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absorbing reactance into series LC

Discussion in 'Electronic Design' started by Phil Newman, Feb 19, 2007.

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  1. Phil Newman

    Phil Newman Guest

    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

    If you can help me make sense of this apparently easier algebra in
    which i'm missing something, that would be great!

  2. 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.
  3. Phil Newman

    Phil Newman Guest

    thanks for your answer.

    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.

  4. Guest

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

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

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

    * LTspice Simulation...
    v1 1 0 ac 1 sin V(3) V1 Rout=1 Rin=1
    C1 1 2 0.18
    L1 2 3 5.406
    R1 3 0 1 lin 1000 120m 200m
    ..plot ac S21(v3)
  6. John Fields

    John Fields Guest


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

    | |
    [GEN] |
    | |

    if you have a frequency invariant component it can't be reactive,
    thus it must be resistive.

    Note that for the circuit to be resonant the capacitive and
    inductive reactances must be equal in magnitude and opposite in
    sign, and at resonance they will cancel.

    Note also that in:

    f = --------------
    2pi sqrt(LC)

    there is no R.

    All the R will do will be to limit the current which will flow
    through the LC, without affecting the location of the resonant peak
    in the slightest. That is, neglecting the stray capacitance and
    inductance associated with the resistor.
  7. John Fields

    John Fields Guest

    This is bandpass: (View in Courier)

    | |
    [GEN [R]
    | |

    This is bandstop:

    | | |
    | [L] |
    [GEN] | [R]
    | [C] |
    | | |
  8. On a sunny day (Mon, 19 Feb 2007 13:06:01 -0600) it happened John Fields
    You want to add some Ri for the generator in the last one,
    if you measure across R.
  9. Guest

    Series RLC in reference to gen.
    Series LC shunt R in reference to gen.
  10. john jardine

    john jardine Guest

    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.
  11. Guest

    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.
  12. Gibbo

    Gibbo Guest

    Or pi
  13. John Fields

    John Fields Guest

    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
    that way. ;)

    But you're right. Thanks.
  14. John Fields

    John Fields Guest

  15. Rich Grise

    Rich Grise Guest

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

    Good Luck!
  16. Rich Grise

    Rich Grise Guest

    This is another bandpass: (View in Courier)

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

    This is another bandstop:

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

  17. john jardine

    john jardine Guest

    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)
  18. Guest

    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.
  19. Phil Newman

    Phil Newman Guest

    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?

  20. Phil Newman

    Phil Newman Guest

    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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