LabVIEW
identify the disturbances in a signal
Dear rahulmore,
You have first to find out the frequency range of the minor disturbances and then create a Filter with his range to filter out the disturbances.
For generating a filter you can use the Filter.vi's located in the signal processing Palette:
Regards,
Oleg Scherling, M.Eng | Applications Engineering | National Instruments | NIG |
Oleg,
Thanks for the suggestion. It works fine but the waveform profile of the signal is still observable.
I tried to work with Bandpass filter in the range of 49 - 51 Hz with IIR, It gives somewhat good results with smooth signal but it takes some time. Below are the snapshots of the option I tried.
I will like to get the good signa filtered out instantly at the start. Below are the snapshots of the option I tried. Can you suggest some other option?
- Rahul
Lynn,
Yes, you are right, the negative spikes are the real aprt of signal.
I am attaching the image mentioning the real signal and disturbance in it.
The negative spike is the only way through which I can analyze and deteremine the disturbances in signal.
- Rahul
Rahul,
I have looked at your data and have several observations and suggestions.
1. I doubt you will be able to use a filter to eliminate the power line frequency components. I will discuss this in more detail below.
2. A baseline correction process may work better.
3. Determining what to use as the "baseline" may not be trivial.
4. After correction or adjustment, automatically locating and analyzing the disturbances may depend on how much you know about the drilling process.
Filter discussion: The desired signal is at about 180 Hz with harmonics out to about 5 kHz at -50 dB. The undesired power line frequency-linked interference has components at 50, 100, 200, out to about 1000 Hz before they drop into the background noise. The overlap makes any simple filter unsuited for removal of the power line frequency related components. In this image the large peak is the desired signal (fundamental component). Peaks at 50, 100, 200, and 300 Hz are all likely linked to the power line frequency.
Another factor is that all filters have transient effects which cannot easily be removed. This image shows the result of a 2nd order Butterworth highpass filter with a 125 Hz cutoff frequency. The transient effect is easily seen as the curve of the top of the envelope. The transient has not completely decayed at the end of the 2 second data segment. And this filter has minimal effect on the power line frequency component of the baseline. Higher order filters have worse transients.
I extracted a baseline on a segment by segment basis. The segments were defined as starting 50 samples after the signal rose above -0.5 until 50 samples before it dropped below -0.5 the next time. The "real signal" part was replaced by the mean of the baseline segment preceeding it. As you can see the replacement at 0.018 to 0.02 in this image is not very close. A refinement might take the average of the baseline segments before and after, but that would add considerable complexity to the VI. The spectrum of the adjusted baseline still has strong components at 50, 100, and 200 Hz but the component at the 185 Hz signal frequency is barely detectable.
Subtracting the adjusted baseline from the original data produces these time and frequency domain images:
Note that the power line frequency components are now lost in the nosie while the main signal components are nearly unchanged.
I have not yet looked at how to detect the distrubed signals, although a threshold comparison may be worth investigating. Note that the baseline adjust process generates arrays containing indexes of the beginning and ending of the spikes. You may be able to use this to find spikes which do not go as negative as others such as the one at 0.133 s. A threshold at -5.5 to -6.0 would find this one.
I have attached my VI. Note that I removed most of your code except for the read file portion. I did not document how things work. If you cannot figure out why I did something, ask. The >0? comparison in the left for loop skips a segment which is probably due to some noise peak which exceeded the -0.5 threshold (which was entirely arbitrary).
Lynn
