NoiseFigure new
= sqrt ( ( NF1^2 - 1 ) + ( NF2^2 - 1 ) + ( NF3^2 - 1 ) + ... + 1 )
where all noise figures are ratios and referenced to a
single location.
A few points:
1 Definition:
Noise Figure is the ratio of increased noise in a system above the expected
level of Johnson noise. For our 50 ohm system that will give a voltage of
sqrt(4KTRBw) where K is Boltzmann's constant, T is temperature in absolute
Kelvin, R is the resistance of the system (for our case 25 ohms {50 ohms in
parallel with 50 ohms}), and Bw is the bandwidth of interest in Hertz. For
a Bw of 1 MHz that yields a noise floor of around 0.641uV, or -3.9dBuV
2 Relate noise figure to some place in your system.
For simplicity (and ease of using specs) relate to the "front end" Noise
figure specs relate to the front end of a system block. The noise is
amplified, or attenuated, along with the signal and therefore track
together. As you go from the output to the input of each system block, gain
subtracts from the NF and attenuation adds to the NF. Modify the Noise
Figure by each block you must go through to get to the input. For example,
through cable loss, add the few dB. For gain, subtract the gain.
3 Make up a list (Use Excell spread sheets) You will end up with
contributions from every block now referenced to the front end. then...
4 Remember that uncorrelated noise does not add, but adds as the square
root of the sum of squares. However! you must only take into consideration
the noise contribution from each "additional" noise source. You cannot keep
adding in the contribution from the 50 ohm source impedance. Therefore,
each Noise Figure ratio must be squared and then have 1 subtracted from it.
After combining all the contributions, you will add the 1 back. Simply take
the square root of that total sum and find the 20log value and you will have
your total system's new NoiseFigure.
For example, let's find the noise figure for a receiving system that
attaches to an antenna consisting of long cable, amp, cable, and spectrum
analyzer (SA).
Passive devices have 0 dB noise figure (they do not add any noise)
Antenna is passive, but converts volts per meter into volts in a 50 ohm
system. Since it does not add any noise, there is no difference between
minimizing the NF at the antenna port or at the field that it measures.
So let's find out the NF of our system at 200MHz
long cable 3dB attenuation
amp 24dB 6 dB NF
cable -
SA 32dB
Note: You could have a perfect receiver that contributes no noise located
after the long cable and you would still have a 3 dB NF That's why
amplifiers are placed near the signal source.
Check your particular SA. It can have a Noise Figure from 26dB to 36dB
depending on its design. That means for a 1 MHz bandwidth you can only see
down to around -80dBm.
So let's move the amplifier out to the antenna and change the order of the
list:
cable -
amp gain 24dB with NF = 6 dB
long cable atten 3dB
SA with NF = 32dB
The list would show 6 dB NF at the input and (32-24+3=11) 11dB from the SA.
That is a ratio of 2 and a ratio of 3.55.
New Noise Figure is sqrt( (2^2-1) + (3.55^2-1) + 1 ) = 4 or 12dB
See how the SA still dominates?
To rewrite the equation:
NoiseFigure new
= sqrt ( ( NF1^2 - 1 ) + ( NF2^2 - 1 ) + ( NF3^2 - 1 ) + ... +
1 )
where all noise figures are ratios and referenced to a
single location.
- Robert -
Robert A. Macy, PE m...@california.com
408 286 3985 fx 408 297 9121
AJM International Electronics Consultants
619 North First St, San Jose, CA 95112
-----Original Message-----
From: KC CHAN [PDD] <kcc...@hkpc.org>
To: emc-p...@majordomo.ieee.org <emc-p...@majordomo.ieee.org>
List-Post: emc-pstc@listserv.ieee.org
Date: Friday, December 21, 2001 5:46 PM
Subject: noise figure
>
>Hi all
>
>It may not be purely EMC question, actually it is RF related, but I am sure
the experts here can answer my questions.
>
>We all know that we need to have a pre-amp. that is as lower noise figure
as possible, but how low it is enough or how it is related to the noise
floor viewed by a receiver or spectrum analyzer.
>
>Thank you
>KC Chan
>
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