Hi all,
I'm new to FFT and I want to ask about calculating FFT in Matlab. The following is my coding guidance from Matlab help:-
clc;
Fs = 512; % Sampling frequency
T = 1/Fs; % Sample time
L = 8100; % Length of signal
t = (0:L-1)*T; % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
y = x + 2*randn(size(t)); % Sinusoids plus noise
plot(Fs*t(1:50),y(1:50));
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('time (milliseconds)');
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2);
% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2)));
title('Single-Sided Amplitude Spectrum of y(t)');
xlabel('Frequency (Hz)');
ylabel('|Y(f)|');
Can someone verify my code because the fft graph unconvincing. Here the figure of the graph
Thank you.

5 Kommentare

Walter Roberson
Walter Roberson am 19 Jun. 2015
What features of the plot are you seeing as "unconvincing" ?
nur yusof
nur yusof am 29 Jun. 2015
the peak.looks not relevant.
Walter Roberson
Walter Roberson am 29 Jun. 2015
I am not sure why you say that? You coded
0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
which calls for a 0.7 peak at 50 Hz and a 1.0 peak at 120 Hz. The peaks are visually shown at the correct place, 50 Hz and 120 Hz. When I extract the value of the peak locations I get 0.683776526075202 and 0.986050681728091 . Why not exactly 0.7 and 1.0? Because of the normally distributed random noise with standard deviation 2 that you added, a standard deviation that exceeds the maximum value of the two sines added together (0.7+1). The peaks look relevant to me.
Sag
Sag am 26 Jan. 2016
Why to multiply by 2 in step: plot(f,2*abs(Y(1:NFFT/2)));
Afshin Loghmani M Toussi
Afshin Loghmani M Toussi am 1 Dez. 2020
Because you are plotting half of the frequency range( because of the symmetry). However, to show the same amount of energy you are multiplying the values by 2.

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Antworten (1)

Andreas Goser
Andreas Goser am 19 Jun. 2015

0 Stimmen

Frequent puzzlese with interpreting FFT:
2 peaks instead of 1 peak
Frequency "off" by a factor of 6.
Both effects are in the nature of the algorithm and "6" is actually "2*Pi". Next steps depend on whether you want to just use the results, then this will help. If you want to understand, read a text book about signal processing.

Gefragt:

am 19 Jun. 2015

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