Elements of Electromagnetics (The Oxford Series in Electrical and Computer Engineering)
Elements of Electromagnetics (The Oxford Series in Electrical and Computer Engineering)
6th Edition
ISBN: 9780199321384
Author: Matthew Sadiku
Publisher: Oxford University Press
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Chapter 14, Problem 28P
To determine

Calculate the global coefficient matrix of the two element region.

Expert Solution & Answer
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Answer to Problem 28P

The global coefficient matrix is [100.50.500.50.500.50.51.50.50.500.51]_.

Explanation of Solution

Calculation:

Refer to Figure 14-61 in the textbook for the two-element region.

Modify Figure 14-61 by mentioning global number. The modified figure as shown in Figure 1.

Elements of Electromagnetics (The Oxford Series in Electrical and Computer Engineering), Chapter 14, Problem 28P

For element 1, local numbering 1-2-3 corresponds to global numbering 1-3-4.

From Figure 14.61, the given values are,

x1=0,x2=2,x3=0,y1=0,y2=2,andy3=2.

Calculate the value of P1.

P1=y2y3=22=0

Calculate the value of P2.

P2=y3y1=20=2

Calculate the value of P3.

P3=y1y2=02=2

Calculate the value of Q1.

Q1=x3x2=02=2

Calculate the value of Q2.

Q2=x1x3=00=0

Calculate the value of Q3.

Q3=x2x1=20=2

Consider the expression for the area A.

A=12(P2Q3P3Q2)        (1)

Substitute 2 for P2, 2 for P3, 0 for Q2, and 2 for Q3 in Equation (1).

A=12[((2)(2)(2)(0))]=2

Write the expression for co-efficient Cij.

Cij=14A[PiPj+QiQj]        (2)

By using Equation (2), the coefficient values C11,C12,C13,C21,C22,C23,C31,C32,andC33 are calculated as 0.5, 0, 0.5, 0, 0.5, 0.5, 0.5, 0.5, and 1 respectively.

Write the expression of the element coefficient matrix C.

C(1)=[C11C12C13C21C22C23C31C32C33]

Substitute 0.5 for C11, 0 for C12, 0.5 for C13, 0 for C21, 0.5 for C22, 0.5 for C23, 0.5 for C31, 0.5 for C32, and 1 for C33.

C(1)=[0.500.500.50.50.50.51]

For element 2, local numbering 1-2-3 corresponds to global numbering 1-2-3.

From Figure 14.61, the given values are,

x1=0,x2=4,x3=2,y1=0,y2=0,andy3=2.

Calculate the value of P1.

P1=y2y3=02=2

Calculate the value of P2.

P2=y3y1=20=2

Calculate the value of P3.

P3=y1y2=00=0

Calculate the value of Q1.

Q1=x3x2=24=2

Calculate the value of Q2.

Q2=x1x3=02=2

Calculate the value of Q3.

Q3=x2x1=40=4

Substitute 2 for P2, 0 for P3, 2 for Q2, and 4 for Q3 in Equation (1) to find area A.

A=12[((2)(4)(0)(2))]=4

By using Equation (2), the coefficient values C11,C12,C13,C21,C22,C23,C31,C32,andC33 are calculated as 0.5, 0, 0.5, 0, 0.5, 0.5, 0.5, 0.5, and 1 respectively.

Write the expression of the element coefficient matrix C(2).

C(2)=[C11C12C13C21C22C23C31C32C33]

Substitute 0.5 for C11, 0 for C12, 0.5 for C13, 0 for C21, 0.5 for C22, 0.5 for C23, 0.5 for C31, 0.5 for C32, and 1 for C33.

C(2)=[0.500.500.50.50.50.51]

Calculate the global coefficient matrix.

C=[C11(1)+C11(2)C12(1)C12(1)+C13(2)C13(1)C21(2)C22(2)C23(2)0C21(1)+C31(2)C32(2)C22(1)+C33(2)C23(1)C31(1)0C32(1)C33(1)]=[0.5+0.5000.50.500.50.5000.50.50.5+10.50.500.51]=[100.50.500.50.500.50.51.50.50.500.51]

Conclusion:

Thus, the global coefficient matrix is [100.50.500.50.500.50.51.50.50.500.51]_.

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