Electronics function help.

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March 20th 2010, 08:27 AM
Hugger

Electronics function help.

Hello,

Been out of practice so long that all is forgotten.(Headbang)
I was wondering if someone could help me understand and apply the following expression:

$V = \frac{{1/V_{tn}}}2+\int_{t0}^{t=n-1} \frac{V_{1}...V_{n}/n}2$

It was suggested to me to help smooth out a noisy voltage signal.

Hugger
March 21st 2010, 01:18 AM
CaptainBlack

Quote:

Originally Posted by Hugger

Hello,

Been out of practice so long that all is forgotten.(Headbang)
I was wondering if someone could help me understand and apply the following expression:

$V = \frac{{1/V_{tn}}}2+\int_{t0}^{t=n-1} \frac{V_{1}...V_{n}/n}2$

It was suggested to me to help smooth out a noisy voltage signal.

Hugger

Please explain you notation, as it is that is meaningless

CB
March 21st 2010, 07:07 AM
Hugger

Thanks for the response CB.

I will go back and check to see if I goofed the notation.

Again, I am trying to come up with an expression that will smooth out samples taken of voltage that is somewhat noisy by averaging a group of samples to get rid of dips and spikes.

I'll reply when I verify notation.

Hugger
March 21st 2010, 09:16 AM
CaptainBlack

Quote:

Originally Posted by Hugger

Thanks for the response CB.

I will go back and check to see if I goofed the notation.

Again, I am trying to come up with an expression that will smooth out samples taken of voltage that is somewhat noisy by averaging a group of samples to get rid of dips and spikes.

I'll reply when I verify notation.

Hugger

A moving window smoother would be something like:

$V_n=\frac{1}{m} \sum_{i=n-m+1}^n v_i$

where the $v_i$ are the input samples and the $V_n$ are the smoothed outputs.

Though this introduces a lag of $(m-1)/2$ samples. So if m is odd, then:

$V_{n-(m-1)/2}=\frac{1}{m} \sum_{i=n-m+1}^n v_i$

is a moving average smoother with no lag (though there is now latency of $(m-1)/2$ samples).

CB