# digital filter

Discussion in 'Embedded' started by Doug Dotson, Oct 1, 2003.

1. ### Doug DotsonGuest

Hi Group,

About a year and a half ago someone (several someones actually) replied to a
request for a simple digital low pass filter. I boiled it down to a single
line of C, tested it and it worked great. Since then I have lost it. Used it
to filter A/D readings. Anybody have this design
handy?

Thanks!
Doug

Doug Dotson, Oct 1, 2003

2. ### CBFalconerGuest

Try executing, at some interval,

filtered = (filtered + newsample) / 2.

CBFalconer, Oct 1, 2003

3. ### Kelly HallGuest

I like having a time constant:
filtered = (0.9 * filtered + 0.1 * newsample)

Play around with it, it's not hard to get useful filtering.

Kelly

Kelly Hall, Oct 1, 2003
4. ### Ville VoipioGuest

A very slight modification of this may be useful in some cases:

The equation above is in the form:

filtered = (1-k) * filtered + k * newsample

This can be written:

filtered = filtered + k * (newsample - filtered)

Mathematically, these are identical. The second one does have some

1. with small k (long decay time) and limited calculation
precision the round-off errors are smaller
2. with non-constant k there is no need for 1-k

Computationally, the relative cost of these two ways depends on

- Ville

Ville Voipio, Oct 1, 2003
5. ### Doug DotsonGuest

I believe this is the one I remember:
I picked k to be a power of 2 so that a shift instruction could be
used for speed. I try these out.

Thanks to all!
Doug

Doug Dotson, Oct 1, 2003

Yes, it's under "advanced search" on this webpage:

7. ### Doug DotsonGuest

Hummm. That's where I looked first. I try again.

Doug

Doug Dotson, Oct 1, 2003
8. ### Spehro PefhanyGuest

Spehro Pefhany, Oct 1, 2003
9. ### Doug DotsonGuest

Doug Dotson, Oct 1, 2003

Here is the very simplest digital lowpass:

Y += (X-Y)*k

X = input, Y = output

Here is how k is defined:

k = 1 - exp(-Fc/Fdis))

where Fc = cutoff frequency, Fdis = sample rate

DSP and Mixed Signal Design Consultant

http://www.abvolt.com

11. ### Doug DotsonGuest

Right! I picked n to be a power of 2 (8 actually in my application)
and in C it turns into:

y += (x - y) >> n;

Works great!

Doug

Doug Dotson, Oct 1, 2003
12. ### Dave HansenGuest

I presume you mean "y += (x - y) >> p;" where "p" is the power rather
than n (e.g., 3 rather than 8).

But beware...

There are no guarantees that right shifting a signed value will yield
the result you desire. You might get zeros shifted in regardless of
the actual sign of the value being shifted.

Most compilers are smart enough to make the "shift to divide by a
power of two" optimization on their own. The following

y += (x - y) / n;

will be guaranteed to work, and should produce nearly identical code

Regards,

-=Dave

Dave Hansen, Oct 2, 2003
13. ### CBFalconerGuest

If it doesn't, and the speed turns out to be critical, he could
use unsigned ints and a bias, such as INT_MAX. The input y would
have to first be corrected to INT_MAX + yin, where yin is int and
y is unsigned. The filtered y would need to be initialized to
INT_MAX also. This assumes that INT_MAX is roughly 1/2 of
UINT_MAX.

So the result would be, more or less:

#include <limits.h>

unsigned int avg = INT_MAX;
unsigned int sig;
int input;

#define FACTOR 3

.....

sig = (unsigned)input + INT_MAX)
avg += (sig - avg) >> FACTOR;

which also avoids triggering UB via integer overflows.

CBFalconer, Oct 2, 2003
14. ### Doug DotsonGuest

Oops! You ae correct.

I didn't include all the contect of my application. 12 bit positive values,
32 bit words.
I've been doing this programming thing for a while

Doug

Doug Dotson, Oct 2, 2003