在 Spectrum Matalb 中查找峰值
Findpeaks in Spectrum Matalb
我试图在频谱中找到峰值,但我只需要提取基频及其谐波的峰值,红色矩形。如何排除基频之前的任何内容,只包括基频及其 3 次谐波。我使用此代码,但没有帮助。有什么想法吗?
pks = findpeaks(q);
findpeaks(q,'MinPeakDistance',99)
%findpeaks(q,'MinPeakHeight',0.0004)
xlim([0.1 500])
使用时:
Fs = 1000;
t = 0:0.001:1-0.001; % 250-Hz sine wave modulated at 100 Hz
x = [1+cos(2*pi*100*t)].*cos(2*pi*250*t);
%envspectrum(x,Fs)
[ES,F]=envspectrum(x,Fs);
%%
findpeaks(ES,F)
% Now for only > 99 Hz (choose the freq you fancy)
idx = F >= 99; % greater than 99 Hz
findpeaks(ES(idx),F(idx)) % idx only select those F > 99
% Good? Keep the values of amplitud and location (in frequencies) of the ES
[pks,loc] = findpeaks(ES(idx),F(idx));
% If you just want to have the first 5 peaks (or the n? you choose):
% Select only 3 first.
if length(pks) > 3 % Check you didnt get less peaks
pks = pks(1:3);
loc = loc(1:3);
end
% To plot the peaks in the envelope
plot(F(idx),ES(idx),loc,pks,'r*')
我明白了:
结果是:
已编辑以通过示例适应新代码。我们正在应用信号的包络谱 - [ES,F] = envspectrum(sig,Fs);
-,因此我们知道信号及其频率采样 (fs)。
还是一样的流程。您可以仅针对高于 250 Hz 的信号样本计算 findpeaks()
中的值。为此,您为想要的频率值定义一个逻辑数组:
idx = F >= 250; % 250 Hz
并将此逻辑索引应用到信号的包络谱和要应用函数的频率 findpeaks()
:
[pks,loc] = findpeaks(ES(idx),F(idx));
示例:
% Let's create a signal with fs = 2500 Hz
% This is just to create an example, don't worry about these lines
fs = 2500;
f0 = 25;
n = 8;
d = 0.02;
p = 0.12;
t = 0:1/fs:1-1/fs;
z = [1 0.5 0.2 0.1 0.05]*sin(2*pi*f0*[1 2 3 4 5]'.*t)/5;
% z: our signal
% t: time array of the signal (to plot)
% How does the signal look?
plot(t,z)
% Calculating the envelope signal and its spectrum
[ES,F]=envspectrum(z,fs);
% Plot the envelop if you want:
envspectrum(z,fs)
% ES: Envelope spectrum of the signal
% F: array of frequencies used for the spectrum (half our fs for Nyquist Theorem)
% Find the peaks of the envelope spectrum
% Let's see how it is for all the envelope
findpeaks(ES,F)
% Now for only > 250 Hz (choose the freq you fancy)
idx = F >= 250; % greater than 250 Hz
findpeaks(ES(idx),F(idx)) % idx only select those F > 250
% Good? Keep the values of amplitud and location (in frequencies) of the ES
% Use 'NPeaks' input to tell function to select only 5 first peaks it can find.
[pks,loc] = findpeaks(ES(idx),F(idx), 'NPeaks', 5);
% To plot the peaks in the envelope
plot(F(idx),ES(idx),loc,pks,'r*')
% Note in this example the first peak (in 250 Hz) is not selected
% because is not a local maximum when we narrow the envelope.
% If the function select false peaks somehow, you can use
%'MinPeakHeight' to specify an absolute amplitude, 'MinPeakProminence '
%for a relative amplitude or 'Threshold' inputs for a better fit.
https://uk.mathworks.com/matlabcentral/answers/768537-use-findpeaks-on-specific-frequency-range
https://uk.mathworks.com/help/signal/ref/findpeaks.html#responsive_offcanvas
我试图在频谱中找到峰值,但我只需要提取基频及其谐波的峰值,红色矩形。如何排除基频之前的任何内容,只包括基频及其 3 次谐波。我使用此代码,但没有帮助。有什么想法吗?
pks = findpeaks(q);
findpeaks(q,'MinPeakDistance',99)
%findpeaks(q,'MinPeakHeight',0.0004)
xlim([0.1 500])
使用时:
Fs = 1000;
t = 0:0.001:1-0.001; % 250-Hz sine wave modulated at 100 Hz
x = [1+cos(2*pi*100*t)].*cos(2*pi*250*t);
%envspectrum(x,Fs)
[ES,F]=envspectrum(x,Fs);
%%
findpeaks(ES,F)
% Now for only > 99 Hz (choose the freq you fancy)
idx = F >= 99; % greater than 99 Hz
findpeaks(ES(idx),F(idx)) % idx only select those F > 99
% Good? Keep the values of amplitud and location (in frequencies) of the ES
[pks,loc] = findpeaks(ES(idx),F(idx));
% If you just want to have the first 5 peaks (or the n? you choose):
% Select only 3 first.
if length(pks) > 3 % Check you didnt get less peaks
pks = pks(1:3);
loc = loc(1:3);
end
% To plot the peaks in the envelope
plot(F(idx),ES(idx),loc,pks,'r*')
我明白了:
结果是:
已编辑以通过示例适应新代码。我们正在应用信号的包络谱 - [ES,F] = envspectrum(sig,Fs);
-,因此我们知道信号及其频率采样 (fs)。
还是一样的流程。您可以仅针对高于 250 Hz 的信号样本计算 findpeaks()
中的值。为此,您为想要的频率值定义一个逻辑数组:
idx = F >= 250; % 250 Hz
并将此逻辑索引应用到信号的包络谱和要应用函数的频率 findpeaks()
:
[pks,loc] = findpeaks(ES(idx),F(idx));
示例:
% Let's create a signal with fs = 2500 Hz
% This is just to create an example, don't worry about these lines
fs = 2500;
f0 = 25;
n = 8;
d = 0.02;
p = 0.12;
t = 0:1/fs:1-1/fs;
z = [1 0.5 0.2 0.1 0.05]*sin(2*pi*f0*[1 2 3 4 5]'.*t)/5;
% z: our signal
% t: time array of the signal (to plot)
% How does the signal look?
plot(t,z)
% Calculating the envelope signal and its spectrum
[ES,F]=envspectrum(z,fs);
% Plot the envelop if you want:
envspectrum(z,fs)
% ES: Envelope spectrum of the signal
% F: array of frequencies used for the spectrum (half our fs for Nyquist Theorem)
% Find the peaks of the envelope spectrum
% Let's see how it is for all the envelope
findpeaks(ES,F)
% Now for only > 250 Hz (choose the freq you fancy)
idx = F >= 250; % greater than 250 Hz
findpeaks(ES(idx),F(idx)) % idx only select those F > 250
% Good? Keep the values of amplitud and location (in frequencies) of the ES
% Use 'NPeaks' input to tell function to select only 5 first peaks it can find.
[pks,loc] = findpeaks(ES(idx),F(idx), 'NPeaks', 5);
% To plot the peaks in the envelope
plot(F(idx),ES(idx),loc,pks,'r*')
% Note in this example the first peak (in 250 Hz) is not selected
% because is not a local maximum when we narrow the envelope.
% If the function select false peaks somehow, you can use
%'MinPeakHeight' to specify an absolute amplitude, 'MinPeakProminence '
%for a relative amplitude or 'Threshold' inputs for a better fit.
https://uk.mathworks.com/matlabcentral/answers/768537-use-findpeaks-on-specific-frequency-range
https://uk.mathworks.com/help/signal/ref/findpeaks.html#responsive_offcanvas