I have the following SISO (Single Input, Single Output) system:

$$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} u $$ $$ y = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$ I know how to calculate the step response of that system (open loop and closed loop).

Now, I have the same system but another control variable: $$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} u + \begin{bmatrix} -40 \\ 0 \end{bmatrix} \omega $$ $$ y = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$ And I have to calculate the step response for open loop and closed loop (I suppose I have to use integral action, but maybe another technique is better). That's a MISO system and I don't know how to do it.

EDIT: I know that the system is linear so maybe I can calculate assuming that first B = [0 1] (transpose) and then B = [-40 0] (transpose) and then applying superposition or something like that


1 Answer 1


The given system is linear time invariant (LTI), therefore the superposition principle can be used.

The system is denoted as: $\dot{x}(t) = Ax(t)+ B_1 u(t) + B_2 \omega (t)$.

Due to the superposition principle the stepresponce (depending on the initial condition and the input) is given by $x(t) = e^{A t} x(0) + \int_0^t e^{A(t-\tau)} B_1 u(\tau)d\tau + \int_0^t e^{A(t-\tau)} B_2 \omega(\tau)d\tau$.

The closed loop response can be found by substituting the (linear) feedback law in the above equation.


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