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

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

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

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

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

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

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

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

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.

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