3
$\begingroup$

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 $x_1 = 5$.

$\endgroup$
6
$\begingroup$

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

$$ \begin{bmatrix} \dot{x} \\ \dot{\xi} \end{bmatrix} = \begin{bmatrix} A & f \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ \xi \end{bmatrix} + \begin{bmatrix} B \\ 0 \end{bmatrix} u \\ y = \begin{bmatrix} C & g \end{bmatrix} \begin{bmatrix} x \\ \xi \end{bmatrix} + D\,u. \tag{4} $$

$\endgroup$
  • $\begingroup$ Thanks, I have an additional questions: Is there any disadvantage in using the second method right away? Because it seems that the second method always works while the first method requires the system$~(2)$ to be solvable. $\endgroup$ – SampleTime Dec 10 '17 at 10:28
  • 1
    $\begingroup$ @SampleTime Depending on what analysis you plan to do there might be a disadvantage of adding an uncontrollable marginally stable mode to the system. It is always good to have multiple options. And my first method might also be useful in other situations where it is not required that $u=f(y)$ with $f(0)=0$. $\endgroup$ – Kwin van der Veen Dec 10 '17 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.