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I'm trying to understand the resolution of the problem 5.11, but I can't understand how she gets the expression in the rectangle: $$\dot{x}_1=-x_1+2u$$ How can I get the expression in the rectangle?

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    \$\begingroup\$ I don't understand why someone would write the small case s so confusingly then almost do the same when writing an x. \$\endgroup\$ – Andy aka Jan 7 at 16:11
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\$ u \$ is the input signal it might be a real number so \$ \dot{u}=0 \$

so i would expect the result to be : \$ \dot{x}_1=−x_1+3u \$

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  • \$\begingroup\$ Why did she obtain two different expressions for $\dot{x_1}$? What is the correct expression? \$\endgroup\$ – Carmen González Jan 7 at 16:26
  • \$\begingroup\$ Which one is correct? That’s a good question! If you can post the rest of the exercise it might be helpful. \$\endgroup\$ – Delphesk Jan 7 at 19:25
  • \$\begingroup\$ I put the rest of the resolution in the post. I don't understand why she gets two expressions for $\dot{x_1}$. \$\endgroup\$ – Carmen González Jan 7 at 22:32
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    \$\begingroup\$ You have to replace x1 by w1 with the given assimilation that w1=x1+u Then you have (x1+u) *(s+1) =(s+3) *u Finally, this the result you are expecting. \$\endgroup\$ – Delphesk Jan 8 at 9:01

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