Sign in

Search Online Books

Free Online Books

Sponsor

Industry's highest performing at the lowest power DSPs now as low as $5.00*
Start development today!
*volume pricing for 10ku

Chapters

Physical Audio Signal Processing
    Wave Digital Filters
       Wave Digital Elements
          A Physical Derivation of Wave Digital Elements
             Reflectance of a General Lumped Waveguide Termination

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?

Reflectance of a General Lumped Waveguide Termination

Calculate the reflectance of the terminated waveguide. That is, find the Laplace transform of the return wave divided by the Laplace transform of the input wave going into the waveguide. In general, the reflectance of an impedance step for force waves (voltage waves in the electrical case) is

$\displaystyle \fbox{$\displaystyle \hat{\rho}_R(s) \isdef \frac{F^{-}(s)}{F^{+}(s)} = \frac{R(s)-R_0}{R(s)+R_0}$} \protect$

(F.1)

This is easily derived from continuity constraints across the junction. Specifically, referring to Fig.F.1b, let $F_R(s) = F^{+}_R(s) + F^{-}_R(s)$ denote the physical force and its traveling-wave components within the ``pseudo-infinitesimal-generalized-waveguide'' defined by the element impedance

, with the `

' superscript denoting a right-going wave.^F.1 Similarly, let $V(s) = V^{+}(s)+V^{-}(s)$ denote the velocity and its component wave variables on the side of the junction at impedance

, and let $V_R(s) = V^{+}_R(s)+V^{-}_R(s)$ denote the corresponding quantities on the element-side of the junction at impedance

. Again, the `

' superscript denotes travel to the right. Then the physical continuity constraints imply

$\begin{eqnarray*} F(s) &=& F_R(s)\\ 0 &=& V(s) + V_R(s) \end{eqnarray*}$

By the definition of wave impedance in a waveguide, we have

$\begin{eqnarray*} F^{+}(s) &=& \quad\! R_0 V^{+}(s)\\ F^{-}(s) &=& - R_0 V^{-}(... ... &=& \quad\! R(s) V^{+}_R(s)\\ F^{-}_R(s) &=& - R(s) V^{-}_R(s) \end{eqnarray*}$

Thus,

$\begin{eqnarray*} 0 &=& V(s) + V_R(s)\\ &=& \left[V^{+}(s)+V^{-}(s)\right] + ... ...s)}\right] &=& \frac{2}{R_0}F^{+}(s) + \frac{2}{R(s)}F^{+}_R(s) \end{eqnarray*}$

Defining $\Gamma_0 \isdef 1/R_0$ and $\Gamma(s) \isdef 1/R(s)$ , we have

$\displaystyle F(s) = \frac{2\Gamma_0}{\Gamma_0+\Gamma(s)} F^{+}(s) + \frac{2\Ga... ...a(s)} F^{+}_R(s) \isdef {\cal A}(s) F^{+}(s) + {\cal A}_R(s)F^{+}_R(s) \protect$

(F.2)

Now that we've solved for the junction force

, the outgoing waves are simply obtained from the force continuity constraint, $F(s) = F^{+}(s)+F^{-}(s) = F^{+}_R(s)+F^{-}_R(s)$ :

$\displaystyle F^{-}(s)$	$\displaystyle =$	$\displaystyle F(s) - F^{+}(s) \protect$	(F.3)
$\displaystyle F^{-}_R(s)$	$\displaystyle =$	$\displaystyle F(s) - F^{+}_R(s) \protect$	(F.4)

Finally, the force-wave reflectance of an impedance step from

can be found by solving Eq. $\,$ (F.3) and (F.2) for $F^{-}(s)/F^{+}(s)$ with $F^{+}_R(s)$ set to zero:

$\begin{eqnarray*} \hat{\rho}(s) &\isdef & \frac{F^{-}(s)}{F^{+}(s)} = \frac{F(s)... ...mma_0-\Gamma(s)}{\Gamma_0+\Gamma(s)} = \frac{R(s)-R_0}{R(s)+R_0} \end{eqnarray*}$

as claimed.

Previous: A Physical Derivation of Wave Digital Elements
Next: Reflectances of Elementary Impedances

Order a Hardcopy of Physical Audio Signal Processing

About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.

Comments

No comments yet for this page

Add a Comment

You need to login before you can post a comment (best way to prevent spam). ( Not a member? )