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I have an exercise and I am trying to understand the solution. Many steps I do not understand. I am going to ask the questions below in the text with italic font.

Exercise

Consider the homogeneous $ 2 \times 2 $ system $$ \dot x = Ax $$ where the distinct, real eigenvalues of $A$ are $λ_1$ and $λ_2$ with corresponding eigenvectors $t_1$ and $t_2$.

a) Using the Laplace transform $$ sX(s) - x(0) = AX(s) $$ show that $$ X(s) = T \begin{bmatrix} \frac{1}{s-\lambda_1} & 0 \\ 0 & \frac{1}{s-\lambda_2} \\ \end{bmatrix} T^{-1}x(0), ~~~~ T=\begin{bmatrix} t_1 & t_2 \end{bmatrix} $$ where $ t_i$ are the columns of T.

b) Show that with the initial condition $$ x(0) = kt_1$$ we have $$X(s)= \frac{k}{s-\lambda_1}t_1 $$

c) For $$ A= \begin{bmatrix} -1 & 1 \\ -2 & -4 \end{bmatrix}$$ and with the initial condition $$ x(0) = \begin{bmatrix} -1 \\3 \end{bmatrix} $$ use the result of part (b) to derive $x(t)$ analytically. Then, by using MATLAB, plot the behaviour of the system in a phase plane diagram (i.e. sketch $x_2(t)$ over $x_1(t)$ as t goes from zero to infinity).

Solution

a) By eigenvalue decomposition $$ A=T\Lambda T^{-1}, ~~~~ T=\begin{bmatrix} t_1 & t_2 \end{bmatrix}, ~~~~ \Lambda=\begin{bmatrix} \lambda_1 & 0 \\ 0 & t_2 \end{bmatrix}$$

so $$(sI-A)X(s) = x(0), ~~~~T(sI-\Lambda)T^{-1}X(s)=x(0) \\ X(s) = T (sI- \Lambda)^{-1}T^{-1}x(0) = T \begin{bmatrix} \frac{1}{s-\lambda_1} & 0 \\ 0 & \frac{1}{s-\lambda_2} \\ \end{bmatrix} T^{-1}x(0) $$ It is worthy to observe that we could not investigate the dynamic of a system with its transfer function when it is affected only by some non-zero initial condition, while this possibility is provided by state space model.

b) With initial conditions $ x(0) = kt_1 $ we have $$ T^{-1}x(0) =\begin{bmatrix} k \\ 0 \end{bmatrix} ~~~~~ (T^{-1} \begin{bmatrix} t_1 & t_2 \end{bmatrix} = I \Rightarrow \begin{bmatrix} T^{-1}t_1 & T^{-1}t_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ $$ X(s) = k\begin{bmatrix} t_1 & t_2 \end{bmatrix} \begin{bmatrix} \frac{1}{s-\lambda_1} \\ 0 \end{bmatrix} = k \frac{1}{s-\lambda_1}t_1 $$

The solution is in the direction $ t_1 $ and only depends on $ \lambda_1 $.

c) From the result of part (b) for any initial condition of the form $x(0) = k_1 t_1 + k_2 t_2 $, the solution $x$ in frequency domain is $$ X(s) = \frac{k_1}{s-\lambda_1}t_1 + \frac{k_2}{s-\lambda_2}t_2$$ and in time domain $$ x(t) = k_1 e^{\lambda t}t_1 + K_2 e^{\lambda_2 t}t_2$$ By eigenvalue decomposition of $A$ with Matlab command eig, we obtain $$ \begin{bmatrix} t_1 & t_2 \end{bmatrix} = \begin{bmatrix} 0.7071 & -0.4472 \\ -0.7071 & 0.8944 \end{bmatrix}, ~~~~ \lambda_1 =-2,~ \lambda_2 = -3 $$

We choose eigenvectors in a way to have $ \begin{bmatrix} t_1 & t_2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} $; note that the eigenvectors are not unique and only their directions matter.

As $t_1$ and $t_2$ are linearly independent, any arbitrary vector like $ x(0) = \begin{bmatrix} -1 \\ 3 \end{bmatrix} $, can be written as linear combination fof $t_1$ and $t_2$. It is straightforward to calculate $k_1 = 1 $ and $k_2=2$ so that we have $ x(0) = t_1 + 2 t_2 $ and the solution is $$ x(t) = e^{-2t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + 2 e^{-3t} \begin{bmatrix} -1 \\ 2 \end{bmatrix} $$

Questions

a) Why is the Λ enclosed by $T$ and $T^{-1}$?

b) How can T have an inverse matrix if T is not a square matrix? I tried the following commands in Matlab syms t_1 t_2, T = [t_1 t_2], T^(-1). This returned the error Error using symengine. Not a square matrix. The solutions says $\begin{bmatrix} k \\ 0 \end{bmatrix} $ Where did the $t_1$ go, when $T^{-1}x(0)$ was executed?

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Question 1: $\Lambda$ is enclosed by $T$ and $T^{-1}$ per the definition of diagonalization.

Question 2: $t_1$ and $t_2$ are each $2 \times 1$ columns, so the resulting matrix $T$ is indeed square.


To your latest comment: they compute $$ T \pmatrix{\frac{1}{s - \lambda_1}&0\\0& \frac{1}{s - \lambda_2}}(T^{-1}x(0)) = \\ k\;T \underbrace{\pmatrix{\frac{1}{s - \lambda_1}&0\\0& \frac{1}{s - \lambda_2}}\pmatrix{1\\0}} =\\ k T \pmatrix{\frac{1}{s - \lambda_1}\\0} $$

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  • $\begingroup$ Thank you for your help. So I understand that T actually has two column vectors. In a) we have found $\begin{bmatrix}\frac{1}{s-\lambda_1} & 0 \\ 0 & \frac{1}{s-\lambda_2} \end{bmatrix}$ in c) it is shortened to $\begin{bmatrix}\frac{1}{s-\lambda_1} \\ 0 \end{bmatrix}$, why? I would now have assumed $$ X(s) = \begin{bmatrix}k \\ 0 \end{bmatrix} \begin{bmatrix}t_{1a} & t_{2a} \\ t_{1b} & t_{2b}\end{bmatrix}\begin{bmatrix} \frac{1}{s-\lambda_1} & 0 \\ 0 & \frac{1}{s-\lambda_2} \\ \end{bmatrix}$$ Is there anything special (as in "light bulb moment") about this exercise for a control engineer? $\endgroup$ – autoship Dec 8 '16 at 19:52
  • $\begingroup$ See my latest edit. I'll get back to you on that "light bulb moment". $\endgroup$ – Omnomnomnom Dec 8 '16 at 19:58
  • $\begingroup$ It is helpful to remember that $T$ represents a change of variables $\endgroup$ – Omnomnomnom Dec 9 '16 at 2:45
  • $\begingroup$ Looking again at this exercise, I wonder how they get from $T^{-1}x(0)$ to $\begin{bmatrix} k \\ 0 \end{bmatrix}$ I assume I am missing the steps in between. $\endgroup$ – autoship Dec 10 '16 at 16:45

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