Ignition systems accumulating energy in inductance are more commonly used than those accumulating energy in capacitance [1, 2, 3]. In Figure 2a diagram of ignition system used to build a mathematical model is shown.

Figure 2 Model of the ignition system for the simulation studies. U_{b} – battery voltage, R_{1} – resistance of the ignition coil primary winding, L_{1} – inductance of the ignition coil primary winding, L_{2} – inductance of the ignition coil secondary winding, R_{2} – resistance of the ignition coil secondary winding, R_{4} – resistance representing the losses in the coil core, R_{3} – radioelectrical interference resistance, R_{45} – flow resistance of the spark plug, R_{ls} – discharge resistance, C_{2} – self-capacity of the coil, C_{45} – selfcapacity of the spark plug, M – coupling [1, 2]

The equivalent circuit of the ignition system presented in Figure 2 is described by equations (1) for two states of the control block.

$$\begin{array}{}{\displaystyle {U}_{B}-{i}_{1}{R}_{1}-{L}_{1}\frac{d{i}_{1}}{dt}+M\frac{d{i}_{2}}{dt}=0,}\\ {\displaystyle {L}_{2}\frac{d{i}_{2}}{dt}-M\frac{d{i}_{1}}{dt}+{i}_{2}{R}_{2}+{i}_{3}{R}_{3}+{u}_{C45}=0,}\\ {\displaystyle {u}_{C2}={L}_{2}\frac{d{i}_{2}}{dt}-M\frac{d{i}_{1}}{dt}+{i}_{2}{R}_{2},}\\ {\displaystyle {u}_{C2}={i}_{R4}{R}_{4},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{i}_{3}={i}_{2}+{i}_{R4}+{i}_{C2}}\\ {\displaystyle {i}_{3}={i}_{R45}+{i}_{C45},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{u}_{C45}={i}_{R45}{R}_{45}}\\ {\displaystyle {i}_{C2}={C}_{2}\frac{d{u}_{C2}}{dt},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{i}_{C45}={C}_{45}\frac{d{u}_{C45}}{dt}}\end{array}$$(1)

By introducing state variables: *x*_{1}= *i*_{1}, *x*_{2}= *i*_{2}, *x*_{3}= *u*_{C2}, *x*_{4}= *u*_{C45} to the equation (1):

$$\begin{array}{}{\displaystyle {U}_{B}-{x}_{1}{R}_{1}-{L}_{1}\frac{d{x}_{1}}{dt}+M\frac{d{x}_{2}}{dt}=0,}\\ {\displaystyle {L}_{2}\frac{d{x}_{2}}{dt}-M\frac{d{x}_{1}}{dt}+{x}_{2}{R}_{2}+{R}_{3}\frac{{x}_{4}}{{R}_{45}}+{R}_{3}{C}_{45}\frac{d{x}_{4}}{dt}+{x}_{4}=0,}\\ {\displaystyle {x}_{3}={L}_{2}\frac{d{x}_{2}}{dt}-M\frac{d{x}_{1}}{dt}+{x}_{2}{R}_{2}}\\ {\displaystyle \frac{{x}_{4}}{{R}_{45}}+{C}_{45}\frac{d{x}_{4}}{dt}={x}_{2}+\frac{{x}_{3}}{{R}_{4}}+{C}_{2}\frac{d{x}_{3}}{dt}}\end{array}$$(2)

Transforming the system of equations (2), we obtain:

$$\begin{array}{}{\displaystyle \frac{d{x}_{1}}{dt}={A}_{1}{x}_{1}+{B}_{1}{x}_{2}+{C}_{1}{x}_{3}+{D}_{1}{U}_{B}}\\ {\displaystyle \frac{d{x}_{2}}{dt}={A}_{2}{x}_{1}+{B}_{2}{x}_{2}+{C}_{2}{x}_{3}+{D}_{2}{U}_{B}}\\ {\displaystyle \frac{d{x}_{3}}{dt}={A}_{3}{x}_{1}+{B}_{3}{x}_{2}+{C}_{3}{x}_{3}+{E}_{3}{x}_{3}+{D}_{3}{U}_{B}}\\ {\displaystyle \frac{d{x}_{4}}{dt}={A}_{4}{x}_{1}+{B}_{4}{x}_{2}+{C}_{4}{x}_{3}+{E}_{4}{x}_{4}+{D}_{4}{U}_{B}}\end{array}$$(3)

where the parameters are determined using the following relationships:

$$\begin{array}{}{\displaystyle {A}_{1}=\frac{{R}_{1}}{\left(\frac{{M}^{2}}{{L}_{2}}-{L}_{1}\right)},}\\ {\displaystyle {B}_{1}=\frac{M{R}_{2}}{{L}_{2}\left(\frac{{M}^{2}}{{L}_{2}}-{L}_{1}\right)},}\\ {\displaystyle {C}_{1}=-\frac{M}{{L}_{2}\left(\frac{{M}^{2}}{{L}_{2}}-{L}_{1}\right)},}\\ {\displaystyle {D}_{1}=-\frac{1}{\left(\frac{{M}^{2}}{{L}_{2}}-{L}_{1}\right)},}\\ {\displaystyle {A}_{2}=\frac{-M{R}_{1}}{{L}_{1}\left({L}_{2}-\frac{{M}^{2}}{{L}_{1}}\right)},}\\ {\displaystyle {B}_{2}=\frac{-{R}_{2}}{\left({L}_{2}-\frac{{M}^{2}}{{L}_{1}}\right)},}\\ {\displaystyle {C}_{2}=\frac{1}{\left({L}_{2}-\frac{{M}^{2}}{{L}_{1}}\right)},}\\ {\displaystyle {D}_{2}=\frac{M}{{L}_{1}\left({L}_{2}-\frac{{M}^{2}}{{L}_{1}}\right)},}\\ {\displaystyle {A}_{3}=\frac{{C}_{45}{A}_{4}}{{C}_{2}},{B}_{3}=\left(\frac{{C}_{45}{B}_{4}}{{C}_{2}}-\frac{1}{{C}_{2}}\right),{C}_{3}=\left(\frac{{C}_{45}{C}_{4}}{{C}_{2}}-\frac{1}{{C}_{2}}\right),}\\ {\displaystyle {D}_{3}=\frac{{C}_{45}}{{C}_{2}}{D}_{4}{U}_{B},{E}_{3}=\left(\frac{1}{{C}_{2}{R}_{45}}+\frac{{C}_{45}{E}_{4}}{{C}_{2}}\right),}\\ {\displaystyle {A}_{4}=\left(\frac{M{A}_{1}}{2{R}_{3}{C}_{45}-\frac{{L}_{2}{A}_{2}}{2{R}_{3}{C}_{45}}}\right),}\\ {\displaystyle {B}_{4}=\left(\frac{M{B}_{1}}{2{R}_{3}{C}_{45}}-\frac{{L}_{2}{B}_{2}}{2{R}_{3}{C}_{45}}+\frac{{R}_{2}}{2{R}_{3}{C}_{45}}\right),}\\ {\displaystyle {C}_{4}=\left(\frac{M{C}_{1}}{2{R}_{3}{C}_{45}}-\frac{{L}_{2}{C}_{2}}{2{R}_{3}{C}_{45}}\right),}\\ {\displaystyle {D}_{4}=\frac{-{L}_{2}{D}_{2}}{2{R}_{3}{C}_{45}},{E}_{4}=-\left(\frac{{R}_{3}}{2{R}_{3}{C}_{45}{R}_{45}}+\frac{1}{2{R}_{3}{C}_{45}}\right)}\end{array}$$(4)

The solution to Equation (3) has the form:

$$\begin{array}{}{\displaystyle x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right],A=\left[\begin{array}{cccc}{A}_{1}& {B}_{1}& {C}_{1}& 0\\ {A}_{2}& {B}_{2}& {C}_{2}& 0\\ {A}_{3}& {B}_{3}& {C}_{3}& {E}_{3}\\ {A}_{4}& {B}_{4}& {C}_{4}& {E}_{4}\end{array}\right],B={U}_{B}\left[\begin{array}{c}{D}_{1}\\ {D}_{2}\\ {D}_{3}\\ {D}_{4}\end{array}\right]}\end{array}$$(5)

At the first switch-on, the initial conditions are zero and Eq. (5) assumes the form:

$$\begin{array}{}{\displaystyle \frac{d}{dt}x=Ax+B,}\\ {\displaystyle x={\mathrm{e}}^{At}{x}_{0}+\underset{0}{\overset{t}{\int}}{\mathrm{e}}^{A(t-\tau )}Bd\tau ,}\\ {\displaystyle x=\underset{0}{\overset{t}{\int}}{\mathrm{e}}^{A(t-\tau )}Bd\tau}\end{array}$$(6)

The solution of equations for the control block (Figure 2) in the open state was obtained by using the state variable method.

The next stage of the calculation (7) refers to the operation of the system with the closed-loop control block where the initial conditions, i.e., the final conditions of the previous state (1), have to be calculated from the control block formula in the contact state for the time equal to the time of contact.

$$\begin{array}{}{\displaystyle {U}_{B}-{i}_{3}{R}_{1}-{L}_{1}\frac{d{i}_{3}}{dt}+M\frac{d{i}_{2}}{dt}-{L}_{2}\frac{d{i}_{2}}{dt}+M\frac{d{i}_{3}}{dt}-{i}_{3}{R}_{3}-{u}_{C45}=0,}\\ {\displaystyle {u}_{C2}={L}_{2}\frac{d{i}_{2}}{dt}-M\frac{d{i}_{3}}{dt}+{i}_{2}{R}_{2},}\\ {\displaystyle {u}_{C2}={i}_{R4}{R}_{4},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{i}_{3}={i}_{2}+{i}_{R4}+{i}_{C2},}\\ {\displaystyle {i}_{3}={i}_{R45}+{i}_{C45},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{u}_{C45}={i}_{R45}{R}_{45},}\\ {\displaystyle {i}_{C2}={C}_{2}\frac{d{u}_{C2}}{dt},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{i}_{C45}={C}_{45}\frac{d{u}_{C45}}{dt}}\end{array}$$(7)

In this case, the unknowns are: *i*_{2}, *i*_{3}, *i*_{R4}, *i*_{C2}, *i*_{R45}, *i*_{C45}, *u*_{C2}, and *u*_{C45}.

By introducing state variables: *x*_{1} = *i*_{2}, *x*_{2} = *i*_{3}, *x*_{3} = *u*_{C2}, *x*_{4} = *u*_{C45} to the equation (7):

$$\begin{array}{}{\displaystyle {U}_{B}-{x}_{2}({R}_{1}+{R}_{3})+(M-{L}_{1})\frac{d{x}_{2}}{dt}+(M-{L}_{2})\frac{d{x}_{1}}{dt}-{x}_{4}=0}\\ {\displaystyle {x}_{3}={L}_{2}\frac{d{x}_{1}}{dt}-M\frac{d{x}_{2}}{dt}+{x}_{1}{R}_{2}\phantom{\rule{negativethinmathspace}{0ex}}\to \phantom{\rule{negativethinmathspace}{0ex}}\frac{d{x}_{2}}{dt}=\frac{{L}_{2}}{M}\frac{d{x}_{1}}{dt}+{x}_{1}\frac{{R}_{2}}{M}-\frac{1}{M}{x}_{3}}\\ {\displaystyle {x}_{2}={x}_{1}+\frac{{x}_{3}}{{R}_{4}}+{C}_{2}\frac{{x}_{3}}{dt}\to {C}_{2}\frac{{x}_{3}}{dt}=-{x}_{1}+{x}_{2}-\frac{{x}_{3}}{{R}_{4}}}\\ {\displaystyle {x}_{2}=\frac{{x}_{4}}{{R}_{45}}+{C}_{45}\frac{d{x}_{4}}{dt}\to {C}_{45}\frac{d{x}_{4}}{dt}={x}_{2}-\frac{{x}_{4}}{{R}_{45}}}\end{array}$$(8)

Transforming the system of equations (8), we obtain:

$$\begin{array}{}{\displaystyle \frac{d{x}_{1}}{dt}={a}_{1}{x}_{1}+{b}_{1}{x}_{2}+{c}_{1}{x}_{3}+{d}_{1}{x}_{4}+{e}_{1}{U}_{B}}\\ {\displaystyle \frac{d{x}_{2}}{dt}={a}_{2}{x}_{1}+{b}_{2}{x}_{2}+{c}_{2}{x}_{3}+{d}_{2}{x}_{4}+{e}_{2}{U}_{B}}\\ {\displaystyle \frac{d{x}_{3}}{dt}={a}_{3}{x}_{1}+{b}_{3}{x}_{2}+{c}_{3}{x}_{3}}\\ {\displaystyle \frac{d{x}_{4}}{dt}={b}_{4}{x}_{2}+{d}_{4}{x}_{4}}\end{array}$$(9)

For the system of equations (9) the parameters *a*_{1} … *d*_{4} are defined by the relationships:

$$\begin{array}{}{\displaystyle {a}_{1}=-\frac{{R}_{2}(M-{L}_{1})}{[{L}_{2}(M-{L}_{1})+M(M-{L}_{2})]},}\\ {\displaystyle {b}_{1}=\frac{M({R}_{1}+{R}_{3})}{[{L}_{2}(M-{L}_{1})+M(M-{L}_{2})]},}\\ {\displaystyle {c}_{1}=\frac{(M-{L}_{1})}{[{L}_{2}(M-{L}_{1})+M(M-{L}_{2})]},}\\ {\displaystyle {d}_{1}=\frac{M}{[{L}_{2}(M-{L}_{1})+M(M-{L}_{2})]},}\\ {\displaystyle {e}_{1}=-{d}_{1},}\\ {\displaystyle {a}_{2}=\frac{{a}_{1}{L}_{2}+{R}_{2}}{M},{b}_{2}=\frac{{b}_{1}{L}_{2}}{M},{c}_{2}=\frac{{c}_{1}{L}_{2}-1}{M}}\\ {\displaystyle {d}_{2}=\frac{{d}_{1}{L}_{2}}{M},{e}_{2}=-\frac{{e}_{1}{L}_{2}}{M}}\\ {\displaystyle {a}_{3}=-\frac{1}{{C}_{2}},{b}_{3}=\frac{1}{{C}_{2}},{c}_{3}=-\frac{1}{{C}_{2}{R}_{4}},}\\ {\displaystyle {b}_{4}=\frac{1}{{C}_{45}},{d}_{4}=-\frac{1}{{C}_{45}{R}_{45}}}\end{array}$$(10)

The solution to Equation (9) has the form:

$$\begin{array}{}{\displaystyle x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right],A=\left[\begin{array}{cccc}{a}_{1}& {b}_{1}& {c}_{1}& {d}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}& {d}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}& 0\\ 0& {b}_{4}& 0& {d}_{4}\end{array}\right],B={U}_{B}\left[\begin{array}{c}{e}_{1}\\ {e}_{2}\\ 0\\ 0\end{array}\right]}\end{array}$$(11)

Eq. (9) assumes the form:

$$\begin{array}{}{\displaystyle \frac{d}{dt}x=Ax+B,}\\ {\displaystyle x={\mathrm{e}}^{At}{x}_{0}+\underset{0}{\overset{t}{\int}}{\mathrm{e}}^{A(t-\tau )}Bd\tau}\end{array}$$(12)

*x*_{0} – initial conditions, or final conditions from the previous state, to be calculated from the formulae for the control block in the contact state for the time equal to the time of contact [2].

The time *t* will be counted from the moment the block is no longer in the contact state.

The results of the computer simulation shown in Figure 3 show the difference in the voltage waveforms *u*_{1}(*t*) and *u*_{2}(*t*). It is evident that the response of *u*_{2}(*t*) (the output voltage to the new spark plug) is different from the response of the spark plug working for 40 hours with contaminated fuel.

Figure 3 Secondary voltage for new spark plug *u*_{1}(*t*) and voltage *u*_{2}(*t*) for spark plug after 30 hours of working with contaminated fuel

It is worth getting to know even the outline of the waveforms that arise as a result of the load of the ignition wire by components such as the spark plug. We often see the effects of reflections on the ends of cables and multiple overlapping of the reflected waves. It can be assumed that such a match will have a significant impact on the spark discharge and consequently the energy, the reliability of the ignition or, of such importance in the present day, the toxicity of exhaust gases. In summary, it seems that the more appropriate model of the ignition system should be the model (being developed by the authors of this paper), which takes into account, in addition, the shape of the spark plug electrodes and the description of the ignition cable as a distributed system.

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