[K3CAL] more interference info

[Bob Sheskin]

The following information was supplied by Rich WW3ZZ who many of you know.
This is his analysis of the signal and it makes it fairly evident why we are
having so much trouble. Dave has already placed a call to the appropriate
person for us to speak to in Calvert and had to leave a message, as of this
moment there has not been a return call.

33.820MHz
OK here it is..
Example:
Frequency	Order	Name 1	      Name 2
1 . f =   440 Hz	n = 1	fundamental tone 1st harmonic
2 . f =   880 Hz	n = 2	1st overtone	      2nd harmonic
3 . f = 1320 Hz	n = 3	2nd overtone	      3rd harmonic
4 . f = 1760 Hz	n = 4	3rd overtone	      4th harmonic

			
33.820	= Fundamental Freq.
67.640 =2nd Harmonic
101.46 =3rd Harmonic
135.28 = 4th Harmonic
169.100 = 5th Harmonic

33.820  + /-  5 MHz for a 10 meg bandwidth = From (-) 28.820Mhz to  (+)
38.820Mhz 
Please note the following frequency falls within the 10Mhz bandwidth
36.74625MHz.
36.74625 X 2 = 73.4925Mhz 1 harmonic
73.4925 + 36.74625=110.23875Mhz 2 harmonic
110.23875+36.74525=146.985MHz 3rd harmonic
146.985 is the 3th harmonic of 36.74625Mhz.
Noted: Your receive frequency 3rd harmonic falls well within the 10 Mhz
overall bandwidth of 33.820 dirty signal output.

73,
Bob N3PPH