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DC Analysis of the WD Mass-Spring Oscillator

Considering the dc case first ($ \rho=1$), we see from Fig.F.35 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

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

The loop on the right in Fig.F.35 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

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

$\displaystyle f_m(n) = f^{{+}}_m(n) + f^{{-}}_m(n) = x_2(n+1) - x_2(n) = 2x_0. \protect$ (F.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 (F.41-F.43) verifies this to be the case.

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WD Mass-Spring Oscillator at Half the Sampling Rate
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Oscillation Frequency
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