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Chapter 3 Sinusoids and Phasors

)
)
)
)
(
A + B = ( 3 j4 + ( 4 j2 = ( 3 + 4 + j 4 – 2 = 7 j2
+
–
+
(
)
)
)
A – B = ( 3 j4+ ) – ( 4 j2 = ( 3 – 4 + j 4 + 2 = – 1 j6
–
+
3.6 Multiplication of Phasors
Phasors are multiplied using the rules of elementary algebra, and making use of the fact that
j 2 = – 1 . Thus, if A = a + jb and B = c + jd , then
⋅
)
2
)
AB = ( a + jb ⋅ ( c + jd = ac + jad + jbc + j bd
2
and since j = – 1 , it follows that
(
)
⋅
–
AB = ac + jad + jbc bd = ( ac bd + jad + bc ) (3.71)
–
Example 3.4
⋅
It is given that A = 3 + j4 and B = 4 – j2 . Find AB
Solution:
)
)
⋅
2
AB = ( 3 + j4 ⋅ ( 4 – j2 = 12 – j6 + j16 – j 8 = 20 + j10
The conjugate of a phasor, denoted as A∗ , is another phasor with the same real component, and

–
with an imaginary component of opposite sign. Thus, if A = a + jb , then A∗ = ajb .

Example 3.5

It is given that A = 3 + j5 . Find A∗

Solution:
The conjugate of the phasor has the same real component, but the imaginary component has
A
opposite sign. Then, A∗ = 3j5
–
If a phasor is multiplied by its conjugate, the result is a real number. Thus, if A = a + jb , then
A
2
2
2 2
)
(
⋅
)
AA∗ = ( a + jb a – jb = a – jab + jab – j b = a + b 2
3−12 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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