poor SINR better Sensitivity?

Hello all,

Negative SINR in dB implies more noise than signal. Lets look at two cases

Case 1
Assuming a 5MHz BW, SINR of 17.5 dB and NF of 9 dB
Sensitivity = KTB + SINR(dB) + NF (dB) = -84 dBm
Case2
If SINR = -1 dB,NF = 9 and BW of 5 MHz
Sensitivity= -99.5 dBm

Looking at both cases tells me sensitivity improves with a negative SNR which I dont understand.
Would appreciate if someone could please clarify.

Thanks

Read the equation the other way round: if your system can operate with smaller signal/noise ratio, then smaller input signal is sufficient.

Could someone give more clarification? I still dont get this..

You can not consider the sensitivity with negative SNR.Negative SNR means there is no signal but noise.So, this equation is not valid anymore..
Which sensitivity will you consider with "pure noise" ? It's nonsense...

I think that the problem is his interpretation, not the equation.

The SNR value in the equation is the required S/N ratio. If we reduce the requirement, and allow a really small SNR, this leads to increased sensitivity.

A negative signal to noise ratio does not imply there is no signal. It implies the signal is less than the noise. There are techniques, such as the use of a lock-in amplifier, that allow signals to be measured well below the noise - I believe more than 100 dB below the noise is possible in some circumstances.

Dave

Negative SNR can appear in Spread Spectrum systems, due to Processing Gain factor: PG[dB] = 10*LOG(BW / BitRate)

http://pdfserv.maxim-ic.com/en/an/AN1140.pdf

Note that, the Processing Gain does not reduce the effects of thermal noise.

If a signal level is below thermal noise level ( so the signal is dropped down under the noise-sea) can you recover this signal again ? I don't think so..

I suggest you look up the lock-in amplifier, as that is one method of signal recovery, and the one I personally know best, having used one extensively during my Ph.D.

http://en.wikipedia.org/wiki/Lock-in_amplifier

The key is that you must have prior knowledge about the signal you want to detect.

There will always be a bandwidth where the thermal noise power is less than the signal. That might mean you have to make the bandwidth of your detector 1 m Hz, but that is possible to do, even if the frequency stability of the signal you want to detect is poor and changes far more than 1 m Hz. The key is to ensure the centre frequency of the detector tracks the signal frequency you want to recover. That is possible to do in a lock-in amplifier, but it does require that you have a clean reference signal, which is at the frequency you want to recover.

Consider what happens if you multiply a large noise voltage by a square wave whose amplitude varies from +1 V to -1 V. Then you low-pass filter the output. The longer the time constant of the filter, the smaller the output will be, since you are just low-pass filtering noise. The fact you have multiplied the noise by a square wave does not change this.

Now consider what happens if you multiply your weak signal (a sine wave) by that same square wave. The frequency of the sine and square wave are the same, but it does not matter what they are. It does not matter if they are constantly changing. As I said before, you must have prior knowledge about the signal you want to detect. In this case, we know the frequency it will be, but we don't know the amplitude. That's what we want to measure.

The easiest to consider is that whenever the sine wave is positive, the square wave is positive, but whenever the sine wave is negative, so is the square wave. Now if you low-pass filter the output, you get a DC voltage which will approach the signal voltage. The longer the time-constant of the filter, the longer it will take the DC voltage to stabilize. (It is normally considered stable after 5 times the time-constant of the filter, though it does depend on the filter characteristics).

It should be clear that if you then put both the signal and the noise into this multiplier, the noise will average to zero, but the signal will average to a steady DC voltage. So you recover the signal. You will need to amplify it to measure it, but that is no big deal. The key is the noise averages to zero, but the signal does not.

The longer the time-constant of your filter, the longer the measurement takes to stabilize, but the better the weak signal recover. I think the Standard lock-in I used had filters with time constants of several thousand seconds, so to measure a really weak signal, you might have to wait hours to measure it. But other filters have time-constants in the μs or ms range.

Of course, if the sine and square wave are 90 degrees out of phase, and you multiply them together, the average of that will also be zero, so you don't get any DC voltage from your signal. However, you get around that problem by doing the multiplication by two square waves, one displaced 90 degrees in phase to the other. So you get two signals, X and Y, then you calculate

sqrt(X^2 + Y^2)

That will always be positive.

Although I've not looked recently, Stanford Research and EG & G were the main manufacturers of lock-in amplifiers. I used a DSP controlled Stanford Research unit extensively, but I've also used analogue instruments from EG & G, and made one myself with a a few op-amps and FET switches. Modern units use DSPs for the filters, but the older analogue units use RC and/or LC filters. The better the unit, the more it is able to recover weak signals.

Lock-in amplifiers are regularly used in research labs. They have many uses.

Dave

Dave, you are right that lock-in amplifiers are great to recover small signals from the noise. As you have described, this depends on time constant for averaging, or in other words: a very small bandwidth. Reducing the bandwidth improves the S/N ratio.

Another thing one can do with lock-in amplifiers, which I did not mention before, it to measure phase shift between the reference and the signal. I was once involved in a project where a laser diode was RF modulated at 200 MHz. This passed into a babies head during labor. The light detected elsewhere on the head was detected on a photomultiplier tube. Changes in blood oxygenation cause changes in both the signal amplitude and phase, both of which were measured on a lock-in amplifier.

There was a complication on this, in that the lock-in worked at 10 kHz, so there were signal generators at 200 and 200.01 MHz, which got mixed. There was a lot of simulatory to a vector network analyser in fact. We basically had a home-brew VNA. Now you can buy lock-ins which work at 200 MHz. But when I spoke to Stanford Research, I found out their RF lock-in worked pretty closely to what we had.

Dave

Take a look at GPS NAVSTAR signal:
The sensivity of the usual cheap receiver is about -150...-160 dBm (acuisition mode), -140 dBm (tracking).
Lets estimate a thermal noise: -174 dBm/Hz for 2 MHz bandwidth means -174+63 = -111 dBm.
The signal is great below thermal noise, but the receiver works good, and that is not a complex system but a low-cost single chip.
Its all about signal accumulation.

@Mityan, why did you calculate with 2 MHz bandwidth?

Its a C/A code bandwidth

I am not an expert in GPS, but did quite some work on noise measurement during my PhD thesis.

Isn't the GPS data rate 50 bit/s? That gives a pretty small effective bandwidth, which is further reduced by averaging.

Yes, 50 bps. Each bit is represented as 20*1023 chips of preudo-random sequence. Modulation is BPSK.
For example, for 10^-2 BER of BPSK you should have about +4 dB SNR. For 20460 spread spectrum ratio this means that we may allow SNR to fall down to +4-43=-41 dB and restore the signal by accumulation of 20460 noisy samples.
Another way to interpret this is to operate effective bandwidth terms.