Line B. of the example should read : log(33)=1.52
Apparently there was a typo in line B that propagated into line C. The log
of 30 is 1.48, that's where the value of 1.48 in line C came from.
Example should read:
A. R=1.0/0.03 = 33
B. log(33)=1.52
C. 20*1.52=30dB
D. MDS= -107dBm - 30dB = -137dBm
Let's see where the value -107dBm came from:
1 dBm means 1 milliwatt of power. In this case the load resistance is
assumed to be 50 ohms, so 1 mW will produce a voltage across the 50 ohm
load that is = SquareRoot( 0.001*50) = 0.224 volt rms.
The XG1 output is 1 microvolt into a 50 ohm load for this test, so the ratio
of it's output to the 1dBm reference signal is:
1 microvolt/0.224volt = 0.000004464.
The log(0.000004464) = -5.35
Multiplying by 20 to convert this voltage ratio to dB, we get 20*(-5.35)
= -107dB
The level of the XG1 test signal is -107dBm, that is, 107dB weaker than one
milliwatt. Since this produces an output from the receiver that is well
above the noise level, then the minimum detectable signal would be even
weaker than -107dBm. Typically for the K2 it could be around 30dB weaker (as
we found in step C above), so the minimum detectable signal is
approximately -137dBm.
73/ Bob - W5BIG
========================================================
----- Original Message -----
From: "Ron D'Eau Claire" <[EMAIL PROTECTED]>
To: <elecraft@mailman.qth.net>
Sent: Thursday, May 12, 2005 1:24 AM
Subject: RE: [Elecraft] Low receiver sensitivity
Bart, PA3GYU wrote:
>Look up "Receiver Sensitivity Testing" (Page 4 of the XG1 manual) and
>"Signal-to-Noise and MDS Calculations (page 5) for instructions on how
>to use the XG1 to measure receiver sensitivity.
I did, and found this procedure and example:
>A. Divide S+N by N; call the resulting ratio R.
>B. Take the base-10 logarithm of R ("log" key on most calculators). C.
>Multiply the result by 20 to obtain the S+N/N ratio at 1 microvolt, in
>dB. D. If the S+N/N is greater than 10 dB, then the MDS is
>approximately equal to the result from (C) subtracted from -107 dBm.
>
>Example: DMM readings of 1.0 Vrms (XG1 on), and 0.030 Vrms (XG1 off).
>
>A. R = 1.0/.03 = 33
>B. log(30) = 1.52
>C. 20 x 1.48 = about 30 dB (this meets the requirement for step D) D.
>MDS = -107 dBm - 30 dB = -137 dBm
Two remarks:
1. I think the second step in the example should read: "B. log(10) = 1.52".
2. I may be missing something here, but if the outcome of B) equals 1.52,
why is "1.48" used in C)? (Not that it changes much, only 0.8 dB, but hey:
it might confuse some people!)
----------------------------------
Sorry for the delay responding Bart! I've been away all day and I see that
none of our other sharp-eyed contributors have answered your question!
Yes, you are right. B. should read log(33) = 1.52 (the base 10 is
"understood" unless a different base is shown and, if so, the base is shown
as a subscript following log, not in parenthesis). The procedure is to take
the result of A. and find the logarithm of that.
I'm really confused about where the 1.48 came from!! That should be 1.52 so
20 x 1.52 = 30 dB.
That instruction sheet was not one if my projects, so I never looked at that
example closely. I'll pass this info along to Wayne.
Ron AC7AC
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