I can't make sense of the amplitude spectrum of a signal. Help please.

Question

MatlabGirl 2014-12-7

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评论： Image Analyst 2014-12-7

采纳的回答： Image Analyst

在 MATLAB Online 中打开

Hi,

I have a sinusoidal signal namely:

signal = sin(2*pi*0.08*t) + sin(2*pi*0.2*t) + sin(2*pi*0.32*t) + sin(2*pi*0.4*t);

so the segment of the code becomes:

freqs = [0.08, 0.2, 0.32, 0.4];
periods = 1./ freqs;
% Let the max value of t be 4 times the largest period
t_max = 4 * periods(1);
% Let us have 500 samples over the interval [0, t_max].
t = linspace(0, t_max, 500);
% The synthesized signal 
signal = sin(2*pi*0.08*t) + sin(2*pi*0.2*t) + sin(2*pi*0.32*t) + sin(2*pi*0.4*t);

Now I wanna view the amplitude spectrum for the first 50 samples.

So I did this:

signal_first_50_samples= signal (1:50);
% Display its magnitude spectrum. 
N = 64;
signal_spect = abs (fft(signal_first_50_samples,N));
signal_spect = fftshift(signal_spect);
F = [-N/2:N/2-1]/N;
plot (F, signal_spect) 
set(gca, 'XTick',[-0.5:0.1:1])
grid on;

I was expecting spikes at each of the frequencies, however, I got this:

can anyone help me understand?

Thanks

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Answer 1

Image Analyst 2014-12-7

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If the signal is 500 samples and has 4 periods, then each period is 125 samples long. If you take less than half that you probably don't have a long enough signal to get a reliable spectrum. You're trying to get a spectrum when some of the frequencies don't even have half a cycle in the signal. Why did you take just 50 and not the whole signal?

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MatlabGirl 2014-12-7

在 MATLAB Online 中打开

Your idea made perfect sense, so I tried doing it for 500 samples by modifying it as follows:

%%Synthesizing a sampled signal that consists of 4 sine waves 
% We are given 4 different frequencies 
freqs = [0.08, 0.2, 0.32, 0.4];
periods = 1./ freqs;
% Let the max value of t be 4 times the largest period
t_max = 4 * periods(1);
% Let us have 500 samples over the interval [0, t_max].
t = linspace(0, t_max, 500);
% The synthesized signal 
signal = sin(2*pi*0.08*t) + sin(2*pi*0.2*t) + sin(2*pi*0.32*t) + sin(2*pi*0.4*t);
% Visualising the synthesized signal 
plot(t, signal, 'b-', 'LineWidth', 1.5);
grid on;
title('The Synthesized Signal', 'FontSize', fontSize);
xlabel('t', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
% Display its magnitude spectrum. 
% i was expecting spikes at each freq 
N = 64;
signal_spect = abs (fft(signal,N));
signal_spect = fftshift(signal_spect);
F = [-N/2:N/2-1]/N;
plot (F, signal_spect) 
%set(gca, 'XTick',[-0.5:0.05:1])
grid on;

and I got this:

kinda the same :(

MatlabGirl 2014-12-7

在 MATLAB Online 中打开

i did something else i had only 50 samples for the singal and took all of them

%%50 samples taking 50
clc;    
clear;  
% We are given 4 different frequencies 
freqs = [0.08, 0.2, 0.32, 0.4];
periods = 1./ freqs;
% Let the max value of t be 4 times the largest period
t_max = 4 * periods(1);
% Let us have 50 samples over the interval [0, t_max].
t = linspace(0, t_max, 50);
% The synthesized signal 
signal = sin(2*pi*0.08*t) + sin(2*pi*0.2*t) + sin(2*pi*0.32*t) + sin(2*pi*0.4*t);
% Determine the first 50 samples of this signal 
% signal_first_50_samples= signal (1:50);
% Display its magnitude spectrum. 
% ??????????????????????????????????????????????????????????????????????
% I was expecting spikes at each freq 
N = 64;
signal_spect = abs (fft(signal,N));
signal_spect = fftshift(signal_spect);
F = [-N/2:N/2-1]/N;
figure
plot (F, signal_spect) 
%set(gca, 'XTick',[-0.5:0.05:1])
grid on;

I got this

it looks about right

but I wonder why it didn't work when i had 500 samples and i took all 500 of them

any ideas?

Image Analyst 2014-12-7

Looks like your sampling frequency is not enough to follow the curves exactly. But at least you're getting all 4 spikes now like you should. It appears that there is some effect of the overall, total sample length because I can see side lobes caused by the sinc function, which is what you will get when you truncate your signal by a rect function. If you had higher resolution and took more cycles, that effect would be less.

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