Editing Discretization (section)

=== Derivation ===
Starting with the continuous model
<math display=block>\mathbf{\dot{x}}(t) = \mathbf{Ax}(t) + \mathbf{Bu}(t)</math>
we know that the [[matrix exponential]] is
<math display=block>\frac{d}{dt}e^{\mathbf{A}t} = \mathbf{A}e^{\mathbf{A}t} = e^{\mathbf{A}t} \mathbf A</math>
and by premultiplying the model we get
<math display=block>e^{-\mathbf{A}t} \mathbf{\dot{x}}(t) = e^{-\mathbf{A}t} \mathbf{Ax}(t) + e^{-\mathbf{A}t} \mathbf{Bu}(t)</math>
which we recognize as
<math display=block>\frac{d}{dt}\Bigl[e^{-\mathbf{A}t}\mathbf x(t) \Bigr] = e^{-\mathbf{A}t} \mathbf{Bu}(t)</math>
and by integrating,
<math display=block>\begin{align}
  e^{-\mathbf{A}t}\mathbf{x}(t) - e^0\mathbf{x}(0) &= \int_0^t e^{-\mathbf{A}\tau} \mathbf{Bu}(\tau) d\tau \\[2pt]
  \mathbf{x}(t) &= e^{\mathbf{A}t}\mathbf{x}(0) + \int_0^t e^{\mathbf{A}(t-\tau)} \mathbf{Bu}(\tau) d\tau 
\end{align}</math>
which is an analytical solution to the continuous model.

Now we want to discretise the above expression. We assume that {{mvar|u}} is [[mathematical constant|constant]] during each timestep.
<math display=block>\begin{align}
  \mathbf x[k] &\, \stackrel{\mathrm{def}}{=}\ \mathbf x(kT) \\[6pt]
  \mathbf x[k] &= e^{\mathbf{A}kT}\mathbf x(0) + \int_0^{kT} e^{\mathbf A(kT-\tau)} \mathbf{Bu}(\tau) d\tau \\[4pt]
  \mathbf x[k+1] &= e^{\mathbf A(k+1)T}\mathbf x(0) + \int_0^{(k+1)T} e^{\mathbf A[(k+1)T-\tau]} \mathbf{Bu}(\tau) d \tau \\[2pt]
  \mathbf x[k+1] &= e^{\mathbf{A}T} \left[  e^{\mathbf{A}kT}\mathbf x(0) + \int_0^{kT} e^{\mathbf A(kT-\tau)} \mathbf{Bu}(\tau) d \tau \right]+ \int_{kT}^{(k+1)T} e^{\mathbf A(kT+T-\tau)} \mathbf B\mathbf u(\tau) d\tau
\end{align}</math>
We recognize the bracketed expression as <math>\mathbf x[k]</math>, and the second term can be simplified by substituting with the function <math>v(\tau) = kT + T - \tau</math>. Note that <math>d\tau=-dv</math>. We also assume that {{math|'''u'''}} is constant during the [[integral]], which in turn yields

<math display=block>\begin{align}
  \mathbf x[k+1] &= e^{\mathbf{A}T}\mathbf x[k] - \left( \int_{v(kT)}^{v((k+1)T)} e^{\mathbf{A}v} dv \right) \mathbf{Bu}[k] \\[2pt]
  &= e^{\mathbf{A}T}\mathbf x[k] - \left( \int_T^0 e^{\mathbf{A}v} dv \right) \mathbf{Bu}[k] \\[2pt]
  &= e^{\mathbf{A}T}\mathbf x[k] + \left( \int_0^T e^{\mathbf{A}v} dv \right) \mathbf{Bu}[k] \\[4pt]
  &= e^{\mathbf{A}T}\mathbf x[k] + \mathbf A^{-1}\left(e^{\mathbf{A}T} - \mathbf I \right) \mathbf{Bu}[k]
\end{align}</math>

which is an exact solution to the discretization problem.

When {{math|'''A'''}} is singular, the latter expression can still be used by replacing <math> e^{\mathbf{A}T} </math> by its [[Taylor series|Taylor expansion]],
<math display=block> e^{\mathbf{A}T} = \sum_{k=0}^{\infty} \frac{1}{k!} (\mathbf{A}T)^k  .</math>
This yields
<math display=block>\begin{align} 
  \mathbf x[k+1] &= e^{\mathbf{A}T}\mathbf x[k] + \left( \int_0^T e^{\mathbf{A}v} dv \right) \mathbf{Bu}[k] \\[2pt]
  &= \left(\sum_{k=0}^{\infty} \frac{1}{k!} (\mathbf{A}T)^k\right) \mathbf x[k] + \left(\sum_{k=1}^{\infty} \frac{1}{k!} \mathbf{A}^{k-1} T^k\right) \mathbf{Bu}[k], 
\end{align}</math>
which is the form used in practice.