using Matlab to solve the linear equations and its application in electrical systems

1. 11
Application of Arrays and
Matrices in Electrical Systems
Lecture – 3
by
Shameer Koya

2. 2
Sets of Linear Equations
 x1+2x2+3x3 = 366
 4x1+5x2+6x3 = 804
 7x1+8x2+0x3 = 351
 In matrix form
1 2 3 x1 366
4 5 6 x2 = 804
7 8 0 x3 351
 This is in the form
A * x = y
Hence x = A-1 y

3. Solving Linear Equations
 The problem have a solution if rank of A and rank of
augmented matrix, ( [ A y] ), are equal.
 rank (A)
 rank ( [ A y] )
 Then test condition number of A, which should be a small
number (close to 1 is better)
( the condition number associated with the linear equation Ax = b gives a bound on how
inaccurate the solution x will be after approximation)
 cond (A)
 Now the solution is
 x= inv(A)*y or x = Ay
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rank of a matrix A is the maximum number
of linearly independent row vectors of A

4. D C Loop Analysis
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5. D C Loop Analysis …
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6. Example Circuit
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7. Example Circuit …..
 In matrix form:
 Matlab Script:
 Z = [ 26 -20 0; -16 41 -6; -4 -6 24]
 V = [10; 5; 0]
 I = ZV
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8. Node Analysis
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9.  Y = [ 0.15 -0.1 -0.05;
-0.1 0.145 -0.025;
-0.05 -0.025 0.075];
 I = [5; 0; 2];
 V = inv(Y)*I
 Or V=YI
 V =
404.2857
350.0000
412.8571
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10. AC Circuit Analysis
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11.  Z = [10-7.5*j -6+5*j;
-6+5*j 16+3*j];
 b = -2*exp(j*pi*75/180);
 V = [5; b]; % voltage vector in column form
 I = inv(Z)*V; % solve for loop currents
 Or I = ZV
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