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I have a 1kN force transducer which can output (through an amplifier) +/-10V for +/-1kN nominal force range and I sample it with a DAQ device which is 24-bit and has +/-10V range. The overall noise floor is 20mV rms.

Now my question is, using the same DAQ device settings (+/-10V input range & 24-bit) but now if I halve the amplification and match the 1kN force transducer range to -5/+5V instead and assuming the noise floor is also halved to 10mV rms, how can I quantify and compare SNRs of both cases? Is one case anyhow better than the other one?

JRE
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GNZ
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  • Unlikely that the DAQ is effective any where near 1.2 uV/bit. And 20mV is 14 bits already at that resolution. Or is this a pretty good laboratory setting? – Paul Uszak May 04 '20 at 23:15
  • The SNR seems to be limited by all sources of noise & not the ADC resolution or supply voltage. Focus on filtering out the noise from the signal. – Tony Stewart EE75 May 05 '20 at 00:45

3 Answers3

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As has been mentioned in some of the comments, your overall noise floor of 20 mV RMS is far larger than your quantization noise, so without doing any calculations, you can be sure that halving your dynamic range while using the same DAQ device will have negligible impact on your SNR. However, you requested a quantitative analysis, so I will start by quantifying the ratio of the transducer noise power to the quantization noise power.

The quantization error is modeled as being equally likely to take on any value within a range of one least significant bit. This is a good assumption as long as your signal is much larger than your least significant bit. With this assumption, one arrives at the well-known result for the time-averaged quantization noise power \$P_{quant} = \frac{LSb^2}{12}\$, where \$LSb\$ is the voltage associated with the least significant bit (reference here). In this case, the LSb value is \$LSb = \frac{20 mV}{2^{24}}\ = 1.2 \mu V\$. Therefore

$$P_{quant} = \frac{(1.2 \mu V)^2}{12} = 0.12 (\mu V)^2$$

Your transducer noise power in the first case (full scale \$\pm 10V\$ output, rms noise voltage 20 mV) is \$(20 mV)^2\ = 400\cdot 10^6(\mu V)^2\$. So your transducer noise power \$P_{tn}\$ is larger than your quantization noise by

$$ 10log(\frac{P_{tn}}{P_{quant}})=10log(\frac{400\cdot 10^6 (\mu V)^2}{0.12 (\mu V)^2}) = 95 dB $$

If you halve the output of your transducer signal so that the noise is now 10 mV rms, you lose just 6 dB from this ratio so that your transducer noise is 89 dB larger than your quantization noise. Since your transducer noise overwhelms your quantization noise to such a high degree, at this point I should stop writing, or you should stop reading and be done. But I just had some coffee, so I'll carry it through and calculate the SNR for a full-scale sine wave output from your transducer.

SNR

A full-scale sine wave would be a sine wave of peak-to-peak amplitude of 20 V in your first scenario, and 10 V in your second scenario. Using this full-scale value leads to the maximum SNR you could have for a sine-wave signal.

The SNR is the ratio of the signal power to the noise power. For a pure sine wave of peak-to-peak amplitude \$V_{pp}\$, the signal power \$P_{sig}\$ is simply the time average of \$V(t)^2\$, which is

$$P_{sig} = \frac{V_{pp}^2}{8}$$

The total noise power \$P_{Noise}\$ is the sum of the transducer noise power and the quantization noise power, or

$$P_{Noise} = P_{tn}+P_{quant}$$

and from that you can calculate your signal-to-noise ratio

$$ SNR = 10log(\frac{P_{sig}}{P_{tn}+P_{quant}}) $$

Let's evaluate this expression to ridiculous precision in each case just to illustrate the negligible effect the quantization noise will have on the SNR. The quantization noise power \$P_{quant}\$ will be the same \$0.12 (\mu V)^2\$ in both cases.

Case 1: \$V_{pp} = 20 V\$, \$V_{noiseRMS} = 20 mV\$

\$P_{sig} = (20 V)^2/8 = 50 V^2\$

\$P_{tn} = (20 mV)^2 = 400\cdot 10^{-6} V^2\$

\$SNR = 50.9691001287 \$ dB

Case 2: \$V_{pp} = 10 V\$, \$V_{noiseRMS} = 10 mV\$

\$P_{sig} = (10 V)^2/8 = 12.5 V^2\$

\$P_{tn} = (10 mV)^2 = 100\cdot 10^{-6} V^2\$

\$SNR = 50.9691001248 \$ dB

The SNR has been kept to absurd precision so you can see just how far out we need to go to see the difference. You can see that the quantization noise is completely and utterly negligible here, and your focus should be on reducing the transducer noise through filtering and averaging.

This is the maximum possible SNR of your system for a full-scale sine wave transducer output. If your output signal is only 1/10th the full-scale voltage, you would subtract 20 dB from your SNR.

As a side note, I will make the same point already made by Paul Uszak in his comment that your DAQ is unlikely valid all the way to 24 bits, and so this 89 dB ratio of transducer noise to quantization noise is actually a high estimate. But 89 dB still gives you so much headroom, you can remove many bits of real resolution and still have your quantization noise absolutely swamped by your transducer noise.

So you asked if one scheme was better than the other. I think the answer is that you are free to take the quantization noise out of that decision, and decide based on other system considerations.

rpm2718
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  • This is a very detailed great answer. Could you also add the minimum forces(in Newton) can be revolved for both cases?(I see that the difference will be so small) I have an idea but since you are very into it the answer would be 100% complete and correct. – GNZ May 07 '20 at 12:04
  • The reason I requested that it would be great to see how it affects out aim which is the minimum force we can resolve in both cases.. – GNZ May 07 '20 at 12:18
  • Thank you. The minimum forces that can be resolved depends entirely on how you set up the filtering and averaging, and what the signal looks like. You can keep improving your minimum resolution by averaging longer and longer, until you eventually run into your quantization limit. You might consider a separate question on that subject. – rpm2718 May 07 '20 at 12:20
  • So in this case, the 51dB you calculated corresponds to a signal voltage to noise voltage ratio of 317 for both cases. And since the transducer has 2kN range, can we say for full scale the minimum force can be resolved is 2000/317 = 6.3N. So I will have at least 6.3N uncertainty correct? – GNZ May 07 '20 at 12:38
  • I'm afraid there's really much more to it than that. To answer that, you have to get into things like your noise spectral density, and the character of your signal. For example, if it is a dc force, you could just sit there and average longer and longer and get better resolution. To estimate how long, you need to know what your noise spectrum looks like. In this dc example, you would want to know the noise spectral density down at very low frequencies. – rpm2718 May 07 '20 at 12:52
  • I think the resolution cannot be better than 6.3N(limited by the noise floor), waiting more would only improve the precision error only (standard deviation). But thats what I think – GNZ May 07 '20 at 12:59
  • We probably should not be carrying the discussion onward in a chat, but just one last comment to try to help illustrate this last point. Imagine a dc signal significantly smaller than the noise so that you can't really see it in real time with no averaging. You sit there and just keep averaging. The noise, being random, continues averaging its way to smaller and smaller values until eventually you can see the dc signal. – rpm2718 May 07 '20 at 13:08
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    I think I see your point. The longer the time the smaller the random noise's rms value which means a bigger SNR. – GNZ May 07 '20 at 13:11
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example:

enter image description here Rated output 1.5mV/V (R.O.) 5~12V excitation Zero balance ±2% of R.O.
Nonlinearity ±0.5% of R.O.
Hysteresis ±0.5% of R.O.
Nonrepeatability ±0.2% of R.O.
Creep(5min) ±0.1% of R.O.

enter image description here

Avoid ground loops, proximity coupling to interference and use ground planes for signals.

You may reduce Bandwidth of the signal by Decimation which also reduces the sampling rate and improves the error-free resolution.

Tony Stewart EE75
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  • Thanks for the answer. In your example wiring is the sensor ground and the ADC ground connected by a dedicated wire? How are they connected physically? The twisted wire in the illustration has negative wire(-), positive wire(+) and a shield wire. How about the analog ground wire? I mean does the analog sensor ground(symbolised with solid reverse triangle) is connected to the other ground(symbolised with parallel lines)? – GNZ May 10 '20 at 18:07
  • Earth bond shields the sensor ,the twisted wires as a shield, and the ADC enclosure. It is isolated from all signal grounds by a clean power supply. SMPS can have noisy leakage currents which might alias with sampling rate or saturate CM signal, . So impedance >> 100k to earth ground. The earth ground shunts stray noise that is common mode at both ends but does not conduct on cable – Tony Stewart EE75 May 10 '20 at 19:54
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Simply put, you cannot know for sure before you try. Because you don't know where does the noise appear. The 24 bit ADC is probably not the source, even the simpler of those give 18-19 bits before noise. So it may be either the sensor, or the amplifier, or the transmission line, or the amplifier inside the DAQ.

So when you scale your amplifier, you can either reduce the noise as well (and get the same SNR) or keep the noise as is (and reduce the SNR).

Good DAQs are generally better than most things you connect to them. I used to work with a medical device with noise lower than 4uVpp. So my bet is that your noise comes from your amplifier. It woule be interesting to see the design and try catching the source from the schematics.