Question

A system is represented in state-space form as follows (\(u\): input, \(\mathbf{x}\): state vector, \(y\): output)
\[ \dot{\mathbf{x}}=\begin{bmatrix}1&2\\-3&0\end{bmatrix}\mathbf{x}+\begin{bmatrix}1\\2\end{bmatrix}u \]
\[ y=\begin{bmatrix}1&2\end{bmatrix}\mathbf{x} \]
Consider the new state vector \(\mathbf{z}=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\mathbf{x}\)
What is the state-space representation of the system in terms of the new state vector \(\mathbf{z}\)?

• A. \(\dot{\mathbf{z}}=\begin{bmatrix}-1&4\\-1&-2\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u\)
  \(y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}\)
• B. \(\dot{\mathbf{z}}=\begin{bmatrix}2&3\\0&3\end{bmatrix}\mathbf{z}+\begin{bmatrix}3\\5\end{bmatrix}u\)
  \(y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}\)
• C. \(\dot{\mathbf{z}}=\begin{bmatrix}4&9\\-2&-3\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u\)
  \(y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}\)
• D. \(\dot{\mathbf{z}}=\begin{bmatrix}2&1\\-4&1\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u\)
  \(y=\begin{bmatrix}4&-1\end{bmatrix}\mathbf{z}\)

Hint:

Use the similarity transform rules A' = TAT^-1, B' = TB, C' = CT^-1 with T being the given transformation matrix.

Answer 1

The Correct Option is C

Step 1: Set up the transformation.
Let \(T=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\), so the new state is \(\mathbf{z}=T\mathbf{x}\), which means \(\mathbf{x}=T^{-1}\mathbf{z}\). Substituting into the original state equation gives the standard similarity transform rules
\[ A'=TAT^{-1},\qquad B'=TB,\qquad C'=CT^{-1} \]

Step 2: Find the determinant of \(T\).
\[ \det(T)=2(0)-1(-1)=1 \]

Step 3: Find \(T^{-1}\).
For a \(2\times2\) matrix \(\begin{bmatrix}p&q\\r&s\end{bmatrix}\), the inverse is \(\dfrac{1}{\det}\begin{bmatrix}s&-q\\-r&p\end{bmatrix}\). Here \(p=2,q=1,r=-1,s=0\), so
\[ T^{-1}=\begin{bmatrix}0&-1\\1&2\end{bmatrix} \]

Step 4: Compute \(AT^{-1}\).
With \(A=\begin{bmatrix}1&2\\-3&0\end{bmatrix}\),
\[ AT^{-1}=\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}0&-1\\1&2\end{bmatrix}=\begin{bmatrix}2&3\\0&3\end{bmatrix} \]

Step 5: Compute \(A'=T(AT^{-1})\).
\[ A'=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\begin{bmatrix}2&3\\0&3\end{bmatrix}=\begin{bmatrix}4&9\\-2&-3\end{bmatrix} \]

Step 6: Compute \(B'=TB\).
With \(B=\begin{bmatrix}1\\2\end{bmatrix}\),
\[ B'=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}4\\-1\end{bmatrix} \]

Step 7: Compute \(C'=CT^{-1}\).
With \(C=\begin{bmatrix}1&2\end{bmatrix}\),
\[ C'=\begin{bmatrix}1&2\end{bmatrix}\begin{bmatrix}0&-1\\1&2\end{bmatrix}=\begin{bmatrix}2&3\end{bmatrix} \]

Step 8: Assemble the final representation.
\[ \boxed{\dot{\mathbf{z}}=\begin{bmatrix}4&9\\-2&-3\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u,\qquad y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}} \]
Hence, the correct option is (C).