How to get rid of the DC-offset from a signal?

[madgreek]

Hello

i have a speech signal which i need to process. I have estimated its DC-offset (using the AC-DC estimate vi) to be approximately 126 V. I have read some posts here that some reccomend to just subtract this value from the array of data or use a filter to do it. What is the difference, which one should perform better? I have also broke down my initial array of 50000 samples into sub-arrays of 512. Where should be the best place to get rid of the DC? Once and out from the initial array or later on from the subarrays?

When someone says subtact that value (DC value) from the signal, what exactly has to be done?

When using a filter, what shall i be looking for to do?Specifications of it?

Any help is appreciated

[johnsold]

Whether to subtract the mean of the whole from the whole or the means of the pieces from the pieces depends upon how much the means of the pieces differ from the mean of the whole. If your process drifts very little over time and none of the subarrays has a DC offset which is much different from any of the others, then I would do it once at the beginning.

Just use the subtract function from the numerical palette. It is polymophic and will happily subtract a scalar from an array. The result is an array with the scalar subtracted from each element. This is so simple, I would try it first before using any filters. Given what you are trying to do in your other thread, I would avoid filters. The delays and transient effects of filters could really complicate your other processing.

Lynn

[madgreek]

Hello Lynn

So you are just saying, get the DC offset since it stays almost constant, from the AC/DC Estimate VI and just subtract it from my initial array at the beginning once and out and further on move with the rest of my process? That's it as it shows in the pic below?

[johnsold]

That is about as easy as it gets. Yes, I would try that first. If the results are not satisfactory, then go back and look at the more complex methods.

Lynn

[madgreek]

Nope it will do i think ....Thank you for your help