constructing discrete time system

I consturcted the loop for disturbance rejection for discrete time system.

But I wonder, why two steps are different?

Gdtest case: discretization and then constructed closed loop

Gdd2 case: constructed closed loop and then discretized the closed loop

I'm confused which one is right.

And I wonder how to check output of controller which is input of system Gd.

Thank you in advance

s = tf('s');
G = 4000/(s*(s+10)*(s+20));
st = 0.02;
Gd= c2d(G,st,'foh');

Cdr = 20 + 100/s + 2*s/(s/1000 + 1);
Gdr0= c2d(Cdr,st,'foh');

Gdtest = feedback(Gd,Gdr0);
Gdd2= c2d(feedback(G, Cdr),st,'foh');

figure;
step(Gdtest)

figure;
step(Gdd2)

Answer 1

[Paul]

Hi Junhwi,

The short answer is that the sample time, st, is too large for discretizing Cdr before forming the closed-loop system.

First, I think it's easier to see what's happening by using the zpk form instead of the tf form.

s       = zpk('s');

Now form G and discretize it.

G       = 4000/(s*(s+10)*(s+20))

G =
 
       4000
  ---------------
  s (s+10) (s+20)
 
Continuous-time zero/pole/gain model.

st      = 0.02;
Gd      = c2d(G,st,'foh')

Gd =
 
  0.0011851 (z+8.814) (z+0.887) (z+0.08924)
  -----------------------------------------
         (z-1) (z-0.8187) (z-0.6703)
 
Sample time: 0.02 seconds
Discrete-time zero/pole/gain model.

We see that all the poles have been mapped in accordance with z = exp(s*T).

exp(pole(G)*st)

ans = 3×1

1.0000 0.8187 0.6703

Note that the sampling time, 0.02, is more 2x smaller than the smallest time constant in G, which is 1/20 = 0.05.

Form Cdr

Cdr    = 20 + 100/s + 2*s/(s/1000 + 1)

Cdr =
 
  2020 (s^2 + 9.95s + 49.5)
  -------------------------
         s (s+1000)
 
Continuous-time zero/pole/gain model.

The sample time, 0.02, is 20x larger than the smallest time constant in Cdr, which is 1/1000 = 0.001. Hence, the discretization will not be what we want.

Gdr0   = c2d(Cdr,st,'foh')

Gdr0 =
 
  121 (z^2 - 1.81z + 0.8264)
  --------------------------
           z (z-1)
 
Sample time: 0.02 seconds
Discrete-time zero/pole/gain model.

The pole at the origin s=0 in Cdr maps to z = 1, and the pole at s = -1000 maps to exp(-1000*st), which is effectively at the origin. This pole at the origin is an open-loop pole.

Now we form the closed-loop system, and because that open-loop pole at the origin isn't really in the correct location, the feedback is wrong and we get an unstable system (poles outside the unit circle)

Gdtest = feedback(Gd,Gdr0)

Gdtest =
 
       0.0010364 z (z+8.814) (z+0.887) (z-1) (z+0.08924)
  -----------------------------------------------------------
  (z+0.05193) (z^2 - 1.811z + 0.8274) (z^2 - 0.2919z + 1.683)
 
Sample time: 0.02 seconds
Discrete-time zero/pole/gain model.

abs(pole(Gdtest))

ans = 5×1

1.2973 1.2973 0.0519 0.9096 0.9096

Here, we form the continuous-time closed-loop system

feedback(G,Cdr)

ans =
 
                   4000 s (s+1000)
  --------------------------------------------------
  (s+1008) (s^2 + 9.887s + 49.81) (s^2 + 12s + 7966)
 
Continuous-time zero/pole/gain model.

Looking at the time constants associated with the poles

-1./real(pole(ans))

ans = 5×1

0.0010 0.1667 0.1667 0.2023 0.2023

we see that the st is considerably smaller than all of them, except the time constant associated with the pole at s = -1008. So that pole will incorrectly map to the origin after discretization. BUT, that pole is a closed-loop pole (not an open-loop pole) and in this case it's nearly cancelled by the zero at s = -1000, which will also result in a near cancellation after discretization

Gdd2   = c2d(feedback(G, Cdr),st,'foh')

Gdd2 =
 
  0.0010929 (z+7.365) (z+0.9108) (z-1) (z+0.1141) (z-1.221e-06)
  -------------------------------------------------------------
       z (z^2 - 1.803z + 0.8206) (z^2 + 0.3702z + 0.7866)
 
Sample time: 0.02 seconds
Discrete-time zero/pole/gain model.

As expected, we have a closed-loop pole essentially at the origin and a zero there as well at z = 1.221e-6. So the fact that st is too large doesn't really have an impact when discretizing the closed-loop system.

The solution for the first case is to use a much smaller sample time

st = st/100;
Gd      = c2d(G,st,'foh');
Gdr0    = c2d(Cdr,st,'foh');
Gdtest  = feedback(Gd,Gdr0);
Gdd2    = c2d(feedback(G, Cdr),st,'foh');
figure;  % put all three systems on the same plot
step(feedback(G,Cdr),Gdtest,Gdd2)

[Lucas]

Thank u so much, man! Great explanation!