Sign in

username:

password:



Not a member?

Search Online Books



Search tips

Free Online Books

Sponsor

NEW! TMS320C6474: 3x the performance. 1/3 the cost. Three 1 GHz cores on 1 chip.

Chapters

Physical Audio Signal Processing
    Elementary String Instruments
       Ideal String Struck by a Mass
          Mass Termination Model

Chapter Contents:

Search Physical Audio Signal Processing

  

Book Index | Global Index

Would you like to be notified by email when Julius Orion Smith III publishes a new entry into his blog?

  

Mass Termination Model

The previous discussion solved for the motion of an ideal mass after hitting an ideal string of infinite length. We now investigate the model from the string's point of view. As before, we will be interested in a digital waveguide model (sampled traveling-wave model) of the string, for efficiency's sake, and we therefore will need to know what the mass ``looks like'' at the end of each string segment.

Let's number the string segments to the left and right of the mass by 1 and 2, respectively. Then Eq.$ \,$(4.17) above may be written

$\displaystyle f_m(t) + f_{1m}(t) + f_{2m}(t) = 0, \protect$ (5.19)

where $ f_{1m}(t)$ denotes the force applied by string-segment 1 to the mass (defined as positive in the ``up'', or positive-$ y$ direction), $ f_{2m}(t)$ is the force applied by string-segment 2 to the mass (again positive upwards), and $ f_m(t)$ denotes the inertial force applied by the mass to both string endpoints (where again, a positive force points up).

To derive the traveling-wave relations in a digital waveguide model, we want to use the force-wave variables $ f_1=f^{{+}}_1+f^{{-}}_1$ and $ f_2=f^{{+}}_2+f^{{-}}_2$ that we defined for vibrating strings in §4.1.5; i.e., we defined $ f\isdeftext -Ky'$, where $ K$ is the string tension and $ y'$ is the string slope.

A moment's thought (or careful reading of §H.7.2) reveals that under our definition, string force acts to the right, irrespective of how the sloped string segment might be moving. That is, as shown in Fig.4.22, a negative string slope pulls ``up'' to the right. The same sloped string segment pulls ``down'' when viewed as acting to the left.

Figure 4.22: Depiction of a string segment with negative slope (center), pulling up to the right and down to the left. (Horizontal force components are neglected.)
\begin{figure}\input fig/stringslope.pstex_t \end{figure}

In the present problem, considering the physical sum of forces at the mass-string junction in Fig.4.20, we must have

$\displaystyle m\dot v(t) + K\,y'_1(t,0) - K\, y'_2(t,0) = 0, \protect$