How to apply bandpass filter to a single data type?

We have recorded signals but when i load the signal into matlab, the data type is single. I want to apply bandpass filter and notch filter to these single data type signals. I have applied both the filters to the signals after converting them to double but it alters the results. Thats why i was curious if we can use band pass and notch filters on single data types.

Answer 1

[Star Strider]

Yes. You can use single.

See the documentation section for x . This specifically refers to the Signal Processing Toolbox bandpass documentation, however it applies to all the related functions.

I do not understand how converting them to double would alter the results, however I do not have your signals to experiment with. If you are using something other than the Signal Processing Toolbox, it would be helpful to know that.

[Rida Alfonso]

I am not using signal processing toolbox, I am doing it through coding.I have two frequencies that are 20HZ-150HZ which i want to use in bandpass filter . but the function (bandpass(x1,[20 150],2048) ;) gives error with the single data type. I would have gladly shared the signal but the signal is bigger the allowed size. Sadly I dont understand the link you provided, Can you kindly explain it.

x — Input signal

vector | matrix

Input signal, specified as a vector or matrix.

Example: sin(2*pi*(0:127)/16)+randn(1,128)/100 specifies a noisy sinusoid

Example: [2 1].*sin(2*pi*(0:127)'./[16 64]) specifies a two-channel sinusoid.

Data Types: single | double

Complex Number Support: Yes

[Star Strider]

I have no idea what the problem is, or what the error is, since you did not share that.

This works when I run it:

t = linspace(0, 1, 5E+3);
x1 = randn(1, 5E+3, 'single');
y = bandpass(x1,[20 150],2048);

figure
plot(t, x1,    t, y)
grid

While you are certainly free to code your won filter functions if you want to, the Signal Processing Toolbox functions make filter design much easier.

.

[Rida Alfonso]

You are right, i am having problem in notch filter ,not the band pass. I am attaching a portion of signal. This is my code

%Desiging Notch filter

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

'HalfPowerFrequency1',48,'HalfPowerFrequency2',52, ...

'DesignMethod','butter','SampleRate',2048); % special case of bandstop filter..

Nfltroutput = filtfilt(d, x1);....And the following is the error

[Star Strider]

This works for me:

nhFilt = designfilt('bandstopiir','StopbandFrequency1',48,'StopbandFrequency2',52, 'PassbandFrequency1',46, 'PassbandFrequency2',54,'DesignMethod','ellip', 'SampleRate',2048);

D = load('x1.mat');
x1 = D.x1;
t = linspace(0,2048,size(x1,1))/2048;                                   % Create Time Vector
dataOut = filtfilt(nhFilt,double(x1));

figure
nsp = size(x1,2);
for k = 1:nsp
    subplot(nsp,1,k)
    plot(t, x1(:,k),    t, dataOut(:,k))
    grid
    title(sprintf('x1(:,%d)',k))
    ylabel('Amplitude')
end
xlabel('Time')

The effect of the filter is obvious, although I did not see 50 Hz noise in the original signal, so I do not understand what the filter is for. I checked the filter with freqz (that launched fvtool), and it is performing correctly.

.

[Rida Alfonso]

Thanks for your guidance. Its strange , when i extract the feature 'cardinality' from EMG data without applying bandpass and band stop filters, The features data makes sense.....But i extract the same feature from the same signal but only pre-processed with bandpass and band stop filters, The resulting data of the feature cardinality is a matrix of n by n size and the only value is 614. I dont know how to resolve this semantics error.

[Star Strider]

As always, my pleasure!

I am not at all certain that I understand what you are describing. The only mention I can find in the MATLAB documentation online about ‘cardinality’ is in the Financial Toolbox (that I do not have).

[Rida Alfonso]

I am attaching a file that would explain it.

[Star Strider]

I cannot comment on the applicability of that paper (saved to my clip file) with respect to what you are doing. I doubt that eliminating line/mains frequency interference would have any specific effect on it, however bandpass filtering the data first certainly could. I do not see that the paper mentions filtering the signal.

I was not previously aware of that paper or its approach to EMG signal classification, since I have not done any EMG work in decades. The only classifiers I ever used were LDA (and only later some nonlinear classifiers on occasion), and only then with short-time Fourier transform (STFT, such as provided in the spectrogram function), and not on the signals themselves in the time domain.

I woud apply the bandstop filter only if a Fourier transform of the original signal demonstrates significant 50 Hz energy (something that appropriate instrumentation and a reference electrode with difference amplifiers could mitigate if not eliminate). I would not use bandpass (or other frequency-selective) filters at all, unless the same Fourier transform analysis demonstrated significant high-frequency noise, above the usual spectrum of the signals you are studying, and if your signals have a significant d-c offset. Most physiological signals have bandwidths of 500 Hz or less, so a 1 kHz sampling frequency should work. A 2 kHz sampling frequency (such as the paper uses) is appropriate, however it could result in unnecessarily large signal records. That is simply a choice you will need to make.

[Rida Alfonso]

Thanks a lot for your detailed reply, i clearly understand your point, i have analysed the signal , the portion i attached is a minor portion of the signal. However there are some minor disturbances from 48-52 Hz as i have analysed the signal using signal analyser. And i am using bandpass because the useful contents of EMG signals lies between 20-100 Hz (From Litrature). I am just unsable to understand how could frequency filtering affect the data in such a way.

[Star Strider]

As always, my pleasure!

Filters by definition remove signal energy, so the waveform will change after the filter is applied. The change depends on the filter characteristics.