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Determine the stability of the systemdx→dt=(000001-1-1-2)x→

Short Answer

Expert verified

The stability of the system dx→dt=000001-1-1-2x→is stable

Step by step solution

01

Step 1:Explanation for the continuous dynamical systems with eigen values p±iq

Consider the linear system dx→dt=Ax→, where A is the real 2×2matrix with complex eigenvaluesp±iq and q≠0.

Consider an eigenvector v→+iw→with eigenvaluep±iq . Then:

x→t=eptScosqt-sinqtsinqtcosqtS-1x→0, where S=w→   v→

Recall that S-1x→0is the coordinate vector of x0with respect to the basisw→,v→

02

Explanation of the stability of a continuous dynamical system.

For a system, dx→dt=Ax→here A is the matrix form.

The zero state is an asymptotically stable equilibrium solution if and only if the real parts of all eigen values of A are negative.

03

Solution for the stability of the system dx→dt=(000001-1-1-2)x→

Consider the stability of the system isdx→dt=000001-1-1-2x→

Here A is the real matrix:

A=010001-1-1-2

dx→dt=010001-1-1-2x→A=010001-1-1-2

Find A-λIas:

A-λI=0010001-1-1-1-λ100010001=0010001-1-1-1-λ000λ000λ=0-λ100-λ1-1-1-2-λ=0

Find characteristics polynomial as:

-λ-λ-2-λ--1-10--1=0-λ(2λ+λ2+1)-(1)=0-2λ2-λ3-λ-1=0λ3+2λ2+λ+1=0λ   is  negative

The stability of the system is negative for all eigen values.

Hence the system is asymptotically stable.

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