Filter signal with 2 different frequencies

Question

Sam Hurrell 2022-9-14

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1805235-filter-signal-with-2-different-frequencies

评论： Star Strider 2022-9-15

I have an equation which gives 2 gaussian-sinc pulses of different frequencies, in dataset wrt time (t = 0:0.001:100). When put into an fft the two frequencies are found, but I want to design a filter for the signal to remove one of the pulses. How can I do this to the signal below for different values for f1 and f2?

f1 = 2.5, f2 = 1.5, s1 = 1.2/f1, s2 = 1.2/f2

D = cos(2*pi*f1*(t-10))*exp(-(t-10).^2/(2*s1^2)) + cos(2*pi*f2*(t-20))*exp(-(t-20).^2/(2*s2^2))

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Answer 1

Star Strider 2022-9-14

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https://ww2.mathworks.cn/matlabcentral/answers/1805235-filter-signal-with-2-different-frequencies#answer_1053365

Use either the bandpass or bandstop functions, depending on the result you want.

For best results, use the 'ImpulseResponse','iir' name-value pair.

.

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Star Strider 2022-9-14

在 MATLAB Online 中打开

I have no idea what the problems with the bandpass function is, since I had problems getting it to work online here as well. I was able to get it to work with my command-line filters (that likely produce the same filters) —

Fs = 1000;

t = linspace(0, 50*Fs, 150*Fs)/Fs;

f1 = 2.5; f2 = 1.5; s1 = 1.2/f1; s2 = 1.2/f2;

D = cos(2*pi*f1*(t-10)).*exp(-(t-10).^2/(2*s1^2)) + cos(2*pi*f2*(t-20)).*exp(-(t-20).^2/(2*s2^2))

D = 1×150000

1.0e+00 * 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

figure

plot(t, D)

grid

xlabel('Time')

ylabel('Amplitude')

title('Original')

L = numel(t);

Fn = Fs/2;

NFFT = 2^nextpow2(L); % For Efficiency

FTD = fft(D - mean(D),NFFT)/L; % Fourier Transform

Fv = linspace(0, 1, NFFT/2-1)*Fn; % Frequency Vector (For Plots)

Iv = 1:numel(Fv); % Index Vector (For Plots)

figure

plot(Fv, abs(FTD(Iv))*2)

grid

xlim([0 1.5])

xlabel('Frequency')

ylabel('Magnitude')

Wp = [0.65 1.3]/Fn; % Define Passband In Hz, Normalise To (0,pi)

Ws = Wp.*[0.95 1.05]; % Stopband (Normalised)

Rs = 50; % Stopband Ripple (Attenuation) dB

Rp = 1; % Passband Ripple dB

[n,Wn] = ellipord(Wp,Ws,Rp,Rs); % Calculate Filter Order

[z,p,k] = ellip(n,Rp,Rs,Wp); % Design Filter

[sosH,gH] = zp2sos(z,p,k); % Implement Filter As Second-Order-Section Representation

figure

freqz(sosH, 2^16, Fs) % Filter Bode Plot

set(subplot(2,1,1), 'XLim',[0 5])

set(subplot(2,1,2), 'XLim',[0 5])

Wp = [0.2 0.65]/Fn; % Define Passband In Hz, Normalise To (0,pi)

Ws = Wp.*[0.95 1.05]; % Stopband (Normalised)

Rs = 50; % Stopband Ripple (Attenuation) dB

Rp = 1; % Passband Ripple dB

[n,Wn] = ellipord(Wp,Ws,Rp,Rs); % Calculate Filter Order

[z,p,k] = ellip(n,Rp,Rs,Wp); % Design Filter

[sosL,gL] = zp2sos(z,p,k); % Implement Filter As Second-Order-Section Representation

figure

freqz(sosL, 2^16, Fs) % Filter Bode Plot

set(subplot(2,1,1), 'XLim',[0 5])

set(subplot(2,1,2), 'XLim',[0 5])

D_FiltH = filtfilt(sosH,gH,D);

D_FiltL = filtfilt(sosL,gL,D);

D_Filt = [D_FiltL; D_FiltH]

D_Filt = 2×150000

-0.0012 -0.0011 -0.0011 -0.0011 -0.0011 -0.0011 -0.0011 -0.0011 -0.0011 -0.0010 -0.0010 -0.0010 -0.0010 -0.0010 -0.0010 -0.0010 -0.0010 -0.0009 -0.0009 -0.0009 -0.0009 -0.0009 -0.0009 -0.0009 -0.0009 -0.0009 -0.0008 -0.0008 -0.0008 -0.0008 -0.0004 -0.0004 -0.0004 -0.0005 -0.0005 -0.0005 -0.0005 -0.0005 -0.0005 -0.0005 -0.0005 -0.0005 -0.0005 -0.0005 -0.0006 -0.0006 -0.0006 -0.0006 -0.0006 -0.0006 -0.0006 -0.0006 -0.0006 -0.0006 -0.0006 -0.0007 -0.0007 -0.0007 -0.0007 -0.0007

figure

plot(t, D_Filt)

grid

xlabel('Time')

ylabel('Amplitude')

title('Filtered')

FTD_Filt = fft(D_Filt - mean(D_Filt,2),NFFT, 2)/L; % Fourier Transform

% D_FiltL = lowpass(D, 0.2, Fs, 'ImpulseResponse','iir')

% D_FiltH = highpass(D, 0.7, Fs, 'ImpulseResponse','iir')

% FTD_Filt = fft([D_FiltH; D_FiltL],NFFT,2)/L % Fourier Transform

figure

plot(Fv, abs(FTD_Filt(:,Iv))*2)

grid

xlim([0 1.5])

xlabel('Frequency')

ylabel('Magnitude')

There is some distortion in the results because some frequencies comomon to both functions were filtered out in each filter operation (seen most easily in the time-domain plot of the filtered signals).

.

Sam Hurrell 2022-9-15

That seems to have fixed it, thank you

Star Strider 2022-9-15

As always, my pleasure!

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Filter signal with 2 different frequencies

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