Is this correct...
I have a complex expression Numerator/Denominator where Numerator is the complex conjugate of the Denominator.
Am I correct that the net phase angle is twice that of the Numerator?
That would simplify my life if I didn't have to make the Denominator real, which turns into an Algebraic mess :-( ...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et |
That doesn't seem right. if N = A+iB then D = A-iB if I multiply top and bottom by A+iB I get (A+iB)^2/(A^2+B^2)
(Unless I didn't understand)
George H
Sure. Just convert to polar coordinates and it's clear.
A exp(j phi)
----------- == exp(j 2 phi) A exp(-j phi)
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant
Duh, (Scuffs shoe in dirt, hangs head in shame.) George H.
Don't feel bad. I just proved it to myself by backing up to a sane expression....
(A-jB)/(A+jB)
and checking against the TAN(2x) expression.
I got myself balled-up with a messy complex expression (fitting an all-pass to specific angles).
...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et |
Well, the net phase angle _is_ twice the numerator's.
-- Tim Wescott Wescott Design Services
If it was simple, they wouldn't have called it "complex".
-- Tim Wescott Wescott Design Services
:-D ...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et |
Hello? Conjugate is just the number rotated 180o on the Argand diagram, angles subtract upon division, so quotient has angle twice the numerator. We don't need no stinking algebra to figure that one out.
I assume you mean Complex as in sqrt(-1), not Complex as in difficult. Rotating, charged colliding black holes are difficult, sums are not...
Kevin Aylward B.Sc.
Yes
:-D
...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et |
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