清空筛选条件
清空筛选条件

Request for review of FFT conversion results

2 次查看（过去 30 天）
J
J 2023-12-6
回答： Sulaymon Eshkabilov 2023-12-6
Although it is not data related to noise and vibration, I would like to do FFT on resistance values that change with a specific speed (10 rpm).
But as you can see in the picture, the maximum comes out at 0 Hz. I think the Peak value would occur in another area(Hz), so if you can look at the attachment and code and advise on how to solve it, I would appreciate it.
clear all
clc
A = xlsread('2F100.xlsx','0.2m','D3:F812');
t = A(:,1);
r = A(:,3);
subplot(2,1,1), plot(t, r, 'linewidth', 2)
xlim([0 80])
ylim([0 30])
xlabel('t (seconds)','fontsize',16,'FontName','Arial')
ylabel('Resistance (\Omega)','fontsize',16,'FontName','Arial')
set(gca,'FontName','Arial','FontSize',14,'FontWeight','bold','XTick',...
[0 10 20 30 40 50 60 70 80]);
Fs = 1000;
Y = fft(r);
L = length(r);
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;
subplot(2,1,2), plot(f, P1, 'linewidth', 2) %semilogx(f,P1)
xlim([0 500])
ylim([0 10])

请先登录，再回答此问题。

采纳的回答

Sulaymon Eshkabilov
编辑：Sulaymon Eshkabilov 2023-12-6
Ran in:
One option is zeroing the DC component by removing the mean of the signal and/or setting DC component equal to zero:
A = xlsread('2F100.xlsx','0.2m','D3:F812');
t = A(:,1);
r = A(:,3);
% Remove mean of r
r=r-mean(r);
subplot(2,1,1), plot(t, r, 'linewidth', 2)
xlim([0 80])
ylim([0 30])
xlabel('t (seconds)','fontsize',16,'FontName','Arial')
ylabel('Resistance (\Omega)','fontsize',16,'FontName','Arial')
set(gca,'FontName','Arial','FontSize',14,'FontWeight','bold','XTick',...
[0 10 20 30 40 50 60 70 80]);
Fs = 1000;
Y = fft(r);
L = length(r);
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;
% Set equal value 1 equal to "0"
%P1(1)=0;
subplot(2,1,2), plot(f, P1, 'linewidth', 2) %semilogx(f,P1)
xlim([0 500])
% Chnage the y-axis limits
ylim([0 3.5])
  1 个评论
J
J 2023-12-6
Thank you for your answer. In the time region(upper graph), the y value was shifted by about 10, and in the frequency region(lower graph), the y value decreased by one-third. Can I see this as the effect of zeroing through the removal of the average value?

请先登录，再进行评论。

更多回答（1 个）

Sulaymon Eshkabilov
Ran in:
The two can be also compared:
A = xlsread('2F100.xlsx','0.2m','D3:F812');
t = A(:,1);
r = A(:,3);
% Remove mean of r
r1=r-mean(r);
subplot(2,1,1),
plot(t, r, 'linewidth', 2)
hold on
plot(t, r1, 'r-', 'linewidth', 1)
legend('Original', 'Mean removed')
xlim([0 80])
ylim([0 30])
xlabel('t (seconds)','fontsize',16,'FontName','Arial')
ylabel('Resistance (\Omega)','fontsize',16,'FontName','Arial')
set(gca,'FontName','Arial','FontSize',14,'FontWeight','bold','XTick',...
[0 10 20 30 40 50 60 70 80]);
Fs = 1000;
Y1 = fft(r1);
L = length(r1);
P2 = abs(Y1/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:(L/2))/L;
subplot(2,1,2),
plot(f, P1, 'linewidth', 2) %semilogx(f,P1)
hold on
Y = fft(r);
L = length(r);
PP2 = abs(Y/L);
PP1 = PP2(1:L/2+1);
PP1(2:end-1) = 2*PP1(2:end-1);
f = Fs*(0:(L/2))/L;
% Set equal value 1 equal to "0"
PP1(1)=0;
plot(f, PP1, 'r-', 'LineWidth', 1)
legend('Mean removed', 'DC = 0')
xlim([0 500])
% Change the y-axis limits
ylim([0 3.5])
% FFT of : Mean removed and DC=0 matches quite well

类别

在 Help Center 和 File Exchange 中查找有关 Fourier Analysis and Filtering 的更多信息

标签

产品


版本

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by