 Login Join Maker Pro

### Network # absorbing reactance into series LC

P

#### Phil Newman

Jan 1, 1970
0
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

J

#### Jan Panteltje

Jan 1, 1970
0
however, with the additional reactance (which is frequency invariant -
i.e constant)

jX = X + jwL - j/wC = 0

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.
http://en.wikipedia.org/wiki/RLC_circuit

P

#### Phil Newman

Jan 1, 1970
0
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.
http://en.wikipedia.org/wiki/RLC_circuit

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

M

#### [email protected]

Jan 1, 1970
0

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

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

M

#### [email protected]

Jan 1, 1970
0
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

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

J

#### John Fields

Jan 1, 1970
0

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.

---

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

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

1
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.

J

#### John Fields

Jan 1, 1970
0
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

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

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

This is bandstop:

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

J

#### Jan Panteltje

Jan 1, 1970
0
On a sunny day (Mon, 19 Feb 2007 13:06:01 -0600) it happened John Fields
This is bandpass: (View in Courier)

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

This is bandstop:

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

You want to add some Ri for the generator in the last one,
if you measure across R.

M

#### [email protected]

Jan 1, 1970
0
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

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

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

Series RLC in reference to gen.
This is bandstop:

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

Series LC shunt R in reference to gen.

J

#### john jardine

Jan 1, 1970
0
Phil Newman said:

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

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

M

#### [email protected]

Jan 1, 1970
0
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
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.

G

#### Gibbo

Jan 1, 1970
0
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.

Or pi

J

#### John Fields

Jan 1, 1970
0
On a sunny day (Mon, 19 Feb 2007 13:06:01 -0600) it happened John Fields
This is bandpass: (View in Courier)

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

This is bandstop:

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

You want to add some Ri for the generator in the last one,
if you measure across R.

---
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.

R

#### Rich Grise

Jan 1, 1970
0
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?

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

Good Luck!
Rich

R

#### Rich Grise

Jan 1, 1970
0
On a sunny day (Mon, 19 Feb 2007 13:06:01 -0600) it happened John Fields
This is bandpass: (View in Courier)

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

This is bandstop:

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

You want to add some Ri for the generator in the last one,
if you measure across R.

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.

This is another bandpass: (View in Courier)

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

This is another bandstop:

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

Cheers!
Rich

J

#### john jardine

Jan 1, 1970
0
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.
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

M

#### [email protected]

Jan 1, 1970
0
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.

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

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.

P

#### Phil Newman

Jan 1, 1970
0
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

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.- Hide quoted text -

- Show quoted text -

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

P

#### Phil Newman

Jan 1, 1970
0
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.

K
Replies
38
Views
2K
keith
K
Replies
24
Views
2K
Replies
1
Views
716
Replies
9
Views
3K
Replies
3
Views
808