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Old 19th July 2009, 11:08 AM   #1
Default when i am alowed to use phasors

Vc''+Vc=3cos(2t)
Vc(0)=2
Vc'(0)=3

i was told that some thing should converge on t->inf

is it the ZIR ?
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Old 21st July 2009, 05:02 PM   #2
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Hi,

Looks like the DC component.
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Old 22nd July 2009, 05:16 AM   #3
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What do you mean when do you use phasors? THey're just another way to write complex numbers.
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Old 24th July 2009, 04:27 AM   #4
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Hi DK,

I was a little confused by this last question too.
We havent heard from transgalactic for a while now, i wonder if he/she
took off to some other galaxy by now :-)
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Old 24th July 2009, 03:35 PM   #5
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I think they allow phasors but have banned disruptors
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Old 29th July 2009, 10:08 PM   #6
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just set them to stun and you will be ok (nobody gets killed just stunned)
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Old 30th July 2009, 07:55 AM   #7
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i was told that when i have a sinus input in m diff equation
than there are case when i am not allowed to use
foorier transformation(phasors)

but i have to solve it like a normal diff equation with a sinus in the right side

what is that case?
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Old 5th August 2009, 01:50 PM   #8
Smile

Phasors can be used in any circuit with a phase shifted power source. They can not be used in underdamped step input circuits because the exponential component adds its own phase component to the ringing component, so:
What equation for Vc can you add to its 2nd derivitive to get 3cos(2t)?

Well, if Vc = Acos(βt) then Vc' = -Aβsin(βt) and Vc'' = -Aβ²cos(βt) so that:

-Aβ²cos(βt) + Acos(βt) = 3cos(2t)
A(-β² + 1)cos(βt) = 3cos(2t)
so: β = 2
and: A(-β² + 1) = 3
A = 3/(-2² + 1)
= -1
so: Vc = -cos(2t) V
I guess the right hand waveform is shifted 180deg from the solution, which makes sense since the equation uses a second derivitive, and therefore shifts the waveform by two lots of 90deg.

Am I right?
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