138 Lesson 1.3.7: Relating discrete-time and continuous-time noise covariances precisely: A little trick

138.1 Setting up the problem

In the prior lesson, you learned that

\Sigma_{\bar{w}} = \int_{0}^{\Delta t} e^{A(\Delta t-\tau)} B_w S_w B_w^T e^{A^T(\Delta t-\tau)} \, d\tau .

But how do we compute this exactly?
You learned that if \Delta t \to 0, then \Sigma_{\bar{w}} \approx B_w S_w B_w^T
But how do we find \Sigma_{\bar{w}} precisely for general \Delta t? Assume that we know S_w and the plant dynamics.
Define

Z = \begin{bmatrix} -A & B_w S_w B_w^T \\ 0 & A^T \end{bmatrix}, \qquad C = e^{Z \Delta t} = \begin{bmatrix} c_{11} & c_{12} \\ 0 & c_{22} \end{bmatrix}.

138.2 Computing C

We seek to compute C = \mathcal{L}^{-1}\!\left[(sI - Z)^{-1}\right]\Big|_{t=\Delta t}
Then if M = sI - Z, recognize the following matrix identity:

M = \begin{bmatrix} sI - z_{11} & -z_{12} \\ 0 & sI - z_{22} \end{bmatrix}; \quad M^{-1} = \begin{bmatrix} (sI-z_{11})^{-1} & (sI-z_{11})^{-1} z_{12} (sI-z_{22})^{-1} \\ 0 & (sI-z_{22})^{-1} \end{bmatrix}.

Then we have that:

\begin{align} c_{11} &= \mathcal{L}^{-1}\!\left[(sI-z_{11})^{-1}\right]\Big|_{t=\Delta t} = \mathcal{L}^{-1}\!\left[(sI+A)^{-1}\right]\Big|_{t=\Delta t} = e^{-A\Delta t},\\ c_{12} &= \mathcal{L}^{-1}\!\left[(sI-z_{11})^{-1} z_{12} (sI-z_{22})^{-1}\right]\Big|_{t=\Delta t} \\ &= \mathcal{L}^{-1}\!\left[(sI+A)^{-1} B_w S_w B_w^T (sI-A^T)^{-1}\right]\Big|_{t=\Delta t},\\ c_{22} &= \mathcal{L}^{-1}\!\left[(sI-z_{22})^{-1}\right]\Big|_{t=\Delta t} = \mathcal{L}^{-1}\!\left[(sI-A^T)^{-1}\right]\Big|_{t=\Delta t} = e^{A^T \Delta t}= A_d^T. \end{align}

138.3 The usefulness of C

So, c_{22}^T = A_d. Also, we will prove that c_{22}^T c_{12} = \Sigma_{\bar{w}}
To show this, first recognize that the transfer function of a state-space model of a system (a,b,c) with d=0 is equal to H(s) = c(sI-a)^{-1}b.
- So (sI+A)^{-1} B_w S_w B_w^T (sI-A^T)^{-1} represents the cascade of two systems:
  - one with c=I, a=-A, b=B_w;
  - the other with c=S_w B_w^T, \quad a=A^T, \quad b=I.
The term c_{12} is the inverse Laplace transform of this cascade, which is the same as the impulse response of the cascade.
The impulse response of the cascade is the convolution of the impulse responses of the two individual systems comprising the cascade.
Note that the impulse response of a state-space model with d=0 is c e^{at} b.
- So, the two systems have impulse responses I e^{-At} B_w \text{ and } S_w B_w^T e^{A^T t}.

138.4 Computing \Sigma_{\bar{w}}

The impulse response of the whole is the convolution of the two cascaded impulse responses: \int_{0}^{\Delta t} I e^{-A\tau} B_w S_w B_w^T e^{A^T(\Delta t-\tau)} \, d\tau .
Multiplying this result on the left by c_{22}^T = A_d = e^{A\Delta t},

c_{22}^T c_{12} = \int_{0}^{\Delta t} e^{A(\Delta t-\tau)} B_w S_w B_w^T e^{A^T(\Delta t-\tau)} \, d\tau = \Sigma_{\bar{w}}.

In conclusion, with one expm(.) command, we can quickly convert dynamics from continuous-time to discrete-time:

\boxed{ A_d = c_{22}^T \qquad \text{and} \qquad \Sigma_{\bar{w}} = c_{22}^T c_{12}}

138.5 A worked example (1)

\dot{x}(t) = \underbrace{\begin{bmatrix}0 & 1 \\ -1 & -1\end{bmatrix}}_{A} x(t) + \underbrace{\begin{bmatrix} 0 \\ 1 \end{bmatrix}}_{B_w} w(t)

with S_w = 0.06 \qquad \text{and} \qquad \Delta t = \frac{2\pi}{16}

Then, B_w S_w B_w^T = \begin{bmatrix}0 & 0 \\0 & 0.06 \end{bmatrix} and:

Z = \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0.06 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix} \qquad C = \begin{bmatrix} 0.91 & -0.47 & -0.006 & -0.0046 \\ 0.47 & 1.38 & 0.0046 & 0.023 \\ 0 & 0 & 0.93 & -0.32 \\ 0 & 0 & 0.32 & 0.62 \end{bmatrix}.

138.6 A worked example (2)

From c_{12} and c_{22}, we can compute the following quantities very quickly:

A_d = \begin{bmatrix} 0.93 & 0.32 \\ -0.32 & 0.62 \end{bmatrix} \quad \Sigma_{\bar{w}} = \begin{bmatrix} 0.0009 & 0.0030 \\ 0.0030 & 0.0156 \end{bmatrix} \quad B_w S_w B_w^T = \begin{bmatrix} 0 & 0 \\ 0 & 0.06 \end{bmatrix}.
Exact \Sigma_{\bar{w}} and approximate \Sigma_{\bar{w}} are very different due to large \Delta t.
Compare predictions of steady-state performance.
- Continuous: A \Sigma_{\tilde{x},ss}+ \Sigma_{\tilde{x},ss} A^T+ B_w S_w B_w^T = 0, \quad \Sigma_{\tilde{x},ss} = \begin{bmatrix}0.03 & 0 \\0 & 0.03 \end{bmatrix}
- Discrete: \Sigma_{\tilde{x},ss} = A_d \Sigma_{\tilde{x},ss} A_d^T + \Sigma_{\bar{w}}, \quad \Sigma_{\tilde{x},ss} = \begin{bmatrix} 0.03 & 0 \\ 0 & 0.03 \end{bmatrix}
Because we used discrete white noise scaled to be equivalent to the continuous noise, the steady-state predictions are the same.

138.7 What about converting S_v to \Sigma_{\tilde{v}}

This is probably a good time to introduce converting S_v to \bar{\Sigma}_v.
In continuous time, the sensor noise v(t) is assumed to be zero mean and have E[v(t_1) v(t_2)^T] = S_v \, \delta(t_1 - t_2) where S_v > 0.
Approach to converting to discrete time: analyze the discrete case as \Delta t \to 0:

v_k = \frac{1}{\Delta t} \int_0^{\Delta t} v(t)\, dt \qquad \text{(average value)}.
Then,

\bar{\Sigma}_v = E[v_k v_k^T] = \frac{1}{\Delta t^2} \int_0^{\Delta t}\int_0^{\Delta t} E[v(\gamma)v(\tau)^T]\, d\gamma\, d\tau = \frac{1}{\Delta t^2}\cdot \Delta t \cdot S_v = \frac{S_v}{\Delta t}.
As \Delta t \to 0, the discrete noise tends to an infinite impulse-like term having weight S_v, similar to S_v \delta(t).

138.8 Summary

The conversion between continuous-time and discrete-time is now complete.

Key takeaways

Discrete A_d, \Sigma_{\tilde{w}}, B_d;

\begin{align} \text{Given } \Sigma_{\tilde{x},0} &:\Sigma_{\tilde{x},k}= A_d \Sigma_{\tilde{x},k-1} A_d^T + \Sigma_{\bar{w}} \\ \text{Given } \bar{x}_0 &: \bar{x}_k = A_d \bar{x}_{k-1} + B_d u_{k-1} \end{align}

Continuous A, S_w, B_u, B_w:

\begin{align} \text{Given } \Sigma_{\tilde{x}}(0) &: \dot{\Sigma}_{\tilde{x}}(t) = A\Sigma_{\tilde{x}}(t) + \Sigma_{\tilde{x}}(t)A^T + B_w S_w B_w^T \\ \text{Given } \bar{x}(0) &: \dot{\bar{x}}(t) = A\bar{x}(t) + B_u u(t) \end{align}

All matrices may be time varying.
Equivalent \Sigma_{\bar{w}} for given S_w can be computed using the trick of this lesson.
Equivalent \Sigma_{\tilde{v}} for given S_v can also be computed as \Sigma_{\tilde{v}} = \frac{S_v}{\Delta t}.

--- title: "Lesson 1.3.7: Relating discrete-time and continuous-time noise covariances precisely: A little trick" subtitle: "Kalman Filter Boot Camp (and State Estimation)" date: "2027-02-24" description: "In this lesson, we will learn a neat trick to compute the discrete-time noise covariance from the continuous-time noise covariance and the plant dynamics." categories: - "Probability and Statistics" keywords: - "Kalman Filter" - "state estimation" - "linear algebra" --- ## Setting up the problem {#sec-kf-1.3.1-relating-discrete-and-continuous-noise-covariances} - In the prior lesson, you learned that $$ \Sigma_{\bar{w}} = \int_{0}^{\Delta t} e^{A(\Delta t-\tau)} B_w S_w B_w^T e^{A^T(\Delta t-\tau)} \, d\tau . $$ - But how do we compute this exactly? - You learned that if $\Delta t \to 0$, then $\Sigma_{\bar{w}} \approx B_w S_w B_w^T$ - But how do we find $\Sigma_{\bar{w}}$ precisely for general $\Delta t$? Assume that we know $S_w$ and the plant dynamics. - Define $$ Z = \begin{bmatrix} -A & B_w S_w B_w^T \\ 0 & A^T \end{bmatrix}, \qquad C = e^{Z \Delta t} = \begin{bmatrix} c_{11} & c_{12} \\ 0 & c_{22} \end{bmatrix}. $$ ## Computing $C$ {#sec-kf-1.3.7-computing-C} - We seek to compute $C = \mathcal{L}^{-1}\!\left[(sI - Z)^{-1}\right]\Big|_{t=\Delta t}$ - Then if $M = sI - Z$, recognize the following matrix identity: $$ M = \begin{bmatrix} sI - z_{11} & -z_{12} \\ 0 & sI - z_{22} \end{bmatrix}; \quad M^{-1} = \begin{bmatrix} (sI-z_{11})^{-1} & (sI-z_{11})^{-1} z_{12} (sI-z_{22})^{-1} \\ 0 & (sI-z_{22})^{-1} \end{bmatrix}. $$ - Then we have that: $$ \begin{align} c_{11} &= \mathcal{L}^{-1}\!\left[(sI-z_{11})^{-1}\right]\Big|_{t=\Delta t} = \mathcal{L}^{-1}\!\left[(sI+A)^{-1}\right]\Big|_{t=\Delta t} = e^{-A\Delta t},\\ c_{12} &= \mathcal{L}^{-1}\!\left[(sI-z_{11})^{-1} z_{12} (sI-z_{22})^{-1}\right]\Big|_{t=\Delta t} \\ &= \mathcal{L}^{-1}\!\left[(sI+A)^{-1} B_w S_w B_w^T (sI-A^T)^{-1}\right]\Big|_{t=\Delta t},\\ c_{22} &= \mathcal{L}^{-1}\!\left[(sI-z_{22})^{-1}\right]\Big|_{t=\Delta t} = \mathcal{L}^{-1}\!\left[(sI-A^T)^{-1}\right]\Big|_{t=\Delta t} = e^{A^T \Delta t}= A_d^T. \end{align} $$ ## The usefulness of $C$ {#sec-kf-1.3.7-usefulness-of-C} - So, $c_{22}^T = A_d.$ Also, we will prove that $c_{22}^T c_{12} = \Sigma_{\bar{w}}$ - To show this, first recognize that the transfer function of a state-space model of a system $(a,b,c)$ with $d=0$ is equal to $H(s) = c(sI-a)^{-1}b.$ - So $(sI+A)^{-1} B_w S_w B_w^T (sI-A^T)^{-1}$ represents the cascade of two systems: - one with $c=I$, $a=-A$, $b=B_w$; - the other with $c=S_w B_w^T, \quad a=A^T, \quad b=I$. - The term $c_{12}$ is the inverse Laplace transform of this cascade, which is the same as the impulse response of the cascade. - The impulse response of the cascade is the convolution of the impulse responses of the two individual systems comprising the cascade. - Note that the impulse response of a state-space model with $d=0$ is $c e^{at} b.$ - So, the two systems have impulse responses $I e^{-At} B_w \text{ and } S_w B_w^T e^{A^T t}.$ ## Computing $\Sigma_{\bar{w}}$ {#sec-kf-1.3.7-computing-sigma-bar-w} - The impulse response of the whole is the convolution of the two cascaded impulse responses: $$ \int_{0}^{\Delta t} I e^{-A\tau} B_w S_w B_w^T e^{A^T(\Delta t-\tau)} \, d\tau . $$ - Multiplying this result on the left by $c_{22}^T = A_d = e^{A\Delta t},$ $$ c_{22}^T c_{12} = \int_{0}^{\Delta t} e^{A(\Delta t-\tau)} B_w S_w B_w^T e^{A^T(\Delta t-\tau)} \, d\tau = \Sigma_{\bar{w}}. $$ - In conclusion, with one `expm(.)` command, we can quickly convert dynamics from continuous-time to discrete-time: $$ \boxed{ A_d = c_{22}^T \qquad \text{and} \qquad \Sigma_{\bar{w}} = c_{22}^T c_{12}} $$ ## A worked example (1) {#sec-kf-1.3.7-worked-example-1} - Let $$ \dot{x}(t) = \underbrace{\begin{bmatrix}0 & 1 \\ -1 & -1\end{bmatrix}}_{A} x(t) + \underbrace{\begin{bmatrix} 0 \\ 1 \end{bmatrix}}_{B_w} w(t) $$ with $S_w = 0.06 \qquad \text{and} \qquad \Delta t = \frac{2\pi}{16}$ Then, $B_w S_w B_w^T = \begin{bmatrix}0 & 0 \\0 & 0.06 \end{bmatrix}$ and: $$ Z = \begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0.06 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix} \qquad C = \begin{bmatrix} 0.91 & -0.47 & -0.006 & -0.0046 \\ 0.47 & 1.38 & 0.0046 & 0.023 \\ 0 & 0 & 0.93 & -0.32 \\ 0 & 0 & 0.32 & 0.62 \end{bmatrix}. $$ ## A worked example (2) {#sec-kf-1.3.7-worked-example-2} - From $c_{12}$ and $c_{22}$, we can compute the following quantities very quickly: $$ A_d = \begin{bmatrix} 0.93 & 0.32 \\ -0.32 & 0.62 \end{bmatrix} \quad \Sigma_{\bar{w}} = \begin{bmatrix} 0.0009 & 0.0030 \\ 0.0030 & 0.0156 \end{bmatrix} \quad B_w S_w B_w^T = \begin{bmatrix} 0 & 0 \\ 0 & 0.06 \end{bmatrix}. $$ - Exact $\Sigma_{\bar{w}}$ and approximate $\Sigma_{\bar{w}}$ are very different due to large $\Delta t$. - Compare predictions of steady-state performance. - Continuous: $A \Sigma_{\tilde{x},ss}+ \Sigma_{\tilde{x},ss} A^T+ B_w S_w B_w^T = 0, \quad \Sigma_{\tilde{x},ss} = \begin{bmatrix}0.03 & 0 \\0 & 0.03 \end{bmatrix}$ - Discrete: $\Sigma_{\tilde{x},ss} = A_d \Sigma_{\tilde{x},ss} A_d^T + \Sigma_{\bar{w}}, \quad \Sigma_{\tilde{x},ss} = \begin{bmatrix} 0.03 & 0 \\ 0 & 0.03 \end{bmatrix}$ - Because we used discrete white noise scaled to be equivalent to the continuous noise, the steady-state predictions are the same. ## What about converting $S_v$ to $\Sigma_{\tilde{v}}$ {#sec-kf-1.3.7-converting-Sv-to-sigma-bar-v} - This is probably a good time to introduce converting $S_v$ to $\bar{\Sigma}_v$. - In continuous time, the sensor noise $v(t)$ is assumed to be zero mean and have $E[v(t_1) v(t_2)^T] = S_v \, \delta(t_1 - t_2)$ where $S_v > 0$. - Approach to converting to discrete time: analyze the discrete case as $\Delta t \to 0$: $$ v_k = \frac{1}{\Delta t} \int_0^{\Delta t} v(t)\, dt \qquad \text{(average value)}. $$ - Then, $$ \bar{\Sigma}_v = E[v_k v_k^T] = \frac{1}{\Delta t^2} \int_0^{\Delta t}\int_0^{\Delta t} E[v(\gamma)v(\tau)^T]\, d\gamma\, d\tau = \frac{1}{\Delta t^2}\cdot \Delta t \cdot S_v = \frac{S_v}{\Delta t}. $$ - As $\Delta t \to 0$, the discrete noise tends to an infinite impulse-like term having weight $S_v$, similar to $S_v \delta(t)$. ## Summary {#sec-kf-1.3.7-summary} - The conversion between continuous-time and discrete-time is now complete. ::: {.callout-tip } ## Key takeaways **Discrete** $A_d$, $\Sigma_{\tilde{w}}$, $B_d$; $$ \begin{align} \text{Given } \Sigma_{\tilde{x},0} &:\Sigma_{\tilde{x},k}= A_d \Sigma_{\tilde{x},k-1} A_d^T + \Sigma_{\bar{w}} \\ \text{Given } \bar{x}_0 &: \bar{x}_k = A_d \bar{x}_{k-1} + B_d u_{k-1} \end{align} $$ **Continuous** $A$, $S_w$, $B_u$, $B_w$: $$ \begin{align} \text{Given } \Sigma_{\tilde{x}}(0) &: \dot{\Sigma}_{\tilde{x}}(t) = A\Sigma_{\tilde{x}}(t) + \Sigma_{\tilde{x}}(t)A^T + B_w S_w B_w^T \\ \text{Given } \bar{x}(0) &: \dot{\bar{x}}(t) = A\bar{x}(t) + B_u u(t) \end{align} $$ - All matrices may be time varying. - Equivalent $\Sigma_{\bar{w}}$ for given $S_w$ can be computed using the trick of this lesson. - Equivalent $\Sigma_{\tilde{v}}$ for given $S_v$ can also be computed as $\Sigma_{\tilde{v}} = \frac{S_v}{\Delta t}.$ :::