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NF of attenuator is same as the attenuation value as the noise floor remains same but signal level gets attenuated? But NF of amplifier depends on the device as it increases both signal and noise?

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The Noise Figure of a passive attenuator is the attenuation value of the attenuator. This is because noise figure is by definition the difference in the SNR of the output and the SNR of the input, with the input terminated with the characteristic impedance of the system (typically 50 ohms). Using this definition, the noise figure of an amplifier similarly is the additional increase in noise beyond the gain of the amplifier when the input of the amplifier is terminated.

The ever-present thermal noise floor is a white noise source with a power spectral density given by kT with k as Boltzmann’s Constant and T as temperature. At room temperature in power units of dBm (dB relative to 1 mW) this ends up being a power spectral density of -174 dBm/Hz. In the case of an attenuator, thermal noise is the only noise source given it is a passive device. If the attenuator is matched to 50 ohms, ans with the input tied to a 50 ohm source resistor only and no other gain at the input, the thermal noise at the input with be -174 dBm/Hz (kTB) at room temperature.

The thermal noise at the output of the 50 ohm attenuator will also still be -174 dBm/Hz! The attenuator cannot somehow attenuate the thermal noise that is present in all parts of the system-- it certainly attenuates any noise that is applied to the input, but in terms of power density, the same level of thermal noise is present at the output as a local independent noise source.

Thus in terms of (SNR out) - (SNR in): Consider a noise-free signal at the input with it's noise floor then limited by thermal noise alone; this sets the input SNR. The signal will certainly be attenuated by the attenuation value, but as we explained above the noise can't get pushed down lower than its theoretical limit (the thermal noise floor). Since the noise is not affected as it is the same value at the output and input in this case, the SNR is and therefore changed by the attenuator value, which by definition is the Noise Figure.

Cascaded Noise Figure is then considering what occurs when you place an amplifier in front of an attenuator. Now in this case you are applying amplified noise to the input, not the thermal noise floor specifically. This amplified noise like any other waveform can of course be attenuated, down to the limit of the thermal noise floor. Consider these two cases and how the overall Noise FIgure compares:

The formula for cascaded Noise Figure is as follows for two components, expandable for any number of components. I'll give the equation first but then demonstrate with two cases where the explanation above becomes very clear.

$$NF_{cascade} = NF_{N-1} + \frac{NF_N-1}{G_{N-1}}$$

This states that the overall noise figure of N cascaded devices is the noise figure of the first component plus the noise figure of the subsequent components reduced by the cummulative gain of to each components input. The gain (G) and noise figure (NF) are given in power ratio (non-dB) quantities.

Consider this with these two components:

Amplifier with 20 dB gain, and 3 dB Noise Figure 10 dB Attenuator

Case 1: Amplifier followed by Attenuator

According to the formula:

$NF_1= 10^{3/10} = 2$

$NF_2 = 10^{10/10} = 10$

$G_1 = 10^{20/10} = 100$

$NF_{cascade} = 2 + \frac{10-1}{100} = 2.09$

The Noise Figure in dB is $10Log_{10}(2.09) = 3.2 \text{ dB} $

Notice the attenuator had very little effect due to the amplifier in front of it! This is the motivation for Low Noise Amplifiers (LNA) at the front-end of the receiver with very low loss components in front of it (typically a necessary filter).

Case 2: Attenuator followed by Amplifier

$NF_1= 10^{10/10} = 10$

$NF_2 = 10^{3/10} = 2$

$G_1 = 10^{-10/10} = 0.1$

$NF_{cascade} = 10 + \frac{2-1}{0.1} = 20$

The Noise Figure in dB is $10Log_{10}(20) = 13 \text{ dB} $

Intuition:

Consider a received signal that has a 30 dB SNR, with the noise limited by thermal noise alone. In case 1, we amplify both the signal and the noise by the amplifier with 20 dB gain. This amplifier has a 3 dB noise figure which means the signal goes up by 20 dB, but the noise goes up by 23 dB! (Due to the added electronics noise that is captured in the Noise Figure metric). So now the SNR at the output of the amplifier is 17 dB, but the thermal noise is amplified and now much higher (23 dB higher!) than the actual thermal noise floor present at the amplifiers output (we have smothered it with an amplified version of the noise). The attenuator can now push both amplified waveforms (signal and noise components down by it's attenuation or 10 dB. Thus the signal that we increased by 20 dB is now reduced by 10 dB, and the noise that we increased by 23 dB is reduced by 10 dB. However the underlying thermal noise is now closer to the amplified noise level, 13 dB away, contributing a small amount to total noise (explaining the additional 0.2 dB in the 3.2 dB Noise Figure result).

Notice the power summation of the amplified thermal noise with the thermal noise 13 dB lower is a total noise increase of 0.2 dB (it all adds up, demystifying cascaded noise figure!):

$$10Log_{10}(1+10^{-13/10}) = 0.2 \text{ dB}$$

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  • $\begingroup$ U gave a great explanation. now i got clarity about noise figure. Its just the formula given in the intuition section i am confused. U mean if my noise floor is -174dBm. With 23 dB(gain +NF) gain it becomes -151 dBm and with attenuation of 10 dB it becomes -161 dBm so now the new noise floor is 13 dBm above -174dBm. But how to analyze the formula u have given? $\endgroup$ – Steve Aug 6 at 10:54
  • $\begingroup$ The amplified noise which indeed in total is -151 dBm gets pushed down 10dB but the underlying thermal noise floor at that part of the system does not go down resulting in a noise floor that will be higher than -161 dBm. The resulting noise floor is the power sum of the noise at -161 dBm and the noise at -174 dBm, and that is what the formula is giving you. $\endgroup$ – Dan Boschen Aug 6 at 20:47
  • $\begingroup$ Now i got it and I did the math for the case 1: amplifier followed by attenuator ( 10log10(2)=3.0103dB for the first amplifier As total noise figure is 10log10(2.045)= 3.1069dB So the second stage contributes 3.1069-3.0103=0.0966dB but the formula for sum of powers at second stage gives 0.2124dB which should be rather 0.0966dB? So am i missing something? $\endgroup$ – Steve Aug 7 at 0:18
  • $\begingroup$ @Steve yes my silly mistake of saying 10^(20/10) = 200, that should be 100 of course! I fixed that, sorry for the confusion. $\endgroup$ – Dan Boschen Aug 7 at 2:19

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