Signal amplitude decreases after FIR filter(windowing method)

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Amy Lg 2022-3-10

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编辑： Amy Lg 2022-3-21

I am applying a low pass filter with the windowing method to my signal in order to truncate it in the frequency domain. The bandwidth of the filter should be 3GHz. I guess that works fine. The filter, however, seems to attenuate the amplitude of the signal. What do I need to do to get the same amplitude as the original signal after filtering?

I am using filters for the first time. I do appreciate any help or explanation.

sig = MY SIGNAL;
fs = 4000;                    % sampling freq. (GHz)
M = 400001;              % signal length
% Filter parameters:
L = M;                         % filter length 
fcut = 1.5;                    % cutoff frequency (GHz) 
% Design the filter using the window method:
hsupp = (-(L-1)/2:(L-1)/2);
hideal = (2*fcut/fs)*sinc(2*fcut*hsupp/fs);
h = hamming(L)' .* hideal;                   % h is our filter
SIG_out = fft(sig);                          % signal
H = fft(h);                                  % filter
FILT_OUT = SIG_out .* H;                   
filt_out = ifft(FILT_OUT);
relrmserr = norm(imag(filt_out))/norm(filt_out) % check... should be zero

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Star Strider 2022-3-10

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I have no idea what the sampling frequency is, so I assume 10 GHz here, with a cutoff of 1.5 GHz. When I simulate it with the freqz function, it shows essentially no attenuation in the passband (as it should not), a stopband frequency of 1.5 GHz (as designed) and an appropriate transition region. Some parameters are not given (specifically ‘M’ and ‘fs’) so the filter behaviour here may be different from what your analysis shows. However on the basis of the freqz results, the problems you are seeing with it may simply be that your analysis — and not the filter design — is incorrect. (A larger value for ‘M’ will give a sharper cutoff and shorter transition region.)

M = 64; % Create 'M'

fs = 1E+10; % Create 'fs'

L = M; % filter length

fcut = 1.5E+9; % cutoff frequency (GHz)

% Design the filter using the window method:

hsupp = (-(L-1)/2:(L-1)/2);

hideal = (2*fcut/fs)*sinc(2*fcut*hsupp/fs);

h = hamming(L)' .* hideal; % h is our filter

figure

freqz(h, 1, 2^16, fs)

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Star Strider 2022-3-10

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My pleasure!

Actually, as designed it has a 1.5 Ghz bandwidth, and due to the filter length, a very short transition region. (I did not change either of those in this analysis.)

Changing those parameters —

M = 400001; % Use Supplied 'M'

fs = 4E+12; % Use Supplied 'fs'

L = M; % filter length

fcut = 1.5E+9; % cutoff frequency (GHz)

% Design the filter using the window method:

hsupp = (-(L-1)/2:(L-1)/2);

hideal = (2*fcut/fs)*sinc(2*fcut*hsupp/fs);

h = hamming(L)' .* hideal; % h is our filter

figure

freqz(h, 1, 2^16, fs)

set(subplot(2,1,1), 'XLim',[0 1E+10]) % Zoom Frequency Axis To Show Detail

set(subplot(2,1,2), 'XLim',[0 1E+10]) % Zoom Frequency Axis To Show Detail

That is a very long filter! Still, it demonstrates that it fulfills the design requirements.

If you are using the Signal Processing Toolbox functions, use the filtfilt function to do the actual filtering. The signal will need to be very long, or the filter length will need to be reduced (possibly significantly so). However it works efficiently as a much shorter filter, so that should not be a problem. If it is, then consider a IIR filter instead, most likely an elliptic design for numerical efficiency.

Star Strider 2022-3-11

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My pleasure!

You are close the the correct code, however it needs a few changes and additions.

I use the ellipord function to calculate the filter order, and generally do something like this to design it:

Fs = 1E+10; % Sampling Frequency

Fn = Fs/2; % Nyquist Frequency

Wp = 1.5E+9/Fn; % Passband Frequency (normalised)

Ws = 1.01*Wp; % Stopband Frequency (Normalised)

Rp = 1; % Passband Ripple (Attenuation)

Rs = 50; % Stopband Ripple (Attenuation)

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

[z,p,k] = ellip(n,Rp,Rs,Wp,'low'); % Zero-Pole-Gain Representation

[sos,g] = zp2sos(z,p,k); % Second-Order-Section Implementation

figure

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

% filteredSignal = filtfilt(sos,g,sig_out); % Filter Signal

I commented the filtfilt call here so it would not execute when I run the rest of the code without the ‘sig_out’ vector. The syntax is correct.

Taking the abs fo ‘filteredSignal’ would simply force the negative parts to be positive. This would not desirable unless it is complex, and I doubt it would be complex. Even if it were complex, it would be better to consider the real and imaginary parts separately, rather than losing that information in combining them with abs.

Also, since the sampling frequency and filter frequency are in the GHz range, they must be specified accurately in order for the filter parameter matrix to be scaled correctly. Using 1.5 when the desired value is 1.5E+9 is not appropriate and will not produce the desired filter.

Star Strider 2022-3-11

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I had forgotten what the sampling frequency was, and use the first value that I saw from my earlier post.

Use this filter instead —

Fs = 4E+12; % Sampling Frequency

Fn = Fs/2; % Nyquist Frequency

Wp = 1.5E+9/Fn; % Passband Frequency (normalised)

Ws = 1.01*Wp; % Stopband Frequency (Normalised)

Rp = 1; % Passband Ripple (Attenuation)

Rs = 50; % Stopband Ripple (Attenuation)

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

[z,p,k] = ellip(n,Rp,Rs,Wp,'low'); % Zero-Pole-Gain Representation

[sos,g] = zp2sos(z,p,k); % Second-Order-Section Implementation

figure

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

set(subplot(2,1,1), 'XLim',[0 5E+9])

set(subplot(2,1,2), 'XLim',[0 5E+9])

% filteredSignal = filtfilt(sos,g,sig_out); % Filter Signal

Did you do a fft on the signal to determine what the frequency components are? If not, that needs to be done in order to design the filter correctly. The filter may have too narrow a passband, or the wrong passband, to filter your signal. I do not have your signal, so I cannot determine this.

Amy Lg 2022-3-19

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I appreciate your help so much.

I still have problems with the result. Increasing my signal length to 4450001 and fs to 10000GHz results in an output similar to what is shown below. I do not know why somewhere the amplitude of the filtered signal increase and somewhere decrease. I think it is wrong to have oscillation in areas where the original signal is zero, because the low pass filter is supposed to smooth the signal and not adding new oscillation. I do appreciate if you could help me to correct the filter for my signal.

sig  = MY SIGNAL;                       % Intensity of output signal(square-law device)
M    = 4450001;                         % signal length
fs   = 10000e9;                         % sampling freq. (10000GHz)
fcut = 3e9;
fn = fs/2;                              % Nyquist Frequency(Hz)
Wp = fcut/fn;                           % Passband Frequency (normalised)-Hz
Ws = 1.05*Wp;                           % Stopband Frequency (Normalised)-Hz
Rp =  1;                                % Passband Ripple (Attenuation)-dB
Rs = 50;                                % Stopband Ripple (Attenuation)-dB
[n,Wp]  = ellipord(Wp,Ws,Rp,Rs);        % Calculate Filter Order
[z,p,k] = ellip(n,Rp,Rs,Wp,'low');      % Zero-Pole-Gain Representation
[sos,g] = zp2sos(z,p,k);                % Second-Order-Section Implementation
filteredSignal = filtfilt(sos,g,sig);
figure
freqz(sos, 2^16, fs)                    % Filter Bode Plot