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One-Multiply Parallel Reflection-Free Three-Port Adaptor

It turns out only one multiply is needed to implement three-port adaptors (scattering junctions) having a reflection-free port. The derivation is very similar to deriving the one-multiply two-port scattering junction from the four-multiply Kelly-Lochbaum scattering junction, as discussed in Fig.[*].

Let's begin with the scattering relations in terms of the alpha parameters introduced in §F.2.2 for a parallel junction of any number of ports:

$\displaystyle f_J$ $\displaystyle =$ $\displaystyle \sum_{i=1}^N \alpha_i \, f^{{+}}_i,$ (F.23)
$\displaystyle f^{{-}}_i(n)$ $\displaystyle =$ $\displaystyle f_J(n) - f^{{+}}_i(n)$ (F.24)

where $ f_J$ denotes the junction force (or voltage), the $ \alpha$ parameters are defined for parallel junctions by

$\displaystyle \alpha_i \isdefs \frac{2\Gamma _i}{\sum_{j=1}^N \Gamma _j},

and $ \Gamma _i=1/R_i$ is the wave admittance on port $ i$ .

Here we wish to focus on the three-port case $ N=3$ . To connect with our previous example, set $ R_1=R_A$ , $ R_2=R_B$ , and $ R_3=R_C$ . When port A is reflection-free, we have the constraint that $ \Gamma _A=\Gamma _B+\Gamma _B$ ( $ R_A=R_B\Vert R_C=R_BR_C/(R_B+R_C)$ ). Scattering relations are unchanged when all port impedances are divided by the same positive number, so divide through by $ R_A$ to obtain $ R_1=\Gamma _1=1$ , and $ \Gamma _2+\Gamma _3=1$ . With $ \Gamma _1=1$ , we have $ \Gamma _2\in[0,1]$ , and $ \Gamma _3=1-\Gamma _2$ . In other words, there is only one degree of freedom, $ \Gamma _2\in[0,1]$ for the three-port parallel junction with a reflection-free port.

The alpha parameters become $ \alpha_1=1$ , $ \alpha_2=\Gamma _2\in[0,1]$ , and $ \alpha_3=\Gamma _3=1-\Gamma _2\in[0,1]$ , so that the scattering formulas simplify to

$\displaystyle f_J$ $\displaystyle =$ $\displaystyle f^{{+}}_1 + \Gamma _2f^{{+}}_2 + (1-\Gamma _2)f^{{+}}_3 \eqsp
f^{{+}}_1 + f^{{+}}_3 + \Gamma _2(f^{{+}}_2 - f^{{+}}_3),$ (F.25)
$\displaystyle f^{{-}}_i$ $\displaystyle =$ $\displaystyle f_J - f^{{+}}_i.$ (F.26)

Thus, only one multiply is necessary to compute the junction force (or voltage). In this form we have six additions, but this can be brought down to four. Expand $ f_J$ in the last equation to get
$\displaystyle f_\delta$ $\displaystyle \isdef$ $\displaystyle f^{{+}}_2 - f^{{+}}_3$ (F.27)
$\displaystyle g_\delta$ $\displaystyle \isdef$ $\displaystyle \Gamma _2 f_\delta$ (F.28)
$\displaystyle f^{{-}}_1$ $\displaystyle =$ $\displaystyle f^{{+}}_3 + g_\delta$ (F.29)
$\displaystyle f^{{-}}_3$ $\displaystyle =$ $\displaystyle f^{{+}}_1 + g_\delta$ (F.30)
$\displaystyle f^{{-}}_2$ $\displaystyle =$ $\displaystyle f^{{+}}_1 + f^{{+}}_3 + g_\delta - f^{{+}}_2
\eqsp f^{{+}}_1 + g_\delta - f_\delta$ (F.31)
  $\displaystyle =$ $\displaystyle f^{{-}}_3 - f_\delta$ (F.32)

Thus, one multiply and four additions suffice for the three-port parallel junction with reflection-free port.
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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4
Copyright © 2023-08-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University