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}