Basic Sampling question involving sampling rate and graphing

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Zachariah Brown
Zachariah Brown el 2 de Abr. de 2021
Comentada: Zachariah Brown el 3 de Abr. de 2021
For a labI was given this code
t=0:1/2000:.02;
x=sin(2*pi*60*t);
t240=0:1/240:.02;
n240=0:length(t240)-1;
x240=sin(2*pi*60/240*n240);
stem(n240,x240,'filled','b','LineWidth',2);
axis([0 4.8 -1 1]);
and was to put that into matlab and display the sampling and function which works fine. We then needed to make this into a function of x=Asin(omega*t) where f = 10000Hz, and find the sampling at rates of 5k, 10k, and 20, 100k. However changing the values of the sampling frequency and the function frequency gives me a graph that is not at all similar and I have no idea what I need to fix or change to get the graph to look like the initial one.
=======================================
t=0:1/3300:.02;
x=sin(2*pi*10000*t);
t5000=0:1/5000:.02;
n5000=0:length(t5000)-1;
x5000=sin(2*pi*10000/5000*n5000);
stem(n5000,x5000,'filled','b','LineWidth',2);
=======================================
t=0:1/3300:.02;
x=sin(2*pi*10000*t);
t10000=0:1/10000:.02;
n10000=0:length(t10000)-1;
x10000=sin(2*pi*10000/10000*n10000);
stem(n10000,x10000,'filled','b','LineWidth',2);
========================================
that is what I changed the code too for 5k and 10k, i know those are both less than nyquist's sampling so it will result in not a good graph, but even with 20k+ the graph isnt similar to what it was before.
what I specifically dont understand is
how
t=0:1/2000:.02
; is time
what
n240=0:length(t240)-1; is
this
stem(n240,x240,'filled','b','LineWidth',2);
part of the code and what it does
how to get a proper sampling graph of 5k,10k,20k and100k, and what to change to get that
  3 comentarios
dpb
dpb el 2 de Abr. de 2021
ADDENDUM
"That's fine when the signal being generated with that time series is one of 60 Hz as the first example; that's 2000/60 or a sample rate of 33X times the signal. You can get by with anything >2 by Nyquist, 3-4X is better with shorter time periods."
That is, >Nyquist by any amount will be enough to find the frequency content via FFT with a suitably-long record; that does NOT mean that the time series graph will look smooth; a sample rate of only slightly over Nyquist will still look like a triangle wave mostly, not a smooth sine even though there is enough frequency content in the resulting signal to compute the appropriate frequencies without aliasing.
Zachariah Brown
Zachariah Brown el 3 de Abr. de 2021
Thank you for the help, I understand it now

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Respuestas (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov el 3 de Abr. de 2021
Your code has a problem with the end time value to take into calculations and display of the calculated data points. Here are some corrected codes:
f1=240;
t=0:1/2000:.02;
%x=sin(2*pi*60*t);
t240=0:1/f1:.02;
n240=0:length(t240)-1;
x240=sin(2*pi*60/f1*n240);
figure
subplot(311)
stem(n240,x240,'filled','b','LineWidth',2);
axis([0 4.8 -1 1]);
%%
%t=0:1/2000:.02;
%x=sin(2*pi*60*t);
f1=240;
t240=0:1/f1:.02;
n240=0:length(t240)-1;
x240=sin(2*pi*f1*n240);
figure
subplot(311)
stem(n240,x240,'filled','b','LineWidth',2);
axis([0 4.8 -1 1]);
subplot(312)
f2=5000; % 5k
t5k=0:1/(f2*10):.0005;
n5k=0:length(t5k)-1;
x5k=sin(2*pi*10000/f2*n5k);
stem(n5k,x5k,'filled','b','LineWidth',2); shg
f3=1e4; % 10k
% t=0:1/(10*f3):.02;
% x=sin(2*pi*f3*t);
t10k=0:1/(10*f3):.0002;
n10k=0:length(t10k)-1;
x10k=sin(2*pi*1*n10k);
subplot(313)
stem(n10k,x10k,'filled','b','LineWidth',2); shg
Good luck.

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