Jtdylktuy

I have a system:

$A$, $B$ and $C$ forms the state space representation of the system. The system has a state feed back controller with an integral controller. I want to draw bode plot of controller, sensitivity function and complementary sensitivity function. How can I create these plots from a system like this? I know that sensitivity function is defined as $1/(1+PC)$ but I can not define $P$ and $C$ in this system; it is same in other two plots.

The open loop $P,C$ can not be defined when you are using cascading control and you want the open loop to be a SISO system. You could define a MIMO open loop using

begin{align}
begin{bmatrix}
dot{x} \ dot{x}_i
end{bmatrix} &=
begin{bmatrix}
A & -B,K_i \ 0 & 0
end{bmatrix}
begin{bmatrix}
x \ x_i
end{bmatrix} +
begin{bmatrix}
B,K & 0 \ 0 & I
end{bmatrix}v \
y &= begin{bmatrix}
I & 0 \ C & 0
end{bmatrix}
begin{bmatrix}
x \ x_i
end{bmatrix},
end{align}

with in closed loop $v = vec{r} - y$ where $vec{r} = begin{bmatrix}r_x^top & r^top end{bmatrix}^top$, so in order to get the same closed loop you would have to set $r_x=0$. I do have to note that the order of multiplication matters when calculating the transfer functions that you are interested, since you are dealing with a MIMO open loop which is a (transfer function) matrix and does not commute.

Another approach is to add an observer and use the error $e=r-y$ instead of the output to estimate the state. In that case only the output of the system will be used in the feedback loop and is $e$ the only input to the controller. Normally the observer dynamics is defined as

$$
dot{hat{x}} = A,hat{x} + B,u + L(y - C,hat{x}),
$$

such that the combined closed loop dynamics can be written as

$$
begin{bmatrix}
dot{x} \ dot{hat{x}} \ dot{x}_i
end{bmatrix} =
begin{bmatrix}
A & -B,K & -B,K_i \
L,C & A-B,K-L,C & -B,K_i \
-C & 0 & 0
end{bmatrix}
begin{bmatrix}
x \ hat{x} \ x_i
end{bmatrix} +
begin{bmatrix}
0 \ 0 \ I
end{bmatrix} r.
$$

But if $-e=y-r$ is used instead of $y$ in $dot{hat{x}}$ then the closed loop dynamics becomes

$$
begin{bmatrix}
dot{x} \ dot{hat{x}} \ dot{x}_i
end{bmatrix} =
begin{bmatrix}
A & -B,K & -B,K_i \
L,C & A-B,K-L,C & -B,K_i \
-C & 0 & 0
end{bmatrix}
begin{bmatrix}
x \ hat{x} \ x_i
end{bmatrix} +
begin{bmatrix}
0 \ -L \ I
end{bmatrix} r,
$$

so the same system matrix but a different input matrix, therefore the closed loop poles will be the same, but the zeros might be different. The open loop then becomes

begin{align}
begin{bmatrix}
dot{x} \ dot{hat{x}} \ dot{x}_i
end{bmatrix} &=
begin{bmatrix}
A & -B,K & -B,K_i \
0 & A-B,K-L,C & -B,K_i \
0 & 0 & 0
end{bmatrix}
begin{bmatrix}
x \ hat{x} \ x_i
end{bmatrix} +
begin{bmatrix}
0 \ -L \ I
end{bmatrix} e \
y &= begin{bmatrix}
C & 0 & 0
end{bmatrix}
begin{bmatrix}
x \ hat{x} \ x_i
end{bmatrix}.
end{align}