4  Time response of state space models

Author

Affiliation

Aditya Mahajan

McGill University

4.1 Solution to vector differential equations

Recall that for a scalar differential equation (i.e., $x(t)\in \mathbb{R}$): $$\dot{x}(t)=ax(t),\text{initial condition }x(0)=x_0$$ 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 vector differential equation (i.e., $x(t)\in {\mathbb{R}}^n$): $$\dot{x}(t)=Ax(t),\text{initial condition }x(0)=x_0$$ where $A\in {\mathbb{R}}^{n\times n}$ 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 $$\begin{array}{rl}{e}^{At}& =I+At+\frac{(At{)}^2}{2!}+\frac{(At{)}^3}{3!}+\cdots \\ & =I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots \end{array}$$ where the second equation uses the fact that $(At{)}^2=AtAt=A^2{t}^2$ because $t$ is a scalar.

With this definition, we have $$\begin{array}{rl}\frac{d}{dt}{e}^{At}& =A+A^{2}t+\frac{A^3{t}^2}{2!}+\cdots \\ & =A[I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots ]\\ & =A{e}^{At}.\end{array}$$

Thus, if we take the candidate solution $x(t)=e^{At}{x}_0$, we have that $$\frac{d}{dt}x(t)=\left[\frac{d}{dt}{e}^{At}\right]x_0=A{e}^{At}{x}_0=Ax(t).$$ Thus, our candidate solution satisfies the vector differential equation!

If we define $e^{At}$ as $$\begin{array}{}\text{(1)}& e^{At}=I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots \end{array}$$

Then, we can write the solution of the vector differential equation (i.e., $x(t)\in {\mathbb{R}}^n$) $$\dot{x}(t)=Ax(t),\text{initial condition }x(0)=x_0$$ where $A\in {\mathbb{R}}^{n\times n}$ as $$x(t)=e^{At}{x}_0.$$

We now present some examples where the matrix exponential can be computed easily.

Example 4.1 Compute $e^{At}$ for $$A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right].$$

Based on the solution, solve the vector differential equation $$\dot{x}(t)=Ax(t),x_0=\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\end{array}\right].$$

Solution

We compute the terms of $\text{(1)}$:

• $A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$.

which means that $A^k=0$ for $k\ge 2$. Thus, the right hand side of $\text{(1)}$ contains only a finite number of nonzero terms: $$\begin{array}{rl}{e}^{At}& =I+At\\ & =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& t\\ 0& 0\end{array}\right]\\ & =\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right].\end{array}$$

Thus, the solution of the vector differential equation $\dot{x}(t)=Ax(t)$ is given by $$x(t)=e^{At}{x}_0=\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\end{array}\right]=\left[\begin{array}{c}{\alpha }_1+{\alpha }_{2}t\\ {\alpha }_2\end{array}\right].$$

Example 4.2 Compute $e^{At}$ for $$A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right].$$

Based on the solution, solve the vector differential equation $$\dot{x}(t)=Ax(t),x_0=\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\\ {\alpha }_3\end{array}\right].$$

Solution

We compute the terms of $\text{(1)}$:

• $A^2=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$.
• $A^3=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$.

which means that $A^k=0$ for $k\ge 3$. Thus, the right hand side of $\text{(1)}$ contains only a finite number of nonzero terms: $$\begin{array}{rl}{e}^{At}& =I+At+\frac{1}{2}{A}^2{t}^2\\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}0& t& 0\\ 0& 0& t\\ 0& 0& 0\end{array}\right]+\frac{1}{2}\left[\begin{array}{ccc}0& 0& t^2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}1& t& \frac{1}{2}{t}^2\\ 0& 1& t\\ 0& 0& 1\end{array}\right].\end{array}$$

Thus, the solution of the vector differential equation $\dot{x}(t)=Ax(t)$ is given by $$x(t)=e^{At}{x}_0=\left[\begin{array}{ccc}1& t& \frac{1}{2}{t}^2\\ 0& 1& t\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\\ {\alpha }_3\end{array}\right]=\left[\begin{array}{c}{\alpha }_1+{\alpha }_{2}t+\frac{1}{2}{\alpha }_3^2\\ {\alpha }_2+{\alpha }_{3}t\\ {\alpha }_3\end{array}\right].$$

Example 4.3 Compute $e^{At}$ for $$A=\left[\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& {\lambda }_2& 0\\ 0& 0& {\lambda }_3\end{array}\right].$$

Based on the solution, solve the vector differential equation $$\dot{x}(t)=Ax(t),x_0=\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\\ {\alpha }_3\end{array}\right].$$

Solution

We compute the terms of $\text{(1)}$:

• $A^2=\left[\begin{array}{ccc}{\lambda }_1^2& 0& 0\\ 0& {\lambda }_2^2& 0\\ 0& 0& {\lambda }_3^2\end{array}\right].$
• $A^3=\left[\begin{array}{ccc}{\lambda }_1^3& 0& 0\\ 0& {\lambda }_2^3& 0\\ 0& 0& {\lambda }_3^3\end{array}\right].$
• and so on.

Thus, we have $$\begin{array}{rl}{e}^{At}& =I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots \\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}{\lambda }_{1}t& 0& 0\\ 0& {\lambda }_{2}t& 0\\ 0& 0& {\lambda }_{3}t\end{array}\right]+\frac{1}{2}\left[\begin{array}{ccc}{\lambda }_1^2{t}^2& 0& 0\\ 0& {\lambda }_2^2{t}^2& 0\\ 0& 0& {\lambda }_3^2{t}^2\end{array}\right]+\cdots \\ & =\left[\begin{array}{ccc}1+{\lambda }_{1}t+\frac{1}{2}{\lambda }_1^2{t}^2+\cdots & 0& 0\\ 0& 1+{\lambda }_{2}t+\frac{1}{2}{\lambda }_2^2{t}^2+\cdots & 0\\ 0& 0& 1+{\lambda }_{3}t+\frac{1}{2}{\lambda }_3^2{t}^2+\cdots \end{array}\right]\\ & =\left[\begin{array}{ccc}{e}^{{\lambda }_{1}t}& 0& 0\\ 0& e^{{\lambda }_{2}t}& 0\\ 0& 0& e^{{\lambda }_{3}t}\end{array}\right]\end{array}$$

Thus, the solution of the vector differential equation $\dot{x}(t)=Ax(t)$ is given by $$x(t)=e^{At}{x}_0=\left[\begin{array}{ccc}{e}^{{\lambda }_{1}t}& 0& 0\\ 0& e^{{\lambda }_{2}t}& 0\\ 0& 0& e^{{\lambda }_{3}t}\end{array}\right]\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\\ {\alpha }_3\end{array}\right]=\left[\begin{array}{c}{\alpha }_1{e}^{{\lambda }_{1}t}\\ {\alpha }_2{e}^{{\lambda }_{2}t}\\ {\alpha }_3{e}^{{\lambda }_{3}t}\end{array}\right].$$

But outside of such few special cases, computing $e^{At}$ via definition $\text{(1)}$ is not computationally feasible. We now present computationally efficient methods to compute the matrix exponential.

4.2 Computing matrix exponential

4.2.1 Method 1: Eigenvalue diagonalization method

As illustrated by Example 4.3, 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 ${\lambda }_1,\dots ,{\lambda }_n$ and $v_1,\dots ,v_n$ are the corresponding eigvenvectors. Thus, $A{v}_k={\lambda }_k{v}_k$ for all $k\in \{1,\dots ,n\}$. Writing this in matrix form, we have $$\begin{array}{rl}A\left[\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\end{array}\right]& =\left[\begin{array}{cccc}A{v}_1& A{v}_2& \cdots & A{v}_n\end{array}\right]\\ & =\left[\begin{array}{cccc}{\lambda }_1{v}_1& {\lambda }_2{v}_2& \cdots & {\lambda }_n{v}_n\end{array}\right]\\ & =\left[\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\end{array}\right]\left[\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \ddots & 0\\ 0& \cdots & \ddots & 0\\ 0& \cdots & \cdots & {\lambda }_n\end{array}\right]\end{array}$$ Now define $$T=\left[\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\end{array}\right]\text{and}\mathrm{\Lambda }=\left[\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \ddots & 0\\ 0& \cdots & \ddots & 0\\ 0& \cdots & \cdots & {\lambda }_n\end{array}\right].$$ So, the above equation can be writen as $AT=T\mathrm{\Lambda }$ or $$T^{-1}AT=\mathrm{\Lambda }$$ Observe that

• $A=T\mathrm{\Lambda }{T}^{-1}$
• $A^2=T{\mathrm{\Lambda }}^2{T}^{-1}$
• $A^3=T{\mathrm{\Lambda }}^3{T}^{-1}$
• and so on.

Therefore, $$\begin{array}{rl}{e}^{At}& =I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots \\ & =I+T\mathrm{\Lambda }{T}^{-1}t+\frac{T{\mathrm{\Lambda }}^2{T}^{-1}{t}^2}{2!}+\frac{T{\mathrm{\Lambda }}^3{T}^{-1}{t}^3}{3!}+\cdots \\ & =T(I+\mathrm{\Lambda }t+\frac{{\mathrm{\Lambda }}^2{t}^2}{2!}+\frac{{\mathrm{\Lambda }}^3{t}^3}{3!}+\cdots )T^{-1}\\ & =T{e}^{\mathrm{\Lambda }t}{T}^{-1}\end{array}$$

Thus, once we know the eigenvalues and eigenvectors of $A$ (and if all eigenvalues are distinct), then $$e^{At}=T{e}^{\mathrm{\Lambda }t}{T}^{-1}$$

Exercise 4.1 Use the eigenvalue diagonalizable method to compute $e^{At}$ for $A=\left[\begin{array}{cc}-6& 4\\ -5& 3\end{array}\right]$.

Solution

We start by computing the eigenvalues and eigenvectors of $A$.

To compute the eigenvalues: $$sI-A=\left[\begin{array}{cc}s+6& -4\\ 5& s-3\end{array}\right]$$ Therefore, the characteristic equation is $$det(sI-A)=s^2+3s+2=(s+1)(s+2)$$ Hence, the eigenvalues of $A$ are ${\lambda }_1=-1$ and ${\lambda }_2=-2$.

We now compute the eigenvectors. Recall that for any eigenvalue $\lambda$, the eigen-vector satisfies $(\lambda I-A)v=0$. We start wtih ${\lambda }_1=-1$. Then, $$\left[\begin{array}{cc}5& -4\\ 5& -4\end{array}\right]\left[\begin{array}{c}{v}_{11}\\ v_{12}\end{array}\right]=0$$ We set $v_{11}=4$. Then, $v_{12}=5$. Thus, the eigenvector $$v_1=\left[\begin{array}{c}4\\ 5\end{array}\right].$$

Similarly, for ${\lambda }_2=-2$, we have $$\left[\begin{array}{cc}4& -4\\ 5& -5\end{array}\right]\left[\begin{array}{c}{v}_{21}\\ v_{22}\end{array}\right]=0$$ We set $v_{21}=1$. Then, $v_{22}=1$. Thus, the eigenvector $$v_2=\left[\begin{array}{c}1\\ 1\end{array}\right].$$

Thus, $T=\left[\begin{array}{cc}4& 1\\ 5& 1\end{array}\right]$ and therefore $T^{-1}=\left[\begin{array}{cc}-1& 1\\ 5& -4\end{array}\right]$. Hence, $$e^{At}=\left[\begin{array}{cc}4& 1\\ 5& 1\end{array}\right]\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 5& -4\end{array}\right]=\left[\begin{array}{cc}-4e^{-t}+5e^{-2t}& 4e^{-t}-4e^{-2t}\\ -5e^{-t}+5e^{-2t}& 5e^{-t}-4e^{-2t}\end{array}\right]$$

Exercise 4.2 Use the eigenvalue diagonalizable method to compute $e^{At}$ for $A=\left[\begin{array}{cc}1& 2\\ -1& 4\end{array}\right]$.

Solution

We start by computing the eigenvalues and eigenvectors of $A$.

To compute the eigenvalues: $$sI-A=\left[\begin{array}{cc}s-1& -2\\ 1& s-4\end{array}\right]$$ Therefore, the characteristic equation is $$det(sI-A)=s^2-5s+6=(s-2)(s-3).$$ Hence, the eigenvalues of $A$ are ${\lambda }_1=2$ and ${\lambda }_2=3$.

We now compute the eigenvectors. Recall that for any eigenvalue $\lambda$, the eigen-vector satisfies $(\lambda I-A)v=0$. We start wtih ${\lambda }_1=2$. Then, $$\left[\begin{array}{cc}1& -2\\ 1& -2\end{array}\right]\left[\begin{array}{c}{v}_{11}\\ v_{12}\end{array}\right]=0$$ We set $v_{11}=2$. Then, $v_{12}=1$. Thus, the eigenvector $$v_1=\left[\begin{array}{c}2\\ 1\end{array}\right].$$

Similarly, for ${\lambda }_2=3$, we have $$\left[\begin{array}{cc}2& -2\\ 1& -1\end{array}\right]\left[\begin{array}{c}{v}_{21}\\ v_{22}\end{array}\right]=0$$ We set $v_{21}=1$. Then, $v_{22}=1$. Thus, the eigenvector $$v_2=\left[\begin{array}{c}1\\ 1\end{array}\right].$$

Thus, $T=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$ and therefore $T^{-1}=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$. Hence, $$e^{At}=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}{e}^{2t}& 0\\ 0& e^{3t}\end{array}\right]\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}2e^{2t}-e^{3t}& -2e^{2t}+2e^{3t}\\ e^{2t}-e^{3t}& -e^{2t}+2e^{3t}\end{array}\right]$$

4.2.2 Method 2: Laplace transform method

Recall that we have $$\dot{x}(t)=Ax(t).$$ Taking (unilateral) Laplace transforms of both sides gives $$sX(s)-x_0=AX(s).$$ Rearranging terms, we get $$(sI-A)X(s)=x_0⟹X(s)=(sI-A{)}^{-1}{x}_0.$$ 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 $$e^{At}={\mathcal{L}}^{-1}((sI-A{)}^{-1}).$$ In the above, we interpret the inverse Laplace transform of a matrix to mean the inverse Laplace transform of each entry.

Exercise 4.3 Use the Laplace transform method to compute $e^{At}$ for $A=\left[\begin{array}{cc}-6& 4\\ -5& 3\end{array}\right]$.

Solution

We first compute $$sI-A=\left[\begin{array}{cc}s+6& -4\\ 5& s-3\end{array}\right]$$ Therefore, $$det(sI-A)=s^2+3s+2=(s+1)(s+2)$$ and $$(sI-A{)}^{-1}=\left[\begin{array}{cc}{\displaystyle \frac{s-3}{(s+1)(s+2)}}& {\displaystyle \frac{4}{(s+1)(s+2)}}\\ {\displaystyle \frac{-5}{(s+1)(s+2)}}& {\displaystyle \frac{s+6}{(s+1)(s+2)}}\end{array}\right]$$ We now do partial fraction expansion of each term: $$(sI-A{)}^{-1}=\left[\begin{array}{cc}{\displaystyle \frac{-4}{s+1}}+{\displaystyle \frac{5}{s+2}}& {\displaystyle \frac{4}{s+1}}-{\displaystyle \frac{4}{s+2}}\\ {\displaystyle \frac{-5}{s+1}}+{\displaystyle \frac{5}{s+2}}& {\displaystyle \frac{5}{s+1}}-{\displaystyle \frac{4}{s+2}}\end{array}\right]$$ Taking the inverse Laplace transform of each term, we get $$e^{At}={\mathcal{L}}^{-1}(sI-A{)}^{-1}=\left[\begin{array}{cc}-4e^{-t}+5e^{-2t}& 4e^{-t}-4e^{-2t}\\ -5e^{-t}+5e^{-2t}& 5e^{-t}-4e^{-2t}\end{array}\right]$$

4.3 Internal stability

As stated in the beginning of the lecture, matrix exponential allows us to solve the vector differential equation $$\dot{x}(t)=Ax(t),\text{initial condition }x(0)=x_0$$ in the same manner as a scalar linear differential equation. The solution is given by $$x(t)=e^{At}{x}_0.$$

Suppose $A$ has distinct eigenvalues $({\lambda }_1,\dots ,{\lambda }_n)$ with corresponding (linearly independent) eigenvalues $(v_1,\dots ,v_n)$.

Recall that for any eigenvector $v_j$, $A^k{v}_j={\lambda }_j^k{v}_j$. Thus, if the system starts from the intial condition $x_0=v_j$, then $$\begin{array}{rl}x(t)& =e^{At}{x}_0=e^{At}{v}_j\\ & =[I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots ]v_j\\ & =[v_j+{\lambda }_{j}t{v}_j+\frac{{\lambda }_j^2{t}^2}{2!}{v}_j+\frac{{\lambda }_j^3{t}^3}{3!}{v}_j+\cdots ]\\ & =e^{{\lambda }_{j}t}{v}_j\end{array}$$ Therefore, if we start along an eigenvector of $A$, the system trajectory $x(t)$ remains along that direction, with the length scaling as $e^{{\lambda }_{j}t}$.

In general, since the eigenvectors $(v_1,\dots ,v_n)$ are linearly independent, an arbitrary initial condition $x_0$ can be written as $$x_0={\alpha }_1{v}_1+\cdots +{\alpha }_n{v}_n.$$ Therefore, $$\begin{array}{rl}x(t)& =e^{At}{x}_0=e^{At}[{\alpha }_1{v}_1+\cdots +{\alpha }_n{v}_n]\\ & ={\alpha }_1{e}^{At}{v}_1+\cdots +{\alpha }_n{e}^{At}{v}_n\\ & ={\alpha }_1{e}^{{\lambda }_{1}t}{v}_1+\cdots +{\alpha }_n{e}^{{\lambda }_{n}t}{v}_n\end{array}$$

Thus, the response of the dynamical system $\dot{x}(t)=Ax(t)$ is a combination of motions along the eigenvectors of the matrix $A$. Each eigendirection is called a mode of the system. A particular mode is excited by choosing an initial condition to have a component ${\alpha }_i$ along that eigendirection.

An implication of the above is the following: the state $x(t)\to 0$ as $t\to \infty$ for all initial states $x_0$ if and only if all eigenvalues of $A$ lie in the open left-hand plane.

A SSM where all eigenvalues of the $A$ matrix lie in the open-left hand plane is called internally stable.

Internal stability vs BIBO stability

The TF of a SSM is given by $$G(s)=C(sI-A{)}^{-1}B=\frac{C\text{adj}(sI-A)B}{det(sI-A)}.$$ The elements of adjoint of $(sI-A)$ are all polynomials in $s$. Thus, the numerator is a polynomial in $s$. So, is the denominator. Moreover, the denominator equals the characteristic equation of $A$; thus, the roots of the denominator are the eigenvalues of $A$.

The numerator and denominator may have common roots that cancel each other. So, in general, $$\{\text{eigenvalues of }A\}\supseteq \{\text{poles of }G(s)\}.$$ Hence, if $A$ is internally stable (i.e., all its eigenvalues are in the open left-hand plane) then $G(s)$ is BIBO stable (i.e., all its poles are in the open left-hand plane).

However, the converse is not true there may be pole-zero cancellations. For example, consider $A=\left[\begin{array}{cc}-1& 0\\ 0& 2\end{array}\right]$, $B=\left[\begin{array}{c}1\\ 0\end{array}\right]$, $C=\left[\begin{array}{cc}1& 0\end{array}\right]$. Then, $$G(s)=\frac{s-2}{(s+1)(s-2)}=\frac{1}{s+1}.$$ Notice that the TF is BIBO stable but the SSM is not internally stable! Any initial condition that activates the mode corresponding to eigenvalue $2$ will cause $x(t)$ to diverge to infinity.

4.4 Time response of state space models

Now consider a SSM given by $$\begin{array}{rl}\dot{x}(t)& =Ax(t)+Bu(t)\\ y(t)& =Cx(t)\end{array}$$

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: $$\begin{array}{rl}sX(s)-x_0& =AX(s)+BU(s)\\ Y(s)& =CX(s)\end{array}$$ Solving for $X(s)$, we get $$X(s)=(sI-A{)}^{-1}{x}_0+(sI-A{)}^{-1}BU(s).$$ Substituting in $Y(s)=CX(s)$, we get $$Y(s)=\underset{\text{zero-input response}}{\underbrace{C(sI-A{)}^{-1}{x}_0}}+\underset{\text{zero-state response}}{\underbrace{C(sI-A{)}^{-1}BU(s)}}$$ 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 $$y(t)=\underset{\text{zero-input response}}{\underbrace{C{e}^{At}{x}_0}}+\underset{\text{zero-state response}}{\underbrace{{\int }_0^{t}C{e}^{A(t-\tau )}Bu(\tau )d\tau }}$$