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9

Your amplifier has average noise characteristics, the problem is that your signal is very, very weak. The amplifier is responsible only of the noise part of the SNR, so it has not a "poor SNR", but a "poor input referred noise". To obtain a better SNR you can either amplify your signal, without adding noise, or reduce the noise. Since amplifying without ...


7

Common mode changes slowly, signal changes fast. What you thus need is a high-pass filter, which filters out the DC component. In the easiest case: that's a capacitor in series with your signal source, with a resistor to ground to "short" low frequency content. here's an easy-to-use RC high-pass design tool. Start with something like C=10nF. You seem to be ...


6

There's two things to consider with a preamp which explains why it improves SNR. The first, and most important, is that the pre-amp is closer to the source (such as a microphone). There's fewer noise-generating components between the microphone and the pre-amp than there would be if you waited all the way to the main amplifier. For example, quite often ...


5

The main ones to consider are \$V_{IH}\$ and \$V_{IL}\$. These are the logic HIGH and logic LOW input voltage thresholds. Anything above \$V_{IH}\$ is considered a logic HIGH, and anything below \$V_{IL}\$ is considered a logic LOW. The space between \$V_{IH}\$ and \$V_{IL}\$ is called the noise margin and any signal within this region is ignored. The ...


5

"How is it even possible for a single fixed measurer to determine both RSSI and Noise?" - very good question. The noise they are talking about is receiver noise and not interfering signal. At very low powers, the noise is mostly the thermal noise of the receiver: ie, if you were to disconnect the antenna and replace it with a 50 Ohm load (most RF systems are ...


5

o does this mean our final objective is to always try to make our SNR(out) as high as we can Yes, we want the output SNR as high as possible, because that means the best chance of recovering the message signal accurately. so that our signal is amplified much more than our noise? This is not possible. The input noise will be amplified just as much as ...


5

Is the noise not amplified the same amount as the signal? From a different but related perspective than Cort Ammon's answer, in a low-noise multi-stage design, one puts most of the overall gain in the 1st stage which might be called a preamplifier. It isn't so much that the preamp amplifies the noise along with the signal, it's that any noise added by the ...


4

35 dBmV means "35 decibels stronger than 1 mV", or about 10^(35/20) = 56 mV RMS. 56 mV is a pretty small signal, so it's unlikely to damage any input that wouldn't be completely destroyed by the ESD of casual handling. There is still the possibility (though unlikely) that the input signal may be too strong for the receiver and would cause clipping. This may ...


4

Yes, that's a reasonable approach, except that you'll also want to calculate the mean (DC bias) of your 10000 samples, and subtract that from the individual samples before you square them for the RMS calculation. This is equivalent to using a high-pass filter to block DC.


4

This analysis assumes that the circuit is a standard transimpedance amplifier and you are varying its gain by changing the feedback resistor value. The signal at the output of the amplifier is proportional to the feedback resistor. However, the feedback resistor also contributes its own thermal noise, which scales with the square-root of the feedback ...


4

I'm going to take a stab at this though I don't know if it applies to sonar. If you have a signal received on a transducer there will be both signal and noise. If you have two transducers (in an array) it can be presumed that the same signal is received on both but the noise on each is likely gaussian and not coherent. The upshot of this is that when you ...


4

Measuring the SNR of an ADSL signal is not a simple proposition. ADSL uses hundreds of subcarriers, each carrying a 16-QAM signal, as shown in the constellation diagram shown below, courtesy Wikimedia Commons. The ADSL modem performs adaptive equalization to get the best possible constellation for each carrier, where "best possible" means that the ...


4

A possible solution, is to use a slew-rate-limited amplifier on this signal, in combination with AC coupling. The step size, then, would be diminished while slower-slewing ripple is passed through. Some external-compensation-capacitor op amps can do this with an oversized compensation capacitor. Another would be synchronous gating, so that a series of ...


3

In accordance with your calculations, the power of the noise would also be zero. And 0/0 is indeterminate. In any case, the SNR is only important for the duration of the signal. That determines the detectability of the signal. SNR has no meaning if there is no signal.Also, by dividing by T, you are calculating average power. Again, the meaningful SNR is ...


3

Less than one ohm with leads- I imagine that represents a big thermal load on the cryostat. Yes, a room-temperature transformer can make a tremendous improvement. The important thing is that the Johnson-Nyquist noise in the windings does not contribute too much noise, so there needs to be enough copper. There are commercial products that go up to the 10s ...


3

SNR as you know means Signal-to-Noise ratio. So it's \$20 log(\$\$V_S\over V_N\$), where \$V_S\$ is signal voltage and \$V_N\$ is noise voltage. So, if noise (\$V_N\$) is really small (compared to signal), then SNR gets really big. It's quite possible to have systems with negative SNR (where the signal level is less than the noise), where you need ...


3

No, the RTL2832u will not be damaged by the cable signal. After all, that's the application it was originally designed for! You'll probably have to reduce the gain settings in your SDR software to get useful results, though.


3

If \$N\$ is the number of bits then your input signal range is divided into quantization intervals of size $$q = \frac{V_{ref}}{2^N}$$ The maximum quantization error is \$q/2\$ and it is usually assumed that the quantization error is uniformly distributed between \$-q/2\$ and \$q/2\$. So the PDF of the quantization error is constant between \$-q/2\$ and \$...


3

Spehro's comment is basically the answer. Imagine a link which subtracts a random, varying noise value between 0 and 0.5V from the input. You measure 4.5V at one end, but that could mean either a 4.5V input with no noise or 5V minus noise. You can't tell, and have lost your signal entirely. With a digital signal varying between 5V and 0V over the same link, ...


3

You're substantially correct on everything you've mentioned. Bigger cable has lower losses. Low loss is important in two areas 1) Noise The attenuation of a feeder is what adds Johnson noise corresponding to its temperature onto the signal. A feeder of near zero length has near zero attenuation and so near zero noise figure. Up to a meter or several (...


3

For most folks posting answers on this particular stack, the answer to optimum cable size generally has a lot to do with economics, service life, ease of use and such. Each individual problem has its own set of defining parameters, which in turn will be used to create a specification that will be met or exceeded. This is an important step to take, because ...


3

The SNR is the ratio between the power of all the signal and the power of all the noise. If I'm not mistaken, the FFT in your diagram shows you the dB in volt, not in W. In the rest of this answer, every time I say dB, it's in voltage. So 0 dB = 1 volt, 6 dB ≃ 2 volt. Let's assume you are sending a pure sine wave that looks like this: \$x(t) = A×\sin(\...


3

I see quite a few misstatements and possible misconceptions both in the answers and in the question, so let’s break down what is meant by “noise.” Analog noise within the same bandwidth as the signal. Analog noise outside of the bandwidth of the signal. Quantization noise introduced by the ADC. I interpreted your question as implying option 1. ...


2

The work Friis did on developing a simple formula for received power makes a basic assumption about distance - all bets are off if the transmitter and receiver are up-close. This is called the near-field and the standard equation of: - LinkLoss (dB) = \$ 32.45 + 20 log_{10}(F) + 20 log_{10}(D)\$ ..... doesn't work up-close because you are not really ...


2

They're negative because they are really small. The dB scale is a logarithmic scale, with 0 dBm referenced to 1 mW. Negative values are smaller and positive values are larger. Like you said -60 dBm is 1 nanowatt and -90 dBm is 1 picowatt. I'm actually not sure where the noise measurement is coming from offhand. The radio receiver does generate some ...


2

"Signal Strength": I believe you are referring to the RSSI value, which is the Received Signal Strength Indicator this value is typically shown as a negative dBm value. RSSI is the measurement of power in an RF signal, the more power in an RF signal the better the connection quality is. It’s typically best practice to have the SNR value 20 to 25 dB’s away ...


2

Generally for radio receivers there is the following "rule of thumb" formula that describes the minimum received antenna power needed to operate at a low bit error rate: - Receive power (dBm) = -154 dBm + 10log\$_{10}\$(data rate) So if you have a 1Mbps data stream, you need to ensure the received signal is: - -154 dBm + 60 dBm = -94 dBm. I call it a ...


2

87.85 db below 2.76 dbV reference level at 0.08 % THD + N Let's start with the 2.76 dBV - this is the voltage level measured by the analyzer and is a sinewave of ~1.374 Vrms plus a bit of noise and distortion. This I believe is what the analyzer is measuring at your amplifier's output i.e. it is the total of: - Output sinewave (at some frequency possibly 1 ...


2

Infinite SNR means no noise. Higher SNR is better from a quality standpoint - it doesn't affect a circuit until noise gets quite close to the signal level then things such as FM demodulators start to get in a muddle as does data transmission systems.


2

I think you have a pretty good test setup. When making your noise measurement in step 1, I would recommend connecting the two differential inputs of your amplifier using an impedance that matches the output impedance of your detector. Also, I agree that you should subtract the DC-offset before computing Vrms as the other user recommended. You can also ...


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