Analysis of coupling characteristics of coaxial line parallel to lightning current channel

Abstract

Base on the analysis of the coaxial coupling theory, the experimental model of the coaxial cable short-range coupling lightning electromagnetic pulse is established to study the problem of coaxial cable short-range coupling lightning electromagnetic pulse on the ship mast and high tower. The combination wave generator is used to simulate the 8/20 μs lightning current waveform, and the metal bar with the length of 1 m and diameter of 0.016 m is used to simulate the lightning channel as the radiation source. The length of the coaxial cable and the distance from the radiation source are changed for the test. The conclusion is as follows: when the coaxial cable is coupled with lightning wave in a short distance, the voltage waveform of cable terminal is still double exponential wave, with strong damping oscillation at 0.1 ∼0.4 μs. When the cable terminal is open circuit or connected with matching resistance, the rising time is shorter, about 1 μs; when the cable terminal shielding layer is grounded, the rising time is longer, about 10 μs. The peak voltage and energy coupled to the coaxial cable at short distance are positively correlated with the cable length, and inversely correlated with the distance from the radiation source. The energy that the cable is coupled to mainly comes from the low-frequency part, and there are multiple energy peaks below 100 Hz. Using fitting curve and experimental data, when the coaxial cable terminal is open circuit, the lightning current reaches 30 kA, the cable length is 30 m, and the distance from the lightning channel is 0.15 m, the peak value of the voltage coupled to the cable terminal is 66 V. It has certain guiding significance and value in the practical application of electromagnetic pulse protection.

Keywords

Coaxial cable 8/20 μs lightning wave coupling peak voltage energy

1. Introduction

Lightning is a common phenomenon in nature, which has large amplitude, high steepness, strong impact and wide spectrum [1–3]. When lightning occurs, the voltage of the return stroke channel is as high as several million volts, and the current can be as high as several hundred thousand amperes, which will produce huge thermal effect, electrodynamic effect, and electromagnetic pulse radiation effect, which is harmful to electronic and electrical equipment, power communication and other aspects [4,5]. With the development of science and technology, coaxial line is more and more indispensable in practical application [6–8]. Coaxial cable has a wide range of transmission frequency, and its attenuation is rising smoothly, so it is widely used [9,10]. Microwave coaxial is widely used in radar, military vehicle, satellite, aerospace, missile, RF ablation and measurement equipment. With the extensive use of coaxial cable, the coupling effect of coaxial cable on strong electromagnetic pulse is more and more worthy of attention [11,12].

Domestic and overseas scholars have done a lot of research on the coupled strong electromagnetic pulse of the overhead cable [13–17], M. Akbari [18] considered the frequency dependence of geoelectric parameters, and a full wave method for calculating induced lightning voltage of overhead lines is proposed which is based on the finite element solution of Maxwell equation. Napolitano et al. [19] found that effect of the nearby line conductors on the induced voltages is noticeable when the ground losses are taken into account in the surge propagation. Li et al. [20] studied the law of the coupled lightning electromagnetic pulse of twisted pair. There are few people studied the situation of the coaxial line coupled lightning electromagnetic pulse parallel to the lightning current channel, and the experimental data was relatively scarce.

Therefore, based on the coaxial coupling theory, the author constructs the experimental model of the coaxial cable coupling lightning electromagnetic pulse wave, analyzes the possible coupling situation of coaxial cable with different distance and length from the field source in various applications, and obtains the voltage waveform of the cable terminal when the coaxial cable is coupled to the lightning wave in a short distance, as well as the relationship between the peak value of the cable terminal voltage and cable length, distance from radiation source. By fitting the curve, the peak value of the coupling voltage of the coaxial terminal in the actual situation is deduced. It has a certain practical significance to the anti-electromagnetic interference of the coaxial line on the mast and tower of the ship.

2. Theoretical analysis

2.1. Transmission line theory

The double transmission lines irradiated by inhomogeneous electromagnetic field are shown in Fig. 1. The transmission line is located in the x-– z plane, the conductor is parallel to the z axis, and the terminal is parallel to the x axis. The length of the transmission line is s, and its diameter is d. Z ₁ and Z ₂ are the terminal impedances on the left and right respectively, and Z ₀ is the characteristic impedances.

Fig. 1.

Double transmission lines irradiated by inhomogeneous electromagnetic field.

The current at any point z ^′ of the cable is satisfy the following equation: $\begin{eqnarray}\displaystyle I(z^{\prime },{\omega})\text{} & =\text{} & \displaystyle {\displaystyle \frac{Z_{0}\cos {\beta}(s-z^{\prime })+jZ_{2}\sin {\beta}(s-z^{\prime })}{Z_{0}D}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad \times \displaystyle \int _{0}^{z^{\prime }}K(z,{\omega})[Z_{0}\cos {\beta}z+jZ_{2}\sin {\beta}z]dz\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad +∼{\displaystyle \frac{Z_{0}\cos {\beta}z^{\prime }+jZ_{2}\sin {\beta}z^{\prime }}{Z_{0}D}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad \times \displaystyle \int _{z^{\prime }}^{s}K(z,{\omega})Z_{0}\cos {\beta}(s-z^{\prime })+jZ_{2}\sin {\beta}(s-z^{\prime })dz\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad +∼{\displaystyle \frac{1}{D}}[Z_{0}\cos {\beta}(s-z^{\prime })+jZ_{2}\sin {\beta}(s-z^{\prime })]\displaystyle \int _{0}^{b}E_{x}^{i}(x,0,{\omega})dx\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad -∼{\displaystyle \frac{1}{D}}[Z_{0}\cos {\beta}z^{\prime }+jZ_{1}\sin {\beta}z^{\prime }]\displaystyle \int _{0}^{b}E_{x}^{i}(x,s,{\omega})dx.\end{eqnarray}$ (1) Where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle D=(Z_{0}Z_{1}+Z_{0}Z_{2})\cos {\beta}s+j({Z_{0}}^{2}+Z_{1}Z_{2})\sin {\beta}s\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\omega}=2{\pi}f\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\beta}={\displaystyle \frac{2{\pi}}{{\lambda}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle K(z,{\omega})=E_{x}^{i}(b,z,{\omega})-E_{x}^{i}(0,z,{\omega}).\nonumber\end{eqnarray}$ Here $E_{x}^{i}(b,z,{\omega})$ is the incident field strength of the upper conductor in the z direction, $E_{x}^{i}(0,z,{\omega})$ is the incident field strength of the lower conductor in the z direction, $E_{x}^{i}(x,0,{\omega})$ is the incident field strength in x direction above the left terminal, $E_{x}^{i}(x,s,{\omega})$ is the incident field strength in x direction above the right terminal.

2.2. Coaxial coupling theory

When the external electromagnetic field irradiates the coaxial cable, it will cause the current distribution on the coaxial shield. Because the shield is not completely shielded, the shield current will pass through the shield and generate voltage inside the cable. This voltage distribution will generate current in the internal load impedance, as shown in Fig. 2.

Fig. 2.

Geometry of coaxial cable.

Here E is the electric field of the irradiated cable, I (z) is the current distribution of the shielding layer, s is the length of the cable, b∕2 is the height of the cable from the ground, and d is the outer diameter of the cable. Z ₁∕2 and Z ₂∕2 refer to the impedance of the cable shield layer as the transmission line on the ground, Z _a and Z _b refer to the internal load impedance, Z ₀∕2 refers to the characteristic impedance of the cable shield layer as the single line on the ground, Z _c refers to the characteristic impedance of the cable internal, and I _L refers to the current in the internal load impedance Z _b.

According to formula (1), the current of cable terminal can be expressed as: $\begin{eqnarray}\displaystyle I(s,{\omega})\text{} & =\text{} & \displaystyle {\displaystyle \frac{1}{D}}\int _{0}^{s}K(z,{\omega})[Z_{0}\cos {\beta}z+jZ_{1}\sin {\beta}z]dz\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad +{\displaystyle \frac{Z_{0}}{D}}\displaystyle \int _{0}^{b}E_{x}^{i}(x,0,{\omega})dx\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad -{\displaystyle \frac{1}{D}}[Z_{0}\cos {\beta}s+jZ_{1}\sin {\beta}s]\int _{0}^{b}E_{x}^{i}(x,s,{\omega})dx.\end{eqnarray}$ (2) And because of this K (z, ω) = Z _T I (z, ω), Z _T is the transfer impedance of coaxial cable, and E _X = 0, so the current in internal load impedance Z _B, I _L(ω) can be presented as: $\begin{eqnarray}\displaystyle I_{L}({\omega})={\displaystyle \frac{Z_{T}}{P}}\displaystyle \int _{0}^{s}I(z,{\omega})[Z_{c}\cos {\beta}z+jZ_{a}\sin {\beta}z]dz. & & \displaystyle\end{eqnarray}$ (3) Where $\begin{eqnarray}\displaystyle P=(Z_{c}Z_{a}+Z_{c}Z_{b})\cos {\beta}s+j({Z_{c}}^{2}+Z_{a}Z_{b})\sin {\beta}s. & & \displaystyle\end{eqnarray}$ (4) The terminal voltage on the cable core can be expressed as: $\begin{eqnarray}\displaystyle V(s,{\omega})=I_{L}({\omega})Z_{b}. & & \displaystyle\end{eqnarray}$ (5)

The current distribution on the shielding layer of coaxial cable can be deduced from the solution of double transmission lines excited by voltage generators at both ends. The field strength along the z direction is given by: $\begin{eqnarray}\displaystyle E_{x}(z,{\omega})=E_{x}^{U}({\omega})e^{-j{\beta}z}. & & \displaystyle\end{eqnarray}$ (6) If the position is taken as the phase reference point, the field strengths of the left and right terminals of the cable are respectively expressed as: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle E_{x}(o,{\omega})=E_{x}^{U}({\omega})\end{eqnarray}$ (7) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle E_{x}(s,{\omega})=E_{x}^{U}({\omega})e^{-j{\beta}z}.\end{eqnarray}$ (8) The voltage generator of the line terminal is given by: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\displaystyle \frac{b}{2}}E_{x}^{U}({\omega})\end{eqnarray}$ (9) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\displaystyle \frac{b}{2}}E_{x}^{U}({\omega})\cdot e^{-j{\beta}z}.\end{eqnarray}$ (10)

According to the solution of the transmission line excited by the voltage generator at both ends, the current distribution on the cable shield can be derived as follows: $\begin{eqnarray}\displaystyle I(z,{\omega})\text{} & =\text{} & \displaystyle {\displaystyle \frac{bE_{x}^{U}}{D}}\big\{(Z_{0}-Z_{1})\sin {\beta}s\sin {\beta}z+j(Z_{0}+Z_{2})\sin {\beta}s\cos {\beta}z\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad -∼j(Z_{1}+Z_{2})\cos {\beta}s\sin {\beta}z\big\}.\end{eqnarray}$ (11) The terminal voltage of the shielding layer can be expressed as: $\begin{eqnarray}\displaystyle V(s,{\omega})={\displaystyle \frac{1}{2}}I(s,{\omega})Z_{2}. & & \displaystyle\end{eqnarray}$ (12)

3. Test and analysis

3.1. Test model

In the experiment, 8/20 μs simulated lightning wave is applied to a metal bar with a height of 1 m and a diameter of 16 mm by a combined wave generator, as shown in Fig. 3. The metal bar can simulate the lightning channel and emit lightning electromagnetic wave. The lightning impulse wave with impulse current from 1 kA to 4 kA is applied at both ends of the metal bar, and the step is 0.5 kA. The oscilloscope collects the test waveform and data of the terminal voltage of the coaxial cable. During the experiment, 0.2 m, 0.6 m and 1 m coaxial cables were placed at 0.15 m, 1 m and 1.5 m away from the metal bar respectively, and the connection mode between the shielding layer and the core wire was adjusted to obtain different experimental data. Finally, the collected data are analyzed to obtain the peak value of the electric voltage of the coupled lightning electromagnetic wave, and the frequency spectrum of the voltage waveform is analyzed to find out the rule. The energy of the coaxial cable coupling lightning wave can be calculated by: $\begin{eqnarray}\displaystyle E=\displaystyle \sum U^{2}{\Delta}t. & & \displaystyle\end{eqnarray}$ (13) In addition, 0.1 m, 0.2 m, 0.3 m to 1 m long cables are respectively placed at 0.15 m away from the metal bar which the cable terminals are open circuited. Statistical data and curve fitting are used to deduce the peak value of voltage coupled to the cable terminal when the cable terminal is open circuit, 30 m in length and 30 kA in impulse current and 0.15 m from the radiation source.

Fig. 3.

Experimental model.

3.2. Test results and analysis

3.2.1. Coaxial cable terminal open circuit

In order to study the protection of ship masts and towers against strong electromagnetic pulse, a close-up coaxial coupling experiment is carried out. The 0.2 m, 0.6 m and 1 m long coaxial cables are respectively placed at 0.15 m, 1 m and 1.5 m away from the metal bar. The cable terminal is open circuit, which is equivalent to infinite terminal load. Use oscilloscope to collect the voltage waveform of cable terminal coupling. When the coaxial line terminal is open circuit, the voltage waveform coupled to the cable terminal is as shown in Fig. 4. The overall waveform is still a double exponential wave, and the rising time is short, almost 1 μs. At the triggering moment, pull the time axis apart, and it can be seen that there is a strong damping oscillation at 0.1—0.4 μs, and the voltage value is large, probably because at the triggering moment of combined wave, the instantaneous change of electromagnetic field is the strongest, and the voltage coupled to the cable is relatively large. The maximum voltage in the damping vibration is set as the peak value of the coupling voltage waveform. The open circuit of the 1 m long cable terminal is placed 0.15 m away from the impact source. When the impact current is 1 kA, 2 kA and 3 kA, the peak value of the coupling voltage of the cable terminal is 6.375 V, 10.73 V and 15.14 V respectively.

The 1 m long coaxial line terminal is open circuited, and the voltage peak value data which is placed at different distance from the radiation source are collected and plotted. It can be seen that with the gradual increase of impulse current, the voltage peak value coupled to the coaxial line also gradually increases, which has a positive correlation, as shown in Fig. 5. The closer the coaxial line is to the radiation source, the higher the voltage peak value is. When the distance is 0.15 m, the voltage peak value is much higher than that when the distance is 1 m and 1.5 m.When the terminal is open circuit, the voltage peak value of the 1 m long coaxial cable at 0.15 m away from the radiation source is the largest, which can reach 18.3 V. Figure 6 shows the voltage peak value of cables with different lengths at a distance of 15 cm from the radiation source. It can be seen that the longer the cable is, the greater the voltage peak value is.

Fig. 4.

Terminal voltage waveform coupled to 1 m coaxial line at 0.15 m away from the radiation source when the terminal is open ((a) Impulse current is 1 kA, (b) Impulse current is 2 kA, (c) Impulse current is 3 kA).

Fig. 5.

Terminal peak voltage of 1 m coaxial cable coupling at different distance from radiation source.

Fig. 6.

Terminal peak voltage of coaxial coupling with different length at 0.15 m from the radiation source.

For 0.1–0.4 μs damped shock wave, Fourier transform is used for spectrum analysis, as shown in Fig. 7. It can be seen that the radiated lightning wave is mainly concentrated in the low-frequency part, and the main energy is concentrated below 100 Hz. When the impulse current is 1 kA, there is a small energy peak near 400 Hz, and with the increase of impulse current, there will be a small peak bulge near 600 Hz and 900 Hz. According to the calculation, the coupling energy of 0.2 m and 1 m long cable at different distance from the radiation source when the cable terminal is open circuit is shown in Table 1.The energy of cable coupling is inversely related to the distance. The closer the distance is, the greater the energy of cable coupling is and the longer the cable is, the greater the energy of coupling is.

Fig. 7.

Terminal voltage spectrum coupled to 1 m coaxial line at 0.15 m away from the radiation source when the terminal is open ((a) Impulse current is 1 kA, (b) Impulse current is 2 kA, (c) Impulse current is 3 kA).

Table 1

The coupling energy of different length cable and different distance of radiation source when the terminal is open

Impulse current (kA)	1 m Coaxial line ∑ U ²Δt (V ² ⋅ s)			0.2 m Coaxial line ∑ U ²Δt (V ² ⋅ s)
	Distance 0.15 m	Distance 1 m	Distance 1.5 m	Distance 0.15 m	Distance 1 m	Distance 1.5 m
0.5	3.02E-06	6.12E-07	4.66E-07	5.60E-07	3.66E-07	1.27E-07
1	6.50E-06	8.95E-07	6.62E-07	2.08E-06	5.68E-07	2.42E-07
1.5	2.31E-05	1.56E-06	1.27E-06	3.20E-06	9.70E-07	3.65E-07
2	2.91E-05	2.44E-06	3.12E-06	4.27E-06	1.19E-06	4.99E-07
2.5	4.12E-05	2.55E-06	4.00E-06	1.86E-05	1.33E-06	7.40E-07
3	4.28E-05	6.22E-06	4.98E-06	2.04E-05	2.58E-06	8.25E-07
3.5	4.59E-05	8.75E-06	6.80E-06	2.33E-05	2.67E-06	9.45E-07
4	5.88E-05	1.56E-05	7.88E-06	5.05E-05	3.32E-06	1.09E-06

3.2.2. Coaxial cable terminal shield grounding

The shielding layer of 1 m long coaxial cable terminal is grounded and placed 0.15 m away from the radiation source. Gradually increase the impulse current to obtain the voltage waveform coupled to the coaxial terminal which is shown in Fig. 8. Compared with the open circuit of the terminal, the rise time of the waveform is longer, about 10 μs, and the voltage peak value increases. When the impact current is 1 kA, 2 kA and 3 kA, the voltage peak value coupled to the cable terminal is respectively 12.33 V, 20.79 V and 28.12 V.

Fig. 8.

Terminal voltage waveform coupled to 1 m coaxial line at 0.15 m away from the radiation source when the terminal shield is grounded ((a) Impulse current is 1 kA, (b) Impulse current is 2 kA, (c) Impulse current is 3 kA).

The shielding layer of the 1 m long coaxial cable terminal is grounded and placed at 0.15 m, 1 m and 1.5 m away from the radiation source respectively. The peak value of the voltage coupled to the cable terminal is shown in Fig. 9. The coupling voltage increases with the increase of impulse current and decreases with the increase of distance from the radiation source. 0.2 m, 0.6 m and 1 m long cables are placed 0.15 m away from the radiation source. When the terminal shielding layer is grounded, the peak value of the coupled voltage is shown in Fig. 10. Basically, the longer the cable is, the greater the voltage it is coupled to. When the cable is 0.2 m long, the voltage peak value it is coupled to is much smaller than that of 0.6 m and 1 m. The peak voltage of the cable with a length of 1 m coupled to the radiation source at 0.15 m is 35.87 V. When the shield is grounded, the voltage amplitude coupled to the core wire is larger than that coupled to the terminal in the open circuit state, which is in line with the reality. Figure 11 is a spectrum analysis of the 0.1–0.4 μs damping oscillation part of the voltage waveform. It can be seen that there are three main energy peaks below 100 Hz, and the energy peak near 100 Hz is the largest, and its amplitude increases with the increase of impulse current.

Fig. 9.

Terminal peak voltage of 1 m coaxial cable coupling different distance from radiation source.

Fig. 10.

Terminal peak voltage of coaxial coupling with different length at 0.15 m from the radiation source.

Fig. 11.

Terminal voltage spectrum coupled to 1 m coaxial line at 0.15 m away from the radiation source when the terminal shield is grounded ((a) Impulse current is 1 kA, (b) Impulse current is 2 kA, (c) Impulse current is 3 kA).

Through calculation, it can be concluded that the coupling energy of cables with different lengths when the terminal shielding layer is grounded is as shown in Table 2. With the increase of distance, the coupling energy decreases gradually. The longer the cable length is, the greater the coupling energy is. When the shield of coaxial terminal is grounded, the coupling energy is one order of magnitude larger than that when the terminal is open circuit.

Table 2

The coupling energy of cable with different length and different distance from radiation source when terminal shield is grounded

Impulse current (kA)	1 m Coaxial line ∑ U ²Δt (V ² ⋅ s)			0.2 m Coaxial line ∑ U ²Δt (V ² ⋅ s)
	Distance 0.15 m	Distance 1 m	Distance 1.5 m	Distance 0.15 m	Distance 1 m	Distance 1.5 m
0.5	1.33E-05	2.44E-06	1.19E-06	6.43E-06	8.86E-07	6.77E-07
1	6.12E-05	3.31E-06	3.10E-06	1.81E-05	2.51E-06	2.44E-06
1.5	1.06E-04	8.50E-06	7.47E-06	4.48E-05	5.73E-06	5.05E-06
2	1.54E-04	1.72E-05	9.62E-06	6.19E-05	9.60E-06	7.21E-06
2.5	1.91E-04	2.34E-05	2.13E-05	1.16E-04	1.16E-05	1.70E-05
3	2.45E-04	2.96E-05	2.89E-05	1.48E-04	2.40E-05	2.13E-05
3.5	3.16E-04	3.53E-05	3.44E-05	1.72E-04	2.48E-05	2.44E-05
4	3.68E-04	4.62E-05	4.07E-05	3.15E-04	3.09E-05	2.73E-05

3.2.3. Coaxial cable terminal connection 51 𝛺 matching resistance

Place a 1 m long coaxial cable at a distance of 0.15 m from the radiation source, and connect a 51 Ω resistor to the terminal, which is equivalent to matching the load impedance of the coaxial cable and the terminal. Increase the impulse current, and the oscilloscope collects the terminal voltage waveform as shown in Fig. 12. The waveform is similar to the voltage waveform collected when the terminal is open circuit, and the rising time is very short. When the impulse current is 1 kA, 2 kA and 3 kA, the peak value of the voltage coupled to the cable terminal is 7.375 V, 10.78 V and 13.59 V.

Fig. 12.

Terminal voltage waveform coupled to 1 m coaxial line at 0.15 m away from the radiation source when the terminal is connected with 51 Ω resistance ((a) Impulse current is 1 kA, (b) Impulse current is 2 kA, (c) Impulse current is 3 kA).

Connect the 1 m long cable terminal to the matching resistance and place it 0.15 m, 1 m and 1.5 m away from the radiation source. The voltage peak value of the cable terminal coupling is shown in Fig. 13. With the increase of impulse current, the voltage peak value increases gradually. The peak value of the voltage coupled to the cable terminal is inversely related to the distance. The peak value of the voltage coupled to 0.15 m away from the radiation source is the largest. When the impact current is 4 kA, the peak value of the voltage can reach 15.65 V. The terminals of 0.2 m, 0.6 m and 1 m long coaxial cables are connected with matching resistors, which are respectively placed at a distance of 0.15 m from the radiation source. The voltage peak value coupled is shown in Fig. 14, which is positively related to the length. As shown in Fig. 15, the main energy is concentrated below 100 Hz, there are four energy peaks, and there is a small energy peak near 400 Hz. 0.2 m and 1 m long coaxial cables are placed 0.15 m, 1 m and 1.5 m away from the radiation source. When the cable terminal is connected with a matching resistance, the coupling energy is shown in Table 3.

Fig. 13.

Terminal peak voltage of 1 m coaxial cable coupling different distance from radiation source.

Fig. 14.

Terminal peak voltage of coaxial coupling with different length at 0.15 m from the radiation source.

Fig. 15.

Terminal voltage spectrum coupled to 1 m coaxial line at 0.15 m away from the radiation source when the terminal is connected with 51 Ω resistance ((a) Impulse current is 1 kA, (b) Impulse current is 2 kA, (c) Impulse current is 3 kA).

Table 3

The coupling energy of cables with different length and different distance from radiation source with 51 Ω resistance added to terminal

Impulse current (kA)	1 m Coaxial line ∑ U ²Δt (V ² ⋅ s)			0.2 m Coaxial line ∑ U ²Δt (V ² ⋅ s)
	Distance 0.15 m	Distance 1 m	Distance 1.5 m	Distance 0.15 m	Distance 1 m	Distance 1.5 m
0.5	2.61E-06	6.23E-07	5.46E-07	4.01E-07	2.66E-07	9.09E-08
1	6.61E-06	2.09E-06	1.46E-06	8.65E-07	5.23E-07	1.56E-07
1.5	2.25E-05	3.43E-06	3.37E-06	1.32E-06	6.77E-07	2.80E-07
2	3.05E-05	5.46E-06	4.72E-06	1.86E-06	7.82E-07	5.05E-07
2.5	3.49E-05	1.67E-05	4.78E-06	2.89E-06	1.14E-06	5.56E-07
3	4.45E-05	1.97E-05	8.06E-06	3.23E-06	1.46E-06	8.24E-07
3.5	5.26E-05	2.15E-05	1.84E-05	4.29E-06	2.17E-06	9.47E-07
4	5.66E-05	2.16E-05	1.97E-05	6.08E-06	2.88E-06	1.17E-06

3.2.4. Simulation

In this paper, CST CABLE STUDIO is used to model, coaxial cable is placed in the air, along the X direction. The electromagnetic pulse waveform is 8/20 pulse waveform specified by IEC, the peak current is 30 kA, the transmission direction is (0, 0, −1), the electric field vector is (1, 0, 0), and the simulation model is shown in the Fig. 16. The coaxial cable is RG58 coaxial cable model which is provided by the software. The length of the cable is 3 M. Its shielding layer is a single-layer shielding braided structure with a coverage of 93%. The number of strands is 7. The number of strands per share is 16. The wire diameter of the copper wire is 0.122 mm. The radius of the shielded cable is 1.8 mm and the braiding angle is 26°. The cable shield terminal is open. The voltage of the core wire is calculated by CST simulation.

Fig. 16.

Electromagnetic pulse simulation model of coaxial cable.

The simulation result is shown in the Fig. 17. The obtained waveform is similar to the measured one, and the peak voltage is close to the measured one, but the voltage coupled to the main energy peak is much larger than the measured one. The simulation results of this paper are compared with Wang’s research [21]. Wang’s coupled waveform is damped shock wave, while the waveform in this paper has shock, but it is still double exponential wave. The coupling voltage is also different.

Fig. 17.

Core wire coupling voltage waveform.

The obtained waveform is similar to the measured one, and the peak voltage is close to the measured one, but the voltage coupled to the main energy peak is much larger than the measured one. Compared with the simulation results of T.W. Wang coaxial coupling electromagnetic pulse, the waveform Wang coupled to is damped shock wave, while the waveform in this paper has shock, but it is still a double exponential wave. The coupling voltage is also different.

3.2.5. Practical application

In the actual situation, the mast of a ship is generally 30–40 m high, the height of the communication tower is from 30 m to 100 m, and the lightning current can reach 30 kA. Therefore, the measured data can be used to deduce the peak voltage coupled to the coaxial line terminal near the mast or tower of a ship in the actual situation. The coaxial lines of 0.1 m, 0.2 m, 0.3 m–1 m are placed at a distance of 0.15 m from the radiation source, and the cable terminals are open circuited. When the impulse current is 0.5–4 kA, the voltage peak value coupled to the cable terminals is measured respectively. Considering the actual situation and data convergence, the logarithmic curve is selected. The data point in Fig. 18 is the real data, and the curve is the curve after fitting. The fitting formula is shown in Table 4. When the cable length is 30 m and the impulse current is 0.5–4 kA, the peak voltage coupled to the cable terminal is calculated, and the curve is drawn as Fig. 19. When the cable terminal is open circuit, the voltage peak value coupled to the 30 m long cable terminal changes with the impact current as y = 15.179 ln (x) + 14.933. If x = 30 kA, y = 66.56 V can be obtained, which means that when the impulse current is 30 kA, the peak voltage coupled to the cable is 66.56 V.

Fig. 18.

Voltage peak value of terminal coupling at different impulse current at 0.15 m from radiation source with different length of cable.

Table 4

Fitting curve between cable length and voltage peak under different impulse current

Impulse current (kA)	Fitting curve x is the cable length y is the voltage peak value	x = 30
0.5	y = 1.3932ln(x) + 3.8157	y = 8.55
1	y = 2.0906ln(x) + 5.8109	y = 12.92
1.5	y = 3.0132ln(x) + 8.1029	y = 18.35
2	y = 3.4262ln (x) + 9.9487	y = 21.6
2.5	y = 4.3748ln(x) + 12.112	y = 27
3	y = 5.2611ln(x) + 14.176	y = 32.07
3.5	y = 5.8857ln(x) + 15.981	y = 36
4	y = 6.4397ln(x) + 17.872	y = 39.77

Fig. 19.

Voltage peak value of 30 m long cable coupled to the terminal under different impulse current at 0.15 m from the radiation source.

4. Conclusion

(1) When 8/20 μs simulated lightning wave is applied to both ends of the metal bar with a height of 1 m and a diameter of 0.016 m, the voltage waveform coupled to the coaxial cable at a short distance is still a double exponential wave, but there is a strong damping vibration at 0.1–0.4 μs. When the coaxial cable terminal is open circuited or connected with 51 Ω matching resistance, the rising edge of the double exponential wave is very short, almost 1 μs. When the cable terminal shielding layer is grounded, the rising edge time of the double exponential wave is longer, about 10 μs.

(2) No matter whether the coaxial cable terminal is open, connected to 51 Ω matching resistance or the shield is grounded, the peak voltage and energy coupled to cables of different lengths and positions increase with the increase of impulse current, and the peak voltage and energy coupled to cables are inversely related to the distance from the radiation source, and positively related to the cable length. When the impulse current is 4 kA, the 1 m long cable is 0.15 m away from the radiation source and the terminal is open, the peak value of the voltage coupled to the cable terminal is 18.3 V; when the terminal shielding layer is grounded, the peak value of the voltage coupled to the cable terminal is 35.87 V; when the terminal is connected to 51 Ω, the peak value of the voltage coupled to the cable terminal is 15.65 V.

(3) By Fourier transform of 0.1–0.4 μs damping vibration part of voltage waveform, we can get that the main energy of cable coupling is in the low frequency part, and there are several main energy peaks below 100 Hz, and the amplitude increases with the increase of impulse current.

(4) By fitting the curve with the experimental data, it is deduced that in the actual situation, when the lightning wave current reaches 30 kA, the coaxial cable terminal is open circuit, the cable length is 30 m, and the voltage peak value coupled to the cable terminal is about 66 V.

Footnotes

Acknowledgements

This work were financially supported by National key R & D plan (Grants No. 2017YFC1501505).

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