Fast Fourier Transform Zero Padding

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Hi all,
I am using the code shown below to plot the FFT of some data. My issue is that the "resolution" seems poor, as the x axis is in increments of 0.2. I would like much finer plotting of points, and have recently seen the Zero Padding method. However, everytime I try to implement other solutions on MATLAB answers, I cannot seem to increase the resolution. Could anyone help me with the necessary code for my specific case?
O2_exp = [0.0247
0.2372
1.9171
1.5570
0.8016
0.5572
1.2185
1.3601
1.0067
0.7767
1.0244
1.1619
1.0210
0.8791
0.9595
1.0592
1.0274
0.9507
0.9735
1.0303
1.0286
0.9912
0.9924
1.0137
1.0143
0.9982
0.9996
1.0097
1.0174
1.0062
1.0052
1.0115
1.0177
1.0131
1.0150
1.0117
1.0182
1.0153
1.0206
1.0177
1.0243
1.0200
1.0221
1.0207
1.0235
1.0256
1.0275
1.0237
1.0248
1.0264];
figure
Fs = 1;
L = length(O2_exp);
Y = fft(O2_exp);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:(L/2))/L;
plot(f(2:end),P1(2:end)/max(P1(2:end)),'color','red','linewidth',4)
  2 个评论
Matt J
Matt J 2021-1-23
I cannot see anywhere in your code where you have attempted zero-padding.
kyle mani
kyle mani 2021-1-23
编辑:kyle mani 2021-1-23
Hi Matt! I have attempted to use zero-padding, but I am getting errors. Essentially, I have doubled the O2 vector to length 100, and have kept the first 50 values to be zero. However, it completely distorts the graph that I am seeing without the zero-padding.
The code I pasted above works perfectly in generating the FFT I would like, except the resolution is very poor. Therefore, I am asking if anyone can use the zero-padding method or any other appropriate method with the code I have specifically provided to achieve my goal of increasing the resolution for this graph.

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采纳的回答

Paul
Paul 2021-1-23
Try changing these lines:
L = length(O2_exp);
Y = fft(O2_exp);
to
L = nfft; % select nfft > numel(O2_exp), preferable a power of 2
Y = fft(detrend(O2_exp),nfft)
  4 个评论
Paul
Paul 2021-1-24
I'm not so sure I'd go so far as to say the output isn't physically meaningful, but I understand the sentiment.
I do agree that in order to get finer resolution, more data is needed.
However, I took the OP's question to not really mean frequency domain resolution in the technical sense (i.e, the ability the distinguish among frequency components); rather the OP was trying to get a reasonable interpolation in the frequency domain, of which zero padding is a reasonable approach among others. This link (and the Next pages) are a good discussion.
The results always need to be understood in the context of the underying samples of data.
Walter Roberson
Walter Roberson 2021-1-24
According to the comment at https://www.mathworks.com/matlabcentral/answers/724683-fast-fourier-transform-zero-padding#comment_1280732 the poster wanted to reduce the uncertainty in the interpretation of the frequency values. However, that is not something that you can do using zero padding.

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更多回答(1 个)

Walter Roberson
Walter Roberson 2021-1-23
Pad_factor = 5;
O2_exp = [0.0247
0.2372
1.9171
1.5570
0.8016
0.5572
1.2185
1.3601
1.0067
0.7767
1.0244
1.1619
1.0210
0.8791
0.9595
1.0592
1.0274
0.9507
0.9735
1.0303
1.0286
0.9912
0.9924
1.0137
1.0143
0.9982
0.9996
1.0097
1.0174
1.0062
1.0052
1.0115
1.0177
1.0131
1.0150
1.0117
1.0182
1.0153
1.0206
1.0177
1.0243
1.0200
1.0221
1.0207
1.0235
1.0256
1.0275
1.0237
1.0248
1.0264];
plot(O2_exp); title('original');
nO2 = numel(O2_exp);
reconstructed = ifft(fft(O2_exp,2*nO2));
plot(reconstructed); title('reconstructed nfft')
%caution: the details that follow are only valid when the
%length of the signal is even, and the signal is purely real.
F = fft(O2_exp);
Fpad = [F(1); F(2:end/2+1); zeros(Pad_factor*nO2-1,1); flipud(conj(F(2:end/2+1)))];
reconstructed_center_padded = ifft(Fpad);
plot(reconstructed_center_padded); title('reconstructed center padded')
  3 个评论
kyle mani
kyle mani 2021-1-23
Hi Walter, thank you for your answer. When I run the code I provided I get this image for the FFT, which is how it should look:
I guess my question is: Is there any way to increase the resolution of this image so that the x-axis is not just plotting points by increments of 0.02? The reason why is 1/frequency will give me the periodicity of these oxygen oscillations. In this graph, the peak is at a value of 0.24, corresponding to a periodicity of 4.16, and based on the plotting the increments directly greater and less would result in periods of: 4.54 and 0.386, which seemingly lead to large variation.
Walter Roberson
Walter Roberson 2021-1-23
You can change how you process to get it to plot at most any increment. The problem is that your output stops becoming meaningful. If you need finer resolution then you need more data.

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