Illustrative Example

Consider the following system in deviation form:

\frac{d}{dt} [ \begin{matrix}X_{1} \\ X_{2} \end{matrix} ] = [ \begin{matrix}−1 & 0 \\ 1 & −1 \end{matrix} ] [ \begin{matrix}X_{1} \\ X_{2} \end{matrix} ] + [ \begin{matrix}1 \\ 0 \end{matrix} ] U

Assume that X(0) = 0. Calculate the forced response to the following input:

U = \{ \begin{matrix}0 & t < 0 \\ 0.5 & 0 ≤ t < 1 \\ 0 & t ≥ 1 \end{matrix}

A = [ \begin{matrix}−1 & 0 \\ 1 & −1 \end{matrix} ]

Thus,

−A = [ \begin{matrix}1 & 0 \\ −1 & 1 \end{matrix} ]

Using the analytical formula for calculating e^{At}, we get:

e^{−At} = [ \begin{matrix}e^{t} & 0 \\ −te^{t} & e^{t} \end{matrix} ]

Since

B = [ \begin{matrix}1 \\ 0 \end{matrix} ]

we have

\begin{aligned}e^{−At}B & = & [ \begin{matrix}e^{t} & 0 \\ −te^{t} & e^{t} \end{matrix} ] [ \begin{matrix}1 \\ 0 \end{matrix} ] \\ & = & [ \begin{matrix}e^{t} \\ −te^{t} \end{matrix} ] \end{aligned}

Now,

e^{At} = [ \begin{matrix}e^{−t} & 0 \\ te^{−t} & e^{−t} \end{matrix} ]

Thus:

[ \begin{matrix}X_{1} \\ X_{2} \end{matrix} ] = [ \begin{matrix}e^{−t} & 0 \\ te^{−t} & e^{−t} \end{matrix} ] ∫_{0}^{t} [ \begin{matrix}e^{t} \\ −te^{t} \end{matrix} ] U(t)dt

Solution for 0 ≤ t < 1

X(t) = e^{At}∫_{0}^{t}e^{−At}B \underbrace{U(t)}_{U=0.5} dt

\begin{aligned}∫_{0}^{t}e^{−At}BUdt & = & [ \begin{matrix}∫_{0}^{t}0.5e^{t}dt \\ ∫_{0}^{t} −0.5te^{t}dt \end{matrix} ] \\ & = & [ \begin{matrix}0.5e^{t} − 0.5 \\ −0.5te^{t} + 0.5e^{t} − 0.5 \end{matrix} ] \end{aligned}

Remember to apply the limits of integration.

Thus

[ \begin{matrix}X_{1} \\ X_{2} \end{matrix} ] = e^{At}∫_{0}^{t}e^{−At}BU(t)dt

= [ \begin{matrix}e^{−t} & 0 \\ te^{−t} & e^{−t} \end{matrix} ] [ \begin{matrix}0.5e^{t} − 0.5 \\ −0.5te^{t} + 0.5e^{t} − 0.5 \end{matrix} ]

= [ \begin{matrix}0.5(1 − e^{−t}) \\ 0.5(1 − e^{−t} − te^{−t}) \end{matrix} ]

Thus when 0 ≤ t < 1,

\begin{aligned}X_{1} & = & 0.5(1 − e^{−t}) \\ X_{2} & = & 0.5(1 − e^{−t} − te^{−t}) \end{aligned}  (1)

Solution for t ≥ 1

X(t) = e^{At}∫_{0}^{t}e^{−At}BU(t)dt

The integration is from 0 to t where t ≥ 1. We split the integral into two parts as shown below:

\begin{aligned}∫_{0}^{t}e^{−At}BU(t)dt & = & ∫_{0}^{1}e^{−At}B \underbrace{U(t)}_{U=0.5} dt + ∫_{1}^{t}e^{−At}B \underbrace{U(t)}_{U=0} dt \\ & = & ∫_{0}^{1}e^{−At}B \underbrace{U(t)}_{U=0.5} dt \end{aligned}

∫_{0}^{1}e^{−At}BU(t)dt = [ \begin{matrix}0.5e^{t}] & {1 \atop 0} \\ −0.5(te^{t} − e^{t})] & {1 \atop 0} \end{matrix} ]

= [ \begin{matrix}0.5e − 0.5 \\ −0.5 \end{matrix} ]

= [ \begin{matrix}0.859 \\ −0.5 \end{matrix} ]

Thus,

[ \begin{matrix}X_{1} \\ X_{2} \end{matrix} ] = e^{At}∫_{0}^{t}e^{−At}BU(t)dt

= [ \begin{matrix}e^{−t} & 0 \\ te^{−t} & e^{−t} \end{matrix} ] [ \begin{matrix}0.859 \\ −0.5 \end{matrix} ]

= [ \begin{matrix}0.859e^{−t} \\ 0.859te^{−t} − 0.5e^{−t} \end{matrix} ]

Thus when t ≥ 1,

\begin{aligned}X_{1} & = & 0.859e^{−t} \\ X_{2} & = & 0.859te^{−t} − 0.5e^{−t} \end{aligned}  (2)

What are X_{1} and X_{2} when t = 0.4? Substitute t = 0.4 in equation 1 to get:

\begin{aligned}X_{1} & = & 0.165 \\ X_{2} & = & 0.031 \end{aligned}

What are X_{1} and X_{2} when t = 1.3? Substitute t = 1.3 in equation 2 to get:

\begin{aligned}X_{1} & = & 0.234 \\ X_{2} & = & 0.168 \end{aligned}