Re: absorbing reactance into series LC
- From: "john jardine" <john@xxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 3 Mar 2007 16:00:40 -0000
"Phil Newman" <phillenium2002@xxxxxxxxxxx> wrote in message
news:1172831064.040203.148470@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 23 Feb, 22:35, "john jardine" <j...@xxxxxxxxxxxxxxxxxxxxx> wrote:wrote:
"John Fields" <jfie...@xxxxxxxxxxxxxxxxxxxxx> wrote in message
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On Thu, 22 Feb 2007 03:00:38 -0000, "john jardine"
<j...@xxxxxxxxxxxxxxxxxxxxx> wrote:
<max...@xxxxxxxxxxx> wrote in message
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On Feb 19, 9:37 pm, "john jardine" <j...@xxxxxxxxxxxxxxxxxxxxx>
stages,[...]
Don't think so..., resistor placement is for microwave this is
subsonic (0.160Hz),It's an impedance mismatch problem between
quoted text -just do a simple norton transform to match 'em up.
[post try #3]
I explained it badly.
Just stick a 4ohm resistor across the cap and watch the
resonant frequency.
---
That's cheating, I think. The OP specified that the resistance was
in series with a series LC.
--
JF
Most probably. Like the Quran/Bible, the more it's read the more
interpretations pop out. I gave up on it. As a homework question it was
obtuse, academic masturbation.
john
--
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- Show quoted text -
Hi,
thanks for your anwers. However, it isn't a resistor, as Impedance is
R + jX, and the reactance/susceptance I'm talking about it clearly the
imaginary part of the impedance, not the resistance.
In order to re-calculate the resonant frequency which IS shifted when
a susceptance/reactance is added in series follow this:
X = wL -1/wC + Xo
at resonant frequency, the reactance of the series L C and Xo is zero
therefore, wL - 1/wC + Xo = 0
therefore, w^2 L - 1/C + wXo = 0
finding the roots of this polynomial equation gives the new resonant
frequency (if there is no reactance Xo, then the w would be 1 in a
normalised circuit)
therefore, the new, shifted, resonant frequency has been found, and
this can then be input into the equation
w^2 = 1/LC
as this is a series circuit, it is also important to make sure the LC
ratio is kept the same
therefore, dX/dw must be identicle.
if X = wL - 1/wC + Xo
dX/dw = L + 1/(w^2)C
from W^2 = 1/LC, it can be seen that L = 1/w^2L, and substituting this
into the dX/dw equation, gives dX/dw as 2L
therefore, in order to keep the LC ratio the same (or in other words,
to keep the reactance slope the same) L must be kept constance, and
therefore C is the parameter that must be tuned in order to redefine
the resonant frequency
therefore, C = 1/Lw^2, where w is the calculated resonant frequency.
Note, for shunt resonators, dB/dw = 2C, so C should be kept constant,
and L should be tuned.
Also note, this is not an obtuse homework question, it was a practical
example that I needed to know the answer of.
Also, adding a resistance does sod-all for the resonant frequency,
apart from reducing the amplitude of the resonant peak.
Phil
Guilty as charged!. (I'm blaming the cat).
john
--
Posted via a free Usenet account from http://www.teranews.com
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