[SOLVED] Is there any Upper limit of sampling frequency

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syedshan

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Hi all,


I want to know about if there is some limitation for highest sampling frequency that can be applied to the signal,
like Nyquist theorum said that at-least more than twice the maximum should be the sampling frequency, but what
about the upper limit, I tried to look but could not find the answer.

For example is 30Mhz, 50Mhz, 150 or 250Mhz sampling is good for the signal of 1-2Mhz....
What is the best of all frequencies for sampling this signal.


Bests
Shan
 

Usually going to higher frequencies is harder to do, hence Nyquist's theorem is useful since it allows one to know what is the MINIMUM sampling freq. But if you want to sample your signal at 5000Terahertz... then sure. Go for it.
 
The limitation is computational resources and the given result. The more the frequency the higher the DSP power needed and at last the achived improvement is not worth the time spent or complexity. Not exact values but for example increasing sample rate from 10 to 100 MHz for 1-MHz band signal you may get 1% of improvement. Just good reason and precision of data representation
 

Actually imho Nyquist criteria states that going from 2Mhz to any higher sampling frequency for a max 1Mhz signal will add NO additional information or improvement. Not even 1%
 
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    FvM

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well in this I disagree, since increasing the sampling speed ofcourse helps, e.g. (and can test on matlab simply) if 1 Khz signal is sampled using 3 Khz the result is actually almost no use. while increasing the sampling rate will increase the quality of signal.

Simply put the more the sampling rate, the less is the intermediary difference b.w. two discrete points (converting from continuous wave), hence giving more look-like of the continuous wave itself.
 

A signal with 1 kHz bandwidth won't have unknown intermediate points. I guess when you say "1 kHz signal" you have a signal of considerably higher bandwidth in mind. A Nyquist frequency of 1 kHz would refer to a stationary 1 kHz sine signal, thus no additional information is revealed when sampling it above 2 kHz, as rohitkhanna mentioned. Under practical measurement conditions, it's usually reasonable to use a slightly higher sampling frequnecy, that also recovers finite waveforms of arbitrary phase.

Your consideration about intermediate points might be applied for waveforms without strict bandwidth limit like the waveforms acquired by an oscilloscope. The Nyquist criterion isn't of much use in this case. Today's digital oscilloscopes expose samplig rates considerably above the Nyquist rate when referring to their nominal bandwidth, e.g. by a factor of 2 or 3.
 

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rohitkhanna said:
Actually imho Nyquist criteria states that going from 2Mhz to any higher sampling frequency for a max 1Mhz signal will add NO additional information or improvement. Not even 1%
Click to expand...
The absolutely limited spectrum corresponds only to the signal with INFINITE duration. And vice-versa: if your signal starts and ends ever, it has an infinite band.
The limited spectrum means infinite suppression in the stopband, and in practice there is always some trade between sharpness of spectral representation and filter complexity
 


I think you rather show a theoretical analysis. Have you ever tried to design a digital filter with a passband equal to half of its input sample rate?
As FvM said in practice the oversampling above nyquist frequency is required, usually x2...x3. For 100 kHz example signal - 200 kHz sampling frequency - impossible, 400-600 - good.
It is even impossible to generate the pure sine with no harmonics.
 

I imply the following:
1. Take the pure sine with frequency f.
2. Sample it with the rate fs = 2*f. E.g. at the sine peaks - +1 and -1.
3. This is a square wave - meander.
4. The spectrum of square wave is the sinc function - sin(x)/x - with the harmonics at n*f, where n - odd numbers (1,3,5, etc.)
5. To get the initial sine wave from this (restore your signal) you should filter - which removes harmonics above 1st.
6. You cannot implement such a filter at all having a sample rate 2*f (or, which is equivalent, the filter order is infinite).
Only having an oversampled signal you may restore the signal in practical application.
So FvM told that 2-3 nyquist frequencies is a reasonable value. There is always some trade between quality of signal representation and processing resources (i.e. sampling frequency), but nyquist is not enough.
 

I think the sequence of +1/-1 doesnt look like sine wave, and you?
I may imagine a lot of functions that have values +1/-1 spaced by 1/Fmax period. Why you say that we "should and must" assume only the sine wave at Fmax? Prove it.
 

2. Sample it with the rate fs = 2*f. E.g. at the sine peaks - +1 and -1.
3. This is a square wave - meander.
Click to expand...
This is no wave. In my DSP text books, there are just data points with nothing in-between. You get into indissoluble confusion when associating a sampled signal sequence with a particular waveform. You can get the same samples from different continuous waveforms, after sampling the extra information is lost, just a trivial conclusion from Nyquist's theorem.

Restoring a sampled waveform is a matter of reconstruction filters. Some possible reconstruction filters exist only in theory or under specific conditions in digital signal processing, e.g. an exact sin(x)/x filter is non-causal, it can be only approximated in a real design, similar for other "ideal" filters.

I'm under the impression that a part of the discussion seems to hold theoretical considerations and properties of ideal systems against technical implementations. I try to live on both sides.
 

This is square wave. When process the sampled data from ADC you always assume that between samples the value of discrete function remains the same as that sampled at the last moment. So +1/-1 sequence is a square wave.
But Mr. Rohitkhanna seems to be unfamiliar with fourier series and sampling theory.
 

this is very interesting thread. one thing is not clear. if we already know that our input has frequency components which are definitely less than f, then how can we assume the opposite also that samples taken at 2f which are +/- 1 must have frequency more than f ? any signal which is not a precise sine wave but shows samples at +/- 1 at f must have other components greater than f, and this does not have to be only a square wave. and anyway, this breaks the original assumption that our signal does not have components greater than f.

using this same reasoning, even if we over sample at 10x the max frequency, or 100x, we will still get discrete samples - which are exact rectangular steps according to you - and so must have frequency components going to infinity ?

in short, it is hopeless trying to do a fourier on discrete samples, because there will always be step changes in the input ?
this sounds incorrect someplace
 

kripacharya said:
using this same reasoning, even if we over sample at 10x the max frequency, or 100x, we will still get discrete samples - which are exact rectangular steps according to you - and so must have frequency components going to infinity ?
Click to expand...

Sure for discrete samples frequency components are going to infinity. But for +1/-1 (the step of 2 between two consequent samples) these components are essential, but when sampling a sine wave at much higher rates this step will be much less - for example, 0.01. So for this case higher frequency components will be negligible. We must stop whenever.
 


i understand.
so to make error negligible, like less than 0.01, then we must sample at rates which is more than 500x of highest frequency. then fourier calculation for signal at even 100khz, we must have system to sample at more than 50mhz ... yes ? And for signal at 10mhz we must sample at 5ghz ?
it must be very difficult to make digital filters and analysis for frequency higher than this. How is it done ?
 

that was just an abstract example.
for more precise result lets make some fiters.
Take a 400 kHz sample rate. To build a low pass filter which passes all frequencies up to 100 kHz with bandpass ripple 1 dB and bandstop attenuation 100 dB for 200 kHz (it is not 0.01 but 0.00001!) the filter order will be 8 (8+1=9 samples oof impulse response length). The filter is FIR equiripple.
But for better presicion - 0.1 dB bandpass ripple and 100dB attenuation for frequencies above 120 kHz the filter order will be 79 - 10 times the delay at output.
When implementing this filter algorithm by direct convolution formula the processing operations needed will grow 10^2 = 100 times.
Can we reduce complexity?
For 250 kHz sampling rate this filter order (for 100 dB bandstop at 120 kHz) will be 49 - good! But for 242 kHz sample rate - almost nyquist - this filter order will grow up to 476!

There are a lot of points for optimization. Try fdatool in matlab.
 

it is very complicated. still i am not understanding why we have to build a low pass filter with cutoff 100khz for fourier analysis when we already know that signal does not have frequency components higher than 100khz ? does not the windowing function work properly ?

or do you mean that because of discrete sampling of max 100khz signal, many new frequency higher than 100khz is getting generated, and so we have to also do filtering ?
 

Sorry I dont understand how the windowing function can help.

kripacharya said:
or do you mean that because of discrete sampling of max 100khz signal, many new frequency higher than 100khz is getting generated, and so we have to also do filtering ?
Click to expand...
Sure.
 

Mityan said:
Sorry I dont understand how the windowing function can help.
Click to expand...

i think for doing FFT on a limited time samples, windowing will make less the effect of signal truncation at Tstart and Tend. otherwise we will get anomalous high frequencies due to sudden cut-in and cut-off of the signal. i think it is similar job to your low pass, but much more easy implementation. is it not ?
 

We where talking about time domain. For representing an analog signal as digital time samples we should sample and then filter.

Im not sure but maybe you're right.
Filtering is a convolution of signal time samples with impulse response of a filter. Direct formula requires much resources, so it is frequently implemented through FFT.
FFT of time samples of period T gives the spectrum in band 1/2T.
To cut off higher frequencies we apply some kind of windowing the spectrum. For example, rectangular.
The rectangular is FFT (spectrum) of filter impulse response of sin(x)/x kind.
So FFT-windowing-IFFT (for restoring time samples is equivalent operation to filtering.
 

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