Page 109 - Numerical Analysis Using MATLAB and Excel

Chapter 3 Sinusoids and Phasors

To divide one phasor by another when both are expressed in exponential or polar form, we divide
the magnitude of the dividend by the magnitude of the divisor, and we subtract the phase angle of
the divisor from the phase angle of the dividend, that is, if

∠
∠
A = M θ and B = N φ
then,
jθ
A M θ ∠ φ – ) = Me M j θ ( ----e φ – ) (3.84)
----- (
------------- =
---- =
B N jφ N
Ne

Example 3.11

∠
∠
Divide A = 12.58 74.3° by B = 7.22 – 118.7°
Solution:
Division in polar form yields

∠
A 12.58 74.3°
∠
∠
---- = ---------------------------------- = 1.74 [ 74.3° ( ∠ – – 118.7° ] ) = 1.74 193° = 1.74 – 167°
∠
B 7.22 – 118.7°
Division in exponential form yields
j74.3°
A 12.58e j74.3° j118.7° j193° j – 167°
---- = ----------------------------- = 1.74e e = 1.74e = 1.74e
B 7.22e – j118.7°
Check with MATLAB:

r1=12.58; r2=7.22; deg1=74.3; deg2=−118.7; r=r1/r2, deg=deg1−deg2

r =
1.7424

deg =
193
*
Check with the Simulink model of Figure 3.19 :

* Same comment as on the footnote of the previous page.

3−22 Numerical Analysis Using MATLAB® and Excel®, Third Edition
Copyright © Orchard Publications