How can i know the best sampling rate to use in a ADC using Fourier
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Jose Francisco Actis Danna
on 22 Oct 2021
Commented: Mathieu NOE
on 25 Oct 2021
I want to analize a a biphasic truncated exponential wave, so i generated the wave in matlab
t1=0:0.0001:0.0025;
VS= 5000;
R1=25;
C1=90*10^(-6);
func1= VS*exp(-t1/(R1*C1))/R1;
plot (t1,func1), grid, xlim ([0 0.01])
t2=0:0.0001:0.0015;
VS= 5000;
R1=25;
C2=90*10^(-6);
func2=(-VS*exp(-t2/(R1*C2))/R1);
plot (t2,func2),xlim ([0 0.01]), grid
t=0:0.0001:0.008;
t0=0:0.0001:0.0038;
zero=t0*0;
final= [func1 func2 zero];
plot (t,final),xlim ([0 0.01]), grid
Now i want to use the FFT but i don't know how to use it because it's for a discrete waveform.
Doing some research i found a code and i modified it a bit to this problem:
fourier=fft (final);
L= length(t)
P2 = abs(fourier/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
Fs=25000
f = Fs*(0:(L/2))/L;
plot(f,P1)
grid
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
I get a razonable plot, but the problem is that i dont know wich sample rate (Fs) should i use, because if i switch this value the values of the max frequencies change too, so i'm not able to choose an adc sampling rate
Thanks
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Accepted Answer
Mathieu NOE
on 25 Oct 2021
hello
Fs is the inverse of the time increment dt
clc; clear all; close all;
dt = 0.0001;
t1=0:dt:0.0025;
VS= 5000;
R1=25;
C1=90*10^(-6);
func1= VS*exp(-t1/(R1*C1))/R1;
% plot (t1,func1), grid, xlim ([0 0.01])
t2=0:dt:0.0015;
VS= 5000;
R1=25;
C2=90*10^(-6);
func2=(-VS*exp(-t2/(R1*C2))/R1);
% plot (t2,func2),xlim ([0 0.01]), grid
t =0:dt:0.008;
t0=0:dt:0.0038;
zero=t0*0;
final= [func1 func2 zero];
plot (t,final),xlim ([0 0.01]), grid
fourier=fft (final);
L= length(t)
P2 = abs(fourier/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
Fs=1/dt;
f = Fs*(0:(L/2))/L;
plot(f,P1)
grid
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
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