# Attenuator noise figure and loss

My question is with regard to how attenuators effect noise. From what I understand $$\text{NF} = 10\log \left( \frac{S_{in}/N_{in}}{S_{out}/N_{out}} \right) = 10\log \left( L \frac{N_{out}}{N_{in}} \right) = L_{dB} + 10\log\left(\frac{N_{out}}{N_{in}}\right)$$ Now since (for an attenuator) $$\text{NF} = L_{dB}$$ that would imply that $$\frac{N_{out}}{N_{in}} = 1$$

From what I can tell, this means that the noise on the input passes through the attenuation unaffected by the attenuator. In other words, the attenuator minimises the signal but not the noise. I read that this is only true at 290°K and that the attenuator can not reduce the noise below the noise floor. I don't really understand this though. If the temperature were 50°K, would the noise still pass though the attenuator unaffected? It'd make sense to me that it would, unless there is something special about 290°K. As a bit of an educated guess, I would've thought that if the room temperature was, say, 50°K and the noise at the input of the attenuator was also 50°K (so $N_{in} = kB \cdot 50$) then the attenuator would not reduce the noise; however, I'm not exactly sure why.

I would've actually thought that the attenuator might actually introduce noise to the output because it is made passively. I can see that perhaps the resistors in the attenuator introduce noise, and the configuration of them reduce noise and the two may cancel and result in no net noise being added, though I'm not sure about this.

If someone could shed some light on this to help me understand this better, that would be great.

Thank you.

Edit: Specifically what I don't get

The scenario above is often a source of confusion for RF engineers, because it is commonly known that the noise figure of any passive component is equal to its loss. Noise figure (NF) and noise factor (F), are commonly described in terms of the ratio of input SNR to output SNR at the specific temperature of 290 K. As described in the equations below, noise figure is simply the logarithmic equivalent of the noise factor measurement.

Based on the equations above, it might SEEM that use of an attenuator on the output of a vector signal generator would attenuate the signal strength without attenuating the noise power. However, it is important to recognize the equation for noise figure ONLY relevant when the noise level is equal to the thermal noise density at -174 dBm/Hz. In general, terms such as noise figure and noise factor should only be applied to wireless receivers, since the noise power from an antenna at 290K will be approximately -174 dBm/Hz.

For example, suppose a receiver observes a signal with a SNR of 60 dB at a power level of -114 dBm. In this scenario, applying a 20 dB attenuator would attenuate the signal power by 20 dB to -134 dBm. However, the noise level would remain unchanged, since the noise power is already at the thermal noise level. Thus, when Pnoise = -174 dBm/Hz, the noise figure of a passive attenuator is equivalent to its loss.

On the other hand, output noise density from an RF generator is caused by active components and will result at levels that are sometimes significantly above the theoretical noise floor. In this scenario, noise power can be attenuated to the level of the thermal noise floor. As a result, high sensitivity receivers can be tested with a stimulus which has a noise power that is well below that of the signal generator.

The above for paragraphs are taken straight from: http://www.ni.com/white-paper/6810/en

I don't understand why 290°K seems special and what the noise power is acting on. I don't understand why in the third paragraph, it says that it won't attenuate the noise.

Reading the question and the comments, there may be a conceptual misunderstanding : the attenuator WILL attenuate any noise presented on its input (even from just a 50 ohm source impedance), to the same extent it attenuates the signal.

However it also generates noise of its own, which may be represented as the noise from a perfect resistor equal to its own output impedance, and this is added at the output to the (attenuated) input signal and noise. So if input and output Z are both 50 ohms, the net result is attenuated signal + marginally increased noise (i.e. NF = attenuation).

But if its output impedance is lower, the added noise is also lower, thus improving the noise voltage as Andy states.

So represent the attenuator as a perfect attenuator (attenuating noise) in series with a Johnson noise voltage source equal to the output impedance. The rest is just applying the formulae.

EDIT: re: updated question.

(1) There is nothing special about 290K except that it's a realistic temperature for the operation of a passive circuit. The reason they chose it is that the article quotes a noise floor ( -174dBm/Hz) which is correct for a specific temperature : yes, 290k.

(2) While any resistance in the attenuator will contribute noise, I realise that it is not a satisfactory explanation as to why you get the same noise out of an attenuator, because (as Andy says) you could make a capacitive attenuator which is not a Johnson noise generator. So we have to look a little deeper, and remember these noise sources are the statistics of the individual electrons that make up the current.

So, let's say we build a (50 ohm in, 50 ohm out) attenuator, and attempt to cheat Johnson by using a capacitive divider. That implies a node within the attenuator which conducts some of the input current to ground. At this node, we have two current paths; a fraction of the current flows to output, the rest to ground. What determines which path an individual electron will take? Essentially, chance. Collectively? Statistics. So this is a noise source.

Or let's just add series capacitance to provide enough attenuation : we thereby avoid dividing the current flow and eliminate the noise source, right? At the cost of reducing the signal current; our statistics now operate with a smaller sample size and consequently greater variance : more noise.

These results are the best you can do, there is no way round them.

• Thanks for the answer. I've added the paragraphs the I don't understand. It says that an attenuator can't attenuate the noise below the noise floor basically; however, I don't understand where this noise comes from... Does it come from the resistors in the attenuator or? – user968243 Apr 1 '13 at 13:15
• Isn't the reason why the 50 ohm attenuator adds the equivalent noise of a 50 ohm resistor due to the fact that it "looks" like that impedance at the output? Hence there has to be some real part; it couldn't be made out of capacitors or inductors only. As far as electron paths, I don't think this explains the noise. You could conduct similar thought experiments at 0 k. Still, some number of electrons will flow in a particular direction but the noise contribution of that passive part will be 0. – Terrabits Mar 22 '18 at 23:43

Firstly, there is nothing special at 290K. Resistor noise increases from zero at -273ºC (absolute zero) to a value governed by actual temperature (in Kelvin) and resistance.

Secondly, attenuators don't have to be resistive - they can be capacitive potential dividers and these will not introduce extra noise. Ditto with inductors.

Thirdly, a resistor attenuator will introduce noise but it may easily be small compared to the noise that is being attenuated.

Fourthly, if the noise into an attenuator is X, that noise will be reduced by the same factor as any signal would be. If the attenuator is 2:1 then the output noise will be X/2.

• Actually, the attenuator does not necessarily reduce noise. Underneath Figure 5 here: ni.com/white-paper/6810/en it explains this. – user968243 Apr 1 '13 at 9:18
• @User968243 The theoretical noise floor of -174dBm per Hz is for 50 ohm systems at room temperature. If the output of the attenuator is 50 ohm then it can't reduce below this figure but if the output is 10 ohms then it can. Ditto if the temperature is lower. – Andy aka Apr 1 '13 at 9:57
• I don't really understand 'what' the noise floor is. Like where does it come from? and why does 50ohm -> 10ohm change anything? – user968243 Apr 1 '13 at 10:08
• @user968243 any resistor at a temperature higher than absolute zero inherently produces noise. Look up Boltzmann constant of Johnson noise. – Andy aka Apr 1 '13 at 10:33
• @user968243 I made a mistake in the above comment about 50 ohms and 10 ohms - the noise floor of -174dBm/Hz is irrespective of resistance. The noise voltage will reduce with lowering resistance but "R" factors out when looking at the power - my apologies. – Andy aka Apr 1 '13 at 15:34