save in current directory

Compile and Run the program For the output see command window\ Figure window




Exp - 1 2 of 3

PROGRAM -1:

clc;

% Creating a row vector A=[1 2 3]

% by using "," between the each element

B= [ 2.3, 4.5, 7.8 ]

% Creating a column vector

X =[1; 2; 3]

% Creating a matrix Q=[1 2 3; 4 5 6;7 8 9]

% Creating a Long array ( series vector) t = 1:10

% series vector with increments k =-2:0.5:1

t =linspace(1,10,4)

% Creating a matrix with series vector

B = [1:4; 5:8]

PROGRAM -2:

% Extracting an elements from the matrix.

A = [1,2,3;4,5,6;7,8,9]

x = A( 3 , 1)

% Extracting a column from the matrix.

y= A( : , 2)

% Extracting a row from the matrix.




Exp - 1 3 of 3

y= A(3 ,: )

% Extracting a sub matrix from a Matrix. B=A(2 : 2 , 1 : 3)

PROGRAM -3:

% addition of two matrices

clc; A=[1 2 3; 4 5 6;7 8 9]

B= [ 5 3 2; -2 -4 0; -9 3 2 ]

C= A+B

% multiplications of two matrices

D= A*B

% Division of two matrices

E=A/B

% Inverse of Matrix

inv(B)

F= A*inv(B)

% Element byelement Multiplication;

A=[1 2 3];

B= [ 10 10 10 ] ; C= A.*B

% Element by element division;

A=[1 2 3]; B= [ 10 10 10 ] ;

D= A./B

Results :

Lab In –Charge HOD –EEE




Exp - 2 1 of 7

Generation of various signals and sequences. Aim:

Generate various signals and sequences (Periodic and aperiodic), such as

Unit Impulse, Unit Step, Square, Saw tooth, Triangular, Sinusoidal, Ramp, Sinc.

Software Required: Matlab software.

Theory:

Unit impulse signal: The Ideal impulse function, is a function that is zero

everywhere but infinitely high at the origin. However, the area of the impulse

is finite . It is defined as δ(t) = 1 at t=0

=0 other wise

Unit step signal:The unit step function, also known as the Heaviside

function, is defined as such:

u(t) = 0 if t<0

=1 if t>=0

Sinc signal: The Sinc function is defined in the following manner:

x

xxc

sin)(sin if x≠0 and

sinc(x) = 1 at x=0

Following are some of the MATLAB functions used to plot the graphs.

Plot(Y): Plots the columns of Y versus their index if Y is a real number.

Plot(x,y): Plots all lines defined by vector X versus vector Y pairs. Stem(x,y):

A two-dimensional stem plot displays data as lines extending from a

baseline along the x-axis. A circle (the default) or other marker whose y-position represents the data value terminates each stem.

Subplot(m,n)

subplot divides the current figure into rectangular panes that are numbered row wise.

Subplot(m,n,p)

subplot(m,n,p) breaks the figure window into an m-by-n matrix

of small axes. P is a number that specifies the position of the pane

(subplot).

https://en.wikipedia.org/wiki/Heaviside_step_function

https://en.wikipedia.org/wiki/Heaviside_step_function




Exp - 2 2 of 7

PROGRAM:

% Generation of signals and sequences

clc;

%generation of unit impulse signal

t1=-1:0.02:1

y1=(t1==0); subplot(2,2,1);

plot(t1,y1);

axis([-1 1 -0.5 1.5]);

xlabel('time'); ylabel('amplitude');

title('unit impulse signal');

%generation of impulse sequence subplot(2,2,2);

stem(t1,y1);

axis([-1 1 -0.5 1.5]);

xlabel('n'); ylabel('amplitude');

title('unit impulse sequence');

%generation of unit step signal

t2=-5:0.1:5; y2=(t2>=0);

subplot(2,2,3);

plot(t2,y2);

axis([-5 5 -0.5 1.5]); xlabel('time');

ylabel('amplitude');

title('unit step signal');

%generation of unit step sequence

subplot(2,2,4);

stem(t2,y2);

axis([-5 5 -0.5 1.5]);

xlabel('n'); ylabel('amplitude');

title('unit step sequence');




Exp - 2 3 of 7

%generation of square wave signal t=0:0.2:10;

y3=square(t);

figure;

subplot(2,2,1); plot(t,y3);

axis([0 10 -2 2]);

xlabel('time');

ylabel('amplitude'); title('square wave signal');

%generation of square wave sequence

subplot(2,2,2);

stem(t,y3); axis([0 10 -2 2]);

xlabel('n');


title('square wave sequence');

%generation of sawtooth signal

y4=sawtooth(t);


axis([0 10 -2 2]);

xlabel('time');

ylabel('amplitude'); title('sawtooth wave signal');

%generation of sawtooth sequence

subplot(2,2,4);

stem(t,y4); axis([0 10 -2 2]);

xlabel('n');


title('sawtooth wave sequence');

%generation of triangular wave signal

y5=sawtooth(t,0.5);

figure; subplot(2,2,1);

plot(t,y5);

axis([0 10 -2 2]);

xlabel('time');




Exp - 2 4 of 7

ylabel('amplitude'); title(' triangular wave signal');

%generation of triangular wave sequence

subplot(2,2,2); stem(t,y5);

axis([0 10 -2 2]);

xlabel('n');

ylabel('amplitude'); title('triangular wave sequence');

%generation of sinsoidal wave signal

y6=sin(t);


axis([0 10 -2 2]);

xlabel('time');

ylabel('amplitude'); title(' sinsoidal wave signal');

%generation of sin wave sequence

subplot(2,2,4);

stem(t,y6); axis([0 10 -2 2]);

xlabel('n');


title('sin wave sequence');

%generation of ramp signal

y7=t;


plot(t,y7);

xlabel('time');

ylabel('amplitude'); title('ramp signal');

%generation of ramp sequence

subplot(2,2,2); stem(t,y7);

xlabel('n');


title('ramp sequence');




Exp - 2 5 of 7

%generation of sinc signal

t3=linspace(-5,5);

y8=sinc(t3); subplot(2,2,3);

plot(t3,y8);

xlabel('time');

ylabel('amplitude'); title(' sinc signal');

%generation of sinc sequence

subplot(2,2,4);

stem(y8); xlabel('n');


title('sinc sequence');

output:

-1 -0.5 0 0.5 1-0.5

0

0.5

1

1.5

time

am

plit

ude

unit impulse signal

-1 -0.5 0 0.5 1-0.5

0

0.5

1

1.5

n

am

plit

ude

unit impulse sequence

-5 0 5-0.5

0

0.5

1

1.5

time

am

plit

ude

unit step signal

-5 0 5-0.5

0

0.5

1

1.5

n

am

plit

ude

unit step sequence




Exp - 2 6 of 7

0 5 10-2

-1

0

1

2

time

am

plit

ude

square wave signal

0 5 10-2

-1

0

1

2

nam

plit

ude

square wave sequence

0 5 10-2

-1

0

1

2

time

am

plit

ude

sawtooth wave signal

0 5 10-2

-1

0

1

2

n

am

plit

ude

sawtooth wave sequence

0 5 10-2

-1

0

1

2

time

am

plit

ude

triangular wave signal

0 5 10-2

-1

0

1

2

n

am

plit

ude

triangular wave sequence

0 5 10-2

-1

0

1

2

time

am

plit

ude

sinsoidal wave signal

0 5 10-2

-1

0

1

2

n

am

plit

ude

sin wave sequence




Exp - 2 7 of 7

0 5 100

5

10

time

am

plit

ude

ramp signal

0 5 100

5

10

n

am

plit

ude

ramp sequence

-5 0 5-0.5

0

0.5

1

time

am

plit

ude

sinc signal

0 50 100-0.5

0

0.5

1

n

am

plit

ude

sinc sequence

Results : Lab In –Charge HOD –EEE




Exp - 3 1 of 7

Basic Operations on Signals and sequences Aim:

Perform the operations on signals and sequences such as addition,

multiplication,scaling, shifting, folding and also compute energy and power.

Software Required: Matlab software.

Theory:

Addition: Any two signals can be added to produce a new signal,

z (t) = x (t) + y (t)

Multiplication :

Multiplication of two signals can be obtained by multiplying their values at

every instants. z(t) = x (t).* y (t)

Time reversal/Folding:

Time reversal of a signal x(t) can be obtained by folding the signal about t=0. z(t)= x(-t)

if x(t) is original signal then z(t) is folded signal.

Signal Amplification/Scaling :

Y(n)=a.x(n) if a < 1 ; attenuation

a >1 ;amplification

Time shifting: The time shifting of x(n) obtained by delay or advance the

signal in time by using y(n)=x(n+k)

If k is a positive number, y(n) shifted to the right i e the shifting delays the

signal.

If k is a negative number, y(n ) it gets shifted left. Signal Shifting advances the signal.

Energy:

N

N

nxnE2

)()(

Average power: N

nxN

nP0

2)(

1)(




Exp - 3 2 of 7

PROCEDURE:- Open new m-file and Save in current directory

Type the code

Compile and Run the program.

For the output, see command window\ Figure window

PROGRAM:

clc;

clear all; close all;

% Generating two input signals t=0:.01:1;

x1=sin(2*pi*4*t);

x2=sin(2*pi*8*t);

subplot(2,2,1); plot(t,x1);

xlabel('time');

ylabel('amplitude'); title('input signal 1');

subplot(2,2,2);

plot(t,x2); xlabel('time');


title('input signal 2');

% Addition of signals

y1=x1+x2;


xlabel('time');

ylabel('amplitude'); title('addition of two signals');

% Multiplication of signals y2=x1.*x2;

subplot(2,2,4);

plot(t,y2); xlabel('time');


title('multiplication of two signals');

% Scaling of a signal1

A=2;

y3=A*x1;




Exp - 3 3 of 7

figure;

subplot(2,2,1); plot(t,x1);

xlabel('time');


title('input signal')


xlabel('time');

ylabel('amplitude'); title('amplified input signal');

% Folding of a signal1 h=length(x1);

nx=0:h-1;

subplot(2,2,3);

plot(nx,x1); xlabel('nx');


title('input signal') y4=fliplr(x1);

nf=-fliplr(nx);

subplot(2,2,4); plot(nf,y4);

xlabel('nf');

ylabel('amplitude'); title('folded signal');

%Shifting of a signal 1


plot(t,x1);

xlabel('time t'); ylabel('amplitude');

title('input signal');

subplot(3,1,2); plot(t+2,x1);

xlabel('t+2');

ylabel('amplitude'); title('right shifted signal');

subplot(3,1,3);

plot(t-2,x1);

xlabel('t-2'); ylabel('amplitude');

title('left shifted signal');

%~~~~~~~~~~~~~~~~~~~~~~~~~




Exp - 3 4 of 7

%Operations on sequences n1=1:1:9;

s1=[1 2 3 0 5 8 0 2 4];


stem(n1,s1);

xlabel('n1'); ylabel('amplitude');

title('input sequence1');

s2=[1 1 2 4 6 0 5 3 6]; subplot(2,2,2);

stem(n1,s2);

xlabel('n2');

ylabel('amplitude'); title('input sequence2');

% Addition of sequences s3=s1+s2;

subplot(2,2,3);

stem(n1,s3); xlabel('n1');


title('sum of two sequences');

% Multiplication of sequences

s4=s1.*s2;

subplot(2,2,4); stem(n1,s4);

xlabel('n1');

ylabel('amplitude'); title('product of two sequences');

% Program for energy of a sequence z1=[ 1 3 2 4 1 ];

e1=sum(abs(z1).^2);

disp('energy of given sequence is');e1

% Program for energy of a signal

t=0:1:10

z2=cos(2*pi*50*t).^2; e2=sum(abs(z2).^2);

disp('energy of given signal is'); e2




Exp - 3 5 of 7

% program for power of a sequence

p1= (sum(abs(z1).^2))/length(z1);

disp('power of given sequence is');p1

% program for power of a signal

p2=(sum(abs(z2).^2))/length(z2); disp('power of given signal is');p2

Outputs:

0 0.5 1-1

-0.5

0

0.5

1

time

am

plit

ude

input signal 1

0 0.5 1-1

-0.5

0

0.5

1

time

am

plit

ude

input signal 2

0 0.5 1-2

-1

0

1

2

time

am

plit

ude

addition of two signals

0 0.5 1-1

-0.5

0

0.5

1

time

am

plit

ude

multiplication of two signals




Exp - 3 6 of 7

0 0.5 1-1

-0.5

0

0.5

1

time

am

plit

ude

input signal

0 0.5 1-2

-1

0

1

2

time

am

plit

ude

amplified input signal

0 50 100-1

-0.5

0

0.5

1

nx

am

plit

ude

input signal

-100 -50 0-1

-0.5

0

0.5

1

nf

am

plit

ude

folded signal

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1

0

1

time t

am

plit

ude

input signal

2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3-1

0

1

t+2

am

plit

ude

right shifted signal

-2 -1.9 -1.8 -1.7 -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1-1

0

1

t-2

am

plit

ude

left shifted signal




Exp - 3 7 of 7

0 5 100

2

4

6

8

n1

am

plit

ude

input sequence1

0 5 100

2

4

6

n2

am

plit

ude

input sequence2

0 5 100

5

10

15

n1

am

plit

ude

sum of two sequences

0 5 100

10

20

30

n1

am

plit

ude

product of two sequences

Results :





Exp - 3 1 of 6

Mesh and Nodal analysis Aim:

To determine

a) The node voltages for the circuit shown in figure 1 and b) The mesh currents for the circuit shown in figure 2 by using MATLAB m-files and simulink

Figure 1.0

Figure 2.0

Software Required: Multisim/Matlab software.




Exp - 3 2 of 6

Theory:

Mesh and Nodal analysis provide a general procedure for analyzing circuits using

currents and voltages as the circuit variables.

Analyzing the connected circuit containing ‘n’ nodes will require ‘n-1’ KCL

equations. The KCL equations are obtained from each node with exception of the reference

node or the datum node. Similarly for mesh analysis KVL equations are obtained from each

mesh.

The above circuits can be solved with MATLAB by two methods

1. By using m-files

2. By using MATLAB Simulink

A.Procedure for Nodal analysis

• Express element current as function of the nodal voltage

• Apply KCL to each node with the exception of reference node

• Solve the resulting simultaneous equations to obtain the unknown voltages.

Applying KCL to node 1 of circuit in figure 1.0 we can obtain the following equations:

𝑖1 + 𝑖2 = 𝑖3

𝑣1 − 0

6+

𝑣1 − 𝑣2

10= 2 (1)

Applying KCL to node 2 of circuit in figure 1.0 we can obtain the following equations:

𝑖2 + 𝑖5 = 𝑖3 + 𝑖4

𝑣1 − 𝑣2

10+ 4 = 2 +

𝑣2 − 0

6 (2)

1 and 2 equations can be written as

(1

6+

1

10) v1 − (

1

10) v2 = 2 (3)

(1

10) 𝑣1 − (

1

10+

1

6) 𝑣2− = −2 (4)

Equations (3) and (4) can be represented in the matrix format in the following manner:

[(

1

6+

1

10) − (

1

10)

(1

10) − (

1

10+

1

6)

] × [𝑣1𝑣2

] = [2

−2]

𝑌 ∗ 𝑉 = 𝐼




Exp - 3 3 of 6

B.Procedure for Mesh analysis

• Express element voltages as function of the mesh currents.

• Apply KVL to each mesh

• Solve the resulting simultaneous equations to obtain the unknown currents.

Applying KVL to mesh 1 of circuit in figure 2.0 we can obtain the following

equations:

10(𝑖1 − 𝑖2) + 30((𝑖1 − 𝑖3) − 10 = 0

40 𝑖1 − 10 𝑖2 − 30 𝑖3 = 10 (5)


equations:

10(𝑖2 − 𝑖1) + 15 𝑖2 + 5((𝑖2 − 𝑖3) = 0

−10 𝑖1 + 30 𝑖2 − 5 𝑖3 = 0 (6)


equations:

30(𝑖3 − 𝑖1) + 5((𝑖3 − 𝑖2) + 30 𝑖3 = 0

−30 𝑖1 − 5 𝑖2 + 65 𝑖3 = 0 (7)

Equations (5) ,(6) and (7) can be represented in the matrix format in the following manner:

[40 −10 −30

−10 30 −5−30 −5 65

] ∗ [𝑖1𝑖2𝑖3

] = [1000

]

𝑍 ∗ 𝐼 = 𝑉




Exp - 3 4 of 6

1. MATLAB CODE FOR NODAL ANALYSIS:

Y=[ 1/6+1/10 -1/10 ; 1/10 -(1/10+1/6) ]

I=[ 2;-2];

V=inv(Y)*I;

2. MATLAB CODE FOR MESH ANALYSIS: Z=[ 40 -10 -30 ; -10 30 -5 ; -30 -5 65 ]

V=[ 10; 0; 0];

I=inv(Z)*V;

3. SIMULATION PROCEDURE BY SIMULINK:

1. Open a new MATLAB/SIMULINK model

2. Model the resistors through the Simulink block (SimPowerSystems--→Elements--

→Series RLC Branch. (Set the resistance to the given value, set L=0 and C=inf)

3. Model the current source through the Simulink block (SimPowerSystems--

→Electrical Sources--→Controlled Current Source)

4. Model the constant for the current source through (Simulink--→Sources--

→Constant)

5. Model the Voltage source through the Simulink block (SimPowerSystems--

→Electrical Sources--→Voltages Source)

6. Model the voltage measurement through (SimPowerSystems--→Measurements--

→Voltage Measurement)

7. Model the current measurement through (SimPowerSystems--→Measurements--

→Current Measurement)

8. Block name display can be invoked through (Simulink--→Sinks--→Display)

9. Nodes can be assigned by invoking the neutral through (SimPowerSystems--

→Elements---→Neutral. The node number can be changed by double click on the

neutral.

10. Continuous power GUI must be embedded in the circuit. (SimPowerSystems--

→power GUI 11. Connect each block/element as shown in the figurers

12. Debug and run the circuit and record the currents and voltages.




Exp - 3 5 of 6

Figure 3.0 Model of the circuit of figure 1.0 for the node voltage analysis

Figure 4.0 Model of the circuit of figure 2.0 for the mesh analysis analysis




Exp - 3 6 of 6

Results :





Exp - 5 1 of 5

Verification of Super position and Thevinin’s Theorems

AIM: To verify a) Superposition theorem and b) Thevenin’s theorem for the following

circuit.

Figure 1.0

SOFTWARE USED: MULTISIM / MATLAB Simulink

SUPERPOSITION THEOREM:

“In a linear network with several independent sources which include equivalent

sources due to initial conditions, and linear dependent sources, the overall response in any

part of the network is equal to the sum of individual responses due to each independent

source, considered separately, with all other independent sources reduced to zero”. Procedure:

Step 1:

1. Make the connections as shown in the circuit diagram, figure 2.0 by using MATLAB

Simulink.

2. Measure the response ‘I’ in the load resistor by considering all the sources 10V, 15V and

8V in the network.

Step 2:

1. Replace the sources 15V and 8V with their internal impedances (short circuited) as shown

in figure 3.0.

2. Measure the response ‘I1’ in the load resistor by considering 10V source in the network.

Step 3:


in figure 4.0.





Exp - 5 2 of 5

Step 4:


in figure 5.0.


The responses obtained in step 1 should be equal to the sum of the responses obtained

in step 2, 3 and 4. Verify I=I1+I2+I3

Figure 2: By considering all sources

Figure 3: By considering 10 V source alone




Exp - 5 3 of 5






Exp - 5 4 of 5

THEVENIN’S THEOREM:

“Any two terminal network consisting of linear impedances and generators may be

replaced at the two terminals by a single voltage source acting in series with an impedance.

The voltage of the equivalent source is the open circuit voltage measured at the terminals of

the network and the impedance, known as Thevenin’s equivalent impedance, ZTH, is the

impedance measured at the terminals with all the independent sources in the network reduced

to zero ”.

Procedure:

Step 1:

1. Make the connections as shown in the circuit diagram ,figure 2.0 by using MATLAB

Simulink.

2. Measure the response ‘I’ in the load resistor by considering all the sources in the network.

Step 2: Finding Thevenin’s Resistance(RTH)

1. Open the load terminals and replace all the sources with their internal impedances as

shown in figure 6.

2. Measure the impedance across the open circuited terminal which is known as Thevenin’s

Resistance.

Step 3: Finding Thevenin’s Voltage(VTH)

1. Open the load terminals and measure the voltage across the open circuited terminals

(Figure 7).

2. Measured voltage will be known as Thevenin’s Voltage.

Step 4: Thevenin’s Equivalent Circuit

1. VTH and RTH are connected in series with the load (figure 8)

2. Measure the current through the load resistor. 𝐼𝐿 = 𝑉𝑇𝐻

𝑅𝑇𝐻+𝑅𝐿

Current measured from Thevenin’s Equivalent Circuit should be same as current obtained

from the actual circuit. Verify I = IL.

Figure 6: Finding Thevenin’s Resistance(RTH)




Exp - 5 5 of 5

Figure 7: Finding Thevenin’s Voltage(VTH)

Figure 8: Thevenin’s Equivalent Circuit ( to find Load current)

Exercise: Simulate to verify the Norton’s theorem for the same given circuit.

Results





Exp - 6 1 of 5

Waveform Synthesis Using Laplace Transform

AIM: To obtain the Laplace transform for the a)Pulse wave form and b) the triangular wave

form by using waveform synthesis

Theory:

In waveform synthesis, the unit step function u(t) and other functions serve as

building blocks in constructing other waveforms. Once the waveforms are synthesized in the

form of other functions, Laplace transform is found and simplified.

a) To obtain the Laplace transform of the pulse using waveform synthesis

For example, we may describe a pulse waveform in terms of unit step functions. A

pulse of unit amplitude from t=a to t=b can be formed by taking the difference between the

two step functions

𝑥(𝑡) = 𝑢(𝑡 − 𝑎) − 𝑢(𝑡 − 𝑏)

Hence 𝑋(𝑠) = 𝑒−𝑎𝑠.1

𝑠− 𝑒−𝑏𝑠.

1

𝑠




Exp - 6 2 of 5

b) To obtain the Laplace transform of the triangular Wave using waveform synthesis

As shown in the figure above, a ramp of slope „A‟ starting at t=0 is taken as f1(t). The

function f2(t) is the ramp of slope -2A starting at t=1. When we add f1(t) and f2(t) we get the

signal shown in figure (c). Observe that a negative going ramp of slope (A-2A=-A) starts at

t=1. To cancel the part of this ramp after t≥2, a positive going ramp of slope +A in figure (d)

is added. Then we get the required triangular pulse of Figure (e).

With the help of step functions, the ramp functions in the figure above can be expressed as

follows:

i) 𝑓1(𝑡) = 𝐴. 𝑡. 𝑢(𝑡)

{ The function u(t)=1 for t>0; It indicates that ramp is present only for t≥0}

ii) 𝑓2(𝑡) = −2𝐴. (𝑡 − 1). 𝑢(𝑡 − 1)

{ The function u(t-1)=1 for t>1; It indicates that ramp is present only for t≥1}




Exp - 6 3 of 5

iii) 𝑓3(𝑡) = 𝐴. (𝑡 − 2). 𝑢(𝑡 − 2)

Therefore, 𝑓(𝑡) = 𝑓1(𝑡) + 𝑓2(𝑡) + 𝑓3(𝑡)

= 𝐴. 𝑡. 𝑢(𝑡) − 2𝐴. (𝑡 − 1). 𝑢(𝑡 − 1) + 𝐴. (𝑡 − 2). 𝑢(𝑡 − 2)

Since u(t), u(t-1) and u(t-2) have values of „1‟ and they just represent the time shifts and

directions of ramp functions, they can be dropped in this expression. Laplace transform of

above equation becomes

𝐴.1

𝑠− 2. 𝐴 .

𝑒−𝑠

𝑠2+ 𝐴.

𝑒−2𝑠

𝑠2

1.PROGRAM FOR PULSE WAVE SYNTHISIS

clear all;

close all;

clc

% WAVE FORM SYNTHISIS OF A PULSE WAVE FORM

syms f t s;

a=2;

b=5;

f1=heaviside(t-a) % unit step function starts at 'a

' sec

f2=-heaviside(t-b) % unit step function startts at

'b' sec

f=f1+f2

FS=laplace(f)

2.PROGRAM FOR TRIANGULAR WAVE SYNTHISIS

clear all;

close all;

clc

% WAVE FORM SYNTHISIS OF A PULSE WAVE FORM

syms f t s;

clc




Exp - 6 4 of 5

A=1;

f1=A*t*heaviside(t) % Ramp function starts at 't=0 '

f2=-2*A*(t-1)*heaviside(t-1) % ramp function at 't=2' sec

f3=A*(t-2)*heaviside(t-2)

f=f1+f2+f3

FS=laplace(f)

Figure(a) Simulink diagram for Pulse waveform




Exp - 6 5 of 5

Figure(b) Simulink diagram for Triangular waveform

Results





Exp - 7 1 of 3

Pole zero Map and root locus design

AIM: 1) To obtain the Pole-zero plots for the given transfer functions.

a) 𝐺(𝑠) = 𝑠2+4.𝑠+3

(𝑠+5).(𝑠2+4.𝑠+7)

b) 𝐺(𝑠) = 2(𝑠+8)

𝑠(𝑠+3).(𝑠+6)2

2) To Design the (amplifier gain ) controller for the given system through pole zero map /

root locus such that settling time ts < 5sec and damping ratio = 0.6.

Verify the output response by Simulink.

Figure (1)

Theory:

pzmap(sys):

Plots the pole-zero map of the continuous- or discrete-time LTI model sys. For

SISO systems, pzmap plots the transfer function poles and zeros. For MIMO

systems, it plots the system poles and transmission zeros. The poles are plotted as

x's and the zeros are plotted as o's.

rlocus(sys)

rlocus computes the root locus of a SISO open-loop model. The root locus

gives the closed-loop pole trajectories as a function of the feedback gain k

(assuming negative feedback). Root loci are used to study the effects of varying

feedback gains on closed-loop pole locations.




Exp - 7 2 of 3

1.A) PROGRAM:

num = [1 4 3]

den= conv ([1 5], [3 4 7])

g = tf (num,den)

[z,p,k] = tf2zp(num,den)

pzmap (g)

1.B) PROGRAM:

p = [0 -3 -6 -6 ]

z = [-8]

k = 2

GS= zpk(z,p,k)

Pzmap(GS)

2. PROGRAM FOR DESIGN OF CONTROLLER/Amplifier Gain

% To obtain the Transfer function

num1 = [60]

den1 = conv ([1 6],[1 2 0])

GS = tf ( num1 , den1 )

% To Draw the root locus/Pole zero MAP

figure (1)

rlocus (GS)

axis ([-3 3 -4 4])

sgrid (0.6,10)

% To find the controller gain

[ k , p ] = rlocfind (GS)

% Obtain the closed loop step response

Cloop = feedback (k*GS , 1)

figure(2)

step ( Cloop );

grid




Exp - 7 3 of 3

Figure (2) : Simulink diagram for step response

Assignment:

Obtain Transfer function of the systems.

1. Poles = -1+i, -1-i, -4. Zeros = -2,-5, gain = 1

2. Poles = -1+4i, -1-4i, -5. Zeros = -8,-5, gain = .75

Results





Exp - 8 1 of 3

Harmonic analysis of non sinusoidal wave forms

AIM: To generate a non sinusoidal wave and to measure the total harmonic distortion of

the non sinusoidal wave

Theory:

Harmonic is multiple of the fundamental frequency and it can be voltage and current in

an electric power system are a result of non-linear electric loads.

In a normal alternating current power system, the current varies sinusoidally at a specific

frequency, usually 50 or 60 hertz. When a linear electrical load is connected to the system, it

draws a sinusoidal current at the same frequency as the voltage .

When a non-linear load, such as a rectifier is connected to the system, it draws a current

that is not necessarily sinusoidal. The current waveform can become quite complex,

depending on the type of load and its interaction with other components of the system.

complex the current waveform can be deconstruct it into a series of simple sinusoids by

Fourier series analysis. The sinusoids start at the power system fundamental

frequency and occur at integer multiples of the fundamental frequency.

Fourier seris and Fourier Transform

Fourier discovered that such a complex signal could be decomposed into an infinite

series made up of cosine and sine terms and a whole bunch of coefficients which can

(surprisingly) be readily determined.

)2

sin()2

cos(2

1)(

11

0T

ntb

T

ntaatf

n

n

n

n

=

=

++=

• The coefficients are “readily” determined by integration.

−

−

=

=

2/

2/

2/

2/

)2

sin()(2

)2

cos()(2

T

Tn

T

Tn

dtT

nttf

Tb

dtT

nttf

Ta

fft(x):

Y = fft(X) returns the discrete Fourier transform (DFT) of vector X, computed with a fast Fourier transform (FFT) algorithm. If X is a matrix, fft returns the Fourier transform of each column of the matrix

https://en.wikipedia.org/wiki/Electric_power_system

https://en.wikipedia.org/wiki/Alternating_current

https://en.wikipedia.org/wiki/Sine_wave

https://en.wikipedia.org/wiki/Hertz

https://en.wikipedia.org/wiki/Linear

https://en.wikipedia.org/wiki/Rectifier

https://en.wikipedia.org/wiki/Fundamental_frequency





Exp - 8 2 of 3

Total harmonic distortion

The total harmonic distortion (THD) is a measurement of the harmonic distortion present in a signal and is defined as the ratio of the sum of the powers of all harmonic components to the power of the fundamental frequency

𝑇𝐻𝐷 = √𝑉2

2+𝑉3

2+𝑉4

2+…………….𝑉𝑛

2

𝑉1

where Vn is the RMS voltage of nth harmonic and n = 1 is the fundamental frequency.

1. PROGRAM FOR HORMONIC DISTORTION ANALYSIS

clear all

clc

fs=1000; % sampling frequency signal

Ts=1/fs

TT =3/50; % Total time

f=50;

V=230;

Vm=sqrt(2)*V

t=0:0.001:0.06

v1=Vm*sin(2*pi*f*t) % Fundamental component of signal

v2=(Vm/2)*sin(2*pi*2*f*t) % 2nd harmonic of signal

v3=(Vm/3)*sin(2*pi*3*f*t) % 3 rd harmonic of signal

% Total harmonic signal

vT=v1+v2+v3

figure(1)

plot(t,vT)

% To calculate Total harmonic distortion

V1=230

V2=230/2

V3=230/3

THD = sqrt((V2^2)+(V3^2))/V1

https://en.wikipedia.org/wiki/Harmonic_distortion

https://en.wikipedia.org/wiki/Harmonic_distortion





Exp - 8 3 of 3

% To display Harmonics ( frequency Vs Amplitude)

FVT=abs(fft(vT))

VF1=max(FVT)

FVT=abs(fft(vT))/VF1

k=length(t)

f=0:1000/k:1000-1

figure(2)

plot(f,FVT)

xlim([0 500])

Figure 1. Simulink diagram for total harmonic distortion

Results :


Continuous

powergui

v+

-

Voltage Measurement

V3

V2

V1

ScopeRL branch




Exp - 9

1 of 4

Transient Response of Series RLC Circuit

Aim: To find time response of current through series RLC circuit for step input voltages.

Theory:

The series RLC circuit is shown below. v(t) is the voltage source. The relation between the source voltage and current is given by equation

=++ )(1

)( tvidtCdt

diLtRi

This is a linear differential equation. Taking Laplace Transformer will get

(CS

LSR1

++ ) I (S) = V (S)

I(S) = V(S) CS /(RCS+LCS2+1)

Values of Parameters to be inserted in Circuit-Diagrams :

Step Input:- V=1 , L=50μH, C= 10μF & R= 1, 4.472 & 8 ohms

V1

L1

50uH

R2

1ohm

C2

10uF

0

21 3




Exp - 9

2 of 4

Theory:- In the RLC circuit shown aside the equation governing the voltage, resistance,

Inductance & Capacitance is as follows:

V(t)=R*I(t)+Ldi(t)/dt+1/C I(t)dt

Taking Laplace Transform ,

V(s)=I(s)[R+sL+1/sC] &

I(s)=V(s) / [R +sL+1/sC]

=V(s) *s/L / {s2 +sR/L+1/LC}

=V(s) *s/L / [(s-1)*(s- 2)] ;

𝛼1 = − 𝑅

2𝐿+ √(

𝑅

2𝐿)

2

−1

𝐿𝐶

𝛼2 = − 𝑅

2𝐿− √(

𝑅

2𝐿)

2

−1

𝐿𝐶

𝛼1 = −𝜎 + 𝑗𝜔𝑑

𝛼2 = −𝜎 − 𝑗𝜔𝑑

The condition for critical resistance Rc is

(𝑅

2𝐿)

2

=1

𝐿𝐶 ; 𝑅𝑐 = 2. √

𝐿

𝐶

If R < Rc , Under Damped & If R > Rc, Over Damped

Current through the RLC circuit:

i) Over damped case:

𝐼(𝑡) = 𝐴1𝑒−𝜎1𝑡 + 𝐴2𝑒−𝜎2𝑡

ii) Critically damped case:

𝐼(𝑡) = (𝐴1 + 𝐴2. 𝑡)𝑒−𝜎𝑡

iii) Under damped case:

𝐼(𝑡) = 𝑒−𝜎𝑡(𝐴1 cos( 𝜔𝑑 . 𝑡) + 𝐴2 sin( 𝜔𝑑 . 𝑡))




Exp - 9

3 of 4

Input Voltage &Current Waveforms

Sample calucultaion:

V=1 , L=50μH, C= 10μF & R= 1 ohms

Rc=2*sqrt(50e-6/10e-6) = 4.472 ,

R=1,

𝜎 =𝑅

2𝐿 = 1/(2*50e-6) = 1e4 ,

𝜔𝑑 = √(𝑅

2𝐿)

2

−1

𝐿𝐶 = √(1e4 ∗ 1e4 − 1/(50e − 6 ∗ 10e − 6)) = j 43589;

R=1




Exp - 9

4 of 4

Figure 1. Simulink diagram for step response of under damped case

Figure 2. Simulink diagram for analysis of step response for different dampings.

Results : Lab In –Charge HOD –EEE

Continuous

powergui

V=3

Scope3R=1 L=1mH

i+-

Current Measurement

C=10 u F3

Breaker2

Continuous

powergui

V=3

V=2

V=1

Scope3

R=8

R=4.472

R=1

L=1mH2

L=1mH1

L=1mH

i+-

Current Measurement2

i+-

Current Measurement1

i+-

Current Measurement

C=10 u F3

C=10 u F2

C=10 u F1

Breaker3

Breaker2

Breaker1




Exp - 10 1 of 6

Three phase power measurement

AIM: To Measure the power by two watt meters method for balance and unbalanced loads

Theory:

Two Wattmeter Method:

In two wattmeter method, a three phase balanced voltage is applied to a balanced three phase load where the current in each phase is assumed lagging or leading by an angle of Ø behind the corresponding phase voltage. The schematic diagram for the measurement of three phase power using two wattmeter method is shown below.

From the figure, it is obvious that current through the Current Coil (CC) of Wattmeter W1 = IR , current though Current Coil of wattmeter W2 = IB whereas the potential difference seen by the Pressure Coil (PC) of wattmeter W1= VRB (Line Voltage) and potential difference seen by Pressure Coil of wattmeter W2 = VYB. The phasor diagram of the above circuit is drawn by taking VR as reference phasor as shown below.

https://electricalbaba.com/differences-between-a-phasor-and-a-vector/




Exp - 10 2 of 6

From the above phasor diagram,

Angle between the current IR and voltage VRB = (30° – Ø)

Angle between current IY and voltage VYB = (30° + Ø)

Therefore, Active power measured by wattmeter W1 = VRB IR Cos (30° – Ø)

Similarly, Active power measured by wattmeter W2 = VYB IY Cos(30° + Ø)

As the load is balanced, therefore magnitude of line voltage will be same irrespective of phase taken i.e. VRY, VYB and VRB all will have same magnitude. Also for Star / Y connection line current and phase current are equal, say IR = IY = IB = I

Let VRY = VYB = VRB = VL

Therefore,

W1 = VRB IR Cos (30° – Ø)

= VL IL Cos(30° – Ø)

In the same manner,

W2 = VL IL Cos(30° + Ø)




Exp - 10 3 of 6

Hence, total power measured by watt meters for the balanced three phase load is given as,

W = W1 + W2

= VL IL Cos(30° – Ø) + VL IL Cos(30° + Ø)

= VL IL [Cos(30° – Ø) + Cos(30° + Ø)]

=√3 VL IL Cos Ø

Therefore, total power measured by watt meters W = √3 VL IL Cos Ø

Case1: When the Load is balanced (star connected )

VL=400 V

Vph = 400/sqrt(3) = 230.94 V

Z1 = 100 +j100 ohm

Z2 =100+j100 ohm

Z3=100+j100 ohm

IL = Vph/Zph = 230.94/(100+j100)

= 230.94/141.42 =1.64 A at -45

P= sqrt(3) Vl.IL. cos(phi) = 803. watts

Case2: When the Load is unbalanced balanced (star connected )

Z1=10 ohms

Z2 = 100+j10 ohms

Z3 = -j20 ohms

VR = 122.4 ∟0 V

VY = 122.4 ∟-120 V

VB = 122.4 ∟+120 V




Exp - 10 4 of 6

IR =10.28 ∟30 A

IB = 9.20 ∟-149.90 A

IY =1.09 ∟-155.7 A

Total power P = (IR )2 *10 + (IY )2 *10

P = (10.28 )2 *10 + (1.09 )2 *100

P = 1056 +118.81 = 1174.81

Simulink procedure:

1. Model the Load resistors through (sim powersystem >> Elements >> Series

RLC branch.(set the values of R,L,C of the load)

2. Model the Voltage sources through (sim powersystem >> Electrical sources. (

set the values of Max Voltage ,frequency)

3. Model the voltage and current measurements through (sim powersystem >>

Measurements.

4. Model the watt meters through (sim powersystem >> Extra Library >>

Measurements.

5. Connect the watt meter-1 , current coil( ‘I’ input of watt meter) to R phase and

voltage input to RY lines

6. Connect the watt meter-2 , current coil( ‘I’ input of watt meter) to B phase and

voltage input to BR lines

7. Connect each block as shown in figure.

8. Model the power gui from simpower system.




Exp - 10 5 of 6




Exp - 10 6 of 6

Results :


basic electrical simulation lab › sites › default › files › pics › eee › eee-e... ·...

Documents