Given a linear state space model as
$$
\begin{split}
\dot{x}_1 &= -50\,x_1 + 5\,x_2 – 0.15\,u + 250 \\
\dot{x}_2 &= 5 – x_1 \\
y &=0.2\,x_1 – 1 \,.
\end{split}
$$
I now would like to analyse this model with harmonic balance given a symetric nonlinear curve
$$
u = f(e)
$$
with $e = w – y$ and $w = 0$, as usual with harmonic balance, so
$$
u = f(-y) = -f(y)\,.
$$
My problem here are the constant terms in the state equations and the output equation. How can I deal with those?
The $f(\cdot)$ curve is point symmetric to the origin, so I cannot just omit the $-1$ in the output equation, otherwise the symmetry required for application of harmonic balance is not given anymore at the equilibrium $x_1 = 5$.
Best Answer
If your system is of the form
$$ \dot{x} = A\,x + B\,u + f \\ y = C\,x + D\,u + g \tag{1} $$
with $f$ and $g$ constant vectors, then you can do a coordinate transformation $z=x+\alpha$ and $v=u+\beta$ with $\alpha$ and $\beta$ constant vectors which satisfies
$$ \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} f \\ g \end{bmatrix}. \tag{2} $$
This can always be solved if the $(A,B,C,D)$ matrix is full rank and if this is not the case when the vector of $(f,g)$ lies in the span of the $(A,B,C,D)$ matrix. After this transformation the dynamics will simply be
$$ \dot{z} = A\,z + B\,v \\ y = C\,z + D\,v. \tag{3} $$
However if it is required that $u=f(y)$, with $f(0)=0$ and $f(-y)=-f(y)$, then this transformation will only work if the value found for $\beta$ is zero. If $\beta\neq0$ or the system of equation in $(2)$ is not solvable, then you could resort to extending the state space by one state $\xi$, whose time derivative is always zero. If the initial condition of $\xi$ is one, then the same dynamics as $(1)$ will be obtained when using the following extended state space model
$$ \begin{bmatrix} \dot{x} \\ \dot{\xi} \end{bmatrix} = \begin{bmatrix} A & f \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ \xi \end{bmatrix} + \begin{bmatrix} B \\ 0 \end{bmatrix} u \\ y = \begin{bmatrix} C & g \end{bmatrix} \begin{bmatrix} x \\ \xi \end{bmatrix} + D\,u. \tag{4} $$