Filtering out spikes in data at n*60 +1

Hello,

I have been working on a spectogram figure for some time now and have been unsuccessful in eliminating spikes in the data that occur at every 60 plus 1 i.e. 61, 121,181, ... etc. I have tried treating my data with a notch filter, filtfilt, medfilt1, and sgolay filt. In the past, I tried whitening it and using an fft to no advantage. Ideally, I would like to use something like butter or fir1 but this is a recorded data set with 64 channels and not a set frequency. What would be the best way to go about filtering this? Aside from the sgolay figure, the results are identical.

Sgolay: third order, framelength of 61

All of the other filters:

Below are the notch filters:

%% Notch Filter: Did not work

% method 1:

wo = 61/(30000/2); %removes 60Hz tome from a signal at 30,000 Hz

bw = wo/35; % Q factor of 35, this can be modified for varied results

[b,a] = iirnotch(wo,bw); % output's num and dom but we are not workign with a function to which they can be applied

% method 2: design butterworth notch filter

% d = designfilt('bandstopiir','FilterOrder',2, ...

% 'HalfPowerFrequency1',60,'HalfPowerFrequency2',62, ...

% 'DesignMethod','butter','SampleRate',samplingRate);

% dataFilt = filtfilt(d,lfp.data);

dataFilt = filtfilt(b,a,lfp.data);

Below are the other filters:

%% Median Filter: did not work

dataFilt = medfilt1(lfp.data,3);

%% Standard Filter: requires the inpouts of numerator and denominator functions 'filter(n,d,X)' where X is the data set, we are not trying to match this data to a filter

% did not work

dataFilt = filter(61,1,lfp.data); % the 1/1 is just a test

%% Savitzky - Golay Filter:

order = 3;

framelength = 61;

dataFilt = sgolayfilt(lfp.data, order, framelength);

%% FIR - Bandpass filter

dataFilt = fir1(lfp.data,60);

Thank you for taking a look at this. There must be a simple way to accomplish this.

Answer 1

[Image Analyst]

If you know the indexes, how about just setting them to zero, or the median of neighboring values.

smoothy = movmedian(y, 5); % Smooth it
index = find(x == 181);
y(index) = smoothy(index); % Replace with median.

[John D'Errico]

Exactly. If you know where the outliers live, then don't use a general filter that will filter the entire set. Just use whatever tool you want, localized at exactly those point. So replace a known outlier at a point with the local mean, or the local median, in both cases, where the known bad point is excluded.

[Emma Walker]

Thank you very much. This would have been a better approach. What do you mean by 'x' here instead of just having the index within the parentheses?

[Emma Walker]

Nevermind, this worked great!