- #1
exidez
- 44
- 0
this comes strait from a textbook:
http://higheredbcs.wiley.com/legacy/college/nise/0471794759/appendices/app_i.pdf
I am looking at how they obtained (I.24) from (I.25) on page 3 and 4.
Firstly we have:
[itex]e^{-\textbf{A}t}x(t)-x(0)=\int{e^{-\textbf{A}t}\textbf{Bu}(\tau)d\tau}[/itex]
Then this is derived to:
[itex]x(t)=e^{-\textbf{A}t}x(0)+\int{e^{-\textbf{A}(t-\tau)}\textbf{Bu}(\tau)d\tau}[/itex]
my question is why does the derived equation have [itex]e^{-\textbf{A}t}[/itex]. I thought it would have been [itex]e^{\textbf{A}t}[/itex] instead. Can someone explain this too me. The text didnt help me.
http://higheredbcs.wiley.com/legacy/college/nise/0471794759/appendices/app_i.pdf
I am looking at how they obtained (I.24) from (I.25) on page 3 and 4.
Firstly we have:
[itex]e^{-\textbf{A}t}x(t)-x(0)=\int{e^{-\textbf{A}t}\textbf{Bu}(\tau)d\tau}[/itex]
Then this is derived to:
[itex]x(t)=e^{-\textbf{A}t}x(0)+\int{e^{-\textbf{A}(t-\tau)}\textbf{Bu}(\tau)d\tau}[/itex]
my question is why does the derived equation have [itex]e^{-\textbf{A}t}[/itex]. I thought it would have been [itex]e^{\textbf{A}t}[/itex] instead. Can someone explain this too me. The text didnt help me.