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Doofy
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#11
Mar31-12, 02:53 PM
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P: 60
Quote Quote by LAHLH View Post
If you look at (#4)^2+(#5)^2, you get an equation constraining just the magnitudes, a1,a2,a3,a4,a5,a6; similarly (#6)^2+(#7)^2 constraints just a1,a2,a3,a7,a8,a9, and (#8)^2+(#9)^2 constraints just a4,a5,a6,a7,a8,a9. How do I get the three equations for just the phases [itex] b_i [/itex] out of these that you mention?
How do you get rid of the [itex] b_i [/itex] when you do (4)^2 + (5)^2 ?

for (4) and (4)^2 I get:
[itex] \alpha_1 \alpha_4 cos(\beta_1 - \beta_4) + \alpha_2 \alpha_5 cos(\beta_2 - \beta_5) = -\alpha_3 \alpha_6 cos(\beta_3 - \beta_6) [/itex]

[itex] (\alpha_1 \alpha_4 )^2 cos^2(\beta_1 - \beta_4) + (\alpha_2 \alpha_5)^2 cos^2(\beta_2 - \beta_5) + 2(\alpha_1 \alpha_4 \alpha_2 \alpha_5)cos(\beta_1 - \beta_4) cos(\beta_2 - \beta_5) = (\alpha_3 \alpha_6)^2 cos^2(\beta_3 - \beta_6) [/itex]


for (5) and (5)^2 I get:
[itex] \alpha_1 \alpha_4 sin(\beta_1 - \beta_4) + \alpha_2 \alpha_5 sin(\beta_2 - \beta_5) = -\alpha_3 \alpha_6 sin(\beta_3 - \beta_6) [/itex]

[itex] (\alpha_1 \alpha_4 )^2 sin^2(\beta_1 - \beta_4) + (\alpha_2 \alpha_5)^2 sin^2(\beta_2 - \beta_5) + 2(\alpha_1 \alpha_4 \alpha_2 \alpha_5)sin(\beta_1 - \beta_4) sin(\beta_2 - \beta_5) = (\alpha_3 \alpha_6)^2 sin^2(\beta_3 - \beta_6) [/itex]


so for (4)^2 + (5)^2 I get:
[itex] (\alpha_1 \alpha_4 )^2 + (\alpha_2 \alpha_5)^2 + 2\alpha_1 \alpha_4 \alpha_2 \alpha_5 (cos(\beta_1 - \beta_4) cos(\beta_2 - \beta_5) + sin(\beta_1 - \beta_4) sin(\beta_2 - \beta_5) ) [/itex]

[itex] = (\alpha_1 \alpha_4 )^2 + (\alpha_2 \alpha_5)^2 + 2\alpha_1 \alpha_4 \alpha_2 \alpha_5 (cos(\beta_1 - \beta_4 - \beta_2 + \beta_5)) = (\alpha_3 \alpha_6)^2 [/itex]


What am I doing wrong here?