DC Analysis of the WD Mass-Spring Oscillator

Considering the dc case first ($\rho=1$ ), we see from Fig.F.37 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

$$f_k(n) = f^{{+}}_k(n) + f^{{-}}_k(n) = 2x_1(n) = 2 x_0. \qquad\hbox{(spring force, dc case)} \protect$$ (F.58)

The loop on the right in Fig.F.37 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

$$x_2(n) = 2 n x_0, \quad n=0,1,2,3,\ldots\,. \protect$$ (F.59)

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:

$$f_m(n) = f^{{+}}_m(n) + f^{{-}}_m(n) = x_2(n+1) - x_2(n) = 2x_0. \protect$$ (F.60)

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 (F.58-F.60) verifies this to be the case.