The graph wasnt showing the appropiate answer

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clc
clear all
Ts=0.1; %Ts=sampling time
Tc=Ts/100; %Tc=continuous time increament
Tr=10; %Tr=record time
A=1; %A=amplitude of vibration
f=1; %f=frequency of signal
%Plot of waveform (continuous and discrete)
t=0.001:Ts:Tr;
x=A*sin(2*pi*1*t)+ A*sin(2*pi*11*t);
tc=0:Tc:Tr;
xc=A*sin(2*pi*1*tc) + A*sin(2*pi*11*tc);
subplot(211)
plot(t,x,'*',tc,xc)
%Find the spectrum of x(kT), so X(iw)=fft(x(k))
Xs=fft(x); %Xs=spectrum of x
%Now we want to plot the magnitude of X(iw), thus |X(iw)|=abs(X(iw))
[k,b]=size(Xs);
N=b;
N2=N/2;
magXs=(2/N)*abs(Xs);
df=1/Tr; %df=Frequency resolution
i=0:(N2-1);
f1=df*i;
subplot(212)
plot(f1,magXs(1,1:N2))
subplot
This is the code and it wont show the right spectrum cause the other spectrum should be at 11 Hz

Answers (1)

Cris LaPierre
Cris LaPierre on 19 Oct 2021
You have specificed a sample frequency of 10 Hz (Ts=0.1). Therefore, according the Nyquist criterion, the highest frequency you can resolve is 5 Hz.
  1 Comment
Cris LaPierre
Cris LaPierre on 19 Oct 2021
Change your sample frequency to something at least double the maximum frequency you want to detect, and it will appear.
% v made Ts 10x smaller. Corresponding Fs is now 100 Hz. Can now resolve up to 50 Hz
Ts=0.01; %Ts=sampling time
Tc=Ts/100; %Tc=continuous time increament
Tr=10; %Tr=record time
A=1; %A=amplitude of vibration
f=1; %f=frequency of signal
%Plot of waveform (continuous and discrete)
t=0.001:Ts:Tr;
x=A*sin(2*pi*1*t)+ A*sin(2*pi*11*t);
tc=0:Tc:Tr;
xc=A*sin(2*pi*1*tc) + A*sin(2*pi*11*tc);
%Find the spectrum of x(kT), so X(iw)=fft(x(k))
Xs=fft(x); %Xs=spectrum of x
% Calculate magnitude of X(iw), thus |X(iw)|=abs(X(iw))
[k,b]=size(Xs);
N=b;
N2=N/2;
magXs=(2/N)*abs(Xs);
df=1/Tr; %df=Frequency resolution
i=0:(N2-1);
f1=df*i;
% Plot
subplot(211)
plot(t,x,'*',tc,xc)
subplot(212)
plot(f1,magXs(1,1:N2))

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