absorbing reactance into series LC

I explained it badly.

Just stick a 4ohm resistor across the cap and watch the resonant frequency. john

On 2 Mar 2007 02:24:24 -0800, "Phil Newman"

--- Not according to your original post:

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.

Notice that you stated that the additional reactance is frequency invariant. That is nonsensical and an impossibility since the reactance is the imaginary part of the impedance and _is_ what changes with frequency since the real part can't.

Then, from your third post, we have:

If i put in my frequency invariant susceptance/reactance which isn't a resistor, it is just constant reactance, then this shifts the resonant freqency from w = 1 to w = 1.15 (eg)

where you seem to still not understand that a frequency invariant susceptance or reactance _must_ be a resistance.

---

--- Oh well...

---

--- Bingo!!! Now, if the resistance doesn't change the resonant frequency it must be because changing the resistance isn't changing the reactance, changing the L or the C is. So, if something _is_ changing the resonance it must be reactance, and of that you've only got two types; inductive and capacitive, neither of which is frequency invariant.

-- JF