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.

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

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

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

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

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

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

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

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

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