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Impedance defined from reflectance

The size of the class of filters called impedances will be seen to be large, because impedances are derived by transformation from an easily specified family of filters called reflectances, say, ct and its Fourier transform $C( \omega )$.To be a reflectance, the time function must be strictly causal, and the frequency function must be strictly less than unity. By strictly causal it is meant that the time function vanishes both at zero time and before. For example, take $-1\ < \ \rho\ < \ +1$ and the reflectance ct to be an impulse of size $\rho$ after a time $\Delta t$.The Fourier transform is  
C \eq \rho\, Z \eq \rho \ e^{ i \omega \Delta t }\end{displaymath} (45)
Obviously, the product of two reflectances is another reflectance.

An impedance has been defined to be a causal filter with a causal inverse and a Fourier transform whose real part is positive. It will be shown that from any reflectance C the expression  
R \eq {1\ -\ C \over 1\ +\ C}\end{displaymath} (46)
generates an impedance. There are three things to show: R is causal, has a causal inverse, and is PR. First because of the assumption that C has a magnitude strictly less than unity, $C \bar C \ < \ 1$,the denominator expands to a convergent $1\,+\,C\,+\,C^2 \,+ \cdots$.Second, the inverse of R is found by simply changing the sign of C. Third, multiply top and bottom by the complex conjugate:
\Re \ R &=& \Re \ {(1\ -\ C)\,(1\ +\ \bar C ) \over \ \rm{posit...
 ... \ +\ \ \ \rm{imaginary} \over \ \rm{positive} }
\ \ \ \ge \ \ \ 0\end{eqnarray} (47)
which shows that R has a real part that is positive.

The expression for R(C) is easily back-solved for C(R), but the converse theorem, that every R generates a reflectance, is harder to show. Nevertheless, it will be proved, along with a deeper theorem. A filter that is both causal and PR is said to be CPR. The deeper theorem is that every CPR has an inverse and hence is an impedance. This will be proved by showing that every CPR--say, $\hat R$, --can be used to construct a reflectance $\hat C$, which, since it is a reflectance, implies that the CPR $\hat R$ is an impedance R. Backsolving gives  
\hat C \eq {1\ -\ \hat R \over 1\ +\ \hat R}\end{displaymath} (49)

Proof requires that two things be shown--first, that the magnitude of $\hat C$ is less than unity. To show this, form the magnitude of the denominator and subtract the magnitude of the numerator. The result is four times the real part of $\hat R$, which is positive. Second, $\hat C$ must be proved causal. This is harder. $(1\ +\ \hat R )^{-1} $ can be expanded into a sum of positive powers of $\hat R$ and hence of positive powers of the delay operator. But the convergence of the series is not assured, because nothing requires $\hat R$ to be less than unity.

To prove that $\hat C$ is causal, we will take advantage of rule 1, namely, that an impedance can be scaled by any real positive number that you like, and it will still be an impedance. Consider a function that is similar to $\hat C$. 
B \eq { 1 \ -\ \epsilon \, \hat R \over 1\ +\ \epsilon \, \hat R }\end{displaymath} (50)
Choose $\epsilon$ small enough that for all $\omega$, $\epsilon \vert \hat R \vert < 1$.This ensures a convergent expansion for the denominator in positive powers of $\hat R$ and hence Z. The expansion contains only positive powers in the delay operator. Thus B is a reflectance, and its corresponding impedance is $\hat \epsilon$.But an impedance can always be scaled by a positive number. Taking the number to be $1/ \epsilon$ shows that $\hat R$ is an impedance. This completes the proof that every CPR is an impedance.

So impedances arise more easily than you might think. It is not necessary to have a reflectance C to insert into the relation R = (1-C)/(1+C). We only need to have a CPR.

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Next: Functional analysis Up: IMPEDANCE Previous: Muir's rules for combining
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