Maheshwari has proposed three differential-voltage current-conveyor configurations for realizing first order all-pass filters only. This paper has exploited these configurations for realizing more complex transfer function T( s) which yield poles and zeros of 1 - T( s) in one of the four admissible patterns. Bilinear and biquadratic functions are dealt in detail. It is shown that only bilinear functions can be realized with all the four passive elements grounded. First order all-pass function is a special case which needs only three elements (2 R, 1 C) or (1 R, 2 C). A biquadratic function requires (2 R, 2 C) elements and has all the capacitor grounded. Design of second order all-pass function is given.
The symbol of a difference voltage current conveyor (DVCC) is shown in
V X = V Y 1 − V Y 2 , I Y 1 = I Y 2 = 0 , I Z + = I X , I Z − = − I X (1)
Maheshwari [
The intent of this paper is to exploit the circuit topologies of
former case, it is shown that only first order functions can have all passive components grounded, and that there are only two possible cases with minimum number of passive elements (1C 2R and 2C 1R) logically, rather than intuitionally in [
Each of the circuit topologies of Figures 2(a)-(c) has the voltage transfer function
T ( s ) = K N ( s ) D ( s ) = 1 − Z 1 Y 2 (2)
From (2),
Z 1 Y 2 = D ( s ) − K N ( s ) D ( s ) . (3)
Note that the poles of Z_{1}Y_{2} are the same as those of T; but zeros are given by D − KN = 0. Impedances Z_{1} and Z_{2} can be identified as RC driving point functions (DPIs) from (3), if the poles and zeros of Z_{1}Y_{2,} arranged in pairs starting from the rightmost pair, each pair consists of a pole and a zero in either order [
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
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 = μ_{1}μ_{2}. The possible canonic realizations of Z_{1} and Z_{2} in Foster and Cauer forms from (7) is given in
Minimum 4 elements (2C, 2R) or (1C, 3R) are required for realizing Z_{1,2} as shown in
which have all three elements (2R, 1C) grounded systematically, and not intuitively as in [
Alternative proof
In
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 R_{1} cannot be equal to R_{2} and C_{1} cannot be equal to C_{2.} To have minimum number of passive components, the choices are C_{2} = 0, R_{1} = 2R_{2} and R_{2} = ∞, C_{2} = 2C_{1}. In the latter choice, it can be seen from (11) that T(s) will be negative. These choices were directly chosen in [
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
Z 1 Y 2 = μ s ( s + p 1 ) ( s + p 2 ) . (16)
where μ = 2(p_{1} + p_{2}).
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 = μ_{1}μ_{2}.
Choosing μ_{1} = μ_{2} = 1, Two realizations of Z_{1,2} given by (17) are shown in
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
Maheshwari [
The author declares no conflicts of interest regarding the publication of this paper.
Rathore, T.S. (2018) Realizations of Voltage Transfer Functions Using DVCCs. Circuits and Systems, 9, 141-147. https://doi.org/10.4236/cs.2018.910015