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(Solved): Nees ASAP Consider the following transfer function \[ \frac{C(s)}{R(s)}=\frac{s+3}{(s+4)(s+6)} \] Fr ...



Nees ASAP

Consider the following transfer function
\[
\frac{C(s)}{R(s)}=\frac{s+3}{(s+4)(s+6)}
\]
From the information given which one
Consider the following transfer function \[ \frac{C(s)}{R(s)}=\frac{s+3}{(s+4)(s+6)} \] From the information given which one of the following is the correct parallel canonical form? a. \[ \dot{x}=\left[\begin{array}{cc} -4 & 0 \\ 0 & -6 \end{array}\right] x+\left[\begin{array}{c} \frac{-1}{2} \\ \frac{3}{2} \end{array}\right] r, \quad y=\left[\begin{array}{ll} 1 & 1 \end{array}\right] x \] b. \[ \dot{x}=\left[\begin{array}{cc} -10 & 0 \\ 1 & -24 \end{array}\right] x+\left[\begin{array}{c} \frac{-1}{2} \\ \frac{3}{2} \end{array}\right] r, \quad y=\left[\begin{array}{ll} -3 & 1 \end{array}\right] x \] c. \( \dot{x}=\left[\begin{array}{cc}1 & 0 \\ -10 & -24\end{array}\right] x+\left[\begin{array}{l}0 \\ 1\end{array}\right] r, \quad y=\left[\begin{array}{ll}3 & 1\end{array}\right] x \) d. \( \dot{x}=\left[\begin{array}{cc}-10 & -24 \\ 1 & 0\end{array}\right] x+\left[\begin{array}{c}\frac{-3}{2} \\ \frac{1}{2}\end{array}\right] r, \quad y=\left[\begin{array}{ll}1 & 3\end{array}\right] x \)


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Solution The transfer function (C(s))/(R(s)) = (s + 3)/((s + 4)(s + 6)) is already in parallel canonical form. The parallel canonical
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