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I built a noise cancellation setup with two microphones and two different microphone preamplifiers that go to two different channels of a stereo recording.
Here is a sample
http://filestore.to/?d=U5FN2IH96K
I tried
char ergebnis[80];
sprintf(ergebnis, "%s.neu.raw", Datei);
FILE* ausgabe = fopen(ergebnis, "wb");
FILE* f = fopen(Datei, "rb");
if (f == NULL)
{
return;
}
int i = -1;
int r1 = 0;
int r2 = 0;
int l1 = 0;
int l2 = 0;
int l = 0;
int r = 0;
int wo = 0;
int dif = 0;
while (wo != EOF)
{
wo = getc(f);
i++;
if (i == 0)
{
r1 = (unsigned)wo;
}
if (i == 1)
{
r2 = (unsigned)wo;
r = (r2 << 8) + r1; //r1 | r2 << 8;
}
if (i == 2)
{
l1 = (unsigned)wo;
}
if (i == 3)
{
l2 = (unsigned)wo;
l = (l2 << 8) + l1; //l1 | l2 << 8;
dif = r - (l * 2);
putc((char)( (unsigned)dif & 0xff), ausgabe);
putc((char)(((unsigned)dif >> 8) & 0xff), ausgabe);
i = -1;
}
}
when the magic happens in
dif = r - (l * 2);
But this does not eliminate the noise surrounding it, all it does is create crackling sounds.
How could I approach this task with my setup instead? I prefer practical solutions over "read this paper only the author of the paper understands".
While we are at it, how do I normalize the final mono output to make it as loud as possible without clipping?
I don't know why you would expect this
dif = r - (l * 2);
to cancel noise, but I can tell you why it "create[s] crackling sounds". The value in dif is often going to be out of range of 16-bit audio. When this happens, your simple conversion function:
putc((char)( (unsigned)dif & 0xff), ausgabe);
putc((char)(((unsigned)dif >> 8) & 0xff), ausgabe);
will fail. Instead of a smooth curve, your audio will jump from large positive to large negative values. If that confuses you, maybe this post will help.
Even if you solve that problem, a few things aren't clear, not the least of which is that for active noise canceling to work you usually assume that one mike provides a source of noise and the other provides signal + noise. Which is which in this case? Did you just place two mikes next to each other and hope to hear some sound source with less ambient noise after some simple arithmetic? That won't work, since they are both hearing different combinations of signal and noise (not just in amplitude, but time as well). So you need to answer 1. which mike is the source of signal and which is the source of noise? 2. what kind of noise are you trying to cancel? 3. what distinguishes the mikes in their ability to hear signal and noise? 4. etc.
Update: I'm still not clear on your setup, but here's something that might help:
You might have a setup where your signal is strong in one mike and weak in the other, and a noise is applied to both mikes. In all likelyhood, there will be signal leak into both mikes. Nevertheless, we will assume
l = noise1
r = signal + noise2
Note that I have not assumed the same noise values for l and r, this reflects the reality that the two mikes will be picking up different noise values due to time delays and other factors. However, it is often the case (and may or may not be the case in your setup) that noise1 and noise2 are correlated at low frequencies. Thus, if we have a low pass filter, lp, we can achieve some noise reduction in the low frequencies as follows:
out = r - lp(l) = signal + noise2 - lp(noise1)
This, of course, assumes that the noise level at l and r is the same, which it may or may not be, depending on your setup. You may want to leave a manual parameter for this purpose for manual tuning at the end:
out = r - g*lp(l)
where g is your tuning parameter and close to 1. I believe in some high-end noise reduction systems, g is constantly tuned automatically.
Selecting a cutoff frequency for your lp filter is all that remains. An approximation you could use is that the highest frequency you can cancel has a wavelength equal to 1/4 the distance between the mikes. Of course, I'm REALLY waving my arms with that, because it depends a lot on where the sound is coming from, how directional your mikes are and so on, but it's a starting point.
Sample calculation for mikes that are 3 inches apart:
Speed of sound = 13 397 inches / sec
desired wavelength = 4*3 inches = 12 inches
frequency = 13,397 / 12 = 1116 Hz
So your filter should have a cutoff frequency of 1116 Hz if the mikes are 3 inches apart.
Expect this setup to cancel a significant amount of your signal below the cutoff frequency as well, if there is bleed.
Related
I am working on embedded programming with written code by other people.
this algorithm be used in calculate average for mic and accelerometer
sound_value_Avg = 0;
sound_value = 0;
memset((char *)soundRaw, 0x00, SOUND_COUNT*2);
for(int i2=0; i2 < SOUND_COUNT; i2++)
{
soundRaw[i2] = analogRead(PIN_ANALOG_IN);
if (i2 == 0)
{
sound_value_Avg = soundRaw[i2];
}
else
{
sound_value_Avg = (sound_value_Avg + soundRaw[i2]) / 2;
}
}
sound_value = sound_value_Avg;
acceleromter is similar to this
n1=p1
(n2+p1)/2 = p2
(n3+p2)/2 = p3
(n4+p3)/2 = p4
...
avg(n1~nx)=px
it not seems to be correct.
can someone explain why he used this algorithm?
is it specific way for sin graph? like noise, vibration?
It appears to be a flawed attempt at maintaining a cumulative mean. The error is in believing that:
An+1 = (An + sn) / 2
when in fact it should be:
An+1 = ((An * n) + s) / (n + 1)
However it is computationally simpler to maintain a running sum and generate an average in the usual manner:
S = S + s
An = S / n
It is possible that the intent was to avoid overflow when the sum grows large, but the attempt is mathematically flawed.
To see how wrong this statement is consider:
True
n s Running Avg. (An + sn) / 2
--------------------------------------
1 20 20 20
2 21 20.5 20.25
3 22 21 20.625
In this case however, nothing is done with the intermediate mean value, so you don'e in fact need to maintain a running mean at all. You simply need to accumulate a running sum and calculate the average at the end. For example:
sum = 0 ;
sound_value = 0 ;
for( int i2 = 0; i2 < SOUND_COUNT; i2++ )
{
soundRaw[i2] = analogRead( PIN_ANALOG_IN ) ;
sum += soundRaw[i2] ;
}
sound_value = sum / SOUND_COUNT ;
In this you do need to make sure that the data type forsum can accommodate a value of the maximum analogRead() return multiplied by SOUND_COUNT.
However you say that this is used for some sort of signal conditioning or processing of both a microphone and an accelerator. These devices have rather dissimilar bandwidth and dynamics, and it seems rather unlikely that the same filter would suit both. Applying robust DSP techniques such as IIR or FIR filters with suitably calculated coefficients would make a great deal more sense. You'd also need a suitable fixed sample rate that I am willing to bet is not achieved by simply reading the ADC in a loop with no specific timing
A microcontroller has the job to sample ADC Values (Analog to Digital Conversion). Since these parts are affected by tolerance and noise, the accuracy can be significantly increased by deleting the 4 worst values. The find and delete does take time, which is not ideal, since it will increase the cycle time.
Imagine a frequency of 100MHz, so each command of software does take 10ns to process, the more commands, the longer the controller is blocked from doing the next set of samples
So my goal is to do the sorting process as fast as possible for this i currently use this code, but this does only delete the two worst!
uint16_t getValue(void){
adcval[8] = {};
uint16_t min = 16383 //14bit full
uint16_t max = 1; //zero is physically almost impossible!
uint32_t sum = 0; //variable for the summing
for(uint8_t i=0; i<8;i++){
if(adc[i] > max) max = adc[i];
if(adc[i] < min) min = adc[i];
sum=sum+adcval[i];
}
uint16_t result = (sum-max-min)/6; //remove two worst and divide by 6
return result;
}
Now I would like to extend this function to delete the 4 worst values out of the 8 samples to get more precision. Any advice on how to do this?
Additionally, it would be wonderful to build an efficient function that finds the most deviating values, instead of the highest and lowest. For example, imagine the this two arrays
uint16_t adc1[8] {5,6,10,11,11,12,20,22};
uint16_t adc2[8] {5,6,7,7,10,11,15,16};
First case would gain precision by the described mechanism (delete the 4 worst). But the second case would have deleted the values 5 and 6 as well as 15 and 16. But this would theoretically make the calculation worse, since deleting 10,11,15,16 would be better. Is there any fast solution of deleting the 4 most deviating?
If your ADC is returning values from 5 to 16 14 bits and the voltage reference 3.3V, the voltage varies from 1mV to 3mV. It is very likely that it is the correct reading. It is very difficult to design good input circuit for 14 bits ADC.
It is better to run the running average. What is the running average? It is software low pass filter.
Blue are readings from the ADC, red -running average
Second signal is the very low amplitude sine wave (9-27mV - assuming 14 bits and 3.3Vref)
The algorithm:
static int average;
int running_average(int val, int level)
{
average -= average / level;
average += val * level;
return average / level;
}
void init_average(int val, int level)
{
average = val * level;
}
if the level is the power of 2. This version needs only 6 instructions (no branches) to calculate the average.
static int average;
int running_average(int val, int level)
{
average -= average >> level;
average += val << level;
return average >> level;
}
void init_average(int val, int level)
{
average = val << level;
}
I assume that average will no overflow. If yes you need to chose larger type
This answer is kinda of topic as it recommends a hardware solution but if performance is required and the MCU can't implement P__J__'s solution than this is your next best thing.
It seems you want to remove noise from your input signal. This can be done in software using DSP (digital signal processing) but it can also be done by configuring your hardware differently.
By adding the proper filter at the proper space before your ADC, it will be possible to remove much (outside) noise from your ADC output. (you can't of course go below a certain amount that is innate in the ADC but alas.)
There are several q&a on electronics.stackexchange.com.
One solution is adding a capacitor to filter some high frequency noise. As noted by DerStorm8
The Photon has another great solution here by suggesting RC, Sallen-Key and a cascade of Sallen-Key filters for a continuous signal filter.
Here (ADN007) is a Analog Design Note from Microchip on "Techniques that Reduce System Noise in ADC Circuits"
It may seem that designing a low noise, 12-bit Analog-to-Digital
Converter (ADC) board or even a 10-bit board is easy. This is
true, unless one ignores the basics of low noise design. For
instance, one would think that most amplifiers and resistors work
effectively in 12-bit or 10-bit environments. However, poor device
selection becomes a major factor in the success or failure of the
circuit. Another, often ignored, area that contributes a great deal
of noise, is conducted noise. Conducted noise is already in the
circuit board by the time the signal arrives at the input of the
ADC. The most effective way to remove this noise is by using a
low-pass (anti-aliasing) filter prior to the ADC. Including by-pass
capacitors and using a ground plane will also eliminate this type
of noise. A third source of noise is radiated noise. The major
sources of this type of noise are Electromagnetic Interference
(EMI) or capacitive coupling of signals from trace-to-trace.
If all three of these issues are addressed, then it is true that
designing a low noise 12-bit ADC board is easy.
And their recommended solution path:
It is easy to design a true 12-bit ADC system by using a few
key low noise guidelines. First, examine your devices (resistors
and amplifiers) to make sure they are low noise. Second, use a
ground plane whenever possible. Third, include a low-pass filter
in the signal path if you are changing the signal from analog to
digital. Finally, and always, include by-pass capacitors. These
capacitors not only remove noise but also foster circuit stability.
Here is a good paper by Analog Devices on input noise. They note in here that "there are some instances where input noise can actually be helpful in achieving higher resolution."
All analog-to-digital converters (ADCs) have a certain amount of input-referred noise—modeled as a noise source connected in series with the input of a noise-free ADC. Input-referred noise is not to be confused with quantization noise, which is only of interest when an ADC is processing time-varying signals. In most cases, less input noise is better; however, there are some instances where input noise can actually be helpful in achieving higher resolution. If this doesn’t seem to make sense right now, read on to find out how some noise can be good noise.
Given that you have a fixed size array, a hard-coded sorting network should be able to correctly sort the entire array with only 19 comparisons. Currently you have 8+2*8=24 comparisons already, although it is possible that the compiler unrolls the loop, leaving you with 16 comparisons. It is conceivable that, depending on the microcontroller hardware, a sorting network can be implemented with some degree of parallelism -- perhaps you also have to query the adc values sequentially which would give you opportunity to pre-sort them, while waiting for the comparison.
An optimal sorting network should be searchable online. Wikipedia has some pointers.
So, you would end up with some code like this:
sort_data(adcval);
return (adcval[2]+adcval[3]+adcval[4]+adcval[5])/4;
Update:
As you can take from this picture (source) of optimal sorting networks, a complete sort takes 19 comparisons. However 3 of those are not strictly needed if you only want to extract the middle 4 values. So you get down to 16 comparisons.
to delete the 4 worst values out of the 8 samples
The methods are described on geeksforgeeks k largest(or smallest) elements in an array and you can implement the best method that suits you.
I decided to use this good site to generate best sorting algorithm with SWAP() macros needed to sort the array of 8 elements. Then I created a small C program that will test any combination of 8 element array on my sorting function. Then, because we only care of groups of 4 elements, I did something bruteforce - for each of the SWAP() macros I tried to comment the macro and see if the program still succeeds. I could comment 5 SWAP macros, leaving 14 comparisons needed to identify the smallest 4 elements in the array of 8 samples.
/**
* Sorts the array, but only so that groups of 4 matter.
* So group of 4 smallest elements and 4 biggest elements
* will be sorted ok.
* s[0]...s[3] will have lowest 4 elements
* so they have to be "deleted"
* s[4]...s[7] will have the highest 4 values
*/
void sort_but_4_matter(int s[8]) {
#define SWAP(x, y) do { \
if (s[x] > s[y]) { \
const int t = s[x]; \
s[x] = s[y]; \
s[y] = t; \
} \
} while(0)
SWAP(0, 1);
//SWAP(2, 3);
SWAP(0, 2);
//SWAP(1, 3);
//SWAP(1, 2);
SWAP(4, 5);
SWAP(6, 7);
SWAP(4, 6);
SWAP(5, 7);
//SWAP(5, 6);
SWAP(0, 4);
SWAP(1, 5);
SWAP(1, 4);
SWAP(2, 6);
SWAP(3, 7);
//SWAP(3, 6);
SWAP(2, 4);
SWAP(3, 5);
SWAP(3, 4);
#undef SWAP
}
/* -------- testing code */
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
int cmp_int(const void *a, const void *b) {
return *(const int*)a - *(const int*)b;
}
void printit_arr(const int *arr, size_t n) {
printf("{");
for (size_t i = 0; i < n; ++i) {
printf("%d", arr[i]);
if (i != n - 1) {
printf(" ");
}
}
printf("}");
}
void printit(const char *pre, const int arr[8],
const int in[8], const int res[4]) {
printf("%s: ", pre);
printit_arr(arr, 8);
printf(" ");
printit_arr(in, 8);
printf(" ");
printit_arr(res, 4);
printf("\n");
}
int err = 0;
void test(const int arr[8], const int res[4]) {
int in[8];
memcpy(in, arr, sizeof(int) * 8);
sort_but_4_matter(in);
// sort for memcmp below
qsort(in, 4, sizeof(int), cmp_int);
if (memcmp(in, res, sizeof(int) * 4) != 0) {
printit("T", arr, in, res);
err = 1;
}
}
void test_all_combinations() {
const int result[4] = { 0, 1, 2, 3 }; // sorted
const size_t n = 8;
int num[8] = { 0, 1, 2, 3, 4, 5, 6, 7 };
for (size_t j = 0; j < n; j++) {
for (size_t i = 0; i < n-1; i++) {
int temp = num[i];
num[i] = num[i+1];
num[i+1] = temp;
test(num, result);
}
}
}
int main() {
test_all_combinations();
return err;
}
Tested on godbolt. The sort_but_4_matter with gcc -O2 on x86_64 compiles to less then 100 instruction.
I have implemented simple PI controller, code is as follows:
PI_controller() {
// handling input value and errors
previous_error = current_error;
current_error = 0 - input_value;
// PI regulation
P = current_error //P is proportional value
I += previous_error; //I is integral value
output = Kp*P + Ki*I; //Kp and Ki are coeficients
}
Input value is always between -π and +π.
Output value must be between -4000 and +4000.
My question is - how to configure and (most importantly) limit the PI controller properly.
Too much to comment but not a definitive answer. What is "a simple PI controller"? And "how long is a piece of string"? I don't see why you (effectively) code
P = (current_error = 0 - input_value);
which simply negates the error of -π to π. You then aggregate the error with
I += previous_error;
but haven't stated the cumulative error bounds, and then calculate
output = Kp*P + Ki*I;
which must be -4000 <= output <= 4000. So you are looking for values of Kp and Ki that keep you within bounds, or perhaps don't keep you within bounds except in average conditions.
I suggest an empirical solution. Try a series of runs, filing the results, stepping the values of Kp and Ki by 5 steps each, first from extreme neg to pos values. Limit the output as you stated, counting the number of results that break the limit.
Next, halve the range of one of Kp and Ki and make a further informed choice as to which one to limit. And so on. "Divide and conquer".
As to your requirement "how to limit the PI controller properly", are you sure that 4000 is the limit and not 4096 or even 4095?
if (output < -4000) output = -4000;
if (output > 4000) output = 4000;
To configure your Kp and Ki you really should analyze the frequency response of your system and design your PI to give the desired response. To simply limit the output decide if you need to freeze the integrator, or just limit the immediate output. I'd recommend freezing the integrator.
I_tmp = previous_error + I;
output_tmp = Kp*P + Ki*I_tmp;
if( output_tmp < -4000 )
{
output = -4000;
}
else if( output_tmp > 4000 )
{
output = 4000;
}
else
{
I = I_tmp;
output = output_tmp;
}
That's not a super elegant, vetted algorithm, but it gives you an idea.
If I understand your question correctly you are asking about anti windup for your integrator.
There are more clever ways to to it, but a simple
if ( abs (I) < x)
{
I += previous_error;
}
will prevent windup of the integrator.
Then you need to figure out x, Kp and Ki so that abs(x*Ki) + abs(3.14*Kp) < 4000
[edit] Off cause as macduff states, you first need to analyse your system and choose the korrect Ki and Kp, x is the only really free variable in the above equation.
I have the following setup https://sketchfab.com/show/7e2912f5f8794a7b96ef3ac5930e090a (It's a 3d viewer, use your mouse to view all angles)
The box has two nondirectional electret microphones(black dots). On the ground there are some elements falling down like water or similar(symbolized by the sphere) and creating noises. On top, someone is speaking in the box. Distances are roughly accurate, so the mouth is pretty close.
Inside the box there are two different amplifiers(but the same electret microphones) with two different amplification circuits(the mouth-one is louder in general and has some normalization circuitry integrated. Long story short, I can record this into a raw audio file with 44100 Hz, 16Bit and Stereo, while the left channel is the upper, the right channel is the lower microphone amplifier output.
Goal is to - even though the electret microphones are not directed and even though there are different amplifiers - subtract the lower microphone(facing the ground) from the upper microphone(facing the speaker) to have noise cancellation.
I tried(With Datei being the raw-filename). This includes a high or low pass filter and a routine to put the final result back into a raw mono file (%s.neu.raw)
The problem is - well - undefinable distortion. I can hear my voice but it's not bearable at all. If you need a sample I can upload one.
EDIT: New code.
static void *substractf( char *Datei)
{
char ergebnis[80];
sprintf(ergebnis,"%s.neu.raw",Datei);
FILE* ausgabe = fopen(ergebnis, "wb");
FILE* f = fopen(Datei, "rb");
if (f == NULL)
return;
double g = 0.1;
double RC = 1.0/(1215*2*3.14);
double dt = 1.0/44100;
double alpha = dt/(RC+dt);
double noise_gain = 18.0;
double voice_gain = 1.0;
struct {
uint8_t noise_lsb;
int8_t noise_msb;
uint8_t voice_lsb;
int8_t voice_msb;
} sample;
while (fread(&sample, sizeof sample, 1, f) == 1)
{
int16_t noise_source = sample.noise_msb * 256 + sample.noise_lsb;
int16_t voice_source = sample.voice_msb * 256 + sample.voice_lsb;
double signal, difference_voice_noise;
difference_voice_noise = voice_gain*voice_source - noise_gain*noise_source;
signal = (1.0 - alpha)*signal + alpha*difference_voice_noise;
putc((char) ( (signed)signal & 0xff),ausgabe);
putc((char) (((signed)signal >> 8) & 0xff),ausgabe);
}
fclose(f);
fclose(ausgabe);
char output[300];
sprintf(output,"rm -frv \"%s\"",Datei);
system(output);
}
Your code doesn't take differences of path length into consideration.
The path difference d2 – d1 between the sound source and the two mics corresponds to a time delay of (d2 – d1) / v, where v is the speed of sound (330 m/s).
Suppose d2 – d1 is equal to 10 cm. In this case, any sound wave whose frequency is a multiple of 3300 Hz (i.e., whose period is a multiple of (0.10/330) seconds) will be at exactly the same phase at both microphones. This is how you want things to be at all frequencies.
However, a sound wave at an odd multiple of half that frequency (1650 Hz, 4950 Hz, 8250 Hz, etc.) will have changed in phase by 180° by the time it reaches the second mic. As a result, your subtraction operation will actually have the opposite effect — you'll be boosting these frequencies instead of making them quieter.
The end result will be similar to what you get if you push all the alternate sliders on a graphic equaliser in opposite directions. This is probably what you're experiencing now.
Try estimating the length of this path difference and delaying the samples in one channel by a corresponding amount. At a sampling rate of 44100 Hz, one centimetre corresponds to about 0.75 samples. If the sound source is moving around, then things get a bit complicated. You'll have to find a way of estimating the path difference dynamically from the audio signals themselves.
Ideas too big for a comment.
1) Looks like OP is filtering the l signal jetzt = vorher + (alpha*(l - vorher)) and then subtracting the r with dif = r - g*jetzt. It seems to make more sense to subtract l and r first and apply that difference to the filter.
float signal = 0.0; (outside loop)
...
float dif;
// Differential (with gain adjustments)
dif = gain_l*l - gain_r*r;
// Low pass filter (I may have this backwards)
signal = (1.0 - alpha)*signal + alpha*dif;
// I am not certain if diff or signal should be written
// but testing limit would be useful.
if ((dif > 32767) || (dif < -32767)) report();
int16_t sig = dif;
// I see no reason for the following test
// if (dif != 0)
putc((char) ( (unsigned)dif & 0xff),ausgabe);
putc((char) (((unsigned)dif >> 8) & 0xff),ausgabe);
2) The byte splicing may be off. Suggested simplification
// This assumes incoming data is little endian,
// Maybe data is in big endian and _that_ is OP problem?
struct {
uint8_t l_lsb;
int8_t l_msb;
uint8_t r_lsb;
int8_t r_msb;
} sample;
...
while (fread(&sample, sizeof sample, 1, f) == 1) {
int16_t left = sample.l_msb * 256 + sample.l_lsb;
int16_t right = sample.r_msb * 256 + sample.r_lsb;
3) Use of float vs. double. Usually the more limiting float creates computational noise, but the magnitude of OP's complaint suggest that this issue is unlikely the problem. Still worth considering.
4) Endian of the 16-bit samples may be backwards. Further, depending on A/D encoding the samples may be 16-bit unsigned rather than 16-bit signed.
5) The phase of the 2 signals could be 180 out from each other due to wiring and mic pick-up. Is so try diff = gain_l*l + gain_r*r.
I'm working on an MC68HC11 Microcontroller and have an analogue voltage signal going in that I have sampled. The scenario is a weighing machine, the large peaks are when the object hits the sensor and then it stabilises (which are the samples I want) and then peaks again before the object roles off.
The problem I'm having is figuring out a way for the program to detect this stable point and average it to produce an overall weight but can't figure out how :/. One way I have thought about doing is comparing previous values to see if there is not a large difference between them but I haven't had any success. Below is the C code that I am using:
#include <stdio.h>
#include <stdarg.h>
#include <iof1.h>
void main(void)
{
/* PORTA, DDRA, DDRG etc... are LEDs and switch ports */
unsigned char *paddr, *adctl, *adr1;
unsigned short i = 0;
unsigned short k = 0;
unsigned char switched = 1; /* is char the smallest data type? */
unsigned char data[2000];
DDRA = 0x00; /* All in */
DDRG = 0xff;
adctl = (unsigned char*) 0x30;
adr1 = (unsigned char*) 0x31;
*adctl = 0x20; /* single continuos scan */
while(1)
{
if(*adr1 > 40)
{
if(PORTA == 128) /* Debugging switch */
{
PORTG = 1;
}
else
{
PORTG = 0;
}
if(i < 2000)
{
while(((*adctl) & 0x80) == 0x00);
{
data[i] = *adr1;
}
/* if(i > 10 && (data[(i-10)] - data[i]) < 20) */
i++;
}
if(PORTA == switched)
{
PORTG = 31;
/* Print a delimeter so teemtalk can send to excel */
for(k=0;k<2000;k++)
{
printf("%d,",data[k]);
}
if(switched == 1) /*bitwise manipulation more efficient? */
{
switched = 0;
}
else
{
switched = 1;
}
PORTG = 0;
}
if(i >= 2000)
{
i = 0;
}
}
}
}
Look forward to hearing any suggestions :)
(The graph below shows how these values look, the red box is the area I would like to identify.
As you sample sequence has glitches (short lived transients) try to improve the hardware ie change layout, add decoupling, add filtering etc.
If that approach fails, then a median filter [1] of say five places long, which takes the last five samples, sorts them and outputs the middle one, so two samples of the transient have no effect on it's output. (seven places ...three transient)
Then a computationally efficient exponential averaging lowpass filter [2]
y(n) = y(n–1) + alpha[x(n) – y(n–1)]
choosing alpha (1/2^n, division with right shifts) to yield a time constant [3] of less than the underlying response (~50samples), but still filter out the noise. Increasing the effective fractional bits will avoid the quantizing issues.
With this improved sample sequence, thresholds and cycle count, can be applied to detect quiescent durations.
Additionally if the end of the quiescent period is always followed by a large, abrupt change then using a sample delay "array", enables the detection of the abrupt change but still have available the last of the quiescent samples for logging.
[1] http://en.wikipedia.org/wiki/Median_filter
[2] http://www.dsprelated.com/showarticle/72.php
[3] http://en.wikipedia.org/wiki/Time_constant
Note
Adding code for the above filtering operations will lower the maximum possible sample rate but printf can be substituted for something faster.
Continusously store the current value and the delta from the previous value.
Note when the delta is decreasing as the start of weight application to the scale
Note when the delta is increasing as the end of weight application to the scale
Take the X number of values with the small delta and average them
BTW, I'm sure this has been done 1M times before, I'm thinking that a search for scale PID or weight PID would find a lot of information.
Don't forget using ___delay_ms(XX) function somewhere between the reading values, if you will compare with the previous one. The difference in each step will be obviously small, if the code loop continuously.
Looking at your nice graphs, I would say you should look only for the falling edge, it is much consistent than leading edge.
In other words, let the samples accumulate, calculate the running average all the time with predefined window size, remember the deviation of the previous values just for reference, check for a large negative bump in your values (like absolute value ten times smaller then current running average), your running average is your value. You could go back a little bit (disregarding last few values in your average, and recalculate) to compensate for small positive bump visible in your picture before each negative bump...No need for heavy math here, you could not model the reality better then your picture has shown, just make sure that your code detect the end of each and every sample. You have to be fast enough with sample to make sure no negative bump was missed (or you will have big time error in your data averaging).
And you don't need that large arrays, running average is better based on smaller window size, smaller residual error in your case when you detect the negative bump.