absorbing reactance into series LC
- From: "Phil Newman" <phillenium2002@xxxxxxxxxxx>
- Date: 19 Feb 2007 05:50:53 -0800
Hi there,
In a filter I've designed, I have a series LC with additional
reactance, X, which gives a transmission zero in the filter.
How can I absorb the reactance into either the L or the C or both?
In a simple series LC, the reactance of the product at resonant
frequency is 0, so
jX (reactance) = jwL - j/wC = 0
from which you get
w^2 = 1/LC
however, with the additional reactance (which is frequency invariant -
i.e constant)
jX = X + jwL - j/wC = 0
the value of X, w, L and C are known.
I'm not entirely sure where i'm supposed to go with this though!
the value of reactance/susceptance in the series arm is -0.865
the value of inductance is 5.406H (this is normalised to one radian)
the value of capacitance is 0.18F (again normalised)
resonance is measured at 0.7130/1.404 (normalised, it's a band-pass
filter)
If you can help me make sense of this apparently easier algebra in
which i'm missing something, that would be great!
Phil
.
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