# Kalman Filter: Why $Q$ Discrete Is Defined as $\int_0^Te^{\mathbf{A}\tau} Q e^{\mathbf{A}^T \tau} d\tau$?

I would like to ask, why in the transformation to the discretization, $\mathbf{Q}$ is obtained from the expression containing the integral (image attached), what is the theory behind it?

– Royi
Nov 21, 2022 at 18:55

Somewhere prior to this step (in the State Transition Equations) there was something along the lines of $$\mathbf{e ^{A(t-\tau)}}$$ which probably factors as

$$\mathbf{e ^{A(t-\tau)}}=\mathbf{e ^{A(t)}} \mathbf{e ^{-A(\tau)}}$$

so things like $$\int_0^T \mathbf{e ^{A(t-\tau)}} \mathbf{G}\mathbf{d\tau}$$ which should have a certain familiar form, can under certain conditions as $$\mathbf{e ^{A(t)}}\int_0^T \mathbf{e ^{A(-\tau)}}\mathbf{G}\mathbf{d\tau}$$ There is also a commutative property on $\mathbf{e ^{A(-\tau)}}$ that is important.

The simple answer is that you discretize Differntial Equation.
Actually a linear differential equation.

The question is why the Matrix Exponential?
Well, try to remember what are the solutions of Linear Scalar Differential equations.
The solutions are formed by the expontial function.
Hence what you see is the Exponent of a Matrix (Since in the General Kalman case we have a vector Linear Differential Equation) as solution (Which is the integration) and we discretize this because in real life the model is discrete.

You may have a look at ECE 595 - Discrete-Time Control Systems - Discretization of Continuous Time State Space Systems.
The solution given there for the Differential Equation in a vector form holds for the cases all variables and coefficients are scalars.

• Thanks Royi. Really useful ECE 595 link. I'm gonna find that other thread where the person asks the question directly and link to it there. Jul 18, 2018 at 6:27
• plus oned the "your welcome" comment by accident but don't know how to undo it. Also, put a 1 on the answer. Jul 18, 2018 at 15:46