I have data from an accelerator which is quiet noisy. The manufacturer states the noise spectral density as 45 micro g /(Hz)^0.5.
How do I use this information to remove noise from the time signal. I don't have a signal processing background. So can anyone point me to a source where I can find how this problem can be handled.
Thank you all.
You can start with a low-pass filter. Ask yourself what is the maximum frequency possible (it depends of where the accelerometer is) and filter out all frequencies above that. The cut-off frequency will be a few Hz. You will need a colleague with some signal processing background to help you implement the LP filter.
What accelerometer are you using? Often digital accelerometers implementing on board a digital filters selectable with a software configuration, as for example the MPU-6000 or MPU--6050 of the INVESENSE.
Tell me the sampling rate and I will design a LP filter for you.
Thanks Fernando, my sampling rate is 1000 Hz. You are suggesting to use a low pass filter. But what if my noise is randomly distributed over all the frequencies?
This is a filter I found at a blog. But it doesn't seem to work.
def lpf(x, Fc, Fs, x0 = None):
alpha = 1 - exp(-2.0 * pi * Fc / Fs)
y = zeros_like(x)
yk = x[0] if x0 is None else x0
for k in range(len(y)):
yk += alpha * (x[k]-yk)
y[k] = yk
return y
Hi Guys
Finally found an answer.I had to use spectral subtraction method. The matlab script I used could be found here.
https://nl.mathworks.com/matlabcentral/fileexchange/7675-boll-spectral-subtraction
Dear Nischal Sehrawat
I don't know what frequencies are of interest and I don't know why you sample so fast. I will assume you know what you are doing and you need to sample at 1000 Hz.
I assume you have MATLAB. If you do, here is the design for your filter (using sampling frequency Fs=1000 Hz, cut-off frequency f0=100 Hz, and using a second order Butterworth filter, with pre-warping). The design is as follows:
clear;
clf;
f0=100; %cut-off frequency
w0=2*pi*f0;
N=512;
Fs=1000; % sampling frequency
[NUMs,DENs]=butter(2,w0,'s'); %Butterworth order 2.
[NUMdp,DENdp] = bilinear(NUMs,DENs,Fs,f0) %with prewarping
The result of that is
NUMdp = 0.0675 0.1349 0.0675
DENdp = 1.0000 -1.1430 0.4128
This will produce the following filter:
H(z) = Y(z)/X(z) = (0.0675 z^2 + 0.1349 z 0.0675) / (1.0000 z^2 - 1.1430 z 0.4128)
which can be implemented as this:
y(n)= 1.1430 y(n-1) - 0.4128 y(n-2) + 0.0675 x(n) + 0.1349 x(n-1) + 0.0675 x(n-2)
You can experiment with different cut-off frequencies or with higher orders. For example, for order 4 you change
[NUMs,DENs]=butter(2,w0,'s'); %Butterworth order 2.
to
[NUMs,DENs]=butter(4,w0,'s'); %Butterworth order 4.
Have fun!
Fernando
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