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R: Question from a newcomer




----- 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


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