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Below are all of the forum threads, including all
the responses to the original posts.
Post subject: Noise Figure of Attenuator
Wed May 11, 2005 6:51 am
Can anyone explain why
the noise is not attenuated too as the signal when passing through attenuator?
Namely, why is the Noise Figure of an attenuator is equal to its
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 + KToB(1-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
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)
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.
Unread postPosted: Fri May 13, 2005 6:08 pm
Sorry for my English.
The answer is simply, you are both right
When noise (white noise, -114 dbm/ 1MHz) pathes attn
its not attenuated because noise cannot go lower then -114bbm/1Mhz.
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
I think noise in an attenuator IS attenuated.
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
Now that really clears things up :roll:
Post subject: NO !!!
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 attenuation value
Unread postPosted: Mon May 16,
2005 2:03 pm
All this depend son the physical tempeature of
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.
From my point of view for a perfectly matched
IF the input noise is higher than the physical ambient
IF the attenuator is physicaly maintained at
a physical ambient of Ta
THEN you cannot obtain in a mached system
a available noise power less than KTaB.
This become false IF:
The input noise is lesser than KTaB,
is at a temperature lesser than KTaB.
Post subject: Noise Figure and Attenuation
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
Tue May 17, 2005 6:06 am
You are 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 9:16 am
The capacity of an attenuator to attenuate noise is directly related
to its physical temperature.
If the attenuator with attenuation
"A" (not in decibel) is at room,
It will attenuate input signal
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
+ Nout = KToB*A + KToB*(1-A) = KToB
But if the attenuator and
the input noise generator have different temperatures, then....
Sout+Nout = KT1B*A + KT2B*(1-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) =
"insertion loss" [dB]
Unread postPosted: Thu Jun 09, 2005 1:56 am
are confusing 2 thigs.
1) White noise
2) White noise which
White 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
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.
As 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?
That's not the issue here
Unread postPosted: Wed Jun 15, 2005 5:07
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?"
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.
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
If you replace -174dBm/Hz by kTB, where T is the attenuating
system's physical temperature, then we have an agreement.
is no magic number. It's just a conveinient reference temperature.
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
I respectfully disagree with the assertion that Noise Figure
(or its equivalent, Noise Temperature) cannot be lower than the ambient
temperature in K.
Low Noise 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
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 does.
Unread postPosted: Mon
Jul 11, 2005 4:55 am
Joined: Mon Jun 27, 2005 2:02 pm
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.
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
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:
and BER indicator.
I 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 ??
the noise out the antenna is small.........
thanks for help!!!!!
Unread postPosted: Thu Jul 14, 2005 9:55 am
Na= Noise at your
Ac= Attenuation of your cable from antenna to attenuator
Aa= the attenuation of your attenuator (not in dB but in magnitude)
Tp= The ambient temperature
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)))