Question

1. Complex Numbers and phasor analysis can be used to solve many problems. For example if we want to determine the currents of each of the voltage sources in the following circuit, VI ov then we can use a technique called mesh analysis to write mesh equations that involve the currents of each source. Let libe the current of the voltage source vl = 2 cos(t), I be the current of the voltage source v2 = -2 cos(t), and 13 be the current of the voltage source v3 = cos(t) all in phasors, after applying mesh analysis, the following equations are obtained (2+2)/ - (21)12 - 213 = 220 -(20)/1 +(2+20) 12 – 213 = 22 -211 - 212 +(4+2j) 13 = 12 (a) write the system of equations in At = b. (b) by hand, find the values of 11, 12and Izin polar form using any method you choose. (c) use Matlab or any other software to check your answers. To solve the equations by hand using Gauss elimination method, you can use the following steps: • Multiply the first row by the conjugate of 211 = (2-2)); "by doing this you should get 8 as the new coefficient of 11" • Repeat the previous step on the second equation," multiply it by its conjugate which is 2j". • Now the coefficients q11, 221 and 231 are real numbers so you can eliminate 221 and azithen use back substitution to solve for 11, Land 13. 2. The following matrix A= 6 -11 5 41 has complex eigenvalues, find all eigenvalues and the corresponding eigenvectors of A by hand, then use matlab or any otehr software to check your answer.

Please use MATLAB for the specified parts, I appreciate the help!

Answer 1

Given the equations of Mesh Analysis, ·(2+2j) 1 - (23) 5 - 213 = 240. - (21) I +(2+2;) I - 2 F3 = 2 LIT -251 - 2 Ty +(4+2j) 13 = 1 LT. a> writing the system equations in the form of Ax' = b.. iT22j. -2; -27577 -2j. 273 -2 | Iz = 2/11 where a= [h z Iz] by Finding values of It, an Iz in polar by using Gauss-elimination method. Simpliflying the matrix (1)(b)values. first row is multiplied by (2-2)). We get- |(2721)(2-2) -2;(2-23) -2 (2-2) 260 (2-2) - -4-4.j -4 +4; 4-4; cs Scannegg with 42/-135 452/135 4822-45 CamScanner

second row is multiplied by 25. (3(23) (2+2j)(xj) (2/2) (245)(-) ч -Чtчі -чі -4ј 4 4.522135 4210 42-90. The modified equations will be. How 1-2 -2 . taj || 13 Lt row @ = row 2 row = row + txrowo Foi cywuj -474j]17 Turuj 10 -2 tbj . 2-6j|| 9 = -2-23 10 -3-j 3 +3j J[ 5] -; ssow B) - row ☺ .- 1x row ( 78 -4-4 -4+4075 77 74-417 To -2+6j 2-6j F2 1 -2-23 CS Scane with 0 .2 173 1 7 Cam sanner

-2+bj Expanding, we get an 2j. Ig = t. .5 = = 0.5j. (-2+6j) IT (2-6j) 13 = -2=2j [F; = 0.5j] (216j) I = (-2-23) -- (-6j) { ; = -2-23 j +3 =-5-3j I = -5-3) = -0.2+0.9 ; 20.922/1035 89 + (-4-43). Iz +(4+4j)} = (y4j) : substituting value of Is, Is, we get IP = 0.25 +0.j = 0.223 [26.5 . on solving, we get. .! [ 0.223 226.5" 1 Iz ] = 10.922 [1025 Es JS L015 (90 Scanned with CamScanner

© Given the system matrix ? eigen, values of matrix A is given by. Ada roots. ( A1-A) = 0 MAI-= (-)(x4 +5= 0 5 42:-100 + 24 +5=0 dio da = 5+21, 5-21. eigen vectors are the vectors that satisfy the following equation for a, and 72. ( 9, I- A) V =0, (diz-A) V2 = 0 sr-p+21 11 16,7 Ics Scanned with CamScanner

= = 6 +2j) 0 + Mz = 0. -5 W + (1+2j) 0,22 0. let 2 = 1, Vi= - = 0,2touj -14-23 Similarly 14-2; il valolol las 1-2; 1 veel lol 5) 4-2j) 494 + V32 = 0. 1 -54 + (1-2) Vr2 = 0. Let U l, Van = I =0.2 -0.43 ----23 • Eigen values of the system. : 5+2j, 5-2j. Eigen vectors of the system : (0.27 0.43 0.2-0.4;) Scanned with CamScanner

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MATLAB CODES FOR BOTH QUESTION

%Q1
A=[2+2i -2i -2;-2i 2+2i -2;-2 -2 4+2i]
b=[2;-2;-1]
x=inv(A)*b
%Q2
A=[6 -1;5 4]
[V,D]=eig(A)

MATLAB RESULTS:

A =

2.0000 + 2.0000i 0.0000 - 2.0000i -2.0000 + 0.0000i
0.0000 - 2.0000i 2.0000 + 2.0000i -2.0000 + 0.0000i
-2.0000 + 0.0000i -2.0000 + 0.0000i 4.0000 + 2.0000i

b =

2
-2
-1

x =

0.2000 + 0.1000i
-0.2000 + 0.9000i
0.0000 + 0.5000i

A =

6 -1
5 4

V =

0.1826 + 0.3651i 0.1826 - 0.3651i
0.9129 + 0.0000i 0.9129 + 0.0000i

D =

5.0000 + 2.0000i 0.0000 + 0.0000i
0.0000 + 0.0000i 5.0000 - 2.0000i

The eigen vector V we have normalised the second part (V12,V22) to 1, but in MATLAB RESPONSE it is 0.9129 which is fine. Because the vector direction represented by MATLAB and our solution is same. It is just multiplied with a gain 0.9129.