Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

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.


Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

[How to cite this work]  [Order a printed hardcopy]  [Comment on this page via email]

``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA