Control.com - Nerds in Control

Hi,

I have a plant transfer function Gp(s) = 4/(s*(s+2)) The denominator has a degree 2. When I convert this to discrete time using the residual method and a Zero Order Hold, I multiply the denominator by another s so the degree is now 3. Using the residual method formula in the notes, I get a Gp(z) = 4.2749/(z-0.8187)

I'm supposed to use Ragazzini's method to design a lead compensator for this system. The pole of this Gp(z) is clearly not outside the unit circle. So there is no need to eliminate any poles/zeros in the F(z) function. I'm not sure how to continue with this. In all my previous lecture examples, there has always been an unstable pole/zero in the Gp(z), which we need to cancel out by inserting a new pole/zero in the F(z) to compensate. Now that the only pole is z=0.8187, and it's stable, is it safe to say that we don't need to cancel any poles out?

I suspect that maybe I'm not applying the Residual method correctly to convert from continuous to discrete time. Remember, I have to apply a ZOH first. The way I did it was I multiplied Gp(s) by (1-exp(-s*T))/s . Then I separated the (1-exp(-s*T)) and that becomes (z-1)/z . The extra s is multiplied with the denominator of the original Gp(s), so now I am looking for the z-transform of (4 / (s^2)*(s+2))), but the original Gp(s) denominator was just s*(s+2).

Moin ywk...
moin all,

Z-Transforming a second order continuous system like your
Gp(s) = 4 / ( s*(s+2) )
will always result in a second order discrete system.

Additionally a continuous system with an integral part as yours will
always keep the integral part after Z-transformation, i.e. will have a pole at z=1.

When using 0.1 (sec) as sampling time T a get a discrete transfer
function of
Gp(z) = ( 0.018731*z^1 + 0.017523 ) / ( 1*z^2 - 1.8187*z^1 + 0.81873 )
with
zeroes =
-0.935525 + 0.000000i
poles =
0.818731 + 0.000000i
1.000000 + 0.000000i

This has the pole you mentioned and additionally the required pole at 1 and a zero at z=-0.935525.

BTW, you cannot cancel an unstable pole, neither in the s-domain nor in the z-domain.

Best regards,
Friedrich Haase