Using MATLAB to Determine the Thévenin Equivalent Circuit

5.9 Using MATLAB to Determine the Thévenin Equivalent Circuit

First, the code from Figure 5.9-2

The way the code is written in the book, the code from Figure 5.9-2 would be in a file. To run from the command line without saving a file, you must set R before typing in the coefficient matrix.

>> R = 6;

Enter coefficient matrix and voltage vector from:
(5.9-3)
Note the use of ellipsis (...) to continue a line and the use of the transpose operator (') on V to write a column vector as the transpose of a row vector.

>> Z = [28 -10 -8;... -10 28 -8;... -8, -8, 16+R] Z = 28 -10 -8 -10 28 -8 -8 -8 22 >> V = [12 12 0]' V = 12 12 0

Use left division to solve for the i values in (5.9-3). Left division in this way is like solving inv(Z)*V.

>> Im = Z\V Im = 0.9851 0.9851 0.7164

To get a particular component (for example, i₃), extract that element from Im by giving the index number. In this case, the third element is extraced as shown at right.

>> I3 = Im(3) I3 = 0.7164

To get the second point, change R and run the commands again.

>> R = 12; >> Z = [28 -10 -8;... -10 28 -8;... -8, -8, 16+R]; >> V = [12 12 0]'; >> Im = Z\V; >> I = Im(3) I = 0.5106

Next, code from Figure 5.9-5 can solve for the Thévenin resistance and open-circuit voltage

Set load resistances and the currents through these resistances.

>> Ra = 12; ia = 0.5106; >> Rb = 6; ib = 0.7164;

Set the coefficient matrix and terminal voltage vector (on the right side of the equation) from:
(5.10-6)
>> A = [1 -ia; 1 -ib] A = 1.0000 -0.5106 1.0000 -0.7164 >> b = [Ra*ia; Rb*ib] b = 6.1272 4.2984

Finally, solve for the open circuit voltage and the Thévenin resistance using left-division and extracting the components.

>> X = A\b X = 10.6645 8.8863 >> Vt = X(1) Vt = 10.6645 >> Rt = X(2) Rt = 8.8863

The way the code is written in the book, the code from Figure 5.9-2 would be in a file. To run from the command line without saving a file, you must set R before typing in the coefficient matrix.	>> R = 6;
Enter coefficient matrix and voltage vector from: (5.9-3) Note the use of ellipsis (...) to continue a line and the use of the transpose operator (') on V to write a column vector as the transpose of a row vector.	>> Z = [28 -10 -8;... -10 28 -8;... -8, -8, 16+R] Z = 28 -10 -8 -10 28 -8 -8 -8 22 >> V = [12 12 0]' V = 12 12 0
Use left division to solve for the i values in (5.9-3). Left division in this way is like solving inv(Z)V*.	>> Im = Z\V Im = 0.9851 0.9851 0.7164
To get a particular component (for example, i₃), extract that element from Im by giving the index number. In this case, the third element is extraced as shown at right.	>> I3 = Im(3) I3 = 0.7164
To get the second point, change R and run the commands again.	>> R = 12; >> Z = [28 -10 -8;... -10 28 -8;... -8, -8, 16+R]; >> V = [12 12 0]'; >> Im = Z\V; >> I = Im(3) I = 0.5106

Set load resistances and the currents through these resistances.	>> Ra = 12; ia = 0.5106; >> Rb = 6; ib = 0.7164;
Set the coefficient matrix and terminal voltage vector (on the right side of the equation) from: (5.10-6)	>> A = [1 -ia; 1 -ib] A = 1.0000 -0.5106 1.0000 -0.7164 >> *b = [Raia; Rbib]* b = 6.1272 4.2984
Finally, solve for the open circuit voltage and the Thévenin resistance using left-division and extracting the components.	>> X = A\b X = 10.6645 8.8863 >> Vt = X(1) Vt = 10.6645 >> Rt = X(2) Rt = 8.8863