What is the frequency response of a first-order system of ODEs?

Question

What is the frequency response of a first-order system of ODEs? Specifically, given the differential equation:

$$ y'(t) = Ay(t) + x(t) $$

where $y:[0,\infty]\rightarrow \mathbb{R}^m$ and $A\in\mathbb{R}^{m\times m}$, what is the solution when

$$ x(t)=\begin{bmatrix}0\\\vdots\\\cos(\omega t)\\\vdots\\0\end{bmatrix} = \cos(\omega t) b_i? $$

Essentially, if we have a first-order system of ODEs and we activate a single input with a sine wave of a given frequency, what is the solution? Normally, the frequency response is worked out for a linear time invariant system of higher order and the solution is just a phase shifted and amplitude scaled version of the sine wave. I'm interested in what this result looks like for a first-order system rather than a single equation.

Answer 1

The solution you are probably looking for has the form $$y(t)=u\cos(ωt)+v\sin(ωt).$$ Insert into the DE to get the linear system for the vectors $u,v$. Take care of what happens if $iω$ is an eigenvalue of $A$.

Answer 2

I use $Y(s)$ and $X(s)$ to denote the Laplace transform $\mathcal{L}\{y(t)\}(s)$ and $\mathcal{L}\{y(t)\}(s)$ respectively. Furthermore, it is worth nothing that for $A(s)=\mathcal{L}\{a(t)\}(s)$ it hold that $\mathcal{L}\{a'(t)\}(s)=s\,A(s)-a(0)$, with $a'(t)$ denoting the derivative of $a(t)$ with respect to time ($t$). Combing this with your original system of first order ODEs yields

$$ s\,Y(s) - y(0) = A\,Y(s) + X(s). \tag{1} $$

Solving $(1)$ for $Y(s)$ yields

$$ Y(s) = (s\,I - A)^{-1} (X(s) + y(0)). \tag{2} $$

Often for frequency response analysis it is assumed that $y(0)=0$. And for the steady state frequency response one can substitute in $s=j\,\omega$.

---

This is also related to converting state space models into transfer functions. So when starting with the following state space model

\begin{align} x'(t) &= A\,x(t) + B\,u(t) \\ y(t) &= C\,x(t) + D\,u(t) \end{align}

then the corresponding transfer function can be obtained using

$$ G(s) = C\,(s\,I - A)^{-1} B + D. $$