Realizations of Voltage Transfer Functions Using DVCCs ()

Tejmal S. Rathore^{}

School of Electrical Sciences, IIT Goa, Ponda, India.

**DOI: **10.4236/cs.2018.910015
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School of Electrical Sciences, IIT Goa, Ponda, India.

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.

Keywords

Differential-Voltage Current-Conveyor, All-Pass Filters, Bilinear Functions, Biquadratic Functions

Share and Cite:

Rathore, T. (2018) Realizations of Voltage Transfer Functions Using DVCCs. *Circuits and Systems*, **9**, 141-147. doi: 10.4236/cs.2018.910015.

1. Introduction

The symbol of a difference voltage current conveyor (DVCC) is shown in Figure 1 and its characteristics are summarized by the following port relationships [1]

${V}_{X}={V}_{Y1}-{V}_{Y2},\text{\hspace{0.17em}}{I}_{Y1}={I}_{Y2}=0,\text{\hspace{0.17em}}{I}_{Z+}={I}_{X},\text{\hspace{0.17em}}{I}_{Z-}=-{I}_{X}$ (1)

Maheshwari [2] has proposed three configurations, shown in Figure 2, using DVCC for realizing only first order voltage-mode all-pass filter. The filter has the advantages of high input resistance and minimum number of RC elements and all of them grounded.

The intent of this paper is to exploit the circuit topologies of Figure 2 to realize more general class of functions. Conditions of realization are derived. Then we consider realizations of bilinear and biquadratic functions in detail. In the

Figure 1. Block representation of a differential voltage current conveyor.

Figure 2. Proposed topologies where zi’s in DVCC (1) and (2) are (a) z+, z−; (b) z−, z+, (c) z+, z+, respectively.

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 [2] . In case of biquadratic function only 4 passive elements (2C, 2R) are required with both the capacitors grounded. Desing of a second order all-pass function is given.

2. Realization of a General Voltage Transfer Function

Each of the circuit topologies of Figures 2(a)-(c) has the voltage transfer function

$T\left(s\right)=K\frac{N\left(s\right)}{D\left(s\right)}=1-{Z}_{1}{Y}_{2}$ (2)

From (2),

${Z}_{1}{Y}_{2}=\frac{D\left(s\right)-KN\left(s\right)}{D\left(s\right)}.$ (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 [3] . The four admissible pole-zero patterns are shown in Figure 3. T(s), so also Z_{1}Y_{2}, must have poles distinct negative real. Zero lociof Z_{1}Y_{2} start from the poles when K = 0 and terminate on the zeros of T(s) when K = ∞. Hence it is possible to choose K sufficiently small so that the zeros are negative real. Thus, if the poles and zeros of Z_{1}Y_{2} fit into one of the patterns shown in Figure 3, T is realizable otherwise not.

Figure 3. Admissible pole zero patterns.

2.1. Realization of Bilinear Voltage Transfer Functions

Let the bilinear voltage transfer function be

$T\left(s\right)=\frac{K\left(s+z\right)}{s+p}=1-{Z}_{1}{Y}_{2}$ (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

${Z}_{1}{Y}_{2}=\frac{\left(1-K\right)s+\left(p+Kz\right)}{s+p}\text{.}$ (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 Z_{1}Y_{2} can be made negative real. Thus it will follow the pattern shown in Figure 3(a) and Figure 3(d). Now

${Z}_{1}{Y}_{2}=\mu \frac{s+\alpha}{s+p},\text{}p<\alpha \le \infty .$ (6)

Only possible identifications are

${Z}_{1}=\frac{{\mu}_{1}}{s+p},\text{}{Y}_{2}={\mu}_{2}\left(s+\alpha \right)$ (7)

${Z}_{1}={\mu}_{1}\frac{s+\alpha}{s},\text{}{Y}_{2}={\mu}_{2}\frac{s}{s+p}$ (8)

${Z}_{1}={\mu}_{1}\frac{s+\alpha}{s+p},\text{}{Y}_{2}={\mu}_{2}$ (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 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 Z_{1}Y_{2} 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

Figure 4. Root locus with K as variable.

(a) (b)(c) (d)

Figure 5. Realizations of Z_{1} and Z_{2} given by (a) (7); (b) (8) and (c) and (d) by (9).

(a) (b)

Figure 6. Two realizations of Z_{1},_{2}.

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

Alternative proof

In Figure 2, Z_{1} and Z_{2} 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

${Z}_{i}=\left(\frac{1}{{C}_{i}}\right)\frac{1}{\left(s+\frac{1}{{C}_{i}{R}_{i}}\right)},\text{}i=1,2$ (10)

Now

$T\left(s\right)=1-\frac{{Z}_{1}}{{Z}_{2}}=\left(1-\frac{{C}_{2}}{{C}_{1}}\right)\frac{s-\frac{1}{\left(1-\frac{{C}_{2}}{{C}_{1}}\right)}\left(\frac{{R}_{1}}{{R}_{2}}-1\right)\frac{1}{{C}_{1}{R}_{1}}}{\left\{s+\frac{1}{{C}_{1}{R}_{1}}\right\}}=\mu \frac{\left\{s-\alpha \right\}}{\left\{s+p\right\}}$ (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

${Z}_{i}=\left(\frac{1}{{C}_{i}}\right)\frac{1}{\left(s+\frac{1}{{C}_{i}{R}_{i}}\right)},\text{}i=1,2$ (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 [2] without any logic.

2.2. Realization of Biquadratic Transfer Functions

Let the function be expressed as

$T\left(s\right)=K\frac{\left(s+{z}_{1}\right)\left(s+{z}_{2}\right)}{\left(s+{p}_{1}\right)\left(s+{p}_{2}\right)}$ (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)

${Z}_{1}{Y}_{2}=\frac{\left(1-K\right){s}^{2}+\left[\left({p}_{1}+{p}_{2}\right)-K\left({z}_{1}+{z}_{2}\right)\right]s+\left({p}_{1}{p}_{2}-K{z}_{1}{z}_{2}\right)}{\left(s+{p}_{1}\right)\left(s+{p}_{2}\right)}$ (14)

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

${Z}_{1}{Y}_{2}=\frac{\left[\left({p}_{1}+{p}_{2}\right)-\left({z}_{1}+{z}_{2}\right)\right]s+\left({p}_{1}{p}_{2}-{z}_{1}{z}_{2}\right)}{\left(s+{p}_{1}\right)\left(s+{p}_{2}\right)}=\mu \frac{\left(s+\alpha \right)}{\left(s+{p}_{1}\right)\left(s+{p}_{2}\right)}$ (15)

where
$\mu =\left({p}_{1}+{p}_{2}\right)-\left({z}_{1}+{z}_{2}\right)$ and
$\alpha =\frac{{p}_{1}{p}_{2}-{z}_{1}{z}_{2}}{\left({p}_{1}+{p}_{2}\right)-\left({z}_{1}+{z}_{2}\right)}$ . 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

${Z}_{1}{Y}_{2}=\mu \frac{s}{\left(s+{p}_{1}\right)\left(s+{p}_{2}\right)}.$ (16)

where μ = 2(p_{1} + p_{2}).

Now Z_{1,2} can be identified, in two possible ways as

${Z}_{1}=\frac{{\mu}_{1}}{s+{p}_{1,2}},\text{}{Z}_{2}=\frac{s+{p}_{2,1}}{{\mu}_{2}s}$ (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 Figure 7. As expected, all elements are not grounded. It is interesting to realize an all-pass filter with double poles, i.e., p_{1} = p_{2} = p. In this case (17) reduces to

${Z}_{1}=\frac{{\mu}_{1}}{s+p},\text{}{Z}_{2}=\frac{s+p}{{\mu}_{2}s}$ (18)

Example: Realize a second order all-pass function

$T\left(s\right)=\frac{{s}^{2}-4s+3}{{s}^{2}+4s+3}\text{.}$ (19)

Here

${Z}_{1}{Y}_{2}=1-T\left(s\right)=\frac{8s}{\left(s+1\right)\left(s+3\right)}\text{.}$ (20)

Identifying

${Z}_{1}=\frac{1}{s+1}\text{,}{Z}_{2}=\frac{s+3}{8s}$ (21)

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

3. Conclusion

Maheshwari [1] proposed three DVCC configurations and used them for realizing only first order all-pass filters. This paper has exploited these configurations for realizing transfer function T(s) such that the location of poles and zeros of 1 - T(s) matches with any one shown in Figure 3. It has been proved 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 (2R, 1C) or (1R, 2C). Biquadratic functions have also been considered. They require (2C, 2R) passive elements with both the capacitors grounded. We have not attempted here the non-ideal analysis, simulation and applications as they are very well documented in [1] .

Figure 7. Two realizations of Z_{1,2} given by (17).

Figure 8. Realization of all pass function given by (19).

Conflicts of Interest

The authors declare no conflicts of interest.

[1] | Ibrahim, M.A., Minaei, S. and Kuntman, H. (2005) A 22.5 MHz Biquad Using Differential Voltage Current Conveyor and Grounded Passive Elements. International Journal of Electronics and Communications, 59, 311-318. |

[2] |
Maheshwari, S. (2007) High Input Impedance VM-APSs with Grounded Passive Elements. IET Circuits, Devices and Systems, 1, 72-78.
https://doi.org/10.1049/iet-cds:20060196 |

[3] |
Rathore, T.S. and Khot, U.P. (2005) Single FTFN Realization of Current Transfer Functions. IETE Journal of Research, 51, 193-199.
https://doi.org/10.1080/03772063.2005.11416395 |

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