Transient response analysis method of shielded cable based on minimum phase method

Qinglin Xu; Jie Zhang; Pingan Liu; Jun Yuan

doi:10.11884/HPLPB202133.210177

Journals >High Power Laser and Particle Beams >Volume 33 >Issue 12 >Page 123004 > Article

High Power Laser and Particle Beams
Vol. 33, Issue 12, 123004 (2021)

Transient response analysis method of shielded cable based on minimum phase method

Qinglin Xu, Jie Zhang, Pingan Liu, and Jun Yuan

Author Affiliations

Xi’an Modern Control Technology Research Institute, Xi’an 710065, China

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DOI: 10.11884/HPLPB202133.210177 Cite this Article

Qinglin Xu, Jie Zhang, Pingan Liu, Jun Yuan. Transient response analysis method of shielded cable based on minimum phase method[J]. High Power Laser and Particle Beams, 2021, 33(12): 123004 Copy Citation Text

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Abstract

To analyze the transient electromagnetic response of shielded cables in time domain when they are subjected to electromagnetic pulse, a frequency domain equivalent method based on minimum phase method is studied in this paper. First, by establishing the circuit model of the transfer impedance testing system, the transfer function of the system in frequency domain is derived, and it is verified as the cascade of a minimum phase system and an all-pass system. Secondly, the minimum phase of the measured amplitude-frequency curve is reconstructed by using the minimum phase method, then the phase of the reconstructed transfer function is compensated according to the distribution of zeros and poles. Finally, the time domain response of the shielded cable is obtained after multiplying the compensated transfer function with the transient interference signal in frequency domain and transforming into time domain. The effectiveness of the proposed method is verified by simulation and experiment.

Keywords

phase reconstruction shielding effectiveness transfer function transfer impedance transient response

$ {Z_{\rm{t}}} = \frac{{{V_{{\rm{l}}2}}}}{{{I_{{\rm{l}}1}}l}} = \frac{{{Z_{{\rm{tt}}}}{V_{{\rm{l}}2}}}}{{{V_{{\rm{l}}1}}l}} $

(1)

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$ SE = 20\lg \frac{{{I_{{\rm{l}}1}}}}{{{I_{{\rm{l}}2}}}} = 20\lg \frac{{{V_{{\rm{l}}2}}}}{{{Z_{\rm{t}}}{I_{{\rm{l}}2}}}} = 20\lg \frac{{{Z_{{\rm{l}}2}}}}{{{Z_{\rm{t}}}}} $

(2)

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$ {I_1}\left( x \right) = \frac{{{V_{\rm{T}}}\left( {1 - {{\mathit{\Gamma}} _{\rm{T}}}} \right){{\rm{e}}^{ - s{T_x}}}}}{{2{Z_0}\left[ {1 - {{\mathit{\Gamma}} _{\rm{T}}}{{\mathit{\Gamma}} _{{\rm{l}}1}}{{\rm{e}}^{ - s\left( {2{T_L}} \right)}}} \right]}}\left[ {1 - {{\mathit{\Gamma}} _{{\rm{l}}1}}{{\rm{e}}^{ - s\left( {2{T_L} - 2{T_x}} \right)}}} \right] $

(3)

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$ \left[ {

\begin{matrix} V_{2} (L) \\ I_{2} (L) \end{matrix}

} \right] = {\boldsymbol{\varPhi}} \left( L \right)\left[ {

\begin{matrix} V_{2} (0) \\ I_{2} (0) \end{matrix}

} \right] + \int_0^L {{\boldsymbol{\varPhi}} \left( {L - \tau } \right)\left[ {

\begin{matrix} V_{F} (τ) \\ I_{F} (τ) \end{matrix}

} \right]{\rm{d}}\tau = {\boldsymbol{\varPhi}} \left( L \right)\left[ {

\begin{matrix} V_{2} (0) \\ I_{2} (0) \end{matrix}

} \right] + \left[ {

\begin{matrix} V_{F T} (L) \\ I_{F T} (L) \end{matrix}

} \right]} $

(4)

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\begin{aligned} V_{F T} (L) = & \frac{Z_{t} (1 - Γ_{T})}{4 Z_{0} [1 - Γ_{T} Γ_{l 1} e^{- s (2 T_{L})}]} [\frac{c}{\sqrt{ε_{r}} + 1} e^{s \sqrt{ε_{r}} T_{L}} \frac{1}{s} V_{T} - \frac{c}{\sqrt{ε_{r}} - 1} e^{- s \sqrt{ε_{r}} T_{L}} \frac{1}{s} V_{T} - \\ \frac{Γ_{l 1} c}{\sqrt{ε_{r}} - 1} e^{s (\sqrt{ε_{r}} - 2) T_{L}} \frac{1}{s} V_{T} + \frac{Γ_{l 1} c}{\sqrt{ε_{r}} + 1} e^{- s (\sqrt{ε_{r}} + 2) T_{L}} \frac{1}{s} V_{T} + \frac{2 (1 + Γ_{l 1}) c}{ε_{r} - 1} e^{- s T_{L}} \frac{1}{s} V_{T}] \end{aligned}

(5)

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\begin{aligned} I_{F T} (L) = & \frac{Z_{t} (1 - Γ_{T})}{4 Z_{c} Z_{0} [1 - Γ_{T} Γ_{l 1} e^{- s (2 T_{L})}]} [- \frac{c}{\sqrt{ε_{r}} + 1} e^{s \sqrt{ε_{r}} T_{L}} \frac{1}{s} V_{T} - \frac{c}{\sqrt{ε_{r}} - 1} e^{- s \sqrt{ε_{r}} T_{L}} \frac{1}{s} V_{T} + \\ \frac{Γ_{l 1} c}{\sqrt{ε_{r}} - 1} e^{s (\sqrt{ε_{r}} - 2) T_{L}} \frac{1}{s} V_{T} + \frac{Γ_{l 1} c}{\sqrt{ε_{r}} + 1} e^{- s (\sqrt{ε_{r}} + 2) T_{L}} \frac{1}{s} V_{T} + \frac{2 \sqrt{ε_{r}} (1 - Γ_{l 1}) c}{ε_{r} - 1} e^{- s T_{L}} \frac{1}{s} V_{T}] \end{aligned}

(6)

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$ {{{{\rm{e}}}}^{ - s{T_0}}} \cong \frac{{1 - \left( {{{s{T_0}} \mathord{\left/ {\vphantom {{s{T_0}} 2}} \right. } 2}} \right)}}{{1 + \left( {{{s{T_0}} \mathord{\left/ {\vphantom {{s{T_0}} 2}} \right. } 2}} \right)}} $

(7)

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$ {I_1}\left( L \right) = \frac{{{V_{\rm{T}}}\left( {1 - {{\mathit{\Gamma}} _{\rm{T}}}} \right)\left( {1 - {{\mathit{\Gamma}} _{{\rm{l}}1}}} \right)}}{{2{Z_0}}}\frac{{1 + s{T_L}}}{{\left( {1 - {{\mathit{\Gamma}} _{\rm{T}}}{{\mathit{\Gamma}} _{{\rm{l}}1}}} \right) + s\left( {1 + {{\mathit{\Gamma}} _{\rm{T}}}{{\mathit{\Gamma}} _{{\rm{l}}1}}} \right){T_L}}}\frac{{1 - \left( {{{s{T_L}} \mathord{\left/ {\vphantom {{s{T_L}} 2}} \right. } 2}} \right)}}{{1 + \left( {{{s{T_L}} \mathord{\left/ {\vphantom {{s{T_L}} 2}} \right. } 2}} \right)}} $

(8)

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$ {I_2}\left( L \right) = \frac{{\left[ {{Z_{02}}{I_{{\rm{FT}}}}\left( L \right) + {V_{{\rm{FT}}}}\left( L \right)} \right] + s\left[ {{Z_c}{I_{{\rm{FT}}}}\left( L \right) + {{{Z_{02}}{V_{{\rm{FT}}}}\left( L \right)} \mathord{\left/ {\vphantom {{{Z_{02}}{V_{{\rm{FT}}}}\left( L \right)} {{Z_c}}}} \right. } {{Z_{\rm{c}}}}}} \right]\sqrt {{\varepsilon _{\rm{r}}}} {T_L}}}{{\left( {{Z_{02}} + {Z_{{\rm{l}}2}}} \right) + s\left( {{Z_{\rm{c}}} + {{{Z_{02}}{Z_{{\rm{l}}2}}} \mathord{\left/ {\vphantom {{{Z_{02}}{Z_{{\rm{l}}2}}} {{Z_{\rm{c}}}}}} \right. } {{Z_{\rm{c}}}}}} \right)\sqrt {{\varepsilon _{\rm{r}}}} {T_L}}} $

(9)

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$ {I_2}\left( L \right) = \frac{{{Z_{02}}{I_{{\rm{FT}}}}\left( L \right) + {V_{{\rm{FT}}}}\left( L \right)}}{{{Z_{02}} + {Z_{{\rm{l}}2}}}} $

(10)

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\begin{aligned} Z_{02} I_{F T} (L) + V_{F T} (L) = & \frac{V_{T} Z_{t} (1 - Γ_{T})}{2 Z_{0}} \frac{[(1 - Γ_{l 1}) + s (1 + Γ_{l 1} + \sqrt{ε_{r}} - Γ_{l 1} \sqrt{ε_{r}}) T_{L} / 2] T_{L} c}{(1 - Γ_{T} Γ_{l 1}) + s (1 + Γ_{T} Γ_{l 1}) T_{L}} \times \\ \frac{1 + s T_{L}}{[1 + s (\sqrt{ε_{r}} - 1) T_{L} / 2] [1 + s (\sqrt{ε_{r}} + 1) T_{L} / 2]} \frac{1 - (s T_{L} / 2)}{1 + (s T_{L} / 2)} \end{aligned}

(11)

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$ \hat H\left( {{{{{\rm{e}}}}^{{{{\rm{j}}}}\omega }}} \right) = \ln \left[ {H\left( {{{\rm{{\rm{e}}}}^{{\rm{j}}\omega }}} \right)} \right] = \ln \left| {H\left( {{{\rm{e}}^{{\rm{j}}\omega }}} \right)} \right| + {\rm{j}}\arg \left[ {H\left( {{{\rm{e}}^{{\rm{j}}\omega }}} \right)} \right] $

(12)

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Qinglin Xu, Jie Zhang, Pingan Liu, Jun Yuan. Transient response analysis method of shielded cable based on minimum phase method[J]. High Power Laser and Particle Beams, 2021, 33(12): 123004

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