Time response of state space models
1 Solution to matrix differential equations
Recall that for a scalar differential equation (i.e., \(x(t) \in \reals\)): \[ \dot x(t) = a x(t), \quad \hbox{initial condition } x(0)=x_0 \] has the solution \[ x(t) = e^{at} x_0. \]
Can we say the same for vector systems? That is, can we define matrix exponential \(e^{At}\) in such a way that the matrix differential equation (i.e., \(x(t) \in \reals^n\)): \[ \dot x(t) = A x(t), \quad \hbox{initial condition } x(0) = x_0 \] where \(A \in \reals^{n × n}\) has the solution \[ x(t) = e^{At} x_0? \]
Mathematically, it turns out that this is straight forward. Recall that for a scalar \(a\), we have \[e^{at} = 1 + at + \frac{(at)^2}{2!} + \frac{(at)^3}{3!} + \cdots\] So, a natural choice is to define matrix exponential as \[\begin{align*} e^{At} &= I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + \cdots\\ &= I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!} + \cdots \end{align*}\] where the second equation uses the fact that \((At)^2 = At At = A^2 t^2\) because \(t\) is a scalar.
With this definition, we have \[\begin{align*} \frac{d}{dt} e^{At} &= A + A^2 t + \frac{A^3 t^2}{2!} + \cdots \\ &= A[I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!} + \cdots] \\ &= A e^{At}. \end{align*}\]
Thus, if we take the candidate solution \(x(t) = e^{At} x_0\), we have that \[ \frac{d}{dt} x(t) = \left[ \frac{d}{dt} e^{At} \right] x_0 = A e^{At} x_0 = A x(t). \] Thus, our candidate solution satisfies the matrix differential equation!
If we define \(e^{At}\) as \[\begin{equation}\label{eq:matrix-exponential} e^{At} = I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!} + \cdots \end{equation}\]
Then, we can write the solution of the matrix differential equation (i.e., \(x(t) \in \reals^n\)) \[ \dot x(t) = A x(t), \quad \hbox{initial condition } x(0) = x_0 \] where \(A \in \reals^{n × n}\) as \[ x(t) = e^{At} x_0. \]
We now present some examples where the matrix exponential can be computed easily.
Example 1 Compute \(e^{At}\) for \[A = \MATRIX{0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0}. \]
We compute the terms of \(\eqref{eq:matrix-exponential}\):
- \(A^2 = \MATRIX{0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 }\).
- \(A^3 = \MATRIX{0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 }\).
which means that \(A^k = 0\) for \(k \ge 0\). Thus, the right hand side of \(\eqref{eq:matrix-exponential}\) contains only a finite number of nonzero terms: \[\begin{align*} e^{At} &= I + At + \frac 12 A^2 t^2 \\ &= \MATRIX{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 } + \MATRIX{0 & t & 0 \\ 0 & 0 & t \\ 0 & 0 & 0} + \frac 12 \MATRIX{0 & 0 & t^2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 } \\ &=\MATRIX{1 & t & \frac 12 t^2 \\ 0 & 1 & t \\ 0 & 0 & 1}. \end{align*}\]
Example 2 Compute \(e^{At}\) for \[A = \MATRIX{λ_1 & 0 & 0 \\ 0 & λ_2 & 0 \\ 0 & 0 & λ_3}. \]
We compute the terms of \(\eqref{eq:matrix-exponential}\):
- \(A^2 = \MATRIX{λ_1^2 & 0 & 0 \\ 0 & λ_2^2 & 0 \\ 0 & 0 & λ_3^2}.\)
- \(A^3 = \MATRIX{λ_1^3 & 0 & 0 \\ 0 & λ_2^3 & 0 \\ 0 & 0 & λ_3^3}.\)
- and so on.
Thus, we have \[\begin{align*} e^{At} &= I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!} + \cdots \\ &= \MATRIX{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} + \MATRIX{λ_1 t & 0 & 0 \\ 0 & λ_2 t & 0 \\ 0 & 0 & λ_3 t} + \frac 12 \MATRIX{λ_1^2 t^2 & 0 & 0 \\ 0 & λ_2^2 t^2 & 0 \\ 0 & 0 & λ_3^2 t^2} + \cdots \\[10pt] &= \MATRIX{1 + λ_1 t + \frac 12 λ_1^2 t^2 + \cdots & 0 & 0 \\ 0 & 1 + λ_2 t + \frac 12 λ_2^2 t^2 + \cdots & 0 \\ 0 & 0 & 1 + λ_3 t + \frac 12 λ_3^2 t^2 + \cdots} \\[10pt] &= \MATRIX{e^{λ_1t} & 0 & 0 \\ 0 & e^{λ_2 t} & 0 \\ 0 & 0 & e^{λ_3 t}} \end{align*}\]
But outside of such few special cases, computing \(e^{At}\) via definition \(\eqref{eq:matrix-exponential}\) is not computationally feasible. We now present computationally efficient methods to compute the matrix exponential.
2 Computing matrix exponential
2.1 Method 1: Eigenvalue diagonalization method
As illustrated by Example 2, compute matrix exponential is easy for diagonal matrix. So, if the matrix \(A\) is diagonalizable (i.e., has distinct eigenvalues) we can do a change of coordinates and compute the matrix exponential in the eigen-coordinates.
In particular, suppose \(A\) has distinct eivenvalues \(λ_1, \dots, λ_n\) and \(v_1, \dots, v_n\) are the corresponding eigvenvectors. Thus, $ A v_k = λ_k v_k$ for all \(k \in \{1,\dots, n\}\). Writing this in matrix form, we have \[\begin{align*} A \MATRIX{v_1 & v_2 & \cdots & v_n} &= \MATRIX{A v_1 & A v_2 & \cdots & A v_n} \\ &= \MATRIX{λ_1 v_1 & λ_2 v_2 & \cdots & λ_n v_n} \\ &= \MATRIX{v_1 & v_2 & \cdots & v_n} \MATRIX{ λ_1 & 0 & \cdots & 0 \\ 0 & λ_2 & \ddots & 0 \\ 0 & \cdots & \ddots & 0 \\ 0 & \cdots & \cdots & λ_n} \end{align*}\] Now define \[T = \MATRIX{v_1 & v_2 & \cdots & v_n} \quad\text{and}\quad Λ = \MATRIX{ λ_1 & 0 & \cdots & 0 \\ 0 & λ_2 & \ddots & 0 \\ 0 & \cdots & \ddots & 0 \\ 0 & \cdots & \cdots & λ_n}. \] So, the above equation can be writen as \(AT = T Λ\) or \[ \bbox[5pt,border: 1px solid] {T^{-1} A T = Λ} \] Observe that
- \(A = T Λ T^{-1}\)
- \(A^2 = T Λ^2 T^{-1}\)
- \(A^3 = T Λ^3 T^{-1}\)
- and so on.
Therefore, \[\begin{align*} e^{At} &= I + At + \frac{A^2 t^2}{2!} + \frac{A^3 t^3}{3!} + \cdots \\ &= I + T Λ T^{-1} t + \frac{T Λ^2 T^{-1} t^2}{2!} + \frac{T Λ^3 T^{-1} t^3}{3!} + \cdots \\ &= T\bigl( I + Λt + \frac{Λ^2 t^2}{2!} + \frac{Λ^3 t^3}{3!} + \cdots\bigr) T^{-1} \\ &= T e^{Λ t} T^{-1} \end{align*}\]
Thus, once we know the eigenvalues and eigenvectors of \(A\) (and if all eigenvalues are distinct), then \[ \bbox[5pt,border: 1px solid] {e^{At} = T e^{Λt} T^{-1}} \]
Exercise 1 Use the eigenvalue diagonalizable method to compute \(e^{At}\) for \(A = \MATRIX{-6 & 4 \\ -5 & 3 }\).
We start by computing the eigenvalues and eigenvectors of \(A\).
To compute the eigenvalues: \[sI - A = \MATRIX{s + 6 & -4 \\ 5 & s-3 }\] Therefore, the characteristic equation is \[\det(sI - A) = s^2 + 3s + 2 = (s+1)(s+2) \] Hence, the eigenvalues of \(A\) are \(λ_1 = -1\) and \(λ_2 = -2\).
We now compute the eigenvalues. Recall that for any eigenvalue \(λ\), the eigen-vector satisfies \((λI - A) v = 0\). We start wtih \(λ_1 = -1\). Then, \[ \MATRIX{5 & -4 \\ 5 & -4 } \MATRIX{ v_{11} \\ v_{12} } = 0 \] We set \(v_{11} = 4\). Then, \(v_{12} = 5\). Thus, the eigenvector \[ v_1 = \MATRIX{ 4 \\ 5}. \]
Similarly, for \(λ_2 = -2\), we have \[ \MATRIX{4 & -4 \\ 5 & -5 } \MATRIX{ v_{21} \\ v_{22} } = -2 \MATRIX{ v_{21} \\ v_{22} } \] We set \(v_{21} = 1\). Then, \(v_{22} = 1\). Thus, the eigenvector \[ v_2 = \MATRIX{ 1 \\ 1}. \]
Thus, \(T = \MATRIX{ 4 & 1 \\ 5 & 1 }\) and therefore \(T^{-1} = \MATRIX{-1 & 1 \\ 5 & -4 }\). Hence, \[ e^{At} = \MATRIX{ 4 & 1 \\ 5 & 1 } \MATRIX{e^{-t} & 0 \\ 0 & e^{-2t} } \MATRIX{-1 & 1 \\ 5 & -4 } = \MATRIX{ -4 e^{-t} + 5 e^{-2t} & 4 e^{-t} - 4 e^{-2t} \\ -5 e^{-t} + 5 e^{-2t} & 5 e^{-t} - 4 e^{-2t} } \]
2.2 Method 2: Laplace transform method
Recall that we have \[ \dot x(t) = A x(t). \] Taking (unilateral) Laplace transforms of both sides gives \[ s X(s) - x_0 = A X(s). \] Rearranging terms, we get \[ (sI - A)X(s) = x_0 \implies X(s) = (sI-A)^{-1} x_0. \] Taking inverse Laplace transforms, we get \[x(t) = \mathcal L^{-1}( (sI - A)^{-1} ) x_0. \] Comparing it with the solution obtained earlier, we get \[ \bbox[5pt,border: 1px solid] { e^{At} = \mathcal L^{-1}( (sI - A)^{-1} ).} \] In the above, we interpret the inverse Laplace transform of a matrix to mean the inverse Laplace transform of each entry.
Exercise 2 Use the Laplace transform method to compute \(e^{At}\) for \(A = \MATRIX{-6 & 4 \\ -5 & 3 }\).
We first compute \[sI - A = \MATRIX{s + 6 & -4 \\ 5 & s-3 }\] Therefore, \[\det(sI - A) = s^2 + 3s + 2 = (s+1)(s+2) \] and \[ (sI - A)^{-1} = \MATRIX{ \dfrac{s-3}{(s+1)(s+2)} & \dfrac{4}{(s+1)(s+2)} \\ \dfrac{-5}{(s+1)(s+2)} & \dfrac{s+6}{(s+1)(s+2)} } \] We now do partial fraction expansion of each term: \[ (sI - A)^{-1} = \MATRIX{ \dfrac{-4}{s+1} + \dfrac{5}{s+2}} & \dfrac{ 4}{s+1} - \dfrac{4}{s+2} \\ \dfrac{-5}{s+1} + \dfrac{5}{s+2} & \dfrac{5}{s+1} - \dfrac{4}{s+2} \] Taking the inverse Laplace transform of each term, we get \[ e^{At} = \mathcal L^{-1}(sI - A)^{-1} = \MATRIX{ -4 e^{-t} + 5 e^{-2t} & 4 e^{-t} - 4 e^{-2t} \\ -5 e^{-t} + 5 e^{-2t} & 5 e^{-t} - 4 e^{-2t}} \]
3 Time response of state space models
Now consider a SSM given by \[\begin{align*} \dot x(t) &= A x(t) + B u(t) \\ y(t) &= C x(t) \end{align*}\]
Suppose the system starts at \(t=0\) with an initial state \(x(0) = x_0\) and we apply the input \(u(t)\). How do we find the output?
Taking Laplace transform of the SSM, we get: \[\begin{align*} sX(s) - x_0 &= A X(s) + B U(s) \\ Y(s) &= C X(s) \end{align*}\] Solving for \(X(s)\), we get \[X(s) = (sI - A)^{-1} x_0 + (sI - A)^{-1} B U(s). \] Substituting in \(Y(s) = C X(s)\), we get \[ \bbox[5pt,border: 1px solid] {Y(s) = \underbrace{C (sI - A)^{-1} x_0}_{\text{zero-input response}} + \underbrace{C (sI - A)^{-1} B U(s)}_{\text{zero-state response}}} \] We can the use the above expression compute \(y(t)\) by taking the inverse Laplace transform.
It is sometimes useful to write the expression in time domain (but we will not use this expression for computations). To do so, recall that \(C(sI - A)^{-1}B\) is the transfer function \(G(s)\) of the system. Therefore, its inverse Laplace transform is the impulse response: \[ g(t) = C e^{At} B. \] Then, we can compute the inverse Laplace transform of \(G(s) U(s)\) using the convolution formula and write \[ \bbox[5pt,border: 1px solid] {y(t) = \underbrace{C e^{At} x_0}_{\text{zero-input response}} + \underbrace{\int_{0}^t C e^{A (t-τ) } B u(τ) d τ}_{\text{zero-state response}}} \]