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Old October 15th 03, 07:11 AM
Active8
 
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On Wed, 15 Oct 2003 02:43:27 +0000 (UTC), Reg Edwards, said...

"budgie" wrote in message
...
On Wed, 15 Oct 2003 02:03:05 +0000 (UTC), "Reg Edwards"
wrote:

Multiply top and bottom of 1/jwC by j (This does not change its value)

and
you get 1/jwC = minus j/wC.

Back to school with your algebra.


His algebra looks perfectly fine to me. But as others have pointed
out, he's left the 2pi out.


------------------------------------------

Yes. I apologise for my remark about school. I gained the incorrect
impression from the previous replies.

The w in wC stands for omega = 2*Pi*F, the angular frequency.

now that were all done playing with j...

don't forget

Z = sqrt{R^2 + [(wL) - (1/wC)]^2]}

and

Z(s) = R + Ls + 1/Cs

which is just plain easier to deal with 'til you need to journey back
into time domain land. no need to leave it f(t) for this deal, though.

all that j stuff... that was scary ****. so easy to make a mistake.
swapping w and f is another good one. only works for f/f stuff.

mike
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Old October 15th 03, 03:43 AM
Reg Edwards
 
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"budgie" wrote in message
...
On Wed, 15 Oct 2003 02:03:05 +0000 (UTC), "Reg Edwards"
wrote:

Multiply top and bottom of 1/jwC by j (This does not change its value)

and
you get 1/jwC = minus j/wC.

Back to school with your algebra.


His algebra looks perfectly fine to me. But as others have pointed
out, he's left the 2pi out.


------------------------------------------

Yes. I apologise for my remark about school. I gained the incorrect
impression from the previous replies.

The w in wC stands for omega = 2*Pi*F, the angular frequency.


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Old October 15th 03, 03:12 AM
budgie
 
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On Wed, 15 Oct 2003 02:03:05 +0000 (UTC), "Reg Edwards"
wrote:

Multiply top and bottom of 1/jwC by j (This does not change its value) and
you get 1/jwC = minus j/wC.

Back to school with your algebra.


His algebra looks perfectly fine to me. But as others have pointed
out, he's left the 2pi out.
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Old October 15th 03, 01:51 AM
Tom Bruhns
 
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Which voltage and which current? Clearly the voltage across the
capacitor will always be pi/2 relative to the current in that
capacitor, and the same (but opposite sign) for an inductor. Also,
are you doing an AC or a transient analysis? If transient, are the
transients settled, are you really on the resonance frequency, and are
you simulating with sufficient time resolution? Oh, and I'm not quite
sure what you mean by "lambda/4" phase shift. Is that degrees or
radians, and what's lambda?

Cheers,
Tom

(Diego Stutzer) wrote in message . com...
Hi,
Well, I'm really confused.
I simulate a simple serial R-C-L-Network (all in series).

As far as I know the total (input-)Impedance of the network is:
Z = R + jwL - j/(wC) resp. the resonance frequency (where Zin=R) is
1/sqrt(L*C).
At resonance frequency, the Impedance should be real and therefore in
my hummel opinion Voltage and Current schould be in phase.

The funny thing is, when i build up such a network in Schematics
(Cadence PSD 14.1/Orcad 9.2) and simulate it with the PSpice A/D
Simulator, the current is displaced (relative to the voltage) about
lambda/4 - obviously not in phase!?

Can anyone tell my where I made a mistake?
Or why this Problem is showing up?
Thanks to anyone reading this and especially to those who post
answers.
D. Stutzer

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Old October 15th 03, 03:03 AM
Reg Edwards
 
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Multiply top and bottom of 1/jwC by j (This does not change its value) and
you get 1/jwC = minus j/wC.

Back to school with your algebra.




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