# Discretise differential equation

Given the differential equation with variables x1(t)x_{1}(t), x2(t)x_2(t), u1(t)u_{1}(t):

dx1dt=√k⋅x1+√k1⋅x2+u1\frac{dx_{1}}{dt}=\sqrt{k\cdot{x_1}}+\sqrt{k_{1}\cdot x_{2}}+{u_1}
I have linearised it around the equilibrium point (dh1(t)dt=0\frac{dh_{1}(t)}{dt}=0) via a first order Taylor expansion to get

dx1dt=k2√k⋅x1equilibria⋅Δx1+k12√k1⋅x2equilibria⋅Δx2+Δu1\frac{dx_{1}}{dt}=\frac{k}{2\sqrt{k\cdot x_{1\hspace{1mm}equilibria}}}\cdot \Delta{x_1}+\frac{k_1}{2\sqrt{k_1\cdot x_{2\hspace{1mm}equilibria}}}\cdot \Delta x_{2}+\Delta u_{1}
Now, by making (just for simplicity)
a=k2√k⋅x1equilibria,b=k12√k1⋅x2equilibriaa=\frac{k}{2\sqrt{k\cdot x_{1\hspace{1mm}equilibria}}}, b=\frac{k_1}{2\sqrt{k_1\cdot x_{2\hspace{1mm}equilibria}}}

the equation becomes

dx1dt=a⋅Δx1+b⋅Δx2+Δu1\frac{dx_{1}}{dt}=a\cdot \Delta{x_1}+b\cdot \Delta x_{2}+\Delta u_{1}

Until this point I had no trouble, but I have to discretise the linearised differential equation, in order to apply the ZZ-Transform, with sampling time T=1sT=1s.

My thoughts are that dx1dt\frac{dx_{1}}{dt} can be approximated with finite differences, thus resulting in

dx1dt=x1((n+1)T)−x1(nT)T=Δx1(n)=a⋅Δx1+b⋅Δx2+Δu1\frac{dx_{1}}{dt}=\frac{x_1((n+1)T)-x_1(nT)}{T}=\Delta x_1(n) =a\cdot \Delta{x_1}+b\cdot \Delta x_{2}+\Delta u_{1}

However, I am not sure on how to proceed now when it comes to discretising the other deltas. If it helps, it seems that the expected result is

Δx1(n+1)=a⋅Δx1(n)+b⋅Δx2(n)+Δu1(n)\Delta x_1(n+1) =a\cdot \Delta{x_1}(n)+b\cdot \Delta x_{2}(n)+\Delta u_{1}(n)

Any ideas or thoughts on how can the linearised equation be discretised?

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Normally if you would have some non-linear time-invariant multidimensional differential equation with input vector →u(t)\vec{u}(t) of the form,

ddt→x(t)=→f(→x(t),→u(t)),
\frac{d}{dt}\vec{x}(t) = \vec{f}(\vec{x}(t), \vec{u}(t)), \tag{1}

with,

→x(t)=[x1(t)x2(t)⋯xn(t)]T,
\vec{x}(t) = \begin{bmatrix}
x_1(t) & x_2(t) & \cdots & x_n(t)
\end{bmatrix}^T, \tag{2}

→u(t)=[u1(t)u2(t)⋯um(t)]T,
\vec{u}(t) = \begin{bmatrix}
u_1(t) & u_2(t) & \cdots & u_m(t)
\end{bmatrix}^T, \tag{3}

→f(→x,→u)=[f1(→x,→u)f2(→x,→u)⋯fn(→x,→u)]T,
\vec{f}(\vec{x}, \vec{u}) = \begin{bmatrix}
f_1(\vec{x}, \vec{u}) & f_2(\vec{x}, \vec{u}) & \cdots & f_n(\vec{x}, \vec{u})
\end{bmatrix}^T, \tag{4}

you could linearize it around an equilibrium point →xeq\vec{x}_{eq} (such that →f(→xeq,→0)=→0\vec{f}(\vec{x}_{eq}, \vec{0})=\vec{0}). Such a linearization would yield a differential equation of the following form,

→z(t)=→x(t)−→xeq,
\vec{z}(t) = \vec{x}(t) – \vec{x}_{eq}, \tag{5}

ddt→z(t)=A→z(t)+Bu(t),
\frac{d}{dt}\vec{z}(t) = A\,\vec{z}(t) + B\,u(t), \tag{6}

where AA is a nn by nn matrix and BB a nn by mm matrix. Matrix AA will be equal to the Jacobian evaluated at the linearization point, which is equivalent to your first order Taylor expansion, which can be found with,

A=[∂f1∂x1⋯∂f1∂xn⋮⋱⋮∂fn∂x1⋯∂fn∂xn]|(→x,→u)=(→xeq,→0).
A = \left.\begin{bmatrix}
\frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\
\vdots & \ddots & \vdots \\
\frac{\partial f_n}{\partial x_1} & \cdots & \frac{\partial f_n}{\partial x_n}
\end{bmatrix}\right|_{(\vec{x},\vec{u})=(\vec{x}_{eq},\vec{0})}. \tag{7}

Matrix BB can be found in a similar way,

B=[∂f1∂u1⋯∂f1∂um⋮⋱⋮∂fn∂u1⋯∂fn∂um]|(→x,→u)=(→xeq,→0).
B = \left.\begin{bmatrix}
\frac{\partial f_1}{\partial u_1} & \cdots & \frac{\partial f_1}{\partial u_m} \\
\vdots & \ddots & \vdots \\
\frac{\partial f_n}{\partial u_1} & \cdots & \frac{\partial f_n}{\partial u_m}
\end{bmatrix}\right|_{(\vec{x},\vec{u})=(\vec{x}_{eq},\vec{0})}. \tag{8}

The solution for →z(t)\vec{z}(t) subjected to equation (6)(6) and initial condition →z(t0)=→z0\vec{z}(t_0)=\vec{z}_0 for t≥t0t\geq t_0 can be found to be,

→z(t)=eA(t−t0)→z0+∫tt0eA(t−τ)B→u(τ)dτ.
\vec{z}(t) = e^{A\,(t-t_0)}\,\vec{z}_0 + \int_{t_0}^t e^{A\,(t – \tau)}\,B\,\vec{u}(\tau)\,d\tau. \tag{9}

Now if you would like to simulate this system in steps of TT and assume that,

\vec{u}(t) = \vec{u}(k\,T), \quad k\,T \leq t < (k+1)\,T, \quad \forall\ \ k\in\mathbb{Z}, \tag{10} \vec{u}(t) = \vec{u}(k\,T), \quad k\,T \leq t < (k+1)\,T, \quad \forall\ \ k\in\mathbb{Z}, \tag{10} then for t_0=k\,Tt_0=k\,T and t=(k+1)\,Tt=(k+1)\,T, equation (9)(9) can be written as, \vec{z}((k+1)\,T) = \underbrace{e^{A\,T}}_{A_d}\,\vec{z}(k\,T) + \underbrace{\int_{k\,T}^{(k+1)\,T}\mkern-18mu e^{A\,((k+1)\,T - \tau)}\,d\tau\,B}_{B_d = A^{-1}\left(A_d-I\right)B}\,\vec{u}(k\,T). \tag{11} \vec{z}((k+1)\,T) = \underbrace{e^{A\,T}}_{A_d}\,\vec{z}(k\,T) + \underbrace{\int_{k\,T}^{(k+1)\,T}\mkern-18mu e^{A\,((k+1)\,T - \tau)}\,d\tau\,B}_{B_d = A^{-1}\left(A_d-I\right)B}\,\vec{u}(k\,T). \tag{11} This is equivalent to the discrete representation, \vec{z}[k+1] = A_d\,\vec{z}[k] + B_d\,\vec{u}[k]. \tag{12} \vec{z}[k+1] = A_d\,\vec{z}[k] + B_d\,\vec{u}[k]. \tag{12} For your system you only gave f_1(\vec{x},\vec{u})f_1(\vec{x},\vec{u}), so you would also need f_2(\vec{x},\vec{u})f_2(\vec{x},\vec{u}) in order to complete the total linearization and subsequently the discretization. You could also use your approach, but it will be less accurate. Namely this would only do a first order approximation of the matrix exponentiation and will only yield approximately the same results for very small TT, e^{A\,T} = I + A\,T + \frac12 A^2\,T^2 + \frac{1}{3!} A^3\,T^3 + \cdots. \tag{13} e^{A\,T} = I + A\,T + \frac12 A^2\,T^2 + \frac{1}{3!} A^3\,T^3 + \cdots. \tag{13}