Resontate frequency of parallel L/C
 

What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel


i.e.



|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

Resontate frequency of parallel L/C
 

Is this by any chance an exam question?

Roy Lewallen, W7EL

Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel


i.e.



|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

Resontate frequency of parallel L/C
 

Roy Lewallen wrote:
Is this by any chance an exam question?

No, it is not. I was shown it by a lecturer of mine more than 10 years
ago. The result is quite interesting.

Resontate frequency of parallel L/C
 

On 18 Nov, 06:05, Dave wrote:
Roy Lewallen wrote:
Is this by any chance an exam question?

No, it is not. I was shown it by a lecturer of mine more than 10 years
ago. The result is quite interesting.

Interesting to me is that there is no parallel resistance bypassing
the capacitor inferring a mythical loss less capacitor.
I await developments with interest
Art

Resontate frequency of parallel L/C
 

In message , Dave writes
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel


i.e.



|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ

--
Brian Howie

Resontate frequency of parallel L/C
 

Brian Howie wrote:


|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ



The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Resontate frequency of parallel L/C
 

On Sun, 18 Nov 2007 19:12:48 +0000, Dave wrote:

Brian Howie wrote:


|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ



The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

Resontate frequency of parallel L/C
 

Richard Clark wrote:

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

For a minute, I thought he had abandoned the "new math" and gone over to
the "dark side." (or, "new-new math!") ;-)
JS

Resontate frequency of parallel L/C
 

Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel

Since the two resistances are equal, seems to me the
resonant frequency would be where the two reactances
are equal. Where the 100 uH line crosses the 100 pf
line on the reactance chart in the ARRL Handbook is
in the ballpark of 1.591549431 MHz. :-)
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

Richard Clark wrote:
On Sun, 18 Nov 2007 19:12:48 +0000, Dave wrote:

Brian Howie wrote:

|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ


The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?
73,
Tom Donaly, KA6RUH

Resontate frequency of parallel L/C
 

On 18 Nov, 11:12, Dave wrote:
Brian Howie wrote:

|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ

finally come up The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);

That equation is obviously know from transmission lines too..- Hide quoted text -

- Show quoted text -

David,
After all those discussions where you have been baiting me on my
antenna
you finally come up with what I have been hitting you with.
Rember my comments where a radiator can be any shape, any elefation as
long as the
element is in equilibrium. Finally the penny has dropped with respect
to the LC ratio!
Art

Resontate frequency of parallel L/C
 

Cecil, W5DXP wrote:
"Since the two resistances are equal, seems to me the resonant frequency
would be where the two reactances are equal."

Yes that`s the unity power factor point. There`s a rule that when the
circuit Q`s not less than 10, fo=1/2pi on the sq.rt. of LC. For lower
Qs, the calculation is more laborious. I sure miss my ARRL Lightning
Coil Calculator!

Best regards, Richard Harrison, KB5WZI

Resontate frequency of parallel L/C
 

Tom Donaly wrote:
My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

The actual formula is 1/[2pi*SQRT(L*C)]
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

On 18 Nov, 14:11, Cecil Moore wrote:
Tom Donaly wrote:
My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

The actual formula is 1/[2pi*SQRT(L*C)]
--
73, Cecil http://www.w5dxp.com

I see that somebody intimated a 1000 ohm resistive impedance.
My antenna on 160 is about half of that! The question I have now is
how can we relate the radiation with respect to that high
resistive impedance?
Regards
Art

Resontate frequency of parallel L/C
 

On 18 nov, 13:11, Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel

i.e.

|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

hello Dave,

Normallly the resonant frequency of circuit is the frequency where Zin
is real. The problem with this circuit is that Z is real everywhere
and Q will be zero. So in my opinion it is useless to define a
resonant frequency for this circuit. The only other option you have
is to find the frequency where Im(current left leg) = -Im(current
right leg), 1.600 MHz.

Best regards,

Wim
PA3DJS
www.tetech.nl

Resontate frequency of parallel L/C
 

On Sun, 18 Nov 2007 21:45:54 GMT, "Tom Donaly"
wrote:

My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

Hi Tom,

You didn't, I misread micro for nanohenry.

No resonance as specified.

73's
Richard Clark, KB7QHC

Resontate frequency of parallel L/C
 

Art wrote:
"The question I now have is how can we relate the radiation with respact
to that high resistive impedance?"

Efficiency = radiation resistance / radiation resistance + loss
resistance

Best regards, Richard Harrison, KB5WZI

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel

Since the two resistances are equal, seems to me the
resonant frequency would be where the two reactances
are equal. Where the 100 uH line crosses the 100 pf
line on the reactance chart in the ARRL Handbook is
in the ballpark of 1.591549431 MHz. :-)

But it does not resonate at 1.591549431 MHz - or anywhere else for that
matter. The impedance is 1000 Ohms, purely resistive, at any frequency.

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
Tom Donaly wrote:
My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

The actual formula is 1/[2pi*SQRT(L*C)]

Not in this case. All you have to do, Cecil, is take your formula,
above, find the frequency you think is the resonant frequency, and
then use it to find the impedance across the circuit. Now, try some
other frequency. You can prove that this circuit can be replaced by
a 1000 ohm resistor for all frequencies, using network analysis, but
that's a little more difficult.
73,
Tom Donaly, KA6RUH

Resontate frequency of parallel L/C
 

Dave wrote:
Roy Lewallen wrote:
Is this by any chance an exam question?

No, it is not. I was shown it by a lecturer of mine more than 10 years
ago. The result is quite interesting.

With the given values, it's a constant-impedance network. I've used one
many times in time domain circuit designs. Its impedance is a constant
real value of 1000 ohms at all frequencies. Since "resonance" implies a
single frequency (at which the reactance is zero), this circuit isn't
resonant at any frequency. The circuit is often used in time domain
applications (e.g., oscilloscopes) where it's sometimes necessary to
provide a constant impedance load but you're stuck with a capacitive
device input impedance. In that situation, the C is the input C of the
device. However, the transfer function isn't flat with frequency-- you
end up with a single pole lowpass rolloff, dictated by the R and C values.

For anyone who cares about such matters, "resonate" is a verb,
"resonant" is the adjective, and "resonance" the noun. A resonant
circuit resonates at resonance.

Roy Lewallen, W7EL

Resontate frequency of parallel L/C
 

Wimpie wrote:
don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel

i.e.

|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

hello Dave,

Normallly the resonant frequency of circuit is the frequency where Zin
is real. The problem with this circuit is that Z is real everywhere

I'm not sure if that is a "problem", or a "nice feature" - it depends on
your viewpoint I guess!

and Q will be zero. So in my opinion it is useless to define a
resonant frequency for this circuit. The only other option you have
is to find the frequency where Im(current left leg) = -Im(current
right leg), 1.600 MHz.

Best regards,

Wim
PA3DJS
www.tetech.nl


It will always be real if

R = sqrt(L/C)


If anyone wants to prove it, I will let them. I did in many years ago,
but don't have the inclination to do it any more.

IIRC, the proof is not particularly difficult.

Resontate frequency of parallel L/C
 

On 18 Nov, 14:41, (Richard Harrison) wrote:
Art wrote:

"The question I now have is how can we relate the radiation with respact
to that high resistive impedance?"

Efficiency = radiation resistance / radiation resistance + loss
resistance

Best regards, Richard Harrison, KB5WZI

Nothing spectacular about that Richard, or are you relating to
something I missed?
Ofcourse nothing is real with the circuit that David provided because
the
capacitor is not real without a bypass resistance!
Art

Resontate frequency of parallel L/C
 

Dave wrote:

The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Another interesting thing about this general topology is that, except
for the special case where R^2 = L/C (the constant impedance case), the
resonant frequency is 1 / (2 * pi * sqrt(LC)) if and only if the two
resistors are equal in value. Otherwise it's at some other frequency
depending on the R values.

Roy Lewallen, W7EL

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
Tom Donaly wrote:
My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?

The actual formula is 1/[2pi*SQRT(L*C)]


1/(2pi*sqrt(l*c))

= 1/(2.28318*100)

= 1/628.318

= 0.001591550775244382621538...

Hmmm. Seems as if my calculator is broken also ...

Regards,
JS

Resontate frequency of parallel L/C
 

Tom Donaly wrote:
You can prove that this circuit can be replaced by
a 1000 ohm resistor for all frequencies, using network analysis, but
that's a little more difficult.

That's pretty interesting and not intuitively obvious.
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

John Smith wrote:

w5dxp wrote:
1/(2pi*sqrt(l*c))

= 1/(2.28318*100)

Hmmm. Seems as if my calculator is broken also ...

If your calculator says that pi = SQRT(2), it
is no doubt broken.
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
John Smith wrote:

w5dxp wrote:
1/(2pi*sqrt(l*c))

= 1/(2.28318*100)

Hmmm. Seems as if my calculator is broken also ...

If your calculator says that pi = SQRT(2), it
is no doubt broken.

Well, I don't see that but ...

Ahhh, just a little fun with "paddin' the figures", heck others do it
here! ;-)

Regards,
JS

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
Tom Donaly wrote:
You can prove that this circuit can be replaced by
a 1000 ohm resistor for all frequencies, using network analysis, but
that's a little more difficult.

That's pretty interesting and not intuitively obvious.

Hi Cecil,
Reg Edward's favorite book _Communication Engineering_ by
William L. Everitt discusses constant resistance networks under the
Chapter entitled "Equalizers." He gives credit to two fellows: R. S.
Hoyt (U.S. Patent 1453980) and O. J. Zobel, who wrote an article
entitled "Distortion Correction in Electrical Circuits with Constant
Resistance Recurrent Networks" in the Bell System Tech. Journal in
1928. (Good luck getting your hands on a copy of that!) Google
"Zobel network" for more information - and much disinformation, too.
73,
Tom Donaly, KA6RUH

Resontate frequency of parallel L/C
 

In message , Roy Lewallen
writes
Dave wrote:
Roy Lewallen wrote:
Is this by any chance an exam question?
No, it is not. I was shown it by a lecturer of mine more than 10
years ago. The result is quite interesting.

With the given values, it's a constant-impedance network. I've used one
many times in time domain circuit designs. Its impedance is a constant
real value of 1000 ohms at all frequencies. Since "resonance" implies a
single frequency (at which the reactance is zero), this circuit isn't
resonant at any frequency. The circuit is often used in time domain
applications (e.g., oscilloscopes) where it's sometimes necessary to
provide a constant impedance load but you're stuck with a capacitive
device input impedance. In that situation, the C is the input C of the
device. However, the transfer function isn't flat with frequency-- you
end up with a single pole lowpass rolloff, dictated by the R and C values.

For anyone who cares about such matters, "resonate" is a verb,
"resonant" is the adjective, and "resonance" the noun. A resonant
circuit resonates at resonance.


I think that the principle of this circuit is similar to the
constant-impedance equaliser - such as used to compensate for the loss
of a length of coaxial cable over a wide range of frequencies (very
common in the cable TV world). This is frequency-selective in that it
has essentially zero loss at a pre-determined 'top' frequency (say
870MHz), with progressively increasing loss at lower frequencies (the
inverse of the cable loss). As it has a constant (75 ohm) input/output
impedance, it is therefore resonant at all frequencies from 0 to 870MHz.
--
Ian

Resontate frequency of parallel L/C
 

Ian Jackson wrote:

I think that the principle of this circuit is similar to the
constant-impedance equaliser - such as used to compensate for the loss
of a length of coaxial cable over a wide range of frequencies (very
common in the cable TV world). This is frequency-selective in that it
has essentially zero loss at a pre-determined 'top' frequency (say
870MHz), with progressively increasing loss at lower frequencies (the
inverse of the cable loss). As it has a constant (75 ohm) input/output
impedance, it is therefore resonant at all frequencies from 0 to 870MHz.

I've designed a couple of coax loss compensators, for very high speed
digital oscilloscope delay lines. They had to preserve the fidelity of a
high speed step to within a very few percent, which amounted to very
precise compensation of both the frequency and phase response.
Bandwidths were about 2 and 9 GHz. The dominant loss mechanism in high
quality coax over those frequency ranges is due to conductor skin effect
which is proportional to the square root of frequency, so no single
network will do the compensation. I used a number of bridged tee
networks to do the job, each correcting a different part of the time
response (equivalent to different frequency ranges), in some cases
transforming them to other topologies to accommodate unavoidable stray
impedances due to components and layout. The circuits were used in the
Tektronix 11802 and TDS820 oscilloscopes.

Roy Lewallen, W7EL

Resontate frequency of parallel L/C
 

John Smith wrote:
Cecil Moore wrote:
John Smith wrote:

w5dxp wrote:
1/(2pi*sqrt(l*c))

= 1/(2.28318*100)

Hmmm. Seems as if my calculator is broken also ...

If your calculator says that pi = SQRT(2), it
is no doubt broken.

Well, I don't see that but ...

Ahhh, just a little fun with "paddin' the figures", heck others do it
here! ;-)

Regards,
JS


I prefer

4427007044615115050034854648525685871587
----------------------------------------
1409160108506276783085718440252375099653

as a rough approximation to Pi.

Resontate frequency of parallel L/C
 

Dave wrote:

I prefer

4427007044615115050034854648525685871587
----------------------------------------
1409160108506276783085718440252375099653

as a rough approximation to Pi.





For sqrt(2),


16801980213452902006767691473844047811063360557160 1
---------------------------------------------------
11880794146294742246965551933607978236747301359246 0

is good enough for most engineering applications.

Anyone care to tell me what well known number this is a reasonable
approximation to?

19078001745537814647816786923051094576738662184691 34974601281
-------------------------------------------------------------
70184046208162524076589093068072649877753883914468 2132979719

Resontate frequency of parallel L/C
 

Dave wrote:

Anyone care to tell me what well known number this is a reasonable
approximation to?

19078001745537814647816786923051094576738662184691 34974601281
-------------------------------------------------------------
70184046208162524076589093068072649877753883914468 2132979719






And for a bonus point, how good is the approximation?

Resontate frequency of parallel L/C
 

Tom Donaly wrote:
Cecil Moore wrote:
That's pretty interesting and not intuitively obvious.

Hi Cecil,
Reg Edward's favorite book _Communication Engineering_ by
William L. Everitt discusses constant resistance networks under the
Chapter entitled "Equalizers."

Thanks very much, Tom, I looked in Johnson's "T-lines &
Networks" and "Reference Data for RF Engineers" and didn't
find that particular design described.
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

John Smith wrote:
Cecil Moore wrote:
John Smith wrote:

1/(2pi*sqrt(l*c))

= 1/(2.28318*100)

Hmmm. Seems as if my calculator is broken also ...

If your calculator says that pi = SQRT(2), it
is no doubt broken.

Well, I don't see that but ...

In your above equation pi = 2.28318/2 = 1.4159
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

Dave wrote:
Anyone care to tell me what well known number this is a reasonable
approximation to?

19078001745537814647816786923051094576738662184691 34974601281
-------------------------------------------------------------
70184046208162524076589093068072649877753883914468 2132979719

http://en.wikipedia.org/wiki/Natural_logarithm
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
In your above equation pi = 2.28318/2 = 1.4159

Sorry, that's wrong - I should have used a calculator
instead of my brain before my first cup of java.
--
73, Cecil http://www.w5dxp.com

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
Dave wrote:
Anyone care to tell me what well known number this is a reasonable
approximation to?

19078001745537814647816786923051094576738662184691 34974601281
-------------------------------------------------------------
70184046208162524076589093068072649877753883914468 2132979719

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


Correct, but to what accuracy?

Resontate frequency of parallel L/C
 

Cecil Moore wrote:

Well, I don't see that but ...

In your above equation pi = 2.28318/2 = 1.4159

Hmmm, pi = ~3.14159, 2pi = ~6.28318

Why the redirected output, into thunderbird, came out like that (the
2.28318)--I don't know.

Ruined it, huh? :-(

Regards,
JS

Resontate frequency of parallel L/C
 

Cecil Moore wrote:
Cecil Moore wrote:
In your above equation pi = 2.28318/2 = 1.4159

Sorry, that's wrong - I should have used a calculator
instead of my brain before my first cup of java.

I knew what you meant ... ;-)

Regards,
JS