Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Basic sampling question

joniengr

Full Member level 3
Joined
Nov 3, 2018
Messages
179
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Activity points
1,598
Hi,

I have a analog signal of 50 KHz and I need to sample it using ADC in such a way that I can see the magnitude of all frequencies with the resolution of 1 Hz. What should be the sampling frequency. Some signal tutorials says that at least twice the highest frequency of the analog signal but I guess it might need more than twice the highest frequency on the signal.
 

stenzer

Full Member level 4
Joined
Oct 1, 2012
Messages
216
Helped
27
Reputation
54
Reaction score
24
Trophy points
1,298
Activity points
3,218
Hi,

the sample frequency should be at least two times higher than the expected measured frequency of interest. By using an ADC with a sample frequency of 100 kHz you are able to evaluate the frequency spectrum from DC to 50 kHz.

Usually someone would use an ADC with a higher sample frequency as 50 kHz is pretty low. But the low (100 kHz) sample frequency might also be an cost issue due to a required high amplitude resolution.

I assume you are evaluating the frequency spectrum by a FFT, thus your frequency resolution depends on your sample period (number of consecutive samples) and your sample frequency. For a sample frequency of 100 kHz (sample rate is 10 µs) you would need at least consecutive 10000 samples.

greets
 

wwfeldman

Advanced Member level 2
Joined
Jan 25, 2019
Messages
693
Helped
169
Reputation
338
Reaction score
167
Trophy points
43
Activity points
5,025
... analog signal of 50 KHz and I need to sample it using ADC in such a way that I can see the magnitude of all frequencies with the resolution of 1 Hz.
all what frequencies?
is your signal a square wave and you want to see the constituent frequencies from Fourier analysis?
or is it a sine wave and you want to distinguish between 49,999 Hz and 50,000 Hz and 50,001 Hz?
or do you mean something else?
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,471
Helped
3,947
Reputation
7,892
Reaction score
3,816
Trophy points
113
Activity points
115,832
Hi,

To be exact: "at least" twice is not correct, you need "more than" twice.

But twice "of what"?
Twice of the maximum frequency of interest.
When you say you have a frequency of 50kHz, then usually there are overtones.
Overtone frequencies can be very high...so you have to decide up to which (overtone) frequency you are interested in..

Klaus
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
46,987
Helped
13,938
Reputation
28,124
Reaction score
12,576
Trophy points
1,393
Location
Bochum, Germany
Activity points
273,971
An ADC with a useful 0 - 50 kHz bandwidth and full suppression of possible aliasing signals will e.g. use 120 kHz sampling rate and a steep anti-aliasing filter with a transition betwen 50 and 60 kHz. This also answers the overtone question.

The original question however doesn't clearly specify the input signal. If it doesn't occupy the 0 - 50 kHz baseband but only a small band around 50 kHz, you might use bandpass sampling with a much lower rate. Consider that Shannon/Nyquist rate doesn't refer to signal frequency rather than total bandwidth.

Frequency solution can be estimated as 1/sample length.
 

vGoodtimes

Advanced Member level 4
Joined
Feb 16, 2015
Messages
1,074
Helped
304
Reputation
608
Reaction score
301
Trophy points
83
Activity points
8,571
this is the classic bad interview question. you can justify sampling rates above or below 100ksps. but I think the interviewer wants something like 2.2x. but if the cost of the adc is low, you could justify using higher rates to make an analog filter easier and then use a digital filter.

the time associated with the number of samples analyzed is what determines frequency resolution.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,471
Helped
3,947
Reputation
7,892
Reaction score
3,816
Trophy points
113
Activity points
115,832
Hi,

or is it a sine wave and you want to distinguish between 49,999 Hz and 50,000 Hz and 50,001 Hz?
Interesting idea.
Theoretically ... for a limited range like 49,000Hz up to 50,000Hz .. when you just want to determine the frequency...you could use undersampling. Then even a sampling frequency of 2kHz could be sufficient. Theoretically.

And for 1Hz resolution you need a measurement window of 1/f_res = 1/ 1Hz = 1s. Independent of sampling frequency.

Klaus
 

joniengr

Full Member level 3
Joined
Nov 3, 2018
Messages
179
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Activity points
1,598
Hi,

The frequency of input signal is un-known. There are several frequencies present in the signal and the highest frequency of the signal is 50 kHz. The question is what should be the sampling frequency of the ADC so that from digital output one can get the amplitude (magnitude) of all frequencies between 1 Hz and 50 kHz. The frequency resolution has to be 1 Hz.
 

joniengr

Full Member level 3
Joined
Nov 3, 2018
Messages
179
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Activity points
1,598
Hi again, I am not sure if I describe my question properly.

I am also wondering about the band-width and sampling rate of oscilloscope. I guess if oscilloscope has band-width of 200 MHz then the sampling rate it has is around 1 Gsps which is relation of factor of five between the highest frequency i.e. band-width and the sampling rate. If we apply sampling theory then with 1 Gsps we can capture up to 500 MHz.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,471
Helped
3,947
Reputation
7,892
Reaction score
3,816
Trophy points
113
Activity points
115,832
Hi,

A couple of things:
* There are undersampling systems with much higher analog bandwidth than sampling rate.
* analog bandwidth is determined at it's -3dB border. So if the -3dB is at 200MHz, then there still is signal at 500MHz.

Klaus
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
46,987
Helped
13,938
Reputation
28,124
Reaction score
12,576
Trophy points
1,393
Location
Bochum, Germany
Activity points
273,971
Parameters of a sampling oscillocope don't have much to do with the original question of this thread or Nyquist/Shannon theorem. It's about the requirements for complete reconstruction of a sampled analog signal. The purpose of digital oscilloscope isn't signal reconstruction. It's measurement of an arbitrary signal, mostly in time domain. A steep anti-aliasing filter would be almost unsuitable due to the caused waveform distortion, instead an oscillocope implements a soft bandwidth limiting.

Beyond sampling theory, an excess sampling rate is necessary for single shot recordings with smooth waveform representation.
 

Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top