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:

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


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

Hugger

Quote:

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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:

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


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

Hugger

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Please explain you notation, as it is that is meaningless

CB

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

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

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A moving window smoother would be something like:

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

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

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

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

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

CB