Page 100 - Numerical Analysis Using MATLAB and Excel

Division of Phasors

Example 3.6

⋅
It is given that A = 3 + j5 . Find AA∗
Solution:
2
2
)
⋅
)
(
–
AA∗ = ( 3 + j5 3j5 = 3 + 5 = 9 + 25 = 34
3.7 Division of Phasors
When performing division of phasors, it is desirable to obtain the quotient separated into a real
part and an imaginary part. This procedure is called rationalization of the quotient, and it is done by
multiplying the denominator by its conjugate. Thus, if A = a + jb and B = c + jd , then,

)
(
(
)
–
A a + jb ( a + jb cjd ) A B∗ ( ac + bd + jbc – ad )
⋅
---- = -------------- = ------------------------------------- = ---- ------- = ------------------------------------------------------
)
(
–
B c + jd ( c + jd cjd ) B B∗ c + d 2
2
( ac + bd ) ( bc – ad ) (3.72)
= ----------------------- + j----------------------
2
2
c + d 2 c + d 2
In (3.72), we multiplied both the numerator and denominator by the conjugate of the denomina-
tor to eliminate the operator from the denominator of the quotient. Using this procedure, we
j
see that the quotient is easily separated into a real and an imaginary part.
Example 3.7
⁄
It is given that A = 3 + j4 , and B = 4 + j3 . Find AB
Solution:
Using the procedure of (3.72), we get

–
A 3 j4+ ( 3 j4+ ) ( 4 j3 ) 12 j9– + j16 + 12 24 j7+ 24 7
+
---- = -------------- = -------------------------------------- = -------------------------------------------- = ----------------- = ------ j------ = 0.96 j0.28+
2
B 4 j3+ ( 4 j3+ ) ( 4 j3 ) 4 + 3 2 25 25 25
–
3.8 Exponential and Polar Forms of Phasors

The relations

e jθ = cos θ + jsin θ (3.73)
and

Numerical Analysis Using MATLAB® and Excel®, Third Edition 3−13
Copyright © Orchard Publications