# Tag Info

41

if they wanted the sampling frequency to be 10Hz, why did they not just sample at 10Hz initially? In order to avoid aliasing, the signal has to be lowpass-filtered before sampling. No frequencies above Fs/2 should be present in the analog signal (or, realistically, they should be attenuated enough to be buried in the noise, or to a level low enough to meet ...

33

First of all, let's get rid of the Nyquist rate misconception. People are usually taught that the minimum sampling frequency needs to be twice the frequency of the highest frequency in the signal. This is completely false! What is true is that if you have a "full" spectrum, and by full, I mean that it completely uses up all frequencies between the lower ...

29

The name for the frequency the samples are taken is Sampling Rate (not specific to audio only). It is measured in samples-per-second or in Hz (which is 1/s). The time between two samples is called sampling period, and is given in units of time. So in your case the period would be 1ms and the frequency 1/1ms=1kHz. Or in terms of rate it's 1000 samples per ...

29

Nyquist-Shannon sampling theorem... often mis-used... If you have a signal that is perfectly band limited to a bandwidth of f0 then you can collect all the information there is in that signal by sampling it at discrete times, as long as your sample rate is greater than 2f0 it is very concise and contains within it two very key caveats PERFECTLY ...

25

Chasing this answer down took a few different links, but it appears to boil down to this: 1. 4 differential pairs (8 wires total, but only 4 lanes). 2. 800 Mega Symbols a second. 3. Using PAM16, 16 symbols are used which translates into 4 bits per baud per lane. Given that information you come up with 4 bits*800 Mhz*4 lanes which results in 12800 Mb/s or ...

24

As you decrease the sampling frequency there is less separation between the images in the frequency domain. source Remember that the repetition of the spectrum occurs at the sampling frequency. When the images are closer together you need to achieve more attenuation in your anti aliasing filter. The filter must transition from pass band to stop band before ...

23

Forget sampling rate for a few seconds... Think about sampling period for a second, which is the time interval between two consecutive samples. This time can be an integer or any real number (as long as it’s positive, of course). Sampling rate is simply the inverse of sampling period. Does it make more sense this way?

20

I would call it the sampling period or measurement period. For short periods like 1 ms, specifying the sampling frequency is another option (1 kHz).

19

Taking a derivative (or an integral) is a linear operation — it doesn't create any frequencies that weren't in the original signal (or remove any), it just changes their relative levels. So the Nyquist rate for the derivative is the same as that for the original signal.

17

The firing of neurons in the eye is completely asynchronous, so for all intents and purposes, the process we call "seeing" must be considered continuous-time, not sampled. Nyquist does not apply. If things move too fast, they simply appear blurry. And "too fast" varies from person to person; there is no well-defined cutoff frequency for the general ...

14

10G ethernet (as described by other answers) does not do signal transitions at 10 GHz, it uses multiple level encoding spread across 4 pairs to achieve 10 Gb/s. However, 10+ gigabit serial transceivers are quite common on high speed chips. For instance PCIe, USB3.1, thunderbolt, and similar protocols all use 10 gbit/s serial rate on individual pairs. You ...

14

Yes, the sampling rate can be any number you want. But you obviously would not get partial samples in the end, you just have to round down. In your example the first sample is taken at $\frac{1}{15.5}s$ = 64.5 ms and then at every multiple from that. This means you get your last sample at 6,966 s. That is the 108's sample. So at 7 s you still have taken ...

13

There's a difference between analyzing a signal for information, and displaying it on a scope screen. A scope display is basically a connect the dots, so if you had a 100 MHz sine wave sampled at 200 MHz (every 5 nsec) AND you had the imaginary component being sampled as well you could reconstruct the signal. Since you only have the real part available, 4 ...

13

So, many people, including professors, are confused about what Nyquist rate is: Nyquist rate is the sample rate that you need to have to sample a signal to avoid damaging it by aliasing What that means is that for real-valued signals and real-valued sampling, the sampling rate must be more than two times the bandwidth of the analog signal. That means that ...

13

Some things are always an integer. Samples are always integer. You can take 108 or 109 samples. Sample rate can be a floating point number, or more generally a rational, or even a real. You calculate the sample rate by dividing the number of samples (less one to get the number of periods between samples) by the time it takes to obtain those samples. ...

11

If you are interested only in signals in the range DC to 3kHz, then only signals above 7kHz will alias onto those. This means that you need a filter with ... a passband to 3kHz a transition band from 3kHz to 7kHz a stopband from 7kHz upwards Note this doesn't define the 5kHz attenuation, and doesn't need to. The stopband must have enough attenuation to ...

10

"even a simple UART samples a digital signal at the same speed..." the UART doesn't need to reconstruct the analog square wave signal that carries the digital information, so it doesn't take the theorem into account. The Shannon-Nyquist theorem actually talks about the perfect representation of an analog signal. Perfect representation here means that ...

10

You mentioned the word magnetometers. This expands the scope a little. Magnetometers for those not familiar measure magnetic flux and create a proportional output voltage/ signal according to the flux. It is likely you will also detect a high amount of unwanted "electrical energy", due to the radiated magnetic energy from any electrical cables around. In ...

9

There is a lot of bad information and audio phoolery available on this topic, but if you're doing one channel of digital audio, 96kHz and 192kHz sample rates are silly. Human hearing extends to 20kHz. To satisfy Nyquist at 20kHz, we need a sampling rate greater than 40kHz. CDs are 44.1kHz, and 48kHz is another common sampling frequency. Now, let's recall ...

9

My guess is that these are the (time) points which the circuit simulator has actually solved. Many analog circuit simulators, unlike simulators for digital circuits (logic), do not use a constant timestep. What these analog simulators do is calculate more timepoints when a lot is "happening" in the circuit and less points when things are more static. I'm ...

8

So you need to measure supply current at 10 ksamp/s from 100µA to 30mA, which is 300:1 range. That by itself sounds doable enough. Even a 10 bit A/D built into a microcontroller is enough resolution if the signal is amplified properly. 10 kHz sample rate is also quite doable. In fact, I'd want to sample faster than that and do a little low pass ...

8

The Rigol 1052E as mentioned by Joe is a great example of how to do this efficiently and cheaply. It uses a pile of independent ADCs, all of which having a slower sampling rate, and clocks them out of phase with each-other. This way, samples get pulled from each ADC in turn round-robin style. Obviously your timing has to be extraordinarily precise to do it ...

8

Debouncing is a FAQ. You should be able to find... nearly unlimited numbers of web pages on the topic. Smith commented about Jack Ganssle's widely read PDF on the topic, as well. And with all these answers you've got both hardware and software methods. I'll add to this "literature" just a little bit by mainly talking about ideas that aren't already covered ...

8

Never. But you need to make sure you understand exactly what the "Nyquist rate" actually is. Nyquist stated that you you can reconstruct the signal as long it is sampled at a rate that is more than twice the bandwidth of the signal. That bandwidth may or may not start at DC, but many sources on this topic assume that it always does, and that the highest ...

8

In order to capture all the information you need to sample at at least twice the highest frequency component in the input signal. If you do a PSD plot of the input you'll see that there is significant power at higher than 400kHz. You might have to sample at 8MHz to get most of it. Also, generally you will want to precede the ADC with an ANALOG low pass ...

8

I am not fond of encoding command data as an analog signal in a digital file. I think I would try something like encoding the lighting commands as text blocks in the lyrics block of the ID3 information inside the mp3 file. The lyrics block is before the sound data, so you should be able to decode it quickly before you start playback. Typical libraries for ...

8

To reconstruct a signal in the digital realm from the analogue realm you need at least two samples in each cycle of the highest frequency present in the analogue signal. For instance, on CDs, they use 44.1 kHz to sample a maximum frequency in the audio band of 20 kHz. They could have used 40 kHz but that is right on the limit and the anti alias filter would ...

7

There is a direct, and actually quite simple, relationship between all the figures. Let's start with the sample size. You get half the bins (or "buckets") as you have samples in your set. For instance, if you have 1024 samples, then you get 512 bins. As simple as that. Now for the sample rate. The maximum frequency is half the sample rate (see Nyquist-...

7

It depends on who you ask. Most humans cannot hear beyond 20 kHz and 16 bits, so 96 or 192 kHz should be plenty. As for hearing a difference between 16 and 24 bit converters it depends on your DSP. The key benefit of 24 bit converters is it gives you tons of additional headroom (dynamic range) so you can do a lot of mathematical operations and not add ...

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