126 Lesson 1.2.3: Understanding the time-domain response of a state-space model

126.1 Understanding the time-domain response of a state-space model

126.1.1 Intro to continuous-time state-space models

You have seen that simulation is a valuable tool for understanding state-space models.
But, we would like to develop more insight into the system response by more general analytic means; specifically, by looking at the time-domain solution for x(t).
We solve first for the homogeneous solution (u(t) = 0):
- Start with \dot x(t) = A x(t) and some initial state x(0).
- Take Laplace transform: \begin{aligned} sX(s) - x(0) &= AX(s) && \text{Initial value theorem} \\ (sI - A)X(s) &= x(0) && \text{take care with matrix dimensions!} \\ X(s) &= (sI - A)^{-1}x(0) && \text{Assume invertible} \end{aligned}
  
  Detailed Laplace-transform knowledge will be helpful
So far we have: x(t) = \mathcal{L}^{-1}[(sI - A)^{-1}]x(0) \tag{126.1}

126.1.2 The state-transition matrix

Recapping, we have: Equation 126.1. But, (sI - A)^{-1} = \frac{I}{s}+\frac{A}{s^2}+\frac{A^2}{s^3}+\frac{A^3}{s^4}+\cdots \quad\text{(geometric series)} so, \mathcal{L}^{-1}[sI - A]^{-1} = I + At + \frac{1}{2!}A^2t^2 + \frac{1}{3!}A^3t^3 + \cdots \stackrel{\triangle}{=} e^{At} \quad\text{(matrix exponential)}
Therefore, we write x(t) =e^{At} x(0)
- e^{At} is called the “transition matrix” or “state-transition matrix.”
- Note that e^{At} is not the same as taking the exponential of every element of At.
- e^{(A+B)t}=e^{At}e^{Bt} iff AB=BA: (that is, not in general).
- In Octave, can compute x(t) as: x = expm(A*t)*x0;
Will say more about e^{At} when we discuss the structure of A.

126.1.3 Computing the state-transition matrix by hand

Computing e^{At} = \mathcal{L}^{-1}[sI - A]^{-1} is straightforward for 2 \times 2.
EXAMPLE: Find e^{At} when A = \begin{bmatrix} 0 & 1 \\ 2 & 3 \end{bmatrix}.
- Solve: \begin{aligned} (sI - A)^{-1} &= \begin{bmatrix} s & -1 \\ 2 & s+3 \end{bmatrix}^{-1} \\ &= \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \\ &= \begin{bmatrix} \frac{ 2}{s+1} - \frac{1}{s+2} & \frac{ 1}{s+1} - \frac{1}{s+2} \\ \frac{-2}{s+1} + \frac{2}{s+2} & \frac{-1}{s+1} + \frac{2}{s+2} \end{bmatrix} \\ e^{At} &= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & - e^{-t} + 2e^{-2t} \end{bmatrix} \mathbb{1}(t) \end{aligned}
This is probably the best way to find e^{At} if the A matrix is 2 \times 2 .

126.1.4 Forced state response

Solving for the forced solution (u(t) \ne 0), we find: x(t) = e^{At} x(0) + \underbrace{\int_0^t e^{A(t-\tau)} B u(\tau) d\tau}_{\text{convolution}}
Where did this come from?

{\color{blue} \dot{x}(t) - A x(t)} = {\color{yellow}B u(t)}, simply by rearranging the state equation.
Multiply both sides by e^{At} and rewrite the LHS as the middle expression: e^{-At} {\color{blue}[\dot x(t) - A x(t)]} = {\color{red}\frac{d}{dt} (e^{-At} x(t))} = e^{-At} {\color{yellow}B u(t)} d\tau
Integrate the middle and the RHS; rearrange and keep new middle and RHS: \begin{aligned} \int_0^t {\color{red}\frac{d}{d\tau} [ e^{-A(t-\tau)} x(\tau)]} d\tau &= e^{-At}x(t)-x(0) \\ &= \int_0^t e^{-A(t-\tau)} {\color{yellow} B u(\tau)} d\tau \end{aligned}

126.1.5 Forced output response

So, we have established the following state dynamics: x(t) = e^{At} x(0) + \underbrace{\int_0^t e^{A(t-\tau)}{\color{yellow}B u(\tau)} d\tau}_{\text{convolution}}
Since z(t) = C x(t) + Du(t), the system output is then comprised of three parts: z(t) = \underbrace{\vphantom{\int_0^t} C e^{At} x(0)}_{\text{initial resp}}+ \underbrace{\int_0^t Ce^{A(t-\tau)}{\color{yellow}Bu(\tau)} d\tau}_{\text{convolution}} + \underbrace{\vphantom{\int_0^t} Du(t)}_{\text{feed through}}

126.1.6 Summary

You have learned how to compute the homogeneous and forced response of a state-space model.
The output z(t) comprises of:
- a response due to initial conditions : C e^{At} x(0)
- a dynamic response due to forcing input : \int_0^t Ce^{A(t-\tau)}Bu(\tau) d\tau
- an instantaneous response due to the feed-through : Du(t)
These terms rely on the matrix exponential e^{At}.
- Caution e^{At} is not the term by term exponential operation on elements of At.
  - Don’t compute it as: ~~exp(A*t)~~;
- One way to compute it (e.g., by hand) is via: e^{At} = \mathcal{L}^{-1}[sI - A]^{-1}.
- Or, in Octave, it is correct to compute: expm(A*t);
We will dive deeper into the matrix exponential in Section 127.1

--- title: "Lesson 1.2.3: Understanding the time-domain response of a state-space model" subtitle: "Kalman Filter Boot Camp (and State Estimation)" description: "This appendix explains the Kalman Filter, a mathematical method for estimating the state of a dynamic system from a series of noisy measurements." categories: - "Probability and Statistics" keywords: - "Kalman Filter" - "state estimation" - "linear algebra" --- ## Understanding the time-domain response of a state-space model {#sec-kf-1.2.3-time-domain-response} ### Intro to continuous-time state-space models - You have seen that simulation is a valuable tool for understanding state-space models. - But, we would like to develop more insight into the system response by more general analytic means; specifically, by looking at the time-domain solution for $x(t)$. - We solve first for the homogeneous solution $(u(t) = 0)$: - Start with $\dot x(t) = A x(t)$ and some initial state $x(0)$. - Take Laplace transform: $$ \begin{aligned} sX(s) - x(0) &= AX(s) && \text{Initial value theorem} \\ (sI - A)X(s) &= x(0) && \text{take care with matrix dimensions!} \\ X(s) &= (sI - A)^{-1}x(0) && \text{Assume invertible} \end{aligned} $$ Detailed Laplace-transform knowledge will be helpful - So far we have: $$ x(t) = \mathcal{L}^{-1}[(sI - A)^{-1}]x(0) $$ {#eq-ct-model-homogeneous-solution} ### The state-transition matrix - Recapping, we have: @eq-ct-model-homogeneous-solution. But, $$ (sI - A)^{-1} = \frac{I}{s}+\frac{A}{s^2}+\frac{A^2}{s^3}+\frac{A^3}{s^4}+\cdots \quad\text{(geometric series)} $$ so, $$ \mathcal{L}^{-1}[sI - A]^{-1} = I + At + \frac{1}{2!}A^2t^2 + \frac{1}{3!}A^3t^3 + \cdots \stackrel{\triangle}{=} e^{At} \quad\text{(matrix exponential)} $$ - Therefore, we write $x(t) =e^{At} x(0)$ - $e^{At}$ is called the “transition matrix” or “state-transition matrix.” - Note that $e^{At}$ is not the same as taking the exponential of every element of $At$. - $e^{(A+B)t}=e^{At}e^{Bt}$ iff $AB=BA$: (that is, not in general). - In Octave, can compute $x(t)$ as: `x = expm(A*t)*x0;` - Will say more about $e^{At}$ when we discuss the structure of $A$. ### Computing the state-transition matrix by hand - Computing $e^{At} = \mathcal{L}^{-1}[sI - A]^{-1}$ is straightforward for $2 \times 2$. - EXAMPLE: Find $e^{At}$ when $A = \begin{bmatrix} 0 & 1 \\ 2 & 3 \end{bmatrix}$. - Solve: $$ \begin{aligned} (sI - A)^{-1} &= \begin{bmatrix} s & -1 \\ 2 & s+3 \end{bmatrix}^{-1} \\ &= \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \\ &= \begin{bmatrix} \frac{ 2}{s+1} - \frac{1}{s+2} & \frac{ 1}{s+1} - \frac{1}{s+2} \\ \frac{-2}{s+1} + \frac{2}{s+2} & \frac{-1}{s+1} + \frac{2}{s+2} \end{bmatrix} \\ e^{At} &= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & - e^{-t} + 2e^{-2t} \end{bmatrix} \mathbb{1}(t) \end{aligned} $$ - This is probably the best way to find $e^{At}$ if the $A$ matrix is $2 \times 2$ . ### Forced state response - Solving for the forced solution $(u(t) \ne 0)$, we find: $$ x(t) = e^{At} x(0) + \underbrace{\int_0^t e^{A(t-\tau)} B u(\tau) d\tau}_{\text{convolution}} $$ - Where did this come from? 1. ${\color{blue} \dot{x}(t) - A x(t)} = {\color{yellow}B u(t)}$, simply by rearranging the state equation. 2. Multiply both sides by $e^{At}$ and rewrite the LHS as the middle expression: $$ e^{-At} {\color{blue}[\dot x(t) - A x(t)]} = {\color{red}\frac{d}{dt} (e^{-At} x(t))} = e^{-At} {\color{yellow}B u(t)} d\tau $$ 3. Integrate the middle and the RHS; rearrange and keep new middle and RHS: $$ \begin{aligned} \int_0^t {\color{red}\frac{d}{d\tau} [ e^{-A(t-\tau)} x(\tau)]} d\tau &= e^{-At}x(t)-x(0) \\ &= \int_0^t e^{-A(t-\tau)} {\color{yellow} B u(\tau)} d\tau \end{aligned} $$ ### Forced output response - So, we have established the following state dynamics: $$ x(t) = e^{At} x(0) + \underbrace{\int_0^t e^{A(t-\tau)}{\color{yellow}B u(\tau)} d\tau}_{\text{convolution}} $$ - Since $z(t) = C x(t) + Du(t)$, the system output is then comprised of three parts: $$ z(t) = \underbrace{\vphantom{\int_0^t} C e^{At} x(0)}_{\text{initial resp}}+ \underbrace{\int_0^t Ce^{A(t-\tau)}{\color{yellow}Bu(\tau)} d\tau}_{\text{convolution}} + \underbrace{\vphantom{\int_0^t} Du(t)}_{\text{feed through}} $$ ### Summary - [You have learned how to compute the **homogeneous** and **forced** response of a state-space model.]{.mark} - The output $z(t)$ comprises of: - [a response due to **initial conditions** :]{.mark} $C e^{At} x(0)$ - [a dynamic response due to **forcing input** :]{.mark} $\int_0^t Ce^{A(t-\tau)}Bu(\tau) d\tau$ - [an instantaneous response due to the **feed-through** :]{.mark} $Du(t)$ - These terms rely on the matrix exponential $e^{At}$. - **Caution** $e^{At}$ is not the term by term exponential operation on elements of $At$. - Don't compute it as: ~~`exp(A*t)`~~; - One way to compute it (e.g., by hand) is via: $e^{At} = \mathcal{L}^{-1}[sI - A]^{-1}$. - Or, in Octave, it is correct to compute: `expm(A*t)`; - We will dive deeper into the matrix exponential in [@sec-kf-1.2.4-time-domain-response]