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WD Mass-Spring Oscillator at Half the Sampling Rate

Under the bilinear transform, the $ s=\infty$ maps to $ z=-1$ (half the sampling rate). It is therefore no surprise that given $ \omega_0=\infty$ ($ \rho=-1$), inspection of Fig.F.35 reveals that any alternating sequence (sinusoid sampled at half the sampling rate) will circulate unchanged in the loop on the right, which is now isolated. Let $ x_2(n) = (-1)^n x_0$ denote this alternating sequence. The loop on the left receives $ - 2 x_2(n)$ and adds $ - x_1(n-1)$ to it, i.e., $ x_1(n+1)= - x_1(n) - 2 x_2(n) = -x_1(n) - 2x_2(0)(-1)^n$. If we start out with $ x_1(0)=0$ and $ x_2(0)=x_0$, we obtain $ x_1 = [0,-2x_0, 4x_0, -6x_0, \ldots]$, or

$\displaystyle x_1(n) = (-1)^n 2 n x_0, \quad n=0,1,2,3,\ldots\,. $

However, the physical spring force is well behaved, since

$\displaystyle f_k(n) = f^{{+}}_k(n) + f^{{-}}_k(n) = x_1(n+1) + x_1(n) = (-1)^{n+1}2 x_0 $

As a check, the mass force is found to be

\begin{eqnarray*} f_m(n) &=& x_2(n+1) - x_2(n)\\ &=& (-1)^{n+1}x_0 - (-1)^n x_0\\ &=& (-1)^{n+1}x_0 + (-1)^{n+1}x_0\\ &=& (-1)^{n+1}2 x_0, \end{eqnarray*}

which agrees with the spring, as it must.

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Linearly Growing State Variables in WD Mass-Spring Oscillator
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DC Analysis of the WD Mass-Spring Oscillator