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.37 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

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

However, the physical spring force is well behaved, since

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

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.