ABSTRACT
We next express the transmission coefficient in terms of other specifications. From (77.12), we have
(77.16)
Substituting
V
(
j
ω
) =
z
(
j
ω
)
I
(
j
ω
) in (77.16) gives
(77.17)
In terms of the transfer voltage ratio and transfer impedance, we apply the relation
V
(
j
ω
) = –
I
(
j
ω
)
z
(
j
ω
) and obtain
(77.18)
(77.19)
Similarly, we can derive a relation between the output reflection coefficient
ρ
(
j
ω
) and the transmission coefficient
ρ
(
j
ω
) magnitude squared as
(77.20)
In fact, for the lossless reciprocal two-port network
N
we have
(77.21)
In this part, we show how to design a lossless two-port network operating between a resistive generator with internal resistance
R
and a resistive load with resistance
R
to yield the
n
th-order Butterworth transducer power-gain characteristic
(77.22)
Since for a passive network
G
(
ω
) is bounded between 0 and 1, the DC gain
K
is restricted by
(77.23)
Substituting (77.22) in (77.12) yields the squared magnitude of the input reflection coefficient as
(77.24)
ρ ω ω ω ω ω21
4j r j r j I j
V jg [ ] = ( ) ( ) ( )( )
ρ ω ω ω ω
ω
4 j
r j r j
z j
I j
I jg ( ) = ( ) ( )( )
( ) ( )
ρ ω ω ω ω
ω
4 j
r j r j
z j
V j
V jg ( ) = ( ) ( )( )
( ) ( )
ρ ω ω ω ω ω
ω
4 j
r j r j
z j z j
V j
I jg ( ) = ( ) ( )( ) ( )
( ) ( )
ρ ω ρ ω12 2
2 j j( ) ≡ − ( )1
ρ ω ρ ω ρ ω ρ ω21 2
2 1j j j j( ) = ( ) = − ( ) = − ( )1
G j Kn
nω ρ ω ω ω
( ) = ( ) = + ( )
0 ≤ ≤Kn 1
ρ ω ω ρ ω ω ω
ω ω 11
21 1
1 j G j
n( ) = − ( ) = − ( ) = − + ( ) + ( )1
or
(77.25)
where
(77.26)
Appealing to analytic continuation by substituting ω by –js results in
(77.27)
where
(77.28)
To obtain the input reflection coefficient ρ11(s) from ρ11(s) ρ11(–s), we need to assign the zeros and poles of (77.27). Since ρ11(s) is devoid of poles in the closed RHS, we must assign all the LHS poles to ρ11(s). The zeros of ρ11(s), however, may lie in the RHS, so that in general a number of different numerators are possible. For our purposes, we choose only the LHS zeros for ρ11(s). Define a minimum-phase reflection coefficient to be one that is devoid of zeros in the open RHS. Then, the minimum-phase solution of (77.27) can be written as
(77.29)
where q(x) is the Hurwitz polynomial with unity leading coefficient formed by the LHS roots of the equation 1 + (–1)nx2n = 0. From (77.1a), the input impedance is found to be
(77.30)
Combining this with (77.29) yields
(77.31)
If both R1 and R2 are specified, then the DC gain Kn cannot be chosen independently. In fact, by substituting s = 0 in (77.31) and assuming that Kn ≠ 0 we obtain
(77.32)
where the ± signs are determined, respectively, according to R2 ≥ R1 and R2 ≤ R1. Therefore, if any two of the three quantities R1, R2, and Kn are specified, the third one is fixed.
