# Detection Bandwidth for Noise Power Calculation

I am trying to calculate the $$P_{noise}$$ in an experimental setup, such as given below. The DUT is excited at a predefined frequency, $$f$$ , with power $$P_\text{in}$$. The total power is measured by the RF digitizer. We can assume that the generator and the sample are ideal so that only noise in the system , $$N_{a1}$$, is originating from the Low Noise Amplifier (LNA). Now, I know that $$P_{Na1} = kT_e \Delta f$$ and $$T_e$$ can be found in LNA specs. What confuses me is the choice of bandwidth.

If we look at the RF front end of the detector;

• We see that RF input is first shifted to IF band, then adjusted (to cover ADC full range I suppose) and filtered before ADC.
• ADC has $$f_{sample} = 250\,\text{MHz}$$. Also, the device allows $$f_{sample}$$ down to 10 kHz. In view of these informations, here are my questions:

1. What determines the $$\Delta f$$ for $$P_{noise}$$ calculation ? Is it $$f_{c}$$ of the BPF at the detector front-end ?
2. What is the role of the sampling frequency? Would it be correct to say $$\Delta f = f_s$$ or $$f_s / 2$$ ?
3. What is the role of the measurement time? If I measure during 1 s and average the data, can we say that the detector will be accepting noise from 1 Hz window of noise PSD ?
4. (additionally,) Given the fact that ADC has $$f_s = 250\,\text{MHz}$$ yet it's possible to sample at $$\text{kHz}$$ regime, can we say that data is being averaged in case of downsampling ?
• Specifically what is the purpose for this computation of noise in your application? Once you have the noise measurement, what are you going to do with it? This may change my answer. – Dan Boschen Apr 2 '20 at 20:01
• Note that the two diagrams you've posted probably aren't directly linked – the input to the digitizer board seems to be some IF or baseband signal, whereas your upper diagram clearly expects things to happen at RF; so, there's a mixer stage missing, and aside from the LNA, that's probably defining the noise figure of the system. – Marcus Müller Apr 3 '20 at 7:29
• @MarcusMüller, you're right about the missing stage. I skipped it in my post. You can take a look at the RF board diagram here and full datasheet can be found here 3035C. The problem is that, the datasheet mentions noise spectral density (dBm/Hz) which again requires a choice for bandwidth – Krlngc Apr 3 '20 at 8:00
• @DanBoschen, I need to calculate the noise power for several reasons. First, I need to make sure that signal arriving at LNA ($P_{in}^{dBm} + S_{21}^2$) is greater than $P_{noise}$ to have SNR > 1. Second, physically interesting phenomena in my case is the fluctuations in DUT. In fact, I'm probing it in real time. So, I need to remove the noise coming from LNA. – Krlngc Apr 3 '20 at 8:06

## 1 Answer

The noise of interest would ultimately be over the bandwidth of your signal and specifically all filtering done prior to ultimately making a decision over which symbol was transmitted- which would typically be the bandwidth of your matched filter. So given a noise density in $$dBm/Hz$$ you can translate that to the total noise in the signal to noise ratio of interest. This is only significant at the point where you make your decision unless you are dealing with waveforms that have negative SNR prior to correlation such as GPS--- in this case you would want to watch the total noise power at the input to each amplifier an ADC to ensure you are not causing saturation and clipping (as the result of too much gain).

For example, the thermal noise floor before we do anything with additional electronics at room temperature is -174 dBm/Hz (given by kT). If we had a 1 MHz signal bandwidth, then this noise level without added contribution would be -174 dBm/Hz + 10Log10(1MHz) = =114 dBm (given by kTB). The added noise by an LNA is the "Noise Figure" and specifically it is how much this noise floor is elevated if the LNA was terminated with it's matched load (meaning no gain prior to the LNA). If the noise figure was 3 dB, which means a 2x increase in power, this would mean the LNA is adding just as much noise as thermal noise is. So this means the LNA self-noise, referred to the LNA input is also -174 dBm/Hz If the gain of the LNA was 10 dB, then the noise power at the output of the LNA would be -174 dBm/Hz + 3dB + 10 dB = -161 dBm/Hz. If the noise figure of the LNA was 0 dB (meaning perfect, no added noise) the noise power at the output of the LNA would be -164 dBm/Hz.

In your comments you discuss a concern with fluctuation and "removing the noise coming from the LNA". This would have nothing to do with the noise figure of the LNA but sounds like an oscillation, instability or intermodulation spur issue in your receiver (or an interference signal being received, which you can rule out by disconnecting your antenna and terminating the antenna input to the receiver). The noise of the LNA that is in band cannot be "removed". The only significance of the noise of the LNA is to set the sensitivity of the receiver. For this the LNA is just one contribution (a dominant one!) but the noise figure of the overall receiver would be computed as a cascaded noise figure calculation for EVERYTHING up to the point where the symbol decision is made- this includes the LNA, the noise figure of every analog component factored by its gain, the aliasing effects of the mixer stages and ADC, the phase noise and jitter of the clocks, and the quantization noise and further possible aliasing effects in the digital part of the system.

I would suggest you do the following: Determine with measurement the overall noise figure of your receiver to see if you have an issue with this. First measure the gain of your overall receiver by connecting a calibrated (known power level) test tone to the receiver input. Sweep this tone over the full signal bandwidth to ensure there isn't significant variability across the pass band. Measure this gain at the latest possible point in your digital receiver so that you can capture all effects up to that point in the system. Ensure the input test signal is low enough in power such that the receiver is linear by increasing the input signal 5 dB and ensuring the output increases 5 dB (I would increase to the point where this isn't true and then back-off 10 dB as you want as strong of a signal as possible). This is the gain, $$G$$, in dB and can be in any units of dB you choose (dBm, dBFS, dBc etc). Then terminate the input with a matched load and measure the noise power density in dB/Hz using the same units of dB. This is $$N$$ in the equation below.

Your noise figure in dB can then be estimated using:

$$NF = N- G - 174$$

This would only be as accurate as the gain variation across your bandwidth, which is why this is actually typically done with "white" noise sources that spread the test signal evenly across the bandwidth of interest. But if you confirmed a minimum passband ripple as I suggested, this is valid approach. This noise figure HAS to be higher than the LNA noise figure given the receiver is a cascade of all noise figure computations. A well designed receiver with an interest in maximum sensitivity would keep all additional contributions below an extra 1 dB, but I would not be surprised if a low cost receiver had an extra 2 to 3 dB of degradation.

• Thanks for the comprehensive answer. A lot to extract, so I'll go one by one. In fact, I'm using single tone CW at UHF band to excite the sample. Given that the source is non-ideal, is there a rule of thumb for considering single tone BW ? or maybe it's not important as I should consider matched filter BW's. Second is a tiny detail; when you calculate the noise floor for 1 MHz BW as $-174 dBm/Hz + 10log10(1 MHz)$ I'd expect the result to be in dBm, as checking the units $k_b$ $(J.K^{-1})$ T$(K)$ B $(1/s)$ yields Watt. – Krlngc Apr 3 '20 at 16:36
• The units are in Watts so you use 10Log(kTB). So what you see in my equation is 10Log(kT)+10Log(B). Yes even with a single tone you consider the matched filter bandwidth in that ultimately the receiver will be making some sort of decision--- so the signal to noise ratio within the bandwidth just prior to detection will ultimately be your margin against a bad decision (bit error). – Dan Boschen Apr 3 '20 at 16:39
• Yes, then the result should be -114 dBm not -114 dBm/Hz. Do you agree with it ? – Krlngc Apr 3 '20 at 16:42
• Sounds like you might be using OOK (on off keying), with CW on = 1 with CW off = 0? In that case the receiver needs to decide between 1 and 0 using a threshold between the two. Follow that through and you'll see better what I am getting at above. – Dan Boschen Apr 3 '20 at 16:43
• No the result for kT is -174 dBm/Hz. The result for kTB is -174 dBm if it is 1 Hz bandwidth. Ah I see, I didn't put that correct in my answer- yes agreed fixing that now. Thanks! – Dan Boschen Apr 3 '20 at 16:43