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## Q. 7.EX.8

Time Scaling an Oscillator
The equation for an oscillator was derived in Example 2.5. For a case with a very fast natural frequency $ω_n = 15,000 \text{ rad/sec }$ (about $2 \text{ kHz}$ ), Eq. (2.23) can be rewritten as

$\ddot{θ} + 15000^2 · θ = 10^6 · T_c.$

$\ddot{\theta}+\frac{g}{l} \theta=\frac{T_{c}}{m l^{2}}$                 (2.23)

Determine the time-scaled equation so that the unit of time is milliseconds.

## Verified Solution

In state-variable form with a state vector $\pmb x = \begin{bmatrix} \theta & \dot{\theta } \end{bmatrix} ^T,$ the unscaled matrices are

$\pmb F = \begin{bmatrix} 0 & 1 \\ -15000^2 & 0\end{bmatrix} \ \ \text { and } \ \ \pmb G = \begin{bmatrix} 0 \\ 10^6 \end{bmatrix} .$

Applying Eq. (7.9) results in

$\frac{d \pmb x}{dτ } = \frac{1}{ω_o} \pmb{Fx} + \frac{1}{ω_o} \pmb Gu = \pmb F^{\prime} \pmb x + \pmb G^{\prime} u.$              (7.9)

$\pmb F^{\prime} = \begin{bmatrix} 0 & \frac{1}{1000} \\ -\frac{15000^2}{1000} & 0\end{bmatrix} \ \ \text { and } \ \ \pmb G^{\prime} = \begin{bmatrix} 0 \\ 10^3 \end{bmatrix} ,$

which yields state-variable equations that are scaled.