Noise Figure of Attenuator - RF Cafe Forums
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Post subject: Noise Figure of Attenuator
Unread postPosted: Wed May 11, 2005 6:51 am
Can anyone explain why the noise is not attenuated too as the signal when passing through
Namely, why is the Noise Figure of an attenuator is equal to its attenuation value.
Unread postPosted: Wed May 11, 2005 9:28 am
Unread postPosted: Wed May 11, 2005 10:07 am
Here is the noise equations for an attenuator
alpha being the attenuation.
Nout= Nin*alpha +
If your input noise is already KToB, then you get.
Nout = KToB*alpha+(KToB(1-alpha) =
KToB, whatever is the attenuation.
Now if the noise at the input is not KtoB then, it's another thing and
the equation shows this.
Signal is treated independently.
Sout = a*Sin, whatever is the signal
So therefore if you assume KToB as input noise and Sin as input signal
F = -(Sout/Nout)/(Sin/Nin), in
this case Nout = Nin
F = -Sout/Sin => NF = (SindB - SoutdB)
Now if you change the temperarure of the
attenuator or the level of input noise simplification does not work and you have to carry the whole equations to
the end result.
Post subject: Noise figure
Unread postPosted: Fri May 13,
2005 6:08 pm
Sorry for my English.
The answer is simply, you are both right and wrong.
noise (white noise, -114 dbm/ 1MHz) pathes attn its not attenuated because noise cannot go lower then
But, on the other hand when noise(white noise as well) passes throug active element
(amplifier etc.) increased due to it's gain, and now passes an attn - the noise's power will be lower by the value
of the attn/NF.
Unread postPosted: Mon
May 16, 2005 9:25 am
I think noise in an attenuator IS attenuated.
The subtility here is that an
attenuator will give back a quantity of noise proportional to its attenuation. The end result is the same amount
of noise, but the origin have changed.
If all temperatures are equal, it looks like if the attenuator is
not attenuate the noise, but it realy does even if you dont see it in the end result.
Unread postPosted: Mon May 16, 2005 10:11 am
Now that really clears things up
Post subject: NO !!!
Unread postPosted: Mon May 16, 2005 10:41 am
You right about one thing , attn loss = attn NF
But !! Noise cannot be attenuated below -174/1hz -
It's a fact !
But, if you the attenuation is after amplifier, the noise will be attenuated by the
Unread postPosted: Mon May 16, 2005 2:03
All this depend son the physical tempeature of the attenuator.
Rembember "To" is the reference
temperature, not a universal constant.
Also, since a resistance noise is KToB and an attenuator is made
from resistance, aint the noise throughput have to be proportional to To?
So if you have let say a
cryogenic 3dB pad let say at a temperature of 0.1*To the limit is not anymore KToB but 0.1KToB.
point of view for a perfectly matched attenuator:
IF the input noise is higher than the physical ambient
IF the attenuator is physicaly maintained at a physical ambient of Ta
cannot obtain in a mached system a available noise power less than KTaB.
This become false IF:
input noise is lesser than KTaB,
The attenuator is at a temperature lesser than KTaB.
Post subject: Noise Figure and Attenuation
Unread postPosted: Mon May 16, 2005 8:29 pm
Actually, it is possible to build a circuit with a noise temperature much lower than the ambient
temperature. Those circuits are active, and thus limited in bandwidth. (This is a fascinating area of circuit
design, by the way!) Cryogenic cooling is not required down to a limit).
But the noise figure of a circuit
is determined primarily by the noise generated in the first amplifier stage. If you weaken the incoming signal by
x dB, you've reduced the SNR by x dB, because of the noise generated after the attenuator. So the noise generated
in the attenuator itself only makes things worse than the standard formula.
Hope this helps!
Post subject: nooooooooooooo
Unread postPosted: Tue May 17, 2005 6:06 am
confusing 2 things. The original question was not about the ambient temp. it was about noise atenuation,,,, You
are way of the issue now
Unread postPosted: Tue May 17, 2005
The capacity of an attenuator to attenuate noise is directly related to its physical
If the attenuator with attenuation "A" (not in decibel) is at room,
It will attenuate
input signal like this
Sout = Sin * A
It will also emmit thermal noise in the following way
Nout = KToB*(1-A)
If it happens that the input signal is noise from a resistor at the same temperature than
the attenuator then combined signal and noise output is
Sout + Nout = KToB*A + KToB*(1-A) = KToB
if the attenuator and the input noise generator have different temperatures, then....
Sout+Nout = KT1B*A +
Therefore the capacity of an attenuator to diminish input noise is related to the input noise
level and to the physical temperature of the attenuator.
Unread postPosted: Tue May 31, 2005 3:25 pm
Noise Figure(attenuator) = 10*log(s21) =
Post subject: Noise
Unread postPosted: Thu Jun 09, 2005 1:56 am
Guys, you are confusing 2 thigs.
1) White noise
2) White noise which was amplified
noise physicly cannot be lower then -114@1MHz, that's a fact. It's a random signal.
White noise at
amplifier output (even if G=1 dB) can go lower then -114 dBm, because from this poin you can "look" at this noise
as a signl - It's no more a random signal as at amplifier input, the amplifier "gave it a spreading vector".
So the answer for the original issue is - the noise is not attenuated by the attn value because at this point
it's white noise (random signal) and cannot be lower then -174 dBm@1Hz.
Unread postPosted: Thu Jun 09, 2005 12:21 pm
You are propagating a very damaging
dogma about thermal noise.
The thing with thermal noise is that it is "thermal", reduce the temperature and
the emitted noise is also reduced.
The equation is Na = KTB
Reduce the temperature you get less
noise. If you put 290K in there you get indeed -174dBm per hertz.
An attenuator will generate noise in
inverse proportion with it's attenuation, and in proportion with the attenuator's physical temperature.
an example a 0dB attenuator, a copper trace, generate no noise and a 1 million dB attenuator will generate 100% of
kTB. But if temperature is not To, then this is not -174dBm/Hz but something lower or higher.
Now if indeed
the environment and equiment you are testing is at To=290K and that the noise source at the input is generating
-174dBm/Hz then you will resolve all the equations to get -174dBm whatever is the attenuator value, but this is
the simplification that is the cause for all this misunderstanding.
If this noise floor would be true, how
would you explain LNAs with equivalent input noise of 30K or 185dBm/Hz, and what would we the reason for cryogenic
LNA if it's not to reduce thermal noise in there?
Post subject: That's not
the issue here
Unread postPosted: Wed Jun 15, 2005 5:07 am
You are right, I know that the noise of an
attenuator depends on it's temp.
That's not what I'm arguing about.
Look at the original question "Can
anyone explain why the noise is not attenuated too as the signal when passing through attenuator?"
question is not regarding the noise value (wheather it's lower or higher -174dBm/Hz).
So the answer is
simple - White noise (If the question was originaly about white noise, lets say at 1hz) cannot be attenuated below
-174dBm, becuse it's something random - it's not a signal.
BUT, when pathing an amplifier - you can consider it
as a signal, and this time it can be lower then -174dBm. Good example for it is a radio. The pass (air) loss is
very high... and the noise power at receiver input is very low.
Unread postPosted: Wed Jun 15, 2005 11:59 am
"So the answer is simple - White noise (If the question was
originaly about white noise, lets say at 1hz) cannot be attenuated below -174dBm, becuse it's something random -
it's not a signal"
If you replace -174dBm/Hz by kTB, where T is the attenuating system's physical
temperature, then we have an agreement.
-174dBm/Hz is no magic number. It's just a conveinient reference
I would say, a system, either attenuating or amplifying, cannot, have a noise output of a
lesser value than the thermal noise corresponding to it's physical temperature. That is roughly -174dBm at room
temperature but less a lower temperatures.
Post subject: Noise
Unread postPosted: Thu Jun 16, 2005 1:34 pm
I respectfully disagree with the assertion that Noise Figure
(or its equivalent, Noise Temperature) cannot be lower than the ambient temperature in K.
Amplifiers in everyday use by the millions in VSAT applications (Dish Network, DirecTV) have Noise Temperatures in
the 50-60K range. Likewise, you can look at the Noise Figure for microwave transistors and find numbers in the
less-than-0.5 dB range, which corresponds to under 40 Kelvin. (NE27200 at 12 GHz, see http://www.cel.com/pdf/datasheets/ne27200.pdf
for one example).
There are several issues involved.
One has to do with impedance
matching and (mis)matching. The Thevenin equivalent resistance of the noise generated by a (perfect) resistor of
50 Ohms is, not surprisingly, 50 Ohms. The Thevenin equivalent noise voltage is sqrt(4kTBR). [Note to the
physics-literate: this is the Rayleigh-Jeans approximation, and it's not valid for very low physical temperatures
T, nor for frequencies in the ultraviolet range.;) ]. The maximum power transfer to a load resistor RL occurs at
Rs = RL. It is under this condition only that you can use the classical equation for noise power Pn = kTB.
This explains the fact that a wire or cable with no loss does not add noise to reach or exceed a noise temperature
of the ambient temp (often 290 K or 300 K for convenience), while a resistor equal to the characteristic impedance
Unread postPosted: Mon Jul 11, 2005
Joined: Mon Jun 27, 2005 2:02 pm
think noise in an attenuator IS attenuated.
here is that an attenuator will give back a quantity of noise proportional to its attenuation. The end result is
the same amount of noise, but the origin have changed.
If all temperatures are equal, it looks like if the
attenuator is not attenuate the noise, but it realy does even if you dont see it in the end result.
That is absolutely right, noise is attenuated when passing through an attenuator, and the attenuator "completes"
the noise by adding its self noise, so the total noise at the output remains the same. The attenuator only
subtracts the signal in an amount equals to its loss. So that the SNR from input to output degrades by the amount
equals to the attenuator's loss.
The spectral density of noise at the output of an attenuator, can't be
reduced under the noise floor at the input of the attenuator.
Unread postPosted: Mon Jul 11, 2005 1:21 pm
... unless the attenuator's body is at a lower temperature than the equivalent temperaturr of the white noise
at the input.
Unread postPosted: Mon Jul 11,
2005 2:17 pm
Joined: Mon Jun 27, 2005 2:02 pm
For that there is the Boltzmans' equation, which defines the T/T0 ratio:
Where T is the actual temperature and T0 is the room temperature
Unread postPosted: Wed Jul
13, 2005 6:59 pm
I have this sitation:
Rx antenna--Attenuator--RX+C/N and BER indicator.
change the attenuator value for change the C/N in RX point.
I know that
(C/N INPUT ATTENUATOR dB) - (C/N OUT
ATTENUATOR dB)=ATT dB
For the Noise????? what is the relation of Noise ??
out the antenna is small.........
thanks for help!!!!!
Unread postPosted: Thu Jul 14, 2005 9:55 am
Na= Noise at your antenna output
Ac= Attenuation of your
cable from antenna to attenuator
Aa= the attenuation of your attenuator (not in dB but in magnitude)
BW= The band you are interested in and the one on which all those noises are evaluated.
Nrx= The noise after the attenuator
NRx = (Na*Aa*Ac)+(K*Tp*BW*(1-(Aa*Ac)))