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Old January 24th 11, 03:47 AM posted to rec.radio.amateur.antenna
John - KD5YI[_3_] John - KD5YI[_3_] is offline
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Default Math help characterizing Antenna, Please

On 1/23/2011 7:58 PM, amdx wrote:
"John - wrote in message
...
On 1/23/2011 11:10 AM, amdx wrote:
I put up a long wire antenna, it is an inverted C.
The antenna is resonant at 3.55 Mhz.
I want to characterize it an the AM broadcast band.
I have made a measurement at 500 Khz and I had to install a parallel
capacitor
to get my variable inductor to bring phase to zero.
I don't know how to do the math to find the impedance of the antenna
with the L and C in the circuit.
Can someone look at my drawing and give me the math so I can figure out
the impedance. Then I can get the numbers at other frequencies for the
band
and calculate those impedances.
See drawing here. I want to calculate the Unknown Impedance.
I need the R and the C of the antenna.
http://i395.photobucket.com/albums/p...naat500Khz.jpg

Thank you, Mikek


Work backwards. You have a 55 ohm resistor in series with a 310 ohm
reactance. Get the equivalent parallel combination of that (the complex
reciprocal). Take the reciprocal of the capacitor (just invert the Xc and
change the sign). Add the two complex numbers. Take the reciprocal of the
answer. This is the impedance the antenna sees looking into the network.
The antenna is the complex conjugate of that (just swap the sign of the
imaginary part).

Does this help?

John


I don't know, I not there yet.

55 + i310
complex reciprocal
55 - i310 / 55^2 + 310^2
reduce
55 - i310 / 3025 + 96100
reduce further
55 - i310 / 99125
Take the reciprocal of the capacitor (just invert the Xc and change the
sign).
Xc = -i717 so 1/i717 ???


??? = +.001395

Add the two complex numbers.
(55 - i310 / 99125) + (1/i717)
Any mistakes yet?


Well, no, but you should simplify here.

Going further doesn't get me the answer you have.


I can't tell why because you didn't show your work.

Mikek



Convert the series impedance 55 + j310 to its parallel equivalent
admittance (1/(55+j310) as follows:

1/(55+j310) = (55-j310)/(55+j310)(55-j310)

In other words, multiply the numerator AND denominator of 1/(55+j310) by
the conjugate (55-j310) of the denominator.

This gives (55-j310)/(3025+96100) or (55-j310)/99125 which is an
admittance of 554.85e-6-j3.127e-3.

The capacitor's parallel equivalent admittance is 1/(0-j717). So, when
numerator and denominator are multiplied by the conjugate we get

0+j717)/(0+514.089e3) or j717/514.089e3 which is 0+1.395e-3.

Now the sum of (554.85e-6 - j3.127) and (0 + 1.945e-6) is

554.855e-6 - j1.733e-3

This the parallel equivalent (the admittance) of your components and
what the antenna is seeing.

Following in the steps of my second and third sentences above, the
equivalent series impedance of your network looking back toward the
resistor is the reciprocal of that, or

167.63 + j523.5

This is the impedance the antenna sees. The antenna's equivalent
impedance is the conjugate of that. That is, the antenna looks like an
impedance of 167.63 - j523.5 according to your data.

A calculator that handles complex numbers is invaluable. I have one on
my computer's desktop. I also have an electronic Smith chart which is
usually faster. If you need a link, let me know.

This whole thing may look a bit complicated to you if you are not
familiar with complex numbers. Look up how to do X, /, +, - using
complex numbers and get familiar. It will help.

Cheers,
John