How to calculate passband and stopband coefficients for signal filtering?

Hello Star Strider,

I have a signal, that is a well log Acoustic Impedance log and I need to apply a Antialias filter in coarser samoling that means a bandpass filter.

Sample frequency is 6562 per kilometer (28000 sample/4.27km). I do need to identify cutt off frequencies correctly. I applied your passband code into my dataset but my proffessors says: apply an anti-alias filter that has a flat spectrum over the band of interest. Would you please direct me how I can identify cutoff frequencies? Appreciated!

Answer 1

[Star Strider]

If you are concerned with an anti-aliasing filter for the time-domain signal, you need to go back and design a hardware Bessel (besself) lowpass anti-aliasing filter with a cutoff frequency that is equal to the Nyquist frequency (half the sampling frequency), then do the sampling. Frequencies higher than the Nyquist frequency are already aliased in your signal and now cannot be removed.

The frequency of ‘6562 per kilometer’ is obviously a spatial frequency, and I assume normal signal processing approaches would apply, so use the lowpass function, or design your own elliptic filter (ellip and related functions), each with a cutoff frequency of 3218 samples/km, as the anti-aliasing filter. Use filtfilt to do the actual filtering.

Then do whatever signal processing you want on the remaining signal. Use a highpass or bandpass filter to eliminate baseline offset and baseline drift. Using elliptic filters to do these are also straightforward to design in MATLAB.

[Maria Amr]

Thank you so much Star Strider

Very helpful to me!

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

[Maria Amr]

Star Strider Sure. But "Accept " is not active. Would you plese active it? Thank you again!

[Star Strider]

I cannot Accept my own Answer.

The site is having problems. I have reported this to MathWorks, along with some other problems with MATLAB Answers that have appeared over the last day or so.

Thank you for replying!

[Anusha Sridharan]

Hi Maria Amir, I'm sorry you are having issues accepting Star Strider's answer.

Are you able to see a blue 'Accept this answer' button on the answer, right below Star Strider's name? Here is how that should look:

Please make sure that you are logged in to the site as the button is only available once signed in.

[Maria Amr]

Hello Star Strider;

I truly appreciate if you answer my question. I have a data, sampled in depth, 8086/second or 8086/1.232 km. I want to measure the average value in 1 ms. I am using this code to measer the average value in every n sample in depth but I don't know how do that in a time block for example 1 millisecond. Thank you!

clear all;

data=xlsread('Data');

dep=data(:,1); % Depth

Vel=data(:,2);%velocity

N = length(Vel);

Fs = 8086/1; % Sampling Frequency (Hz)

t = linspace(0, N, N)/Fs; % Time Vector (If One Has Not Been Supplied With Your EEG Record)

N1 = length(t);

Fn = Fs/2;

vavg=mean(reshape(Vel,8,[]));

V3=vavg;

[Star Strider]

I am not exactly certain what you want.

This uses the Signal Processing Toolbox buffer function to segment the signal:

T1 = readtable('Data.xlsx');
Q1 = T1(1:5,:);
d = T1.Depth;
v = T1.Velocity;
Fs = 8086;                                                  % Sampling Frequency (Hz)
Ts = 1/Fs;                                                  % Sampling Interval (sec)
tsec = size(v,1)/8086;                                      % Record Length (sec)
frame = 0.001;                                              % Frame Length (sec)

figure
plot(d, v)
grid
xlabel('Depth')
ylabel('Velocity')
title('Signal Segment')
xlim([5000 5500])

vmtx = buffer(v, round(tsec/frame));                        % Uses ‘buffer’ To Segment Signal
vmtx_mean = mean(vmtx,1);

figure
plot((1:size(vmtx,2)), vmtx_mean)
grid
xlabel('Depth')
ylabel('Velocity')
title('Signal 1 Sec Segment Means')

The buffer function is easier to use than reshape in this situation.

[Maria Amr]

Star Strider

Thank you for your responding. There is an error "dot indexing is not supported for variables of this type" does not allow to run the code.

I mean: I want to measure the average of velocity values in every 1 ms in time and plot it versus time. Please find attached! Appreciated.

[Star Strider]

My code runs without error in R2020b. You should be able to address the ‘T1’ table just as I do here, using the dot-index convention.

I am confused. You stated: ‘8086/second’, and there are 8086 sampes (rows) is the file you posted, so there should be 1000 1 ms segments of about 9 samples each. (The buffer function zero-pads the vector to produce the correct segment lengths, creating a (9000x1) vector that it then segments.)

Changing that part of the code to:

vmtx = buffer(v, round(tsec/frame));                        % Uses ‘buffer’ To Segment Signal
vmtx_mean = mean(vmtx,2);

figure
plot((1:size(vmtx,1)), vmtx_mean)
grid
Ax = gca;
xt = Ax.XTick;
Ax.XTickLabel = xt*size(vmtx,2);
xlabel('Depth Segment')
ylabel('Velocity')
title('Signal 1 ms Segment Means')

produces them. (I reversed the row-column references earllier.)

.

[Maria Amr]

Star Stride

Thank you so much. The plot shows that is totally correct but it is not runing for me. My MATLAB verson is 2019b.

The same error is still remaining.'' Dot indexing is not supported for variables of this type.''

[Star Strider]

As always, my pleasure!

You should be able to run my code without error in R2019b. I have no idea what the problem could be.

My full code is:

T1 = readtable('Data.xlsx');
d = T1.Depth;
v = T1.Velocity;
Fs = 8086;                                                  % Sampling Frequency (Hz)
Ts = 1/Fs;                                                  % Sampling Interval (sec)
tsec = size(v,1)/Fs;                                        % Record Length (sec)
frame = 0.001;                                              % Frame Length (sec)

figure
plot(d, v)
grid
xlabel('Depth')
ylabel('Velocity')
title('Signal Segment')
xlim([5000 5500])

vmtx = buffer(v, round(tsec/frame));                        % Uses ‘buffer’ To Segment Signal
vmtx_mean = mean(vmtx,2);

figure
plot((1:size(vmtx,1)), vmtx_mean)
grid
Ax = gca;
xt = Ax.XTick;
Ax.XTickLabel = xt*size(vmtx,2);
xlabel('Depth Segment')
ylabel('Velocity')
title('Signal 1 ms Segment Means')

What line is throwing the error?

.

[Maria Amr]

Star Strider

Really appreciated!! That is perfect. I have installed a 2020b and it is working very well.

The only one I should change is xlable in the figure "Time Segment". Is that correct?

How may I ''Accept it'' again? Thank you!

[Star Strider]

As always, my pleasure!

Label the axes appropriately. I am not certain what they should be, so I simply provided the code to label them.

I appreciate your wanting to Accept it again! That is not possible, however I would definitely encourage you to Vote for it! That option is above the text of my Answer, to the right of my name, just below the green bar! (I get 2 Reputation Points for a Vote.)

[Maria Amr]

Star Strider I truly appreciate if you can check my code. I have already asked my question and got different answers.

I should measure a parameter of an array and I have wrtten the code but my prof. says it is something wrong.

The equation is:

RC = ½(deltaVp/averageVp + deltaRHO/averageRHO). % reflection coefficients

when I run the first code , the output is half when I run the second one. ( I only replace ".*o.5) instead of "mean" to measure average)

My codes are:

First:

for j=2:l;

rc(j)=1/2.*(((v(j)-v(j-1))/(mean(v(j)+v(j-1))))+((d(j)-d(j-1))/(mean(d(j)+d(j-1)))));% reflection coefficients

end

Second:

for j=2:l;

rc(j)=1/2.*(((v(j)-v(j-1))/(0.5.*(v(j)+v(j-1))))+((d(j)-d(j-1))/(0.5.*(d(j)+d(j-1))))); % reflection coefficients

end

[Star Strider]

I have absolutely noi idea what you are doing.

However, the second one appears to be correct, because in the first one, the mean call is not doing what you believe it is doing. It should be:

mean([v(j) v(j-1)])

instead of:

mean(v(j)+v(j-1))

taking the mean of the vector, rather than taking the mean of the sum of the two elements. The mean of the sum of the two elements will produce only the sum of the two elements, since the mean of a constant is the constant.

[Maria Amr]

Star Strider Thank you so much! Perfect!

[Star Strider]

As always, my pleasure!

Answer 2

[Maria Amr]

Yes. I saw it just now and I done. Thank you!

[Star Strider]

I very much appreciate your Accepting my Answer!