- #1
bsodmike
- 82
- 0
How does one go from the simple substitution to the end result?
[tex]\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)}[/tex]
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
[tex]G(s) = K_P+\dfrac{K_I}{s}+K_Ds[/tex]
If I were to attempt to simplify the above by considering a = Kp, b = Ki, c = Kd
[tex]a+\dfrac{b}{s}+cs = a + \dfrac{b+cs^2}{s}=\dfrac{as+b+cs^2}{s}[/tex]
Translate to the Z-domain by substituting
[tex]s = \dfrac{z-1}{Tz}[/tex]
[tex]\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}}[/tex]
[tex]\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}[/tex]
Multiplying through top and bottom by 'z'
[tex]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)}[/tex]
[tex]\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)}[/tex]
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
[tex]G(s) = K_P+\dfrac{K_I}{s}+K_Ds[/tex]
If I were to attempt to simplify the above by considering a = Kp, b = Ki, c = Kd
[tex]a+\dfrac{b}{s}+cs = a + \dfrac{b+cs^2}{s}=\dfrac{as+b+cs^2}{s}[/tex]
Translate to the Z-domain by substituting
[tex]s = \dfrac{z-1}{Tz}[/tex]
[tex]\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}}[/tex]
[tex]\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}[/tex]
Multiplying through top and bottom by 'z'
[tex]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)}[/tex]
Last edited by a moderator: