# absorbing reactance into series LC

• posted

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

• posted

On a sunny day (19 Feb 2007 05:50:53 -0800) it happened "Phil Newman" wrote in :

If your X is frequency invariant then it is a resistor. If you have L C R in series (where R can be the resistance of the L), then in resonance you will see R (real). R also sets the Q factor of your series circuit, so its bandwidth.

• posted

However, I beleive the job of the susceptance/reactance is to shift the resonant frequency of the LC series from w = 1 to another frequency, which denotes the transmission zero of the filter.

Phil

• posted

Series LC is not a bandpass, it's a bandstop...

• LTspice Simulation... v1 1 0 ac 1 sin .net V(1) V1 Rout=10K Rin=10K C1 1 2 0.18 L1 2 0 5.406 .ac lin 1000 120m 200m .plot ac S21(v1) .end
• posted

But with an "R" (RLC) then it's a bandpass.

• LTspice Simulation... v1 1 0 ac 1 sin .net V(3) V1 Rout=1 Rin=1 C1 1 2 0.18 L1 2 3 5.406 R1 3 0 1 .ac lin 1000 120m 200m .plot ac S21(v3) .end
• posted

```--
As Jan has pointed out already, In a series-resonant circuit like
this: (view in Courier)```
• posted

```--
This is bandpass: (View in Courier)

+--[L]--[C]--+```
• posted

On a sunny day (Mon, 19 Feb 2007 13:06:01 -0600) it happened John Fields wrote in :

```--
>This is bandpass: (View in Courier)
>```
• posted

Series RLC in reference to gen.

Series LC shunt R in reference to gen.

• posted

It's very difficult to understand the wording. It is possible to drop the resonant frequency to say 0.5 Rads, using just a resistance. Maybe that's what's wanted in this case. john

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Posted via a free Usenet account from http://www.teranews.com```
• posted

To drop the res freq you increase the inductance and/or capacitance. Increasing resistance increases the bandwidth not the resonant (center) frequency. Reactance is XL=2*pi*F*L, XC=1/(2*pi*F*C), so changing the reactance requires change L, C, or F.

• posted

Or pi

```--
Gibbo

• posted

```--
At resonance there'll be an infinitely deep notch, so it'll short
the output of the generator and you'll get the bandstop function```
• posted

```--
or 2```
• posted

Figure out if the reactance is inductive or capacitive, and just use the formula for series inductors or series capacitors, depending.

Good Luck! Rich

• posted

This is another bandpass: (View in Courier)

+--------+--------+ | | | | +--+--+ | | | | | [GEN] [L] [C] [R] | | | | | +--+--+ | | | | +--------+--------+

This is another bandstop:

+--[L]--+ | | +---+ +---+ | | | | | +--[C]--+ | [GEN] [R] | | +---------------+

Cheers! Rich

• posted

L),

just a

Yes. But ... It's not the size of that resistor that's the problem, it's knowing where to put it!.

As example, for a parallel tuned circuit we always go ... Fres=1/2*pi*root (L/C). Fine, no problem!.

It's not correct. The true formula is ... Fres=1/2*pi* root(1/L*C-R^2/L^2). [a ballache to use so we don't]

But ... look how that coil resistance "R" has slimed itself in and ingratiated itself with the resonant frequency. (only rears it's head at very very low Q values ). That "R" can become big enough to noticeably poison the reactive effect of the inductor and drop the Fres.

Idle thought could suggest maybe it's not unreasonable that a series tuned circuit, which is a kind of inversion of a parallel one, might for a similar reason also suffer an "R" caused resonant frequency shift. I.e same lousy Q and lower Fres. In this case (as you note), the "R" can have no Fres effect if in series with the inductor, so maybe it needs to go in ... with the ... :)

(Dragged out I know but I'm constantly surprised at the number of ways a few RCL components can be put together yet offer distinct features) john

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Posted via a free Usenet account from http://www.teranews.com```
• posted

Don't think so..., resistor placement is for microwave this is subsonic (0.160Hz),It's an impedance mismatch problem between stages, just do a simple norton transform to match 'em up.

• posted

Thanks, how can I do this? my network theory isn't great.

basically, I've been using the equation Lw^2 + Xw + 1/C to find the resonant frequency, and then using w^2 = 1/LC to calculate the new values of L or C.

This works in terms of shifting the resonant frequency, but in terms of the whole circuit really doesn't work very well.

How can I do these norton transforms?

Phil

• posted

If I have a series LC (in a one-ohm circuit, everything normalised)

then S21 is resonates at w = 1 radian, and had magnitude of 0dB at w =

1. S11 has magnitude of -infinite dB (or -80dB as ADS doesn't interpolate that far!)

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)

S21 still resonates at a level of 0dB and S11 and -infinite dB.

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