*NGC as a Five-Node Network*

In Figure 4 we assume NGC as a five-node network. We start analyzing this network by performing KCL at node *Y* as reported in the following:

$$
\dot{\imath}\_Y = \dot{\imath}\_3 + \dot{\imath}\_6 + \dot{\imath}\_7 = \dot{\imath}\_3 + \Upsilon\_6 V\_Z + \Upsilon\_7 V\_Z. \tag{13}
$$

Since no current is flowing into *Y*<sup>8</sup> and *Y*9, these admittances can be assumed as open circuit (*Y*<sup>8</sup> = *Y*<sup>9</sup> = 0). Routine analysis of Figure 4 results in *i*<sup>3</sup> as:

$$i\_3 = \frac{\chi\_3(\chi\_1 + \chi\_2)}{\chi\_1 + \chi\_2 + \chi\_3 + \chi\_4} V\_Z. \tag{14}$$

Using (13)–(14), we have:

$$\frac{\dot{Y}\_Y}{V\_Z} = \frac{Y\_3(Y\_1 + Y\_2)}{Y\_1 + Y\_2 + Y\_3 + Y\_4} + Y\_6 + Y\_7. \tag{15}$$

Similar analysis for *ix* results:

$$\frac{\dot{q}\_X}{\dot{V}\_Z} = -\frac{\chi\_2(\chi\_3 + \chi\_4)}{\chi\_1 + \chi\_2 + \chi\_3 + \chi\_4} - \chi\_5 - \chi\_7 \tag{16}$$

**Figure 4.** The NGC as a five-node network.

Using (15) and (16) in (7), the CE of the five-node network is found as:

$$(1 - \mathbf{Y}\_2 \mathbf{Y}\_4 + \mathbf{Y}\_3 \mathbf{Y}\_1 + (\mathbf{Y}\_6 - \mathbf{Y}\_5)(\mathbf{Y}\_1 + \mathbf{Y}\_2 + \mathbf{Y}\_3 + \mathbf{Y}\_4) = \mathbf{0}.\tag{17}$$

It can be noticed that CE does not depend on *Y*7, ... ,*Y*<sup>10</sup> which means that these branches can be assumed to be open circuit. For the other branches we can make different choices. If two branches have non-zero admittances, the following CEs are possible:

$$\mathcal{Y}\_1 \mathcal{Y}\_3 \neq 0 \to \mathcal{C} \\
\mathcal{E} : \mathcal{Y}\_1 \mathcal{Y}\_3 = 0 \tag{18a}$$

$$\mathcal{Y}\_1 \mathcal{Y}\_5 \neq 0 \to \mathbb{C}E : \mathcal{Y}\_1 \mathcal{Y}\_5 = 0 \tag{18b}$$

$$\mathcal{Y}\_1 \mathcal{Y}\_6 \neq 0 \to \mathcal{C}E : \mathcal{Y}\_1 \mathcal{Y}\_6 = 0 \tag{18c}$$

$$\mathcal{Y}\_{2}, \mathcal{Y}\_{4} \neq 0 \rightarrow \mathcal{C}E : \mathcal{Y}\_{2}\mathcal{Y}\_{4} = 0 \tag{18d}$$

$$\mathcal{Y}\_{2\prime}\mathcal{Y}\_5 \neq 0 \to \mathbb{C}E : \mathcal{Y}\_2\mathcal{Y}\_5 = 0 \tag{18e}$$

$$
\mathcal{Y}\_2 \mathcal{Y}\_6 \neq 0 \to \mathbb{C}E : \mathcal{Y}\_2 \mathcal{Y}\_6 = 0 \tag{18f}
$$

$$
\mathcal{Y}\_3 \mathcal{Y}\_5 \neq 0 \to \mathcal{C}E : \mathcal{Y}\_3 \mathcal{Y}\_5 = 0 \tag{18g}
$$

$$
\mathcal{Y}\_3 \mathcal{Y}\_6 \neq 0 \to \mathcal{C}E : \mathcal{Y}\_3 \mathcal{Y}\_6 = 0 \tag{18h}
$$

$$\mathcal{Y}\_{4\prime}\mathcal{Y}\_{5} \neq 0 \to \mathcal{C}E : \mathcal{Y}\_{4}\mathcal{Y}\_{5} = 0 \tag{18i}$$

$$
\mathcal{Y}\_{4\prime}\mathcal{Y}\_6 \neq 0 \to \mathbb{C}E : \mathcal{Y}\_4\mathcal{Y}\_6 = 0 \tag{18j}
$$

In the general case, (18) can be expressed as

$$CE: \quad \mathcal{Y}\_a \mathcal{Y}\_b = 0 \tag{19}$$

By assuming *Ya = sCa + Ga* and *Yb = sCb + Gb*, (19) can be written as:

$$s^2\mathcal{C}\_a\mathcal{C}\_b + s\left[\mathcal{C}\_a\mathcal{G}\_b + \mathcal{C}\_b\mathcal{G}\_a\right] + \mathcal{G}\_a\mathcal{G}\_b = 0\tag{20}$$

From (20) it is not possible to have imaginary roots. Therefore, in case of two non-zero branches, no oscillation is possible.

Finally, we investigate the possibility of achieving oscillations from (17) in the case that three branches of NGC present non-zero admittance.

For (*Y*<sup>1</sup> = *Y*<sup>5</sup> = *Y*<sup>6</sup> = 0) or (*Y*<sup>3</sup> = *Y*<sup>5</sup> = *Y*<sup>6</sup> = 0), we have *Y2Y4* = 0, while for (*Y*<sup>2</sup> = *Y*<sup>5</sup> = *Y*<sup>6</sup> = 0) or (*Y*<sup>4</sup> = *Y*<sup>5</sup> = *Y*<sup>6</sup> = 0), we have *Y1Y3* = 0. In both these cases, the CE has the general form of (19).

For (*Y*<sup>1</sup> = *Y*<sup>2</sup> = *Y*<sup>5</sup> = 0), (*Y*<sup>1</sup> = *Y*<sup>2</sup> = *Y*<sup>6</sup> = 0), (*Y*<sup>1</sup> = *Y*<sup>4</sup> = *Y*<sup>5</sup> = 0), (*Y*<sup>1</sup> = *Y*<sup>4</sup> = *Y*<sup>6</sup> = 0), (*Y*<sup>2</sup> = *Y*<sup>3</sup> = *Y*<sup>5</sup> = 0), (*Y*<sup>2</sup> = *Y*<sup>3</sup> = *Y*<sup>6</sup> = 0), (*Y*<sup>3</sup> = *Y*<sup>4</sup> = *Y*<sup>5</sup> = 0), (*Y*<sup>3</sup> = *Y*<sup>4</sup> = *Y*<sup>6</sup> = 0) the CE has the following form:

$$\mathcal{Y}\_{\mathfrak{C}}(\mathcal{Y}\_{\mathfrak{a}} + \mathcal{Y}\_{\mathfrak{b}}) = 0. \tag{21}$$

For (*Y*<sup>1</sup> = *Y*<sup>3</sup> = *Y*<sup>6</sup> = 0) and (*Y*<sup>2</sup> = *Y*<sup>4</sup> = *Y*<sup>5</sup> = 0), the CE is obtained as:

$$
\mathcal{Y}\_a \mathcal{Y}\_b + \mathcal{Y}\_c (\mathcal{Y}\_a + \mathcal{Y}\_b) = 0. \tag{22}
$$

The CEs of (21) and (22) do not result in pure imaginary roots; therefore, these cases cannot give oscillator topologies.

Considering instead the cases (*Y*<sup>1</sup> *= Y*<sup>2</sup> *= Y*<sup>3</sup> = 0), (*Y*<sup>1</sup> *= Y*<sup>2</sup> *= Y*<sup>4</sup> = 0), (*Y*<sup>1</sup> *= Y*<sup>3</sup> *= Y*<sup>4</sup> = 0) and (*Y*<sup>2</sup> *= Y*<sup>3</sup> *= Y*<sup>4</sup> = 0), the CE has the following general form:

$$
\mathcal{Y}\_{\mathcal{C}}(\mathcal{Y}\_{\mathfrak{a}} - \mathcal{Y}\_{\mathfrak{b}}) = 0 \; . \tag{23}
$$

It is easy to verify that the CE in (23) cannot be associated with an oscillator topology if only two capacitors are used (we need three of them at least).

Finally, for (*Y*<sup>1</sup> *= Y*<sup>3</sup> *= Y*<sup>5</sup> = 0) and (*Y*<sup>2</sup> *= Y*<sup>4</sup> *= Y*<sup>6</sup> = 0), the CEs will be given by (24a) and (24b), respectively:

$$
\chi\_2 \chi\_4 - \chi\_6 (\chi\_2 + \chi\_4) = 0 \tag{24a}
$$

$$
\chi\_1 \chi\_3 - \chi\_5 (\chi\_1 + \chi\_3) = 0 \tag{24b}
$$

which are equations with the general form:

$$
\Upsilon\_a \Upsilon\_b - \Upsilon\_c (\Upsilon\_a + \Upsilon\_b) = 0. \tag{25}
$$

In (25), oscillation condition is related to the choice of *Yc* and *Ya* or *Yb* as a capacitance. In order to design an oscillator with the minimum number of components, we now have to verify the choice of the components in (25). It can be demonstrated that it is possible to have a minimum of two capacitors and at least two resistors in order to have a constant term in the constituting equation. In fact, with this choice we obtain a complete polynomial. In this case, having only three branches of the type *sC* + *G* with *C* ≥ 0, *G* ≥ 0, it is a matter of choosing an admittance between *Ya*, *Yb*, *Yc* of the type *sC* + *G*; the two remaining admittances will be a capacitance *sC*, and a conductance *G*. Inserting all possible combinations of options into (25), two sets of CEs which show imaginary roots are obtained.

For *Yc* = *sCc* + *Gc*; *Ya* = *sCa*; *Yb* = *Gb*, the CE becomes

$$-s^2\mathcal{C}\_a\mathcal{C}\_c + s[\mathcal{C}\_a\mathcal{G}\_b - \mathcal{C}\_a\mathcal{G}\_c - \mathcal{C}\_c\mathcal{G}\_b] - \mathcal{G}\_b\mathcal{G}\_c = 0. \tag{26}$$

For *Yc* = *sCc* + *Gc*; *Yb* = *sCb*; *Ya* = *Ga*, the CE becomes

$$-s^2\mathcal{C}\_b\mathcal{C}\_c + s\left[\mathcal{C}\_b\mathcal{G}\_a - \mathcal{C}\_c\mathcal{G}\_a - \mathcal{C}\_b\mathcal{G}\_c\right] - \mathcal{G}\_b\mathcal{G}\_c = 0. \tag{27}$$

From (26) and (27), the oscillation condition (*Co*) and oscillation frequency (*ω*0) for the two cases are obtained respectively as:

$$\mathbf{C}\_{o}: \frac{\mathbf{G}\_{\mathbf{c}}}{\mathbf{G}\_{b}} + \frac{\mathbf{C}\_{\mathbf{c}}}{\mathbf{C}\_{a}} = 1, \ \omega\_{0} = \sqrt{\frac{\mathbf{G}\_{b}\mathbf{G}\_{\mathbf{c}}}{\mathbf{C}\_{a}\mathbf{C}\_{\mathbf{c}}}}\tag{28a}$$

$$\mathcal{C}\_{\mathcal{o}} : \frac{\mathcal{G}\_{\mathcal{C}}}{\mathcal{G}\_{a}} + \frac{\mathcal{C}\_{\mathcal{c}}}{\mathcal{C}\_{b}} = 1, \ \omega\_{0} = \sqrt{\frac{\mathcal{G}\_{\mathcal{C}} \mathcal{G}\_{\mathcal{C}}}{\mathcal{C}\_{b} \mathcal{C}\_{\mathcal{c}}}} \tag{28b}$$

Thus, the minimum number of elements necessary to obtain an oscillator based on the scheme of Figure 2 is four, being two of these capacitors and two resistors. Considering the two cases (*Y*<sup>1</sup> *= Y*<sup>3</sup> *= Y*<sup>5</sup> = 0) and (*Y*<sup>2</sup> *= Y*<sup>4</sup> *= Y*<sup>6</sup> = 0) and the possible choices for *Ya* and *Yb*, we obtain a total of four canonic oscillators, corresponding to the following CEs:

$$-s^2\mathcal{C}\_1\mathcal{C}\_5 + s\left[\mathcal{C}\_1\mathcal{G}\_3 - \mathcal{C}\_1\mathcal{G}\_5 - \mathcal{C}\_5\mathcal{G}\_3\right] - \mathcal{G}\_3\mathcal{G}\_5 = 0\tag{29}$$

$$-s^2\mathcal{C}\_3\mathcal{C}\_5 + s\left[\mathcal{C}\_3\mathcal{G}\_1 - \mathcal{C}\_3\mathcal{G}\_5 - \mathcal{C}\_5\mathcal{G}\_1\right] - \mathcal{G}\_1\mathcal{G}\_5 = 0\tag{30}$$

$$\rm s^2 C\_2 C\_6 + s \left[ C\_6 G\_4 + C\_2 G\_6 - C\_2 G\_4 \right] + G\_4 G\_6 = 0 \tag{31}$$

$$\rm{s}^{2}\rm{C}\_{4}\rm{C}\_{6} + \rm{s} \left[ \rm{C}\_{6}\rm{G}\_{2} + \rm{C}\_{4}\rm{G}\_{6} - \rm{C}\_{4}\rm{G}\_{2} \right] + \rm{G}\_{2}\rm{G}\_{6} = 0 \tag{32}$$

However, this number is reduced again to two if we consider that from each of the cases (*Y*<sup>1</sup> *= Y*<sup>3</sup> *= Y*<sup>5</sup> *=* 0) and (*Y*<sup>2</sup> *= Y*<sup>4</sup> *= Y*<sup>6</sup> *=* 0) we obtain two equal oscillators if we exchange the order of the elements which are connected in series. These two configurations are shown in Figure 5, and the corresponding transfer functions, oscillation frequencies *ω*<sup>0</sup> and oscillation conditions are reported in Table 1. The oscillation frequencies and oscillation conditions in (28) show a strong interdependence since they are functions of the same parameters. Since the oscillation condition requires that the sum of the ratios of the capacitances and of the conductances is constant and equal to 1, a possible strategy for frequency tuning requires varying both resistors or both capacitors, maintaining their ratio constant. For example, a ratio of 2 between *Ca* and *Cc* can be obtained by using two parallel capacitors equal to *Ca* to obtain *Cc*; all three capacitors can be varied together; thus their ratio remains constant unless there are mismatches and the effect of parasitics.

**Figure 5.** VCII-based oscillators obtained in the case NGC is a five-node network. (**a**)Series-parallel impedances configuration; (**b**) parallel-series impedances configuration.

**Table 1.** Main equations of the canonic VCII-based oscillators.

