I played a bit with the code in Matlab and I can only confirm @baggepinnen 's explanation. The PID controller in this case is formed solely by the D term, for which the transfer function is just `278.7369 s`

and no state equations can be formulated to model it. Instead, differential-algebraic (also called descriptor) equations can be formulated. And that is what Matlab does automatically:

```
>> K = 278.7369 * tf([1 0],1)
K =
278.7 s
Continuous-time transfer function.
Model Properties
>> ss(K)
ans =
A =
x1 x2
x1 1 0
x2 0 1
B =
u1
x1 0
x2 -16
C =
x1 x2
y1 17.42 0
D =
u1
y1 0
E =
x1 x2
x1 0 1
x2 0 0
Continuous-time state-space model.
Model Properties
```

Matlab then proceeds with the descriptor description (`dss`

instead of the more familiar `ss`

class) when forming the open- and whater closed-loop transfer functions. All these are then modelled by the five matrices `A`

, `B`

, `C`

, `D`

, and `E`

. It is then incorrect to ignore the `E`

matrix. For example when analysing the stability of the closed-loop system, checking the eigenvalues of the `A`

matrix leads to incorrect conclusions.

In this particular case, since the original plant is strictly proper (more poles than zeros), it seems useful to form the open-loop transfer function in a way that avoids the need to form a standalone model of the derivative term first, as @baggepinnen suggests.

When doing this in Matlab, I first reduced the model of the plant a bit. The original model is rather redundant â most poles can be either exactly or approximately cancelled by zeros (I was just wondering, wasnât the model obtained by some ARX-like identification procedure?).

```
>> G = ss(A,B,C,D); % SISO, order 163
G = minreal(G); % order 146
19 states removed.
>> Gred = reduce(G,3);
>> zpk(Gred)
ans =
-0.0049294 (s+8.172) (s+33.99)
------------------------------
(s+2.323) (s-2.047) (s+13.72)
Continuous-time zero/pole/gain model.
Model Properties
```

It appears that this simple model captures the main aspects of the original model quite well. It is strictly proper, it does contain a single unstable pole, and the Bode plots reasonably agree too. I found it more convenient to play with this simple model.

Now, I obtained the open-loop transfer function by manipulating the transfer functions instead of state-space models. Perhaps this could be done more efficiently (transforming the model description from the state-space format to the transfer function format and back is not particularly recommendable), anyway:

```
>> L = tf(Gred)*K
L =
-1.374 s^3 - 57.93 s^2 - 381.6 s
---------------------------------
s^3 + 13.99 s^2 - 0.9693 s - 65.2
Continuous-time transfer function.
Model Properties
>> zpk(L)
ans =
-1.374 s (s+33.99) (s+8.172)
-----------------------------
(s+13.72) (s+2.323) (s-2.047)
Continuous-time zero/pole/gain model.
Model Properties
```

Obviously, the open-loop transfer function is proper and can be conveniently described by a state-space model. But what we are ultimately after is that one particular closed-loop transfer function

```
>> U = feedback(tf(Gred),K)
U =
0.004929 s^2 + 0.2078 s + 1.369
--------------------------------------
0.374 s^3 + 43.94 s^2 + 382.6 s + 65.2
Continuous-time transfer function.
Model Properties
>> zpk(U)
ans =
0.01318 (s+33.99) (s+8.172)
----------------------------
(s+108) (s+9.282) (s+0.1739)
Continuous-time zero/pole/gain model.
Model Properties
```

All the poles are obviously stable. This can also be seen upon drawing a root locus:

For sufficiently large gain (just by trying it appears the treshold is around 0.75) the close-loop pole that lies on the locus emanating from the only unstable open-loop pole âjumpsâ over the infinity to the left half plane.