Let the bilinear voltage transfer function be

T ( s ) = K ( s + z ) s + p = 1 − Z 1 Y 2 (4)

If z is positive, i.e., it lies on the negative real axis, then T(s) can be realized by RC passive elements. Therefore, we shall consider the case when z is negative i.e., 0 ≤ z ≤ ∞.

Then

Z 1 Y 2 = ( 1 − K ) s + ( p + K z ) s + p . (5)

The zero-locus with K as variable is shown in Figure 4. It starts from pole at ?p when K = 0 and reaches ?∞ when K = 1 and again from +∞ to zero at z from K = 1 to K = ∞. Thus choosing 0 < K ≤ 1, the zero of Z₁Y₂ can be made negative real. Thus it will follow the pattern shown in Figure 3(a) and Figure 3(d). Now

Z 1 Y 2 = μ s + α s + p , p < α ≤ ∞ . (6)

Only possible identifications are

Z 1 = μ 1 s + p , Y 2 = μ 2 ( s + α ) (7)

Z 1 = μ 1 s + α s , Y 2 = μ 2 s s + p (8)

Z 1 = μ 1 s + α s + p , Y 2 = μ 2 (9)

where μ = μ₁μ₂. The possible canonic realizations of Z₁ and Z₂ in Foster and Cauer forms from (7) is given in Figure 5(a), from (8) in Figure 5(b), and from (9) in Figure 5(c) and Figure 5(d), respectively.

Minimum 4 elements (2C, 2R) or (1C, 3R) are required for realizing Z_1,2 as shown in Figure 5. However, one element can be reduced by choosing K = 1 which forces the zero of Z₁Y₂ at ∞ (see (5)). The reduced realizations of Z_1,2 are shown in Figure 6. The complete realizations of T(s) given by (4) is obtained by inserting Z_1,2 of Figure 6 in Figure 2. They reduce to all-pass functions when p = z. Thus we get (2R, 1C) and (1R, 2C) realizations for first order all-pass function. The realization corresponding to (2R, 1C) has all the passive elements grounded. We have thus shown that there are four possible realizations and only one of

which have all three elements (2R, 1C) grounded systematically, and not intuitively as in [2] .

Alternative proof

In Figure 2, Z₁ and Z₂ can have only one R and one C element at the most in parallel for them to be grounded. Any additional resistance (capacitance) in parallel will be absorbed in R (C) already present. Let Z_i (i = 1, 2) be parallel combinations of one R_i and one C_i as shown in the Figure 5(d). Then

Z i = ( 1 C i ) 1 ( s + 1 C i R i ) , i = 1 , 2 (10)

Now

T ( s ) = 1 − Z 1 Z 2 = ( 1 − C 2 C 1 ) s − 1 ( 1 − C 2 C 1 ) ( R 1 R 2 − 1 ) 1 C 1 R 1 { s + 1 C 1 R 1 } = μ { s − α } { s + p } (11)

Since the denominator is of first order, the circuit can realize only bilinear transfer functions with four components (2C and 2R); all grounded. The number of elements can be reduced by 1 under specific conditions. Let us consider the special case of all pass function. From (11), the condition is

Z i = ( 1 C i ) 1 ( s + 1 C i R i ) , i = 1 , 2 (12)

There are infinite number of solutions to satisfy (11). From (11), it is obvious that R₁ cannot be equal to R₂ and C₁ cannot be equal to C_2. To have minimum number of passive components, the choices are C₂ = 0, R₁ = 2R₂ and R₂ = ∞, C₂ = 2C₁. In the latter choice, it can be seen from (11) that T(s) will be negative. These choices were directly chosen in [2] without any logic.

Let the function be expressed as

T ( s ) = K ( s + z 1 ) ( s + z 2 ) ( s + p 1 ) ( s + p 2 ) (13)

where poles p_1,2, as discussed above, have to be negative real and z_1,2 may lie anywhere in the s-plane. Then from (2)

Z 1 Y 2 = ( 1 − K ) s 2 + [ ( p 1 + p 2 ) − K ( z 1 + z 2 ) ] s + ( p 1 p 2 − K z 1 z 2 ) ( s + p 1 ) ( s + p 2 ) (14)

To realize with minimum number of elements, we choose K = 1. Then

Z 1 Y 2 = [ ( p 1 + p 2 ) − ( z 1 + z 2 ) ] s + ( p 1 p 2 − z 1 z 2 ) ( s + p 1 ) ( s + p 2 ) = μ ( s + α ) ( s + p 1 ) ( s + p 2 ) (15)

where μ = ( p 1 + p 2 ) − ( z 1 + z 2 ) and α = p 1 p 2 − z 1 z 2 ( p 1 + p 2 ) − ( z 1 + z 2 ) . If α and p_1,2

satisfy any of the pole-zero patterns shown in Figure 3 then Z_1,2 can be identified as driving point impedances. Since there are many possible locations of z_1,2, we explain the procedure by taking the all-pass function for which z_1,2 = −p_1,2. Then (15) reduces to

Z 1 Y 2 = μ s ( s + p 1 ) ( s + p 2 ) . (16)

where μ = 2(p₁ + p₂).

Now Z_1,2 can be identified, in two possible ways as

Z 1 = μ 1 s + p 1 , 2 , Z 2 = s + p 2 , 1 μ 2 s (17)

where μ = μ₁μ₂.

Choosing μ₁ = μ₂ = 1, Two realizations of Z_1,2 given by (17) are shown in Figure 7. As expected, all elements are not grounded. It is interesting to realize an all-pass filter with double poles, i.e., p₁ = p₂ = p. In this case (17) reduces to

Z 1 = μ 1 s + p , Z 2 = s + p μ 2 s (18)

Example: Realize a second order all-pass function

T ( s ) = s 2 − 4 s + 3 s 2 + 4 s + 3 . (19)

Here

Z 1 Y 2 = 1 − T ( s ) = 8 s ( s + 1 ) ( s + 3 ) . (20)

Identifying

Z 1 = 1 s + 1 , Z 2 = s + 3 8 s (21)

the complete realization of T(s) of (19) is given in Figure 8.