Electronic – arduino – How to implement a discrete transfer function into an arduino

arduinodigital filterfiltermicrocontrollertransfer function

I am here asking how does one transfer a difference equation into a MCU? I have never done it personally and looking into this topic I was never able to find a good answer. Its usually used with Matlab but never a MCU.

I have a continuous transfer function:

$$LPF_A = \frac{1}{6.416\cdot 10^{-11}s^2+1.133\cdot 10^{-5}s+1}$$

Its a 2nd Order Butterworth filter with a \$F_c = 20\text{kHz}\$.

Using the Tustin method I have the discrete equation:

$$LPF_D=\frac{0.1441z^2+0.2881z+0.1441}{z^2-0.6777z+0.254}$$
With a sample of \$T_s = \$ 8e-6 seconds, in scientific E notation.

The difference equation:

\$Y\$ = Output

\$U\$ = Input

$$Y_i =0.1441U_{i}+0.2281U_{i-1}+0.1441U_{i-2}+0.6777Y_{i-1}-0.254Y_{i-2}$$

I know I have to make my ADC output = \$U_i\$ and initialize \$U_{i-1}\$, \$U_{i-2}\$, \$Y_{i-1}\$, \$Y_{i-2} = 0\$, But after that no idea.

I have no idea what to do with this after. I have always used Matlab's embedded ways of doing it but never have transferred a difference equation into a MCU before. Any help would be appreciated.

Edit: Here's the code based on what I think

  unsigned short x;
double Y_i[10] = {0,0,0,0,0,0,0,0,0,0};
double U_i[10] = {0,0,0,0,0,0,0,0,0,0};
double Y_i_1 = 0;
double Y_i_2 = 0;
double U_i_1 = 0;
double U_i_2 = 0;

void setup() {

  ADMUX |= 0xC0;
  ADCSRA |= 0x8F;
  ADCSRA |= 0x40;
  Serial.begin(9600);
}

ISR(ADC_vect) {
 x = ADCL << 2;
 x |= ADCH;
 ADCSRA |= 0x40;
}

void loop() {
  //Yi = 0.1441Ui + 0.2281Ui-1 + 0.1441Ui-2 + 0.6777Yi-1 - 0.254Yi-2 
  for (int i = 1; i < 4; i++){

    if (i == 1) {
       Y_i[i] = 0.144*x + 0.2281*U_i_1 + 0.1441*U_i_2 + 0.6777*Y_i_1 - 0.254*Y_i_2;
       U_i[i] = x;
    } else if (i == 2) {
      Y_i[i] = 0.144*x + 0.2281*U_i[i-1] + 0.1441*U_i[i-2] + 0.6777*Y_i[i-1] - 0.254*Y_i[i-2];
      U_i[i] = x;
    } else {
      Y_i[i] = 0.144*x + 0.2281*U_i[i-1] + 0.1441*U_i[i-2] + 0.6777*Y_i[i-1] - 0.254*Y_i[i-2];
      U_i[i] = x;
    }

  }

}

What happens after the iteration is done? It resets? thats it?

Best Answer

Maybe you are having trouble with the "advance time" step. It helps to separate the logic as follows:

  1. Declare your variables: \$Y_{i}\$, \$Y_{i-1}\$, \$Y_{i-2}\$, \$U_{i}\$, \$U_{i-1}\$ and \$U_{i-2}\$. Let's also initialize a counter \$i = 0\$, which represents the current value of the subscript in your variables (this counter is completely unnecessary, but helps getting across what we are doing). Here is how \$i\$ works: if \$i = 10\$, this means the variable for \$Y_{i-2}\$ currently holds \$Y_{8}\$. Initialize everyone to zero, but keep in mind \$Y_{i}\$ and \$U_{i}\$ are not valid yet - we'll deal with them in the next two steps.
  2. Acquire new data. That is, acquire \$U_{i}\$ from the ADC.
  3. Recalculate your output. You know how to do it. Store it in \$Y_i\$.
  4. Profit. Now all your variables are correct and reflect the system at the timestep \$i\$ (if this is your first pass, now \$i = 0\$). Do whatever you want with them.
  5. Now you have to advance time. This means updating your variables as if time passed. This is necessary for calculating the next timestep. To do this, have \$Y_{i}\$ become \$Y_{i-1}\$, and so on (I won't spoil everything you have to do). You'll notice old values of \$U\$ and \$Y\$ will be lost as time advances. Also increment \$i\$ to keep track of which sample we are talking about.
  6. Now, all your variables except \$Y_i\$ and \$U_i\$ reflect the state of the sistem at the current \$i\$ (if this is your first pass, now \$i = 1\$). \$Y_i\$ and \$U_i\$ are still unknown because you need another sample to calculate them.
  7. Wait for the time to sample again and go to step 2.
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