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Chapter 7

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.

Step-by-Step

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.