Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
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, 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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