Which Is Correct Equation To Convert To C Embedded Code #1240

oneEclipseUser1 · 2026-07-12T00:29:32Z

oneEclipseUser1
Jul 12, 2026

Hello community,

I have a simple question that I hope you can help me with.

# Plant
Gm = ct.tf([1],[1, 14, 40.02])

# PI Compensator
K = 30 # Kp = 30, Ki = 150
Gc = ct.tf([K, 5*K], [1, 0])

# Plant + Compensator:
Ls = ct.series(Gm, Gc)         # Open loop
G_closed = ct.feedback(Ls, 1)  # Closed loop

# 2. Discrete time domain
T = 0.02 # 20 ms; T = 1 / (10xfb) = 1 / (10xWpm/(2xpi)
Gm_z = ct.c2d(Gm, Ts=T, method='zoh')
Gc_z = ct.c2d(Gc, Ts=T, method='bilinear')
Ls_z = ct.series(Gm_z, Gc_z)
Tz = ct.feedback(Ls_z, 1)

print(Gc_z)  # print the compensator transfer function

When I print the compensator discrete transfer function, I get the following result:

  31.5 z - 28.5
  -------------
      z - 1

I now would like to convert it to usable C embedded code. Do I need to divide by z so that I obtain the transfer function in terms of z^-1? For example:

-28.5z^-1 + 31.5
 ---------------
   1 - z^-1

If I assume I can work directly as is from the original printed compensator transfer function, then the difference discrete output function would be:

u[n] = u[n-1] - 31.5e[n - 1] + 28.5e[n]

or, if I divide by z so that I obtain the second equation, then I obtain (that is, so that I have the transfer function in terms of z^-1 and not z:

u[n] = u[n-1] + 31.5e[n] - 28.5e[n - 1]

Can someone please clarify this small issue.

Thank you.

Answered by AMBRA7592

Jul 13, 2026

Yes, the second recurrence is the correct one.

Starting from

U(z)   31.5 z - 28.5
---- = -------------
E(z)       z - 1

divide both numerator and denominator by z. This does not change the transfer function; it only rewrites it using the unit-delay operator:

U(z)   31.5 - 28.5 z^-1
---- = -----------------
E(z)       1 - z^-1

Cross-multiplying gives

U(z)(1 - z^-1) = E(z)(31.5 - 28.5 z^-1)

and therefore

u[n] - u[n-1] = 31.5 e[n] - 28.5 e[n-1]

so the implementation is

u[n] = u[n-1] + 31.5 e[n] - 28.5 e[n-1]

The first recurrence in the question is not equivalent: it swaps the current and previous error coefficients and changes their signs.

A minimal floating-point update would look like:

…

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oneEclipseUser1 · 2026-07-12T04:05:46Z

oneEclipseUser1
Jul 12, 2026
Author

Solved. After a little digging, confirmed that I need to convert to z^-k format as opposed to z only.

0 replies

AMBRA7592 · 2026-07-13T15:15:11Z

AMBRA7592
Jul 13, 2026

Yes, the second recurrence is the correct one.

Starting from

U(z)   31.5 z - 28.5
---- = -------------
E(z)       z - 1

divide both numerator and denominator by z. This does not change the transfer function; it only rewrites it using the unit-delay operator:

U(z)   31.5 - 28.5 z^-1
---- = -----------------
E(z)       1 - z^-1

Cross-multiplying gives

U(z)(1 - z^-1) = E(z)(31.5 - 28.5 z^-1)

and therefore

u[n] - u[n-1] = 31.5 e[n] - 28.5 e[n-1]

so the implementation is

u[n] = u[n-1] + 31.5 e[n] - 28.5 e[n-1]

The first recurrence in the question is not equivalent: it swaps the current and previous error coefficients and changes their signs.

A minimal floating-point update would look like:

double u = u_prev + 31.5 * e - 28.5 * e_prev;
e_prev = e;
u_prev = u;

The stored values must be initialized deliberately. For an actual actuator, also account for output saturation and integrator windup; otherwise the internal u_prev state can continue accumulating while the physical output is clamped. These coefficients are specific to the chosen T = 0.02 and bilinear discretization, so regenerate them if either changes.

0 replies

oneEclipseUser1 · 2026-07-13T19:45:02Z

oneEclipseUser1
Jul 13, 2026
Author

@AMBRA7592,

thank you for following up and confirming the results. Yes, I have accounted for saturation
and the required minimum value. The following code represents the discrete time PI compensator output equation in C . It serves more of an exercise so that I go through all of the required steps when designing a compensator.

/*
   Test ISR source file

   compensator values:
   ----------------------
   u[n]      -> current result sum
   u[n - 1]  -> previous result sum
   e[n]      -> current error
   e[n - 1]  -> previous error

   PI compensator
   u[n] = u[n - 1] + Kp*e[n] - Ki*e[n - 1]

*/

// Clamp max upper value (misc.) place holders
#define UPPER_LIM  0xde35
#define LOWER_LIM 0x2524

// Define the fixed reference
#define VREF 0x9b10

// PI compensator gain constants
#define Kp 31.5
#define Ki 28.5


#pragma vector = PWM1_ISR_VECTOR // Associates ISR to this handler
__interrupt void PWM1_ISR(void)
{
  // local variables
  static float prev_err = 0;
  static float prev_yout = 0;
  float y_out = 0; 
   
  float error = VREF - ADC3; // latest error 

  // PI compensator
  y_out = prev_yout + Kp*error - Ki*prev_err;

  // clamp if outside limit values
  if (y_yout > UPPER_LIM){
    y_out = UPPER_LIM;}
      
  elif(y_out < LOWER_LIM){
    y_out = LOWER_LIM;}

  /*
    Code to update actuator (i.e., PWM) value here - pending
  */

  // Update history variables prior to exiting ISR
  prev_out = y_out;
  prev_err = error;

}

Cheers!

0 replies

AMBRA7592 · 2026-07-18T18:33:02Z

AMBRA7592
Jul 18, 2026

Glad the derivation was useful. Since this is a Q&A discussion and it still appears as unanswered, would you mind marking the explanatory response as the accepted answer? That will make the resolved recurrence easier for future readers to find.

0 replies

oneEclipseUser1 · 2026-07-19T01:31:28Z

oneEclipseUser1
Jul 19, 2026
Author

No problemo.

Thank you @AMBRA7592.

Much appreciated.

0 replies

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