State space formulas

$n$ states, $m$ inputs, $p$ outputs. $A,B,C,D$ are constant matrices because the system is time-invariant.

The denominator must be monic. The last row of $A$ holds the negated denominator coefficients in ascending powers of $s$, and $C$ holds the numerator coefficients in ascending powers. This form is always controllable. Some texts number the states in reverse, which puts the coefficients in the first row; $G(s)$ is the same either way.

Convention used here: $A_o=A_c^T$, $B_o=C_c^T$, $C_o=B_c^T$. This form is always observable. Reversing the state order flips rows and columns without changing $G(s)$.

Only for distinct poles; the $r_i$ are the partial-fraction residues. The residues can sit in $B$ instead of $C$, because only each product $b_i{c}_i=r_i$ is fixed. Complex poles give complex entries.

Assumes zero initial conditions. For MIMO systems $G(s)$ is a $p\times m$ matrix with $G_{ij}=Y_i/U_j$. The order of multiplication is fixed: $C$, then $(sI-A{)}^{-1}$, then $B$.

Its roots are the eigenvalues of $A$. Every pole of $G(s)$ is an eigenvalue. If $C\text{adj}(sI-A)B$ shares a factor with $\det (sI-A)$, that eigenvalue cancels and does not appear as a pole. $\det (A-\lambda I)$ differs from $\det (\lambda I-A)$ only by $(-1{)}^n$, so the two have the same roots.

Swap the diagonal entries, negate the off-diagonal ones and divide by the determinant. You use this in nearly every 2×2 $(sI-A{)}^{-1}$ calculation.

Holds for any square $A$, with eigenvalues counted by multiplicity. The eigenvalues of a triangular or diagonal matrix are its diagonal entries.

$P$ is any invertible $n\times n$ matrix. The transformation does not change the eigenvalues, the characteristic equation, $G(s)$, controllability or observability. With $P$ = modal matrix (eigenvectors as columns), $\bar{A}$ is diagonal. This needs $n$ linearly independent eigenvectors, which is guaranteed when the eigenvalues are distinct.

Valid for any constant square $A$. The series stops after finitely many terms if $A$ is nilpotent, e.g. $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ gives $e^{At}=\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]$.

True for every constant $A$, so $\Phi (t)$ is always invertible. Use $\Phi (0)=I$ and $\dot{\Phi }(0)=A$ to test whether a given matrix can be an STM, and to recover $A$ from a given STM.

The third matrix has eigenvalues $\sigma \pm j\omega$. The last line is the diagonalisation method: $\Lambda$ is the diagonal matrix of eigenvalues and $P$ is the modal matrix, which exists when $A$ has $n$ linearly independent eigenvectors (always the case for distinct eigenvalues).

Initial time 0. For initial time $t_0$: $x(t)=\Phi (t-t_0)x(t_0)+{\int }_{t_0}^t\Phi (t-\tau )Bu(\tau )d\tau$.

Often the quickest route by hand. Take the inverse transform of each entry at the end.

The first formula (unit-step input, zero initial state) needs $A$ invertible. The final values hold, for any initial state, when every eigenvalue of $A$ has a negative real part; $A$ is then automatically invertible, because 0 is not an eigenvalue. You can also get $x(\infty )$ by setting $\dot{x}=0$: $0=Ax+B$.

Here $A,B,C,D$ are the discrete-time matrices, and $A^k$ plays the role of the STM. The system is asymptotically stable exactly when every eigenvalue of $A$ has $|\lambda |<1$. The Kalman matrices $[BAB\cdots A^{n-1}B]$ and $[C;CA;\cdots ;C{A}^{n-1}]$ take the same form: full rank of the first means every state can be reached from the origin, and full rank of the second means $x[0]$ can be recovered from the output. Sampling $\dot{x}=Ax+Bu$ with a zero-order hold of period $T$ gives $A_d=e^{AT}$ and $B_d=\left({\int }_0^T{e}^{A\sigma }d\sigma \right)B$, which equals $A^{-1}(e^{AT}-I)B$ when $A$ is invertible.

Depends only on $(A,B)$. For a single input, $Q_c$ is $n\times n$ and the system is controllable exactly when $\det Q_c\ne 0$. For $m$ inputs, $Q_c$ is $n\times nm$ and you must check its rank. Compute $A^{2}B$ as $A(AB)$.

Depends only on $(A,C)$. An equivalent transposed convention is $[C^T{A}^T{C}^T\cdots (A^T{)}^{n-1}{C}^T]$, which has the same rank. For a single output, the system is observable exactly when $\det Q_o\ne 0$.

Use only after transforming to a diagonal $\bar{A}$ with distinct eigenvalues. A zero row $i$ of $\bar{B}$ means the mode $e^{{\lambda }_{i}t}$ cannot be excited, and a zero column $i$ of $\bar{C}$ means that mode never reaches the output. In Jordan form (one block per distinct eigenvalue), check the row of $\bar{B}$ at the last row of each block and the column of $\bar{C}$ at the first column of each block. The test does not apply to a diagonal $\bar{A}$ with a repeated eigenvalue; see the last tip of this section.

$K$ is $m\times n$ ($1\times n$ for a single input) and $r$ is the new reference input. The eigenvalues of $A-BK$ can be placed at any chosen locations (complex ones in conjugate pairs) if and only if $(A,B)$ is controllable. In the pole-placement formula, ${\alpha }_d(s)=s^n+{\alpha }_{n-1}{s}^{n-1}+\cdots +{\alpha }_0$ is the desired closed-loop characteristic polynomial and ${\alpha }_d(A)=A^n+{\alpha }_{n-1}{A}^{n-1}+\cdots +{\alpha }_{0}I$. State feedback does not change controllability and, for a SISO system, does not move the zeros of the transfer function; it can destroy observability if a closed-loop pole is placed on a zero.