How can be the state-space representation rewritten into differential equation?x˙1=−x1+x2x˙2=−x1+uA(−11−10)xB(01)u

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Lesly Fernandez · Accepted Answer

Step 1For this specific example,x˙1&amp;nbsp;=−x1+x2x˙2&amp;nbsp;=−x1+u&amp;nbsp;⟹&amp;nbsp;x¨1&amp;nbsp;=−x˙1+x˙2x˙2&amp;nbsp;=−x1+u―x¨1=−x˙1+u−x1―⇒x¨1+x˙1+x1=uwhere we substituted x˙2 (the underlined equation) into&amp;nbsp;x¨1. In general, if you have a single input single output system of the form,x˙=Ax+Buy=Cx+DuYou can compute the transfer functionG(s)=C(sI−A)−1B+DLet&amp;nbsp;G(s)=P(s)Q(s)is a polynomial of s in the numerator and&amp;nbsp;Q(s)&amp;nbsp;is a polynomial of s in the denominator of&amp;nbsp;G(s).Then from&amp;nbsp;Y(s)=G(s)U(s)=P(s)Q(s)U(s)&amp;nbsp;we can get&amp;nbsp;Q(s)Y(s)=P(s)U(s)&amp;nbsp;and by taking the inverse Laplace transform we can obtain the differential equation, i.e.,L−1(Q(s)Y(s))=L−1(P(s)U(s)).For your example,A=[−11−10]&amp;nbsp;B=[01]&amp;nbsp;C=[10]&amp;nbsp;D=0Therefore,C(sI-A)-1B+D=10s+1-11s-101=1s2+s+1and from the transfer function we can obtain,Y(s)=1s2+s+1U(s)⇒(s2+s+1)Y(s)=U(s)⇒L−1y¨+y˙+y=uSince&amp;nbsp;y=x1&amp;nbsp;we can write this as&amp;nbsp;x¨1+x˙1+x1=u&amp;nbsp;(same as above).

How can be the state-space representation rewritten into

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