PID Control Equation in Z-Domain

[bsodmike]

How does one go from the simple substitution to the end result?

$$\dfrac{U(z)}{E(z)}=K'_P+\dfrac{K'_ITz}{z-1}+\dfrac{K'_D(z-1)}{Tz}=\dfrac{K_Pz^2+K_Iz+K_D}{z(z-1)}$$

Found the above simplification in http://www.nt.ntnu.no/users/skoge/prost/proceedings/acc04/Papers/0008_WeA02.5.pdf" in Eq 23-24.

Edit: Here's my attempt going back to first principles. In the s-domain, the laplace TF of a PID controller can be written as

$$G(s) = K_P+\dfrac{K_I}{s}+K_Ds$$

If I were to attempt to simplify the above by considering a = Kp, b = Ki, c = Kd

$$a+\dfrac{b}{s}+cs = a + \dfrac{b+cs^2}{s}=\dfrac{as+b+cs^2}{s}$$

Translate to the Z-domain by substituting

$$s = \dfrac{z-1}{Tz}$$

$$\dfrac{a\left(\dfrac{z-1}{Tz}\right)+b+c\left(\dfrac{z-1}{Tz}\right)^2}{\dfrac{z- 1}{Tz}}=\dfrac{a\left(\dfrac{z-1}{Tz}\right)+b+c\left(\dfrac{z^2-2z+1}{T^2z^2}\right)}{\dfrac{z-1}{Tz}}$$

$$\dfrac{a\left(\dfrac{z-1}{Tz}\right)+b+c\left(\dfrac{z^2-2z+1}{T^2z^2}\right)}{\dfrac{z-1}{Tz}} = \dfrac{a(z-1)+b(Tz)+c(\dfrac{z^2-2z+1}{Tz})}{z-1}$$

Multiplying through top and bottom by 'z'

$$z \times \dfrac{a(z-1)+b(Tz)+c(\dfrac{z^2-2z+1}{Tz})}{z-1}=\dfrac{az(z-1)+b(Tz^2)+c(\dfrac{z^2-2z+1}{T})}{z(z-1)}$$

 

[AlephZero]

I'm not sure what you are having problems with. Getting from (22) to (23) to (24) is just algebra.

Did you miss the fact that the constants in (23) and (24) are different? The continuous ones are K'p etc, the discrete ones are Kp etc.

 

[bsodmike]

Thanks Aleph.

Figured this much out; missed out on the primes...

Kp + (Ki T z)/(z - 1) + (Kd (z - 1))/(T z) = (Kd - 2*Kd*z - Kp*T*z + Kd*z^2 + Kp*T*z^2 + Ki*T^2*z^2)/(T*(-1 + z)*z)

Collecting z's gives: Kd/T + ((-2 Kd - Kp T) z)/T + ((Kd + Kp T + Ki T^2) z^2)/T Therefore, K_D = K_D'/T; K_I = -(K_P'T + 2 K_D')/T; K_P = (K_D' + K_P' T + K_I' T^2)/T

 

[AlephZero]

The continuous controller response function is a + b/s + cs

Then he substitutes s =(z-1)/(Tz)

and gets a + bTz/(z-1) + c(z-1)/(Tz)

Putting everything over z(z-1) gives

[ az(z-1) + bTz^2 + (c/T)(z-1)^2 ] / z(z-1)

The numerator is a quadratic expression in z, so this is of the form

[ Az^2 + Bz + C ] / z(z-1)

for some constants A B and C.

You were heading towards working out the exact formulas for A B and C in terms of a b c and T, but he doesn't bother to do that. He is only interested in the form of the expression, not the exact details.

 

[bsodmike]

You were heading towards working out the exact formulas for A B and C in terms of a b c and T, but he doesn't bother to do that. He is only interested in the form of the expression, not the exact details.

Indeed! Much appreciated.

Cheers, Mike.