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Example: A 3 dB pie-pad is said to have a 3 dB noise figure. If the input noise floor is greater than 3 dB above the thermal noise floor, resulting in both signal-in and noise-in to experience the same level of attenuation, then the output SNR has not changed. The question is, does the pad truly have a noise figure? An alternative question is how does one relate the resistors that comprise a pad to the noise figure of said pad?
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That's like saying if the signal into my receiver is really strong, say 80 dB above thermal noise, does my receiver truly have a noise figure. Of course it does. If the 3 dB pad is downstream in a cascade, where the noise and signal will be attenuated by the same amount as in your example, the pad will have very little effect on the overall noise figure of the cascade but the pad itself still has a noise figure of 3 dB. From HP App Note 57-1, page 6, "The specific input noise level for determining noise figure is that associated with a 290K souce temperature (ie: thermal noise). Your error is that you did not assume thermal noise at the input of the pad.
I don't understand your second question. A 50 ohm, 3 dB pi (not pie) pad has, starting at the input, a 292 ohm shunt resistor, a 17.6 ohm series resistor and a 292 ohm shunt resistor. This network has a loss of 3 dB and a noise figure of 3 dB.
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The noise figure is defined to equal (SNR)ip/(SNR)op when the SNR at the input is determined with only the thermal noise from the source impedance contributing. If this wasn't the case you could make F as close to 1 as you wanted by simply increasing the input signal and noise as you suggest. Not only that, but your comment would apply to any device - amplifier etc, as well as attenuators.
Peter.
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Of course a receiver has a noise figure. The point to the question is this: assume the input noise level to the pad, wherever it is in the cascade, is greater than thermal noise. Does the pad truly contribute noise to the system? If so, how much?
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Agree. So given a resistive pad of say 3dB(50 Ohm system), what is its noise contribution? Does it truly add noise to the system? If so, how much? And how do you arrive at said figure?
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Hello Paul,
Yes it had noise, thermal noise, the complete formula to calculate equivalent noise temperature is:
Tout = alpha*Tinput + (1-alpha)Tatt
Where Tout is proportionnal to the avalaible output noise temperature, Tinput is your avalaible input noise temperature, Tatt is the temperature of the attenuator and alpha is the attenuation in magnitude.
As you can see, the noise generated by an infinite attenuation pad is the full thermal noise, and the noise generated by a pad with infinitely small attenuation is O. And you can also see that the pad attenuate the thermal noise from the source.
The thing is what whatever temperature you have, you will always get thermal noise at the output if you had termal noise at the input (alpha + 1-alpha = 1). Consequence S/Nin - S/Nout = 3dB.
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The pad does not add noise in my interpretation. It just takes away your ability to detect a signal. Therefore we would consider the receiver to be more deaf with a pad than without it. This is equivalent to saying that the receiver noise figure has increased.
If you add a 3 dB pad to a receiver (at room temperature) then the new composite receiver will need 3dB more signal to detect. The pad did not add 3 dB more noise, because if it did, you could play very interesting games with the laws of physics and gett something from nothing. However nothing useful ever comes from doing nothing unless you are in politics but even then the merits of such outcomes are debatable.
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Yes, agreed.
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O.K. Help me here with a numerical example.
Noise in = -174 dBm / Hz, Tinput = 300 Kelvin, Attn = 3 dB. Is T(att) dervied from the attenuator value? Is it derived from the physical temperature. Alpha = 10^(3/10)=2, I'll assume Tatt = 300 K also. Then Tout = 2*300 + (1-2)*300 k = 300 k. If this is correct, then the thermal noise floor does not change from pad input to pad output.
Is this incorrect? Why?
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It definitely adds its own noise.
If the resistors in the pad were held at absolute zero then they would not contribute any noise, and both the input noise and signal would be attenuated by 3dB. Ths output S/N would be the same as the input and the pad noise figure would be 0dB.
When the resistors in the pad are at the same temperature as the source, they add noise to the system. One way of representing this is to consider the pad as noiseless and include the pad noise as an equivalent input noise power, which for a 3dB pad is equal to thermal noise. Thus the total effective input noise is twice thermal, and so the output noise is the same as with no pad. The output signal has been halved and so the output S/N is half the input. Thus the noise figure of the 'hot' pad is 3dB. You can change its noise figure to other values by changing its temperature.
Peter
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It definitely adds its own noise.
If the resistors in the pad were held at absolute zero then they would not contribute any noise, and both the input noise and signal would be attenuated by 3dB. Ths output S/N would be the same as the input and the pad noise figure would be 0dB.
When the resistors in the pad are at the same temperature as the source, they add noise to the system. One way of representing this is to consider the pad as noiseless and include the pad noise as an equivalent input noise power, which for a 3dB pad is equal to thermal noise. Thus the total effective input noise is twice thermal, and so the output noise is the same as with no pad. The output signal has been halved and so the output S/N is half the input. Thus the noise figure of the 'hot' pad is 3dB. You can change its noise figure to other values by changing its temperature.
Peter
http://www.radiolab.com.au>click here for Applied Radio Labs
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Is this so complicated?
Assuming a 50 ohm system: If it was a 300 dB attenuator it would add just about as much noise as a 50 Ohm resistor.
Now it is a 3 dB attenuator, so it adds half as much (in power).
If this attenuator is all you have on the input, it is mismatched and the case is uninteresting. At least for this discussion.
If your receiver has a 2.7 dB NF and you put a 3 dB attenuator on the input you now have a receiver with 3 + 2.7 = 5.7 dB.
Simple as that.
All the best,
Carl.
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Yes Carl, it is uncomplicated if viewed superficially. 3dB pad + low noise amp = noise figure larger than 3 dB pad.
The true question touches upon a subtle point. What is the noise power of a 3 dB attenuator?
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Maybe the formula of NF derived from SNR in and out is not accurate or not applicabe in this case? Why? No reference temperature? There must be noise due to bumping of electrons and such.
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The only thing is that alpha is not 10^3/10 but 10^-3/10.
So you must use 0.5 (it is 1/2) instead of 2 into your computation.
And it it true that the thermal noise floor doesn't change if the attenator is at same temperature than temperature reference To.
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The pad always has a noise figure equal to its attenuation. The pad no matter where it is in the cascade will contribute -174 dBm/Hz of added noise due to thermal effects. If the noise level is higher than thermal at a certain point in the cascade the noise added by the pad will have little affect. The other effect of the pad is to reduce the signal and noise at its output closer to the noise floor of the system. Again depending on where you are in the cascade the affect on the overall noise figure will vary. These effects are all taken care of with the cascaded noise figure equation. There is a nice RF cascade program in the free download area of RF Globalnet. Hope this helps.
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Assuming the system is 50 ohms throughout , the basic noise from the source is that of an ideal 50 ohm resistor , adding an attenuator does not change the value of the new source impedance and therefore does not change the noise however it will attenuate both the source signal and noise -- net result is a loss of signal to noise ratio of 3 db . If the source has excess noise above thermal it will also be attenuated .In practice attenuator noise may be larger than this because resistors can exhibit noise above ideal thermal values.
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Ian, are you saying that politics violates the second law of thermodynamics?
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Now, that last statement surprises me. Can you expand on it John?
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I love these noise discussions. There is always more to discuss and and new surprises. Had a few posts on this very topic sometime ago.
The prize goes to the claim that an attenuator does not contribute to noise! Next time you sit in front of that spectrum analyzer, go and put a 20dB pad in front. Not only your signal will go down but your floor will come up too.
Ok, back to physics. Noise out of an attenuator has two components. One that is attanuated coming from the source and one that is generated internally and which can be represented by an equivalent temp, Te. Then the power density at the output is (I can derive it for sceptics!):
No = (k/L)(Tg+(L-1)Te)
where Tg is the equivalent temp of the source, right? Note that since the equivalent noise temp of an attenuator is given by:
Te = (L-1)Tattn
where L is the attn ratio and Tattn is the PHYSICAL temp of the attn. Then
No = k((Tg-Tattn)/L+Tattn)
For very low L values, or very high source noise, the output noise from the attenuator is:
No ~ kTg/L
which is indeed approaching Ni/L. For very high L:
No ~ kTattn
which is approximately the noise power density the attenuator itself is generating. The exact equation given previously is only a special application of the Pierce law which explains the contributions to noise output of a network of resistors with DIFFERENT physical temperatures. Another note here, be aware of the gain definition used in the NF formula, unmatched attenuator will be quite different.
Now, given all this, what can you say about the noise figure of a filter OUTSIDE the pass band?!
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