TOWARD THE THEORY OF PARAMETRIC SYSTEMS

V. M. OSIPOV

(Presented by the scientific seminar of the Department of TOE)

Parametric systems, i.e., systems with periodically varying parameters, possess a whole series of special properties whose technical use is of considerable interest. In this connection, attention to them has noticeably increased in recent years. However, the significant mathematical difficulties encountered in the study of parametric systems naturally complicate their analysis and use. Therefore, the creation of a convenient engineering methodology for the calculation and analysis of such systems continues to remain a timely task. Below, using as an example a complex electrical circuit with one periodically varying parameter, a method is presented for calculating parametric systems that admits various generalizations.

1. Statement of the Problem

Let a multiloop electrical circuit contain, in one of its loops, a periodically varying inductance or capacitance. For simplicity, we shall assume that in the same loop, beginning from the moment \(t = 0\), a voltage \(U(t)\) acts. We shall separate the periodically varying element together with the voltage \(U(t)\), and represent the remaining part of the circuit in the form of a passive two-terminal network.

Consider the transient process in this circuit. In such a formulation the problem was considered by V. A. Taft (1958), who essentially gave a method for constructing the solution based on the use of the Fourier integral and providing for the use of a variable-current calculating table [5, 6]. Below a method for an analytical solution of the problem is presented.

We shall assume that

1. \(U(t)\) is a piecewise-continuous function satisfying the condition

\[ \int_{0}^{\infty} U(t)e^{-at}\,dt < \infty; \qquad a \geqslant 0, \]

i.e., the Laplace transform of this function exists.

1. The inductance (capacitance) varies according to a simple harmonic law:

\[ L(t)=L_0[1+m\cos(\Omega t+\alpha)], \tag{1,1} \]

where \(m\) is the modulation coefficient \((m<1)\).

1. A passive two-terminal network satisfies the conditions of physical realizability, i.e., the input operator impedance of the two-terminal network \(Z_{\mathrm{tw}}(p)\) is a positive real function [2].

2. The integral equation of the circuit. The shift operator

The problem posed is evidently reduced to integrating the following differential equation:

\[ \frac{d[L(t)\cdot i(t)]}{dt}+U_{\mathrm{tw}}(t)=U(t). \tag{2.1} \]

Taking (1.1) into account, the equation assumes the form

\[ L_0\frac{di}{dt}+L_0m\,\frac{d\,[i(t)\cos(\Omega t+\alpha)]}{dt}+U_{\mathrm{tw}}(t)=U(t). \]

Transforming the last equation by Laplace, we obtain

\[ pL_0I(p)+\frac{m}{2}\,pL_0\,[I(p-j\Omega)e^{j\alpha}+I(p+j\Omega)e^{-j\alpha}]+ \]
\[ +Z_{\mathrm{tw}}(p)I(p)=U(p), \tag{2.2} \]

where

\[ Z_{\mathrm{tw}}(p)I(p)=U_{\mathrm{tw}}(p). \]

Introduce the notation

\[ y_0(p)=\frac{1}{pL_0+Z_{\mathrm{tw}}(p)}, \tag{2.3} \]

\[ I_0(p)=U(p)y_0(p). \tag{2.4} \]

We shall have the following relation:

\[ I(p)=I_0(p)-\frac{m}{2}\,pL_0y_0(p)[I(p-j\Omega)e^{j\alpha}+I(p+j\Omega)e^{-j\alpha}]. \tag{2.5} \]

Let us now pass to the originals. First note that the operator admittance \(y_0(p)\) corresponds to the impulse response of the original circuit for \(m=0\), which we denote by \(h(t)\), i.e.,

\[ y_0(p)=\int_{0}^{\infty} h(t)e^{-pt}\,dt. \tag{2.6} \]

Applying the convolution theorem, we find

\[ i(t)=i_0(t)-m\frac{d}{dt}L_0\int_{0}^{t}h(t-x)i(x)\cos(\Omega x+\alpha)\,dx. \tag{2.7} \]

We reduce the obtained integral equation to canonical form. Carrying out the differentiation, we shall have

\[ i(t)=i_0(t)-mL_0h(0)i(t)\cos(\Omega t+\alpha)- \]

\[ -mL_0\int_{0}^{t}h'(t-x)i(x)\cos(\Omega x+\alpha)\,dx. \]

After elementary transformations we obtain:

\[ i(t)=f_0(t)-m\int_{0}^{t}K(t,x)i(x)\,dx, \tag{2.8} \]

where it is denoted

\[ f_0(t)=\frac{i_0(t)}{1+mL_0h(0)\cos(\Omega t+\alpha)}, \tag{2.9} \]

\[ K(t,x)=\frac{L_0h'(t-x)\cos(\Omega x+\alpha)} {1+mL_0h(0)\cos(\Omega t+\alpha)}. \]

Thus we have arrived at a Volterra integral equation of the second kind with free term \(f_0(t)\) and kernel \(K(t,x)\). This equation, as is known, will have one and only one continuous solution if the free term \(f_0(t)\) and the kernel \(K(t,x)\) are bounded and continuous functions for \(t>x\) [3]. Let us note that the solution will also be unique under weaker restrictions, namely: it is sufficient to require square summability of \(f_0(t)\) and \(K(t,x)\) [4].

Let us show that in our case, for \(m\leqslant 1\), both the free term and the kernel will be bounded and continuous. Clearly, the functions \(i_0(t)\) and \(h'(t)\) are continuous and bounded; therefore it is sufficient to show that \(L_0h(0)\leqslant 1\). It is known that the value of a function at \(t=0\) is equal to the value of its Laplace transform multiplied by \(p\) as \(p\to\infty\), i.e.

\[ \lim_{p\to\infty} L_0p\,y_0(p)=L_0h(0). \]

From (2.8) it is seen that the behavior of \(y_0(p)\) at the infinitely distant point is determined by the behavior of \(Z_{\mathrm{tw}}(p)\) as \(p\to\infty\). For physically realizable passive two-terminal networks two cases are possible.

1. The infinitely distant point is a special point and, in particular, a zero, i.e.

\[ \lim_{p\to\infty} Z_{\mathrm{tw}}(p)=\operatorname{const}\ \text{(or zero)} \]

and, consequently,

\[ \lim pL_0y_0(p)=\lim \frac{pL_0}{pL_0+Z_{\mathrm{tw}}(p)}=1=L_0h(0). \]

1. The infinitely distant point is a pole, i.e.

\[ \lim_{p\to\infty} Z_{\mathrm{tw}}(p)=\infty, \]

and for physically realizable two-terminal networks the order of the pole cannot be greater than one [2], i.e. the tendency of \(Z_{\mathrm{tw}}(p)\) to infinity will occur in the same way as the tendency to infinity of an expression of the form \(Lp\), where \(L\) is some positive coefficient. We therefore have:

\[ \lim_{p\to\infty} pL_0y_0(p) = \lim_{p\to\infty}\frac{pL_0}{pL_0+pL} = \frac{L_0}{L_0+L}<1. \]

Thus, for real two-terminal networks the inequality

\[ L_0h(0)\leqslant 1 \]

holds, and consequently, for \(m\leqslant 1\) the integral equation (2.8) or (2.7) has a unique solution.

Let us note that in the second case one may reason differently. Represent \(Z_{\mathrm{tw}}(p)\) as the sum of two components

\[ Z_{\mathrm{tw}}(p)=pL+Z'_{\mathrm{tw}}(p), \]

where \(Z'_{\mathrm{tw}}(p)\) may have at infinity either a removable singular point or a zero. Introduce the operator inductive resistance

\(pL\) into the active circuit, then instead of \(pL_0\) we shall have \(p(L+L_0)\). The modulation coefficient will then change by a factor of \(\dfrac{L_0}{L_0+L}\) and will be equal to

\[ m' = m \frac{L_0}{L_0+L}. \]

Consequently, we shall have

\[ \lim_{p\to\infty} p(L_0+L)y'_0(p) = \lim_{p\to\infty} \frac{p(L_0+L)}{p(L_0+L)+Z'_{\mathrm{in}}(p)} =1=(L_0+L)h(0). \]

The quantity \(mL_0h(0)\) remains unchanged, since

\[ m'(L_0+L)h(0)=mL_0h(0). \]

Since in the image domain the integral equation (2,7) corresponds to an operator equation, by virtue of the uniqueness of the Laplace transform the latter will also have a unique solution. Before seeking this solution, let us introduce the concept of the shift operator. The meaning of this operator is determined by the relation

\[ I(p+j\Omega)=e^{j\Omega D}I(p), \tag{2,11} \]

which formally arises in the following way. Expansion of \(I(p+j\Omega)\) in a Taylor series gives

\[ I(p+j\Omega)=I(p)+\frac{dI(p)}{dp}j\Omega+\frac{d^2I(p)}{dp^2}\frac{(j\Omega)^2}{2!}+\cdots . \]

This series converges for all points of the domain of definition of the function \(I(p+j\Omega)\) that do not coincide with the poles. Let us introduce the operator of differentiation with respect to the variable \(p\),

\[ \frac{d}{dp}=D;\qquad \frac{d^2}{dp^2}=D^2;\ \ldots\ldots\ \frac{d^n}{dp^n}=D^n. \]

Then

\[ I(p+j\Omega)= \left[ 1+\frac{j\Omega D}{1!} +\frac{(j\Omega D)^2}{2!} +\frac{(j\Omega D)^3}{3!} +\cdots \right]I(p), \]

i.e.,

\[ I(p+j\Omega)=e^{j\Omega D}I(p), \]

\[ I(p-j\Omega)=e^{-j\Omega D}I(p), \tag{2,12} \]

and the principal property of the shift operator consists in the following:

\[ e^{\pm j\Omega D}[f(p)\cdot \varphi(p)] = \left[e^{\pm j\Omega D}f(p)\right] \left[e^{\pm j\Omega D}\varphi(p)\right] = \]

\[ = f(p\pm j\Omega)\cdot \varphi(p\pm j\Omega). \tag{2,13} \]

The operator acts under multiplication from the left. The other properties of this operator formally coincide with the properties of the ordinary exponential function.

With the aid of the shift operator, equation (2,5), whose solution we seek, is written in the following form:

\[ \left\{ 1+\frac{m}{2}pL_0y_0(p) \left[ e^{j(\Omega D-\alpha)}+e^{-j(\Omega D-\alpha)} \right] \right\}I(p)=I_0(p). \tag{2,14} \]

Denote

\[ pL_0y_0(p)=N(p), \tag{2,15} \]

\[ 1+\frac{m}{2}N(p) \left[ e^{j(\Omega D-\alpha)}+e^{-j(\Omega D-\alpha)} \right] = B(p,D). \tag{2,16} \]

We shall regard the last expression as a certain generalized operator; then our equation is written symbolically in the form

\[ B(d,D)I(p)=I_0(p)=U(p)y_0(p). \tag{2,17} \]

3. The inverse operator and its realization

Formally, the solution of equation (2,17) has the form

\[ I(p)=B(p,D)^{-1}I_0(p), \tag{3,1} \]

where \(B(p,D)^{-1}\) is the operator inverse to the previously introduced operator \(B(p,D)\). Obviously,

\[ B(p,D)\cdot B(p,D)^{-1}=1. \tag{3,2} \]

It is easy to see that

\[ B(p,D)^{-1}=\frac{1}{1+m\cos(\Omega D-\alpha)}=\frac{1}{B(p,D)}. \tag{3,3} \]

In this form, however, the use of the inverse operator is difficult, since we do not know the rule for division by a generalized operator. It is necessary to represent it in a form that permits realization.

Expression (3,3) may formally be regarded as a periodic function of the argument \(\Omega D-\alpha\); therefore it is natural to seek a representation of the inverse operator in the form of a certain symbolic Fourier series, i.e.,

\[ B(p,D)^{-1}=\sum_{n=-\infty}^{\infty} F_n(p)e^{jn(\Omega D-\alpha)}, \tag{3,4} \]

where \(F_n(p)\) are certain functions to be determined. Using equality (3,2), as well as the basic property of the shift operator, we obtain

\[ \begin{aligned} 1={}&\sum_{n=-\infty}^{\infty}F_n(p)e^{jn(\Omega D-\alpha)} +\frac{m}{2}N(p)\sum_{n=-\infty}^{\infty}F_n(p+j\Omega)e^{j(n+1)(\Omega D-\alpha)} \\ &+\frac{m}{2}N(p)\sum_{n=-\infty}^{\infty}F_n(p-j\Omega)e^{j(n-1)(\Omega D-\alpha)}. \end{aligned} \tag{3,5} \]

Equating the coefficients of shift operators of the same order, we shall have the following system of recurrence equations

\[ \left. \begin{aligned} &\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \\[2mm] 0={}&F_{-k}(p)+\frac{m}{2}N(p)\bigl[F_{-(k+1)}(p+j\Omega)+F_{-(k-1)}(p-j\Omega)\bigr]; \\[2mm] &\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \\[2mm] 0={}&F_{-1}(p)+\frac{m}{2}N(p)\bigl[F_{-2}(p+j\Omega)+F_0(p-j\Omega)\bigr]; \\[2mm] 1={}&F_0(p)+\frac{m}{2}N(p)\bigl[F_{-1}(p+j\Omega)+F_1(p-j\Omega)\bigr]; \\[2mm] 0={}&F_1(p)+\frac{m}{2}N(p)\bigl[F_0(p+j\Omega)+F_2(p-j\Omega)\bigr]; \\[2mm] &\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \\[2mm] 0={}&F_k(p)+\frac{m}{2}N(p)\bigl[F_{k-1}(p+j\Omega)+F_{k+1}(p-j\Omega)\bigr]. \end{aligned} \right\} \tag{3,6} \]

Before solving the resulting system, let us note the following. By virtue of the existence and uniqueness of the solution of equation (2.18), the inverse operator is meaningful and, consequently, the series (3.4) converges; then

\[ \lim_{n\to\infty} |F_n(p)| \to 0 . \tag{3.7} \]

If, starting from some sufficiently large \(n\), we neglect all functions, then the error made in determining all the other functions with indices smaller than \(n\) will be the smaller, the larger \(n\) is. Setting \(F_{\kappa+1}(p) \approx 0\), we obtain

\[ F_\kappa(p) \approx -\frac{m}{2} N(p) F_{\kappa-1}(p+j\Omega). \tag{3.8} \]

Next, bearing in mind that

\[ 0 = F_{\kappa-1}(p) + \frac{m}{2} N(p)\left[F_{\kappa-2}(p+j\Omega)+F_\kappa(p-j\Omega)\right], \]

and substituting the approximate value of \(F_\kappa(p)\), we find

\[ F_{\kappa-1}(p) \approx -\frac{\dfrac{m}{2}N(p)F_{\kappa-2}(p+j\Omega)} {1-\left(\dfrac{m}{2}\right)^2 N(p)N(p-j\Omega)} . \tag{3.9} \]

Substituting again \(F_{\kappa-1}(p)\) into the recurrence relation

\[ 0 = F_{\kappa-2}(p) + \frac{m}{2}N(p)\left[F_{\kappa-3}(p+j\Omega)+F_{\kappa-1}(p-j\Omega)\right], \]

we obtain

\[ F_{\kappa-2}(p) \approx -\frac{\dfrac{m}{2}N(p)F_{\kappa-3}(p+j\Omega)} {1-\dfrac{\left(\dfrac{m}{2}\right)^2 N(p)N(p-j\Omega)} {1-\left(\dfrac{m}{2}\right)^2 N(p-j\Omega)N(p-j2\Omega)}} . \]

Continuing this process \(q\) times, we shall have

\[ F_{\kappa-q}(p) \approx -\frac{\dfrac{m}{2}N(p)F_{\kappa-q-1}(p+j\Omega)} {1-\dfrac{\left(\dfrac{m}{2}\right)^2 N(p)N(p-j\Omega)} {1-\dfrac{\left(\dfrac{m}{2}\right)^2 N(p-j\Omega)N(p-j2\Omega)} {1-\cdots\;-\cdots\;-\cdots\;-\cdots\;-\cdots\;-\cdots\;-\cdots \;-\cdots\;-\cdots\;-\cdots\;-\cdots\;-\cdots\;-\cdots\;-\cdots \over 1-\left(\dfrac{m}{2}\right)^2 N[p-j(q-1)\Omega]N(p-jq\Omega)}}} . \tag{3.10} \]

Put \(q=\kappa-1\); then

\[ F_1(p)\approx \frac{ -\dfrac{m}{2}N(p)F_0(p+j\Omega) }{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2N(p)N(p-j\Omega)} {1-\dfrac{\left(\dfrac{m}{2}\right)^2N(p-j\Omega)N(p-j2\Omega)} {1-\cdots}} } \]

\[ \cdots \frac{1}{ 1-\left(\dfrac{m}{2}\right)^2 N[p-j(\kappa-2)\Omega]\,N[p-j(\kappa-1)\Omega] }. \]

Now letting \(\kappa\) tend to infinity, we obtain the exact value of \(F_1(p)\), expressed in terms of \(F_0(p+j\Omega)\),

\[ F_1(p)=-\frac{m}{2}N(p)A_1(p)F_0(p+j\Omega). \tag{3,11} \]

Here \(A_1(p)\) denotes the infinite continued fraction of the form

\[ A_1(p)= \frac{1}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2N(p)N(p-j\Omega)} {1-\dfrac{\left(\dfrac{m}{2}\right)^2N(p-j\Omega)N(p-j2\Omega)} {1-\cdots}} }. \tag{3,12} \]

Putting \(F_{-(\kappa+1)}(p)\approx 0\) and repeating the same operations, we obtain

\[ F_{-1}(p)=-\frac{m}{2}N(p)A_{-1}(p)F_0(p-j\Omega), \tag{3,13} \]

where

\[ A_{-1}(p)= \frac{1}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2N(p)N(p+j\Omega)} {1-\dfrac{\left(\dfrac{m}{2}\right)^2N(p+j\Omega)N(p+j2\Omega)} {1-\cdots}} }. \tag{3,14} \]

The convergence of the infinite continued fractions (3,12) and (3,14) follows from the theorem on the existence and uniqueness of the solution proved earlier; however, essentially the same result can also be obtained directly.

Let us write the continued fraction (3,12) in the form

\[ A_1(p)=\frac{1}{1+ \dfrac{-\left(\dfrac{m}{2}\right)^2N(p)N(p-j\Omega)}{1} + \cdots + \dfrac{-\left(\dfrac{m}{2}\right)^2N[p-j(\kappa-1)\Omega]N(p-j\kappa\Omega)}{1} +\cdots }. \]

Suppose that \(\displaystyle \lim_{\kappa\to\infty} N[p-j(\kappa-1)\Omega]N(p-j\kappa\Omega)=C\) for any fixed \(p\); then the infinite continued fraction converges for any \(m\), with the exception of values satisfying the inequality

\[ -\infty<-\left(\frac{m}{2}\right)^2<-\frac{1}{4C}\,[7] \]

or

\[ \infty>m>\frac{1}{\sqrt{C}}. \]

Let us find \(C\):

\[ \begin{aligned} \lim_{\kappa\to\pm\infty} N[p-j(\kappa-1)\Omega]N(p-j\kappa\Omega) &= \\ &= \lim_{\kappa\to\pm\infty} \frac{[p-j(\kappa-1)\Omega]L_0\cdot(p-j\kappa\Omega)L_0} {\{[p-j(\kappa-1)\Omega]L_0+Z_{\mathrm{dv}}[p-j(\kappa-1)\Omega]\}\times} \\ &\quad \times \frac{[p-j(\kappa-1)\Omega]L_0\cdot(p-j\kappa\Omega)L_0} {\{p-j\kappa\Omega)L_0+Z_{\mathrm{dv}}(p-j\kappa\Omega)\}} = C. \end{aligned} \]

For fixed \(p\), as \(\kappa\to\pm\infty\), \(p-j\kappa\Omega\to\mp\infty\), but the infinitely remote point for \(Z_{\mathrm{dv}}(j\kappa\Omega)\) is either a removable singular point or a pole of first order. In the first case we immediately obtain \(C=1\). In the second case, as was already noted, put \(Z_{\mathrm{gv}}(p)=Lp+Z_{\mathrm{dv}}(p)\), \((p\to j\kappa\Omega)\), where \(Z'_{\mathrm{dv}}(j\kappa\Omega)\) can have only a removable singular point or zero. The quantity \(L\) may be regarded as a certain inductance, which we can take out into the active circuit and include in \(L_0\); consequently, the second case reduces to the first. Thus, convergence of our continued fractions is ensured for any \(m\) satisfying the inequality

\[ 0\leq m\leq 1. \]

From (3.11) and (3.13) it follows that:

\[ \begin{aligned} F_1(p-j\Omega) &= -\frac{m}{2}N(p-j\Omega)A_1(p-j\Omega)F_0(p),\\ F_{-1}(p+j\Omega) &= -\frac{m}{2}N(p+j\Omega)A_{-1}(p+j\Omega)F_0(p). \end{aligned} \tag{3.15} \]

Substituting these expressions into the only inhomogeneous equation of system (3.6), we find

\[ F_0(p)= \frac{1} {1-\left(\frac{m}{2}\right)^2 N(p)\,[A_{-1}(p+j\Omega)N(p+j\Omega)+A_1(p-j\Omega)N(p-j\Omega)]}. \tag{3.16} \]

With the aid of the recurrence relation

\[ 0=F_k(p)+\frac{m}{2}N(p)[F_{k-1}(p+j\Omega)+F_{k+1}(p-j\Omega)] \]

or

\[ F_{k+1}(p)= -\frac{F_k(p+j\Omega)} {\dfrac{m}{2}N(p+j\Omega)} -F_{k-1}(p+j2\Omega). \tag{3.17} \]

we can express all the unknown functions in terms of the known \(F_0(p)\). For \(F_1(p)\) and \(F_{-1}(p)\) this has already been done [formulas (3.11) and (3.13)]. Next we find

\[ F_2(p)=\bigl[A_1(p+j\Omega)-1\bigr]F_0(p+j2\Omega), \tag{3.18} \]

\[ F_3(p)=-\frac{\dfrac{m}{2}N(p+j2\Omega)\bigl[A_1(p+j\Omega)-1\bigr]F_0(p+j3\Omega)} {1-\left(\dfrac{m}{2}\right)^2 N(p+j\Omega)N(p+j2\Omega)A_1(p+j\Omega)}. \tag{3.19} \]

Let us note that if we use the identity

\[ A_1(p+j\Omega)= \frac{1}{1-\left(\dfrac{m}{2}\right)^2 N(p)N(p+j\Omega)A_1(p)}, \tag{3.20} \]

whose validity is verified directly, then one can ascertain that

\[ |F_k(p)|\equiv \left(\frac{m}{2}\right)^k . \]

As for the functions with negative indices, the corresponding formulas can be obtained by replacing, in (3.16), (3.18), (3.19), \(\Omega\) by \(-\Omega\) and \(A_1(p)\) by \(A_{-1}(p)\). The following relations are also valid:

\[ A_{-1}(p-j\Omega)= \frac{1}{1-\left(\dfrac{m}{2}\right)^2 N(p)N(p-j\Omega)A_{-1}(p)}. \tag{3.21} \]

Thus, the desired solution is written in the form

\[ I(p)=B(p,D)^{-1}I_0(p)= \sum_{n=-\infty}^{\infty} F_n(p)e^{jn(\Omega t-\alpha)}I_0(p)= \]

\[ =\sum_{n=-\infty}^{\infty} F_n(p)I_0(p+jn\Omega)e^{-jn\alpha} =\sum_{n=-\infty}^{\infty}F_n(p)y_0(p+jn\Omega)U(p+jn\Omega)e^{-jn\alpha}\simeq \]

\[ \simeq\sum_{n=-q}^{q}F_n(p)y_0(p+jn\Omega)U(p+jn\Omega)e^{-jn\alpha}, \tag{3.22} \]

where \(q\) is determined by the required accuracy. Since, as \(n\to \pm\infty\),

\[ |y_0(p+jn\Omega)|\to 0, \]

\[ |U(p+jn\Omega)|\to 0, \]

the absolute values of the terms of the series (3.22) and of the corresponding time functions decrease faster than \(\left(\dfrac{m}{2}\right)^n\), and the faster, the larger \(\Omega\) is and the narrower the passband of the frequency response.

4. Forced Component of the Solution

The desired current as a function of time is determined as a result of the inverse Laplace transform (the Bromwich integral)

\[ i(t)=\frac{1}{2\pi j}\int_{a-j\infty}^{a+j\infty} I(p)e^{pt}\,dp. \tag{4.1} \]

According to the residue theorem, the value of the Bromwich integral is equal to the sum of the residues of the integrand with respect to all poles situated to the left of the line \(Re\,p=a\), if \(t>0\) [1], i.e.,

\[ i(t)=\frac{1}{2\pi j}\int_{a-j\infty}^{a+j\infty} I(p)e^{pt}\,dp = Res\,[I(p)e^{pt}]. \tag{4,2} \]

According to (3,22)

\[ I(p)=\sum_{n=-\infty}^{\infty} F_n(p)y_0(p+jn\Omega)\dot U(p+jn\Omega)e^{-jn\alpha} \]

\[ =\sum_{n=-\infty}^{\infty} G_n(p)U(p+jn\Omega)e^{-jn\alpha}, \tag{4,3} \]

where

\[ G_n(p)=F_n(p)y_0(p+jn\Omega). \]

This function characterizes the dynamic properties of the circuit under consideration itself and does not depend on the form of the applied voltage. Let this function have poles at the points \(p=\lambda_i\) \((i=1,2,3\ldots)\), with \(Re\,\lambda_i<a\) for any \(i\). Suppose also that the poles of the function \(U(p+jn\Omega)\) are located at the points \(p=p_\nu\) \((\nu=1,2,3\ldots)\), with, for real inputs, \(Re\,p_\nu\leq 0<a\) for any \(\nu\). Taking this into account, we can write

\[ i(t)=\sum Res\,[I(p)e^{pt}] =\sum_{p=\lambda_i} Res\,[I(p)e^{pt}]+ \]

\[ +\sum_{p=p_\nu} Res\,[I(p)e^{pt}] =i(t)_{\mathrm{sv}}+i(t)_{\mathrm{pr}}. \tag{4,4} \]

Here, by analogy with ordinary linear systems, we denote

\[ i(t)_{\mathrm{sv}}=\sum_{p=\lambda_i} Res\,[I(p)e^{pt}], \tag{4,5} \]

\[ i(t)_{\mathrm{pr}}=\sum_{p=p_\nu} Res\,[I(p)e^{pt}], \tag{4,6} \]

i.e., the free component of the current is equal to the sum of the residues with respect to all poles of the function \(G_n(p)\), while the forced component is equal to the sum of the residues with respect to the poles of the function \(U(p+jn\Omega)\). It should be emphasized, however, that if for any electrical circuit with constant parameters the free component of the current decays with time, tending to zero, then in parametric systems this need not be the case. In a once-excited parametric system the free process may even increase, if the energy generated by the periodically varying inductance exceeds the losses in the active resistances [5, 6].

Let us consider in more detail the forced component, when the applied voltage \(U(t)\) is a periodic function with fundamental frequency \(\omega\). This case is sufficiently general and is of undoubted practical interest. We have

\[ U(t)=\frac{U_0}{2}+\sum_{\nu=1}^{\infty} U_m\cos(\nu\omega t+\beta_\nu) =\sum_{\nu=-\infty}^{\infty} \dot U_{m\nu}e^{j\nu\omega t}, \tag{4,7} \]

where \(\dot U_{m\nu}=\dfrac{1}{2}U_{m\nu}e^{j\beta_\nu}\) is the complex amplitude of the \(\nu\)-th harmonic, with \(\beta_{-\nu}=-\beta_\nu;\ \beta_0=0\), i.e.,
\[ \dot U_{m\nu}=\dot U_{m(-\nu)};\qquad \dot U_{m0}=\frac{1}{2}U_0 . \]

The operator image of the voltage will be equal to
\[ U(p)=\sum_{\nu=-\infty}^{\infty}\frac{\dot U_{m\nu}}{p-j\nu\omega} \tag{4,8} \]
and, consequently,
\[ U(p+jn\Omega)=\sum_{\nu=-\infty}^{\infty}\frac{\dot U_{m\nu}}{p-j(\nu\omega-n\Omega)} . \tag{4,9} \]

Thus, the poles of the function \(U(p+jn\Omega)\) are located at the points \(j(\nu\omega-n\Omega)\) \((\nu=0,\pm1,\pm2,\ldots)\) \((n=0,\pm1,\pm2,\ldots)\), and the forced component will be equal to
\[ i(t)_{\mathrm{pr}}=\sum_{n=-\infty}^{\infty}\sum_{\nu=-\infty}^{\infty} F_n[j(\nu\omega-n\Omega)]\,y_0(j\nu\omega)\,\dot U_{m\nu} e^{j[(\nu\omega-n\Omega)t-n\alpha]} . \tag{4,10} \]

Introduce the notation
\[ \dot I_{\nu n}=\dot U_{m\nu}F_n[j(\nu\omega-n\Omega)]\,y_0(j\nu\omega), \tag{4,11} \]
then
\[ i(t)_{\mathrm{pr}}=\sum_{n=-\infty}^{\infty}\sum_{\nu=-\infty}^{\infty} \dot I_{\nu n}e^{j[(\nu\omega-n\Omega)t-n\alpha]} . \tag{4,12} \]

In practice, in the series (4,12) we shall obtain a finite number of terms, since the frequency response \(G_n(j\omega)\) has a finite passband and, moreover, rapidly decreases in absolute value as \(n\) increases.

For several initial values of \(n\) the complex amplitudes will be equal to:
\[ \dot I_{\nu0}=F_0(j\nu\omega)y_0(j\nu\omega)\dot U_{m\nu}= \]
\[ =\frac{y_0(j\nu\omega)\dot U_{m\nu}} {1-\left(\dfrac{m}{2}\right)^2 N(j\nu\omega)\left[A_{-1}(j\nu\omega+j\Omega)N(j\nu\omega+j\Omega) +A_1(j\nu\omega-j\Omega)N(j\nu\omega-j\Omega)\right]}, \tag{4,13} \]
\[ \dot I_{\nu1}=F_1(j\nu\omega-j\Omega)y_0(j\nu\omega)\dot U_{m\nu}= \]
\[ =-\frac{m}{2}N(j\nu\omega-j\Omega)A_1(j\nu\omega-j\Omega)\dot I_{\nu0}, \tag{4,14} \]
\[ \dot I_{\nu2}=F_2(j\nu\omega-j2\Omega)y_0(j\nu\omega)\dot U_{m\nu}= \]
\[ =[A_1(j\nu\omega-j\Omega)-1]F_0(j\nu\omega)y_0(j\nu\omega)\dot U_{m\nu} =[A_1(j\nu\omega-j\Omega)-1]\dot I_{\nu0}, \tag{4,15} \]
\[ \dot I_{\nu3}=F_3(j\nu\omega-j3\Omega)y_0(j\nu\omega)\dot U_{m\nu}= \]
\[ -\frac{m}{2}N(j\nu\omega-j\Omega)[A_1(j\nu\omega-j2\Omega)-1]\dot I_{\nu0} \]
\[ =\frac{ -\dfrac{m}{2}N(j\nu\omega-j\Omega)[A_1(j\nu\omega-j2\Omega)-1]\dot I_{\nu0} }{ 1-\left(\dfrac{m}{2}\right)^2N(j\nu\omega-j2\Omega)N(j\nu\omega-j\Omega)A_1(j\nu\omega-j2\Omega) }. \tag{4,16} \]

By these formulas the complex amplitudes can be determined with arbitrary accuracy if the infinite continued fractions are replaced by the corresponding convergents. The complex amplitudes for negative \(n\) can be obtained by replacing \(\Omega\) by \(-\Omega\) and \(A_1\) by \(A_{-1}\). Let us note several important special cases. For a sinusoidal applied voltage \((\nu=\pm 1)\) we shall have:

\[ i(t)_{\mathrm{pr}}= \sum_{n=-\infty}^{\infty} \left[ \dot I_{1n} e^{j[(\omega-n\Omega)t-n\alpha]} + \dot I_{-1n} e^{-j[(\omega-n\Omega)t+n\alpha]} \right]. \tag{4.17} \]

The peculiarity in this case will be that the complex amplitudes \(\dot I_{1n}\) and \(\dot I_{-1(-n)}\), as well as \(\dot I_{1(-n)}\) and \(\dot I_{-1n}\), will form pairs of complex-conjugate quantities. If one takes into account that

\[ \dot I_{1n}=\dot I_{1n}e^{j\varphi_n} = I_n e^{j\varphi_n} = \hat I_{-1(-n)}, \]

\[ \dot I_{1(-n)}=\dot I_{1(-n)}e^{j\varphi_{-n}} = I_{-n}e^{j\varphi_{-n}} = \hat I_{-1n}, \tag{4.18} \]

then instead of (4.17) one may write:

\[ i(t)_{\mathrm{pr}} = 2\sum_{n=-\infty}^{\infty} I_n \cos\{(\omega-n\Omega)t-n\alpha+\varphi_n\}. \tag{4.19} \]

The case of a constant applied voltage is obtained for \(\nu=0\). In this case (4.12) is transformed into the ordinary Fourier series:

\[ i(t)_{\mathrm{pr}} = \sum_{n=-\infty}^{\infty} \dot I_{0n} e^{-jn(\Omega t-\alpha)}, \tag{4.20} \]

where formulas (4.13)—(4.16) are substantially simplified, since

\[ N(j\nu\omega)\big|_{\nu=0} = \frac{j\nu\omega L_0}{j\nu\omega L_0+Z_{\mathrm{dv}}(j\nu\omega)\big|_{\nu=0}} = j\nu\omega L_0 y_0(j\nu\omega)\big|_{\nu=0} =0, \]

because for real two-terminal networks \(y_0(0)\ne 0\). Thus, \(N(0)=0\)

\[ \dot I_{00}=y_0(0)\dot U_{m0} = y_0(0)\frac{U_0}{2}. \tag{4.21} \]

Finally let us note the special case when \(m=0\). In this case all the circuit parameters are constant and we obtain the well-known result

\[ i(t)_{\mathrm{pr}} = \sum_{\nu=-\infty}^{\infty} \dot I_{\nu 0} e^{j\nu\omega t}, \tag{4.22} \]

where

\[ \dot I_{\nu 0}=\dot U_{m\nu}y_0(j\nu\omega). \]

It is not difficult to find analogous relations for the case when, instead of an inductance, a variable capacitance varying according to the simple harmonic law is connected in series with the passive two-terminal network,

\[ C(t)=C_0[1+m\cos(\Omega t+\alpha)]. \tag{4.23} \]

Assuming \(U_c(0)=0\), we shall have

\[ U(p)=U_c(p)+U_{\mathrm{dv}}(p), \]

where \(U(p)\) is the operator voltage applied to the circuit. Taking into account that

\[ \frac{d\,[C(t)\cdot U_c(t)]}{dt}\doteq I(p), \]

\[ I(p)=pC_0U_c(p)+\frac{m}{2}pC_0\left[U_c(p+j\Omega)e^{-j\alpha}+U_c(p-j\Omega)e^{j\alpha}\right], \tag{4.24} \]

we obtain, after transformations:

\[ U_c(p)=U_0(p)-\frac{m}{2}M(p)\left[U_c(p+j\Omega)e^{-j\alpha}+U_c(p-j\Omega)e^{j\alpha}\right]. \tag{4.25} \]

This is an equation for the unknown voltage on the capacitance \(U_c(p)\).

\[ \left. \begin{aligned} U_0(p)&=U(p)\frac{Y_{\mathrm{dv}}(p)}{pC_0+Y_{\mathrm{dv}}(p)},\\ M(p)&=\frac{pC_0}{pC_0+Y_{\mathrm{dv}}(p)}. \end{aligned} \right\} \tag{4.26} \]

The solution of equation (4.25) is obtained from (3.22) if one makes the substitution
\(I(p)\to U_c(p);\ I_0(p)\to U_0(p);\ N(p)\to M(p)\).

5. The case \(Z_{\mathrm{dv}}(p)\equiv 0\)

This is the simplest variant of the parametric circuit under consideration. It is characterized by the fact that the differential equation (2.1) can be directly integrated, and we obtain the possibility of checking the proposed method. For simplicity, consider the case when \(U(t)\) is a sinusoidal function of time. For \(U_{\mathrm{dv}}(t)=0\) we have

\[ \frac{d\,[L(t)\cdot i(t)]}{dt}=U_m\sin(\omega t+\beta). \tag{5.1} \]

Integrating and solving for the sought current, we obtain

\[ i(t)=-\frac{U_m}{\omega L_0}\cdot \frac{\cos(\omega t+\beta)}{1+m\cos(\Omega t+\alpha)}. \tag{5.2} \]

The constant of integration is omitted, since the steady-state regime is being considered.

The periodic function

\[ f(t)=\frac{1}{1+m\cos(\Omega t+\alpha)} \]

will be expanded in a Fourier series. Since this function is even with respect to the argument \((\Omega t+\alpha)\), the expansion will contain only cosine terms. The Fourier coefficients will be equal to

\[ \frac{1}{\pi}\int_0^\pi \frac{d(\Omega t+\alpha)}{1+m\cos(\Omega t+\alpha)} = \frac{1}{\sqrt{1-m^2}}, \tag{5.3} \]

\[ \frac{2}{\pi}\int_0^\pi \frac{\cos n(\Omega t+\alpha)}{1+m\cos(\Omega t+\alpha)} \,d(\Omega t+\alpha) = \frac{2}{\sqrt{1-m^2}} \left[ \frac{\sqrt{1-m^2}-1}{m} \right]^n . \tag{5.4} \]

Thus,

\[ f(t)=\frac{1}{\sqrt{1-m^{2}}}+\sum_{n=1}^{\infty}\frac{2}{\sqrt{1-m^{2}}} \left[\frac{\sqrt{1-m^{2}}-1}{m}\right]^{n}\cos n(\Omega t+\alpha). \tag{5.5} \]

Further, bearing in mind that

\[ i(t)=-\frac{U_m}{\omega L_0} f(t)\cos(\omega t+\beta) \]

and

\[ \begin{aligned} \cos(\omega t+\beta)\cdot \cos n(\Omega t+\alpha) &=\frac{1}{2}\{\cos[(\omega+n\Omega)t+n\alpha+\beta] \\ &\quad+\cos[(\omega-n\Omega)t-n\alpha+\beta]\} \\ &=\frac{1}{4}\{e^{j[(\omega+n\Omega)t+n\alpha+\beta]} \\ &\quad+e^{-j[(\omega+n\Omega)t+n\alpha+\beta]} +e^{j[(\omega-n\Omega)t-n\alpha+\beta]} \\ &\quad+e^{-j[(\omega-n\Omega)t-n\alpha-\beta]}\} \\ &=\frac{1}{4}\sum_{n=\pm 1}\{e^{j[(\omega-n\Omega)t-n\alpha+\beta]} +e^{-j[(\omega+n\Omega)t+n\alpha+\beta]}\}, \end{aligned} \]

we finally obtain

\[ i(t)=-\frac{1}{\omega L_0}\sum_{n=-\infty}^{\infty} \frac{1}{\sqrt{1-m^{2}}} \left[\frac{\sqrt{1-m^{2}}-1}{m}\right]^{|n|} \left\{ \dot U_m e^{j[(\omega-n\Omega)t-n\alpha]} +\hat U_m e^{-j[(\omega+n\Omega)t+n\alpha]} \right\}, \tag{5.6} \]

where

\[ \dot U_m=\frac{1}{2}U_m e^{j\beta} \quad\text{and}\quad \hat U_m=\frac{1}{2}U_m e^{-j\beta}. \]

Thus, we have reduced (5.2) to the form (4.17)

\[ i(t)=\sum_{n=-\infty}^{\infty} \left\{ \dot I_{1n} e^{j[(\omega-n\Omega)t-n\alpha]} +\dot I_{-1\,n} e^{-j[(\omega+n\Omega)t+n\alpha]} \right\}, \tag{5.7} \]

where

\[ \dot I_{1(\pm n)} =-\frac{1}{2}\frac{U_m e^{j\beta}}{\omega L_0\sqrt{1-m^{2}}} \left[\frac{\sqrt{1-m^{2}}-1}{m}\right]^{|n|}, \]

\[ \dot I_{-1(\pm n)} =-\frac{1}{2}\frac{U_m e^{-j\beta}}{\omega L_0\sqrt{1-m^{2}}} \left[\frac{\sqrt{1-m^{2}}-1}{m}\right]^{|n|}. \tag{5.8} \]

Moreover,

\[ \dot I_{1(\pm n)}=\hat I_{-1(\pm n)}. \]

Let us show that the same result follows immediately from our general formulas (4.13)—(4.16).

Since \(Z_{\mathrm{in}}(p)=0\), it follows that

\[ N(p)=\frac{pL_0}{pL_0+Z_{\mathrm{in}}(p)}=1, \tag{5.9} \]

\[ y_0(p)=\frac{1}{pL_0}; \qquad y_0(\pm j\omega)=\frac{1}{\pm j\omega}. \]

The continued fractions (3, 12) and (3, 14) take the form

\[ A_1=A_{-1}= \frac{1}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2}{ 1-\cdots }}}. \tag{5,10} \]

The following expansion is known [7]:

\[ \sqrt{x}=a+ \frac{\dfrac{x-a^2}{2a}}{ 1+\dfrac{\dfrac{x-a^2}{4a^2}}{ 1+\cdots }}, \tag{5,11} \]

where \(a\) is an approximate value of the root.

Put \(x=1-m^2\) and, consequently, \(a=1\); then we shall have

\[ \sqrt{1-m^2} = 1- \frac{\dfrac{m^2}{2}}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2}{ 1-\cdots }}} = 1-\frac{m^2}{2}A_1, \tag{5,12} \]

whence

\[ \frac{\sqrt{1-m^2}-1}{m} = -\frac{m}{2}A_1 = - \frac{\dfrac{m}{2}}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2}{ 1-\dfrac{\left(\dfrac{m}{2}\right)^2}{ 1-\cdots }}}. \tag{5,13} \]

For the application of our formulas (4, 13, 14, ....) it is necessary to put

\[ \beta_1=\beta-\frac{\pi}{2}, \]

then

\[ \dot U_{m1} = \frac{1}{2}U_m e^{j\beta}\cdot e^{-j\frac{\pi}{2}} = -j\frac{1}{2}U_m e^{j\beta}, \]

\[ \dot U_{m(-1)} = \frac{1}{2}U_m e^{-j\beta}\cdot e^{j\frac{\pi}{2}} = j\frac{1}{2}U_m e^{-j\beta}. \tag{5,14} \]

Taking into account (5.9), (5.10), (5.13), (5.14), formula (4.13) gives

\[ \dot I_{10} = -\frac{1}{2}\, \frac{U_m e^{j\beta}}{\omega L_0\sqrt{1-m^2}} = \hat I_{-10}. \tag{5.15} \]

Next, from (4.14) we find

\[ \dot I_{11} = -\frac{m}{2} A_1 \dot I_{10} = -\frac{1}{2}\, \frac{U_m e^{j\beta}}{\omega L_0\sqrt{1-m^2}} \left[ \frac{\sqrt{1-m^2}-1}{m} \right] = \hat I_{-11}. \tag{5.16} \]

Bearing in mind that

\[ A_1=\frac{1}{1-\left(\frac{m}{2}\right)^2 A_1}, \]

we obtain from (4.15)

\[ \dot I_{12} = \frac{\left(\frac{m}{2}\right)^2 A_1}{1-\left(\frac{m}{2}\right)^2 A_1}\,\dot I_{10} = \]

\[ = -\frac{1}{2}\, \frac{U_m e^{j\beta}}{\omega L_0\sqrt{1-m^2}} \left[ \frac{\sqrt{1-m^2}-1}{m} \right]^2 = \hat I_{-12}. \tag{5.17} \]

And, in general, in this way one can verify that for \(\dot I_{1n}\) and \(\dot I_{-1n}\) we obtain formulas (5.8), i.e., the result found by us by direct integration of the differential equation for the particular case under consideration.

Conclusion

This article considers a method for the analytical calculation of the forced component of the current in a parametric circuit of a particular type. This method can also be successfully used for solving more complicated problems, as well as for calculating transient processes in parametric systems.

LITERATURE

1. André Ango. Mathematics for Electrical and Radio Engineers. Moscow, “Nauka,” 1964.

2. G. I. Atabekov. Theory of Linear Electrical Circuits. Moscow, “Sovetskoe radio,” 1959.

3. V. Lovitt. Linear Integral Equations. State Publishing House of Technical-Theoretical Literature. Moscow, 1959.

4. S. G. Mikhlin. Lectures on Linear Integral Equations. Moscow, Fizmatgiz, 1959.

5. V. A. Taft. Electrical Circuits with Periodically Varying Parameters and Transient Processes in Synchronous Machines. Moscow, Publishing House of the Academy of Sciences of the USSR, 1958.

6. V. A. Taft. Problems in the Theory of Electrical Circuits with Variable Parameters and the Synthesis of Digital and Pulse Automatic Regulators. Moscow, Publishing House of the Academy of Sciences of the USSR, 1960.

7. A. N. Khovanskii. The Application of Continued Fractions and Their Generalization to Problems of Approximate Analysis. Moscow, Gostekh­teorizdat, 1956.