Final Answer:
For
![the condition \(K=1\), the state-space representation for the system with \(x_1 = \Phi_a(t), x_2 = x_1, x_3 = x_2\) and \(u = \delta_d(t)\) is:\[\begin{align*}\dot{x} &= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & -0.7 \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} u \\y &= \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} x\end{align*}\]](https://img.qammunity.org/2024/formulas/engineering/college/5mg153d7i50k0cn6fi4fs6bs6rlnx6ebog.png)
Step-by-step explanation:
Given the transfer function \(\frac{\Phi_a(s)}{\delta_d(s)} = \frac{K}{(s + 1)(s^2 + 0.7s + 2)}\) for \(K = 1\), the state-space representation is obtained by converting the transfer function into state-space form. The state variables are defined as \
and the input as \(u = \delta_d(t)\).
Using the system's transfer function, we rewrite it in the standard form of a transfer function as \(\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D\). After performing partial fraction decomposition and equating coefficients, we extract matrices \(A, B, C,\) and \(D\). The resulting matrices are:
\[
![A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & -0.7 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}, \quad D = 0\]](https://img.qammunity.org/2024/formulas/engineering/college/pgpvvhsi6qnk6asng5k1qykwetppb79pue.png)
Therefore, the state-space representation for the system, considering \(K = 1\) and the defined state and input variables, is given by the state equation \(\dot{x} = Ax + Bu\) and the output equation \(y = Cx\).