Oscillation Frequency

From Fig. N.32, we can see that the impedance of the parallel combination of the mass and spring is given by

$$R_{m\vert\vert k}(s) \isdef \left.\frac{k}{s} \right\Vert ms = \frac{\frac{k}{s}ms}{\frac{k}{s}+ms} = \frac{ks}{s^2+\frac{k}{m}} \protect$$ (N.38)

(using the product-over-sum rule for combining impedances in parallel). The poles of this impedance are given by the roots of the denominator polynomial in $s$:

$$s = \pm j\sqrt{\frac{k}{m}} \protect$$ (N.39)

The resonance frequency of the mass-spring oscillator is therefore

$$\omega_0 = \sqrt{\frac{k}{m}} \protect$$ (N.40)

Since the poles $s=\pm j\omega_0$ are on the $j\omega$ axis, there is no damping, as we expect.

We can now write reflection coefficient $\rho$ (see Fig. N.34) as

$$\rho = \frac{m-k}{m+k} = \frac{1-\frac{k}{m}}{1+\frac{k}{m}} = \frac{1-\omega_0^2}{1+\omega_0^2}$$

We see that dc ( $\omega_0=0$) corresponds to $\rho=1$, and $\omega_0=\infty$ corresponds to $\rho=-1$.

DC Analysis.

Considering the dc case first ($\rho=1$), we see from Fig. N.34 that the state variable $x_1(n)$ will circulate unchanged in the isolated loop on the left. Let's call this value $x_1(n)\equiv\message{CHANGE eqv TO equiv IN SOURCE} x_0$. Then the physical force on the spring is always equal to

$$f_k(n) = f^{{+}}_k(n) + f^{{-}}_k(n) = 2x_1(n) = 2 x_0. \qquad\hbox{(spring force, dc case)} \protect$$ (N.41)

The loop on the right in Fig. N.34 receives $2 x_0$ and adds $x_2(n)$ to that. Since $x_2(n+1) = 2x_1(n)+x_2(n)$, we see it is linearly growing in amplitude. For example, if $x_2(0)=0$ (with $x_1(0)=x_0$), we obtain $x_2=[0, 2x_0, 4x_0, 6x_0,\ldots]$, or

$$x_2(n) = 2 n x_0, \quad n=0,1,2,3,\ldots\,. \protect$$ (N.42)

At first, this result might appear to contradict conservation of energy, since the state amplitude seems to be growing without bound. However, the physical force is fortunately better behaved:

$$f_m(n) = f^{{+}}_m(n) + f^{{-}}_m(n) = x_2(n+1) - x_2(n) = 2x_0. \protect$$ (N.43)

Since the spring and mass are connected in parallel, it must be the true that they are subjected to the same physical force at all times. Comparing Equations (N.41-N.43) verifies this to be the case.

Analysis 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. N.34 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.