Ton Van Den Bogert

07-26-1999, 03:35 AM

Dan Major wrote:

> >From the replies I received, no one could tell me exactly what the effect

> of a rolling average would be, except that it is a low-pass filter, and the

> mathematical formula that describes it is called an FIR (Finite Impulse

It is not too hard to determine the transfer function in the frequency

domain of such a filter. In fact, I used to demonstrate this for my

undergraduate biomechanics class. If you apply a 3-point moving average

to a signal x(t) = sin(wt), sampled at intervals of T, you get as output:

y(t) = [x(t-T) + x(t) + x(t+T)]/3

After some expansions of the sin() terms, you get:

y(t) = [1 + 2*cos(wT)]*sin(wt)/3

This is the input signal sin(wt) attenuated by a factor:

H(w) = [1 + 2*cos(wT)]/3 (1)

This is the transfer function in the frequency domain. The cut-off

frequency w0 of a filter is defined as the frequency where the transfer

is exactly sqrt(2)/2. For this filter, solve the equation:

[1+2*cos(w0*T)]/3 = sqrt(2)/2

To find: w0 = 0.976/T

The equations become longer with more points averaged, but the procedure

is still the same.

Moving average filters have a transfer function that does not decrease

monotonically to zero as frequency increases. There are secondary peaks

(see equation 1). This is less of a problem, I think, when more points

are averaged.

Ton van den Bogert

--

A.J. (Ton) van den Bogert, PhD

Department of Biomedical Engineering

Cleveland Clinic Foundation

9500 Euclid Avenue (ND-20)

Cleveland, OH 44195, USA

Phone/Fax: (216) 444-5566/9198

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> >From the replies I received, no one could tell me exactly what the effect

> of a rolling average would be, except that it is a low-pass filter, and the

> mathematical formula that describes it is called an FIR (Finite Impulse

It is not too hard to determine the transfer function in the frequency

domain of such a filter. In fact, I used to demonstrate this for my

undergraduate biomechanics class. If you apply a 3-point moving average

to a signal x(t) = sin(wt), sampled at intervals of T, you get as output:

y(t) = [x(t-T) + x(t) + x(t+T)]/3

After some expansions of the sin() terms, you get:

y(t) = [1 + 2*cos(wT)]*sin(wt)/3

This is the input signal sin(wt) attenuated by a factor:

H(w) = [1 + 2*cos(wT)]/3 (1)

This is the transfer function in the frequency domain. The cut-off

frequency w0 of a filter is defined as the frequency where the transfer

is exactly sqrt(2)/2. For this filter, solve the equation:

[1+2*cos(w0*T)]/3 = sqrt(2)/2

To find: w0 = 0.976/T

The equations become longer with more points averaged, but the procedure

is still the same.

Moving average filters have a transfer function that does not decrease

monotonically to zero as frequency increases. There are secondary peaks

(see equation 1). This is less of a problem, I think, when more points

are averaged.

Ton van den Bogert

--

A.J. (Ton) van den Bogert, PhD

Department of Biomedical Engineering

Cleveland Clinic Foundation

9500 Euclid Avenue (ND-20)

Cleveland, OH 44195, USA

Phone/Fax: (216) 444-5566/9198

---------------------------------------------------------------

To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl

For information and archives: http://isb.ri.ccf.org/biomch-l

---------------------------------------------------------------