Discretization of process noise

Numerical evaluation of ${\displaystyle \mathbf {Q} _{d}}$ is a bit trickier due to the matrix exponential integral. It can, however, be computed by first constructing a matrix, and computing the exponential of it^[3]

${\displaystyle \mathbf {F} ={\begin{bmatrix}-\mathbf {A} &\mathbf {Q} \\\mathbf {0} &\mathbf {A} ^{\top }\end{bmatrix}}T}$

${\displaystyle \mathbf {G} =e^{\mathbf {F} }={\begin{bmatrix}\dots &\mathbf {A} _{d}^{-1}\mathbf {Q} _{d}\\\mathbf {0} &\mathbf {A} _{d}^{\top }\end{bmatrix}}.}$

The discretized process noise is then evaluated by multiplying the transpose of the lower-right partition of G with the upper-right partition of G:

${\displaystyle \mathbf {Q} _{d}=(\mathbf {A} _{d}^{\top })^{\top }(\mathbf {A} _{d}^{-1}\mathbf {Q} _{d})=\mathbf {A} _{d}(\mathbf {A} _{d}^{-1}\mathbf {Q} _{d}).}$

Derivation

Starting with the continuous model

${\displaystyle \mathbf {\dot {x}} (t)=\mathbf {A} \mathbf {x} (t)+\mathbf {B} \mathbf {u} (t)}$

we know that the matrix exponential is

${\displaystyle {\frac {d}{dt}}e^{\mathbf {A} t}=\mathbf {A} e^{\mathbf {A} t}=e^{\mathbf {A} t}\mathbf {A} }$

and by premultiplying the model we get

${\displaystyle e^{-\mathbf {A} t}\mathbf {\dot {x}} (t)=e^{-\mathbf {A} t}\mathbf {A} \mathbf {x} (t)+e^{-\mathbf {A} t}\mathbf {B} \mathbf {u} (t)}$

which we recognize as

${\displaystyle {\frac {d}{dt}}(e^{-\mathbf {A} t}\mathbf {x} (t))=e^{-\mathbf {A} t}\mathbf {B} \mathbf {u} (t)}$

and by integrating..

${\displaystyle e^{-\mathbf {A} t}\mathbf {x} (t)-e^{0}\mathbf {x} (0)=\int _{0}^{t}e^{-\mathbf {A} \tau }\mathbf {B} \mathbf {u} (\tau )d\tau }$

${\displaystyle \mathbf {x} (t)=e^{\mathbf {A} t}\mathbf {x} (0)+\int _{0}^{t}e^{\mathbf {A} (t-\tau )}\mathbf {B} \mathbf {u} (\tau )d\tau }$

which is an analytical solution to the continuous model.

Now we want to discretise the above expression. We assume that u is constant during each timestep.

${\displaystyle \mathbf {x} [k]\ {\stackrel {\mathrm {def} }{=}}\ \mathbf {x} (kT)}$

${\displaystyle \mathbf {x} [k]=e^{\mathbf {A} kT}\mathbf {x} (0)+\int _{0}^{kT}e^{\mathbf {A} (kT-\tau )}\mathbf {B} \mathbf {u} (\tau )d\tau }$

${\displaystyle \mathbf {x} [k+1]=e^{\mathbf {A} (k+1)T}\mathbf {x} (0)+\int _{0}^{(k+1)T}e^{\mathbf {A} ((k+1)T-\tau )}\mathbf {B} \mathbf {u} (\tau )d\tau }$

${\displaystyle \mathbf {x} [k+1]=e^{\mathbf {A} T}\left[e^{\mathbf {A} kT}\mathbf {x} (0)+\int _{0}^{kT}e^{\mathbf {A} (kT-\tau )}\mathbf {B} \mathbf {u} (\tau )d\tau \right]+\int _{kT}^{(k+1)T}e^{\mathbf {A} (kT+T-\tau )}\mathbf {B} \mathbf {u} (\tau )d\tau }$

We recognize the bracketed expression as ${\displaystyle \mathbf {x} [k]}$, and the second term can be simplified by substituting with the function ${\displaystyle v(\tau )=kT+T-\tau }$. Note that ${\displaystyle d\tau =-dv}$. We also assume that ${\displaystyle \mathbf {u} }$ is constant during the integral, which in turn yields

${\displaystyle {\begin{matrix}\mathbf {x} [k+1]&=&e^{\mathbf {A} T}\mathbf {x} [k]-\left(\int _{v(kT)}^{v((k+1)T)}e^{\mathbf {A} v}dv\right)\mathbf {B} \mathbf {u} [k]\\&=&e^{\mathbf {A} T}\mathbf {x} [k]-\left(\int _{T}^{0}e^{\mathbf {A} v}dv\right)\mathbf {B} \mathbf {u} [k]\\&=&e^{\mathbf {A} T}\mathbf {x} [k]+\left(\int _{0}^{T}e^{\mathbf {A} v}dv\right)\mathbf {B} \mathbf {u} [k]\\&=&e^{\mathbf {A} T}\mathbf {x} [k]+\mathbf {A} ^{-1}\left(e^{\mathbf {A} T}-\mathbf {I} \right)\mathbf {B} \mathbf {u} [k]\end{matrix}}}$

which is an exact solution to the discretization problem.

When ${\displaystyle \mathbf {A} }$ is singular, the latter expression can still be used by replacing ${\displaystyle e^{\mathbf {A} T}}$ by its Taylor expansion,

${\displaystyle e^{{\mathbf {A} }T}=\sum _{k=0}^{\infty }{\frac {1}{k!}}({\mathbf {A} }T)^{k}.}$

This yields

${\displaystyle {\begin{matrix}\mathbf {x} [k+1]&=&e^{{\mathbf {A} }T}\mathbf {x} [k]+\left(\int _{0}^{T}e^{{\mathbf {A} }v}dv\right)\mathbf {B} \mathbf {u} [k]\\&=&\left(\sum _{k=0}^{\infty }{\frac {1}{k!}}({\mathbf {A} }T)^{k}\right)\mathbf {x} [k]+\left(\sum _{k=1}^{\infty }{\frac {1}{k!}}{\mathbf {A} }^{k-1}T^{k}\right)\mathbf {B} \mathbf {u} [k],\end{matrix}}}$

which is the form used in practice.

Approximations

Exact discretization may sometimes be intractable due to the heavy matrix exponential and integral operations involved. It is much easier to calculate an approximate discrete model, based on that for small timesteps ${\displaystyle e^{\mathbf {A} T}\approx \mathbf {I} +\mathbf {A} T}$. The approximate solution then becomes:

${\displaystyle \mathbf {x} [k+1]\approx (\mathbf {I} +\mathbf {A} T)\mathbf {x} [k]+T\mathbf {B} \mathbf {u} [k]}$

This is also known as the Euler method, which is also known as the forward Euler method. Other possible approximations are ${\displaystyle e^{\mathbf {A} T}\approx \left(\mathbf {I} -\mathbf {A} T\right)^{-1}}$, otherwise known as the backward Euler method and ${\displaystyle e^{\mathbf {A} T}\approx \left(\mathbf {I} +{\frac {1}{2}}\mathbf {A} T\right)\left(\mathbf {I} -{\frac {1}{2}}\mathbf {A} T\right)^{-1}}$, which is known as the bilinear transform, or Tustin transform. Each of these approximations has different stability properties. The bilinear transform preserves the instability of the continuous-time system.