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*Subject*: R: [amsat-bb] Question from a newcomer*From*: "i8cvs" <domenico.i8cvs@xxxxxx>*Date*: Sun, 25 Nov 2001 00:15:18 +0100

----- Original Message ----- From: Dr. Tom Clark <tac@clark.net> To: AMSAT BB <amsat-bb@AMSAT.Org>; <ea4bfk@eresmas.net> Sent: Thursday, November 22, 2001 6:28 PM Subject: RE: [amsat-bb] Question from a newcomer > Alex, EA4BFK asked a question and N0AN gave the answer "don't worry, it's no > problem. Bet let me show how to make the calculation: > > > How can I calculate the NF for my receiver knowing that > > is as TM255 with the following specs: > > Sensitivity 0.13 uV in SSB at 10 db (S+N/N)? Selectivity 2.1 KHz > > 1. Let's figure out how much power the spec means: > > Pspec = e^2/R = (0.13 x 10e-6 volts)^2/(50 ohms) = 3.38e-16 watts. > > which equates to -154.7 dBw (dB below 1 watt) > or =124.7 dBm (dB below 1 milliwatt) > > The spec says that the S/(S+N) noise floor is 10 dB below these numbers. A > little algebra (see comments at end) says that the Noise level is a factor > of 9 below these numbers, so the Noise Floor would then be a level of > > Pfloor = Pspec/9 = 3.76e-17 watts > or -164.25 dBw or -134.25 dBm in dB units > > 2. The physical temperature of a resistor is related to the power that the > resistor generates as white noise by > > Pnoise = kTB where T is the absolute temperature in Kelvin, B is the > Bandwidth of the measurement in Hz, and k is Boltzmann's > constant = 1.38e-23 watts/Hz/Kelvin. > > every Hz of bandwidth has the same noise as every other Hz, so for the > spec's 2.1 kHz bandwidth, you have P/T = 1.38e-13 *2100 = 2.90e-20 > watts/Kelvin of power. > > 3. Therefore the spec says that the noise floor is equivalent to a resistor > at a temperature of > > Tspec = Pfloor/(kB) = 3.76e-17/2.90e-20 = 1296K > > 4. Noise Figure and Noise Factor usually refer to room temperature of 290K, > so the spec corresponds to a Noise Factor F (which has no units) of > > F = (T+290)/290 = (1296/290)+1 = 5.47 > > The noise figure that is normally reported is related to the Noise Factor F > by > > NF = 10*log(F) which in your case is 7.4 dB > > Hope this helped -- 73 de Tom, W3IWI > > > [An aside: Some may be surprised by the divide by 9 in step 1 and assert > that the value is 10. > > Manufacturers have been known to share this confusion. The problem one of > definition -- Just how did the manufacturer measure what he calls S/S+N. If > he simply measured the noise signal that produces a 10 dB rise, then the > receiver has 9 units of added noise plus one original unit of noise. This is > commonly called a Y-factor of 10 and you will usually see Y-1 written in the > textbooks. This is what I assumed. > > In some cases, manufacturers have chosen a different measurement strategy > (or incorrectly piece together other specs from data sheets) resulting in > the values that reflect actual S/N numbers, and then mis-label them as S/S+N > (which what is what users normally expect). If this is the case, then the > factor of 9 becomes 10. > > If the factor = 10 and not 9, then the specs imply a 1166K noise > temperature, Noise Factor F=5.0 and NF=7.00. The NF=7.00 looks suspicious to > me. I always worry when "measured" spec sheet numbers "hit" exact numbers. > The agreement in this case under the =10 speculation is closer than the > chance agreements that PI=sqrt(10) and that there are PI*10e7 seconds in a > year.] > Tom, It was very interesting to read and study your calculations and now i have some questions for you. Please look at the problem from anoder point of view The receiver specs. are : Sensitivity 0,13 uV in SSB at 10 dB (S+N/N) ,Selectivity 2.1 KHz This means that the sum of signal and total noise is 10 dB greater than the noise alone and that the signal is 9.54 dB greater than the noise because (10/10) (S+N)/N =10 dB is equivalent to 10 = 10 in factor and S/N = ((S+N)/N)-1 = 10-1 = 9 in factor equivalent to 10 log 9 = 9.54 dB 10 The result of your calculation in 1) is: Pfloor= -134.25 dBm The result of your calculation in 4) is: NF= 7.4 dB Noise Figure,noise bandwidth and 10 dB sensitivity for SSB/CW are nearly tied togheter by the following formula: S = -174 + 9.54 + NF + 10 log (BW) 10 where S is the 10 dB sensitivity (dBm), -174 is the available thermal noise power density (dBm in 1 Hz at room temperature,9,54 is signal to noise ratio S/N in dB,NF is Noise Figure (dB) and BW is noise power bandwidth (Hz) If i put NF=7.4 dB and 2100 Hz in the above formula i get: S= -174+9,54+7.4+10 log 2100 = -174 + 9.54 + 7.4 + 33.2 = -123.8 dBm 10 -123.8 dBm is the 10 dB Sensitivity and so the noise floor is 10 dB less noisy or -133.8 dBm This figure is very very close to your calculated Pfloor = -134.25 dBm and the 0.45 dBm difference is probably due to the decimals in our calculators. To verify S in terms of voltage E (called "hard" signal level) from a 50 ohm signal generator or electromotive force,EMF(twice the reading indicated on the generator's output meter with the generator operating into a 50 ohm load ) i have used the following formula from Sabin W0IYH article in QST october 1992 page 30 (S/20 E= 0.4467 x 10 And infact if i put in the above formula S= -123.8 dBm wich is the 10 dB sensitivity i get 0.28 uV wich is about two time the applied voltage or the EMF generator voltage with open output ( 0.13 uV on specs.) >From all the above i have derived a formula originally discussed by I.L. McNally,K3WX in HAM RADIO magazine april 1980 page 71 This formula permit to easily calculate the noise figure NF in dB knowing only the RMS voltage effectively applied to the receiver input,its BW and provided that the receiver input impedance is 50 ohm. Without to go now in to math details the derived formula is the following: -6 2 ( uV x 10 ) x 20 NF(dB) = [ 10 log ----------------------------- ] + 174 10 BW x (S/N) Were: uV = effective applied voltage to receiver input over Z=50 0hm BW= bandwidth in Hz S/N= any signal to noise ratio in factor (not dB) (see the above calculations) If i put in the above formula uV=0.13 microvolt BW= 2100 Hz S/N= 9 in factor corresponding to an (S+N)/N ratio of 10 dB Than i get NF= 6.52 dB for the above receiver and the differrence from your calculation is only -0.88 dB Since it is very easy to use this formula for a newcomer,please take a look at it and please let me know wath do you think about If it is necessary i will send a complete math demontration to show how this formula has been derived. Tank you very much for your help and 73 de i8CVS Domenico > ---- > Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. > To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org > ---- Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

**References**:**RE: Question from a newcomer***From:*Dr. Tom Clark

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