3 lABORATORY WORK

TOPIC: Modeling of steady-state electric processes in MATLAB environment.

PURPOSE OF THE WORK: Development and debugging of the program of calculation of currents and voltages in the given electrical circuit and proof of correctness of account.

The task for fulfillment

For the given variant of the electric circuit:

 Develop mathematical model of calculation of branches currents and reactive elements voltages;

 Make the program realizing developed model;

 Include into the program the operators, allowing carry out verification of calculation results;

 Organize a graphical output of the specified functions of currents and voltages;

 Debug the program;

 Save results of the work (the program, listing of calculation, graphics) at your personal folder;

· Draw up report.

Individual tasks

Variant 1:

1. H aving denoted the directions of currents in branches the equations of impedance and EMF in complex form were cjmposed.

2. Directed graph of the scheme:

So the branch here is just one. And there are three chords. It means that 1 equation should be composed according to the 1^st K`s law and three more are to be composed according to the 2^nd one.

3. Table of equations.

Table 1.1 – Table of coefficients of equations

	I1	I2	I3	I4
I	Z1	Z2	0	0	E
II	0	-Z2	Z3	0	0
III	0	0	-Z3	Z4	0
b	1	-1	-1	-1	0

4. Block-scheme of the designed program

5. Program that executes the task

f=50;R1=0;R2=10;R3=5;L4=1e-3;C2=1e-4;C3=2e-4;R4=3;ph=0;

w=2*pi*f;XL4=w*L4;XC2=1/(w*C2);XC3=1/(w*C3);

E1=25*exp(ph*pi/180*i);

Z1=R1;Z2=R2-i*XC2;Z3=R3-i*XC3;Z4=R4+i*XL4;

M=[1 -1 -1 -1;Z1 Z2 0 0;0 -Z2 Z3 0;0 0 -Z3 Z4];

F=[0;E1;0;0];

I=M\F;

disp(['I1=',num2str(I(1,1))]);

disp(['I2=',num2str(I(2,1))]);

disp(['I3=',num2str(I(3,1))]);

disp(['I4=',num2str(I(4,1))]);

I1M=abs(I(1,1))

I2M=abs(I(2,1))

I3M=abs(I(3,1))

I4M=abs(I(4,1))

phI1=(angle(I(1,1))/pi*180)

phI2=(angle(I(2,1))/pi*180)

phI3=(angle(I(3,1))/pi*180)

phI4=(angle(I(3,1))/pi*180)

h=1/(10^2*f);

t(1)=0;

i1(1)=current(I1M,f,phI1,t(1));

i2(1)=current(I2M,f,phI2,t(1));

i3(1)=current(I3M,f,phI3,t(1));

i4(1)=current(I4M,f,phI4,t(1));

for k=2:201

t(k)=t(k-1)+h;

i1(k)=current(I1M,f,phI1,t(k));

i2(k)=current(I2M,f,phI2,t(k));

i3(k)=current(I3M,f,phI3,t(k));

i4(k)=current(I4M,f,phI4,t(k));

end

plot(t,i1,'x',t,i2,'+',t,i3,t,i4,'y');

grid on;

eps1=-I(1,1)+I(2,1)+I(3,1)+I(4,1);

eps21=E1-Z1*I(1,1)-Z4*I(4,1);eps22=E1-Z1*(I(1,1))-Z3*I(3,1);eps23=-I(2,1)*Z2+I(4,1)*Z4;

disp(['error eps1=', num2str(eps1)]);

disp(['error 1st contour eps2=', num2str(eps21)]);

disp(['error 2nd contour eps2=', num2str(eps22)]);

disp(['error 3rd contour eps2=', num2str(eps23)]);

Listing of the program

>> lw3_v1

I1=8.9167+1.2813i

I2=0.22458+0.71485i

I3=0.44915+1.4297i

I4=8.2429-0.8632i

I1M =

9.0083

I2M =

0.7493

I3M =

1.4986

I4M =

8.2880

phI1 =

8.1775

phI2 =

72.5594

phI3 =

72.5594

phI4 =

72.5594

error eps1=0+2.2204e-016i

error 1st contour eps2=1.0658e-014

error 2nd contour eps2=7.1054e-015

error 3rd contour eps2=-7.1054e-015

Plot of the currents:

Figure 1.1 – Plot of found currents

Conclusion: Having done a program that solves a pretty difficult problem of plane circuit with complex components, as the summary it should be said the simulation and programming should be used as frequent as it is possible in the diversity of technical problems.