Finite-Difference Time Domain Method

The FDTD method requires spatial and temporal discretizations of the electric- and magnetic-field components in the Yee grid to calculate the temporal and spatial variations in the electromagnetic field, respectively [14]. The Maxwell rotation equation system in a vacuum is as follows [15]: $｛\begin{array}{l}\nabla \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}\hfill \\ \nabla \times \overrightarrow{H}=\frac{\partial \overrightarrow{D}}{\partial t}+\overrightarrow{J}\hfill \end{array}$ (1) where $\overrightarrow{E}$ is the electric-field intensity vector, in V/m, $\overrightarrow{D}$ is the electric-displacement vector, expressed in C/m², $\overrightarrow{H}$ is the magnetic-field-intensity vector, measured in V/m², $\overrightarrow{B}$ is the magnetic-induction-intensity vector, in units of Wb/m², and $\overrightarrow{J}$ is the current-density vector, measured in A/m². In this study, the current density is caused by photoelectron emission.

In an isotropic medium, we have: $｛\begin{array}{l}\overrightarrow{D}=ϵ\overrightarrow{E}\hfill \\ \overrightarrow{B}=\mu \overrightarrow{H}\hfill \end{array}$ (2) where ϵ is the dielectric constant of the medium, in F/m and μ is the magnetic-permeability coefficient, in H/m. This study assumes vacuum conditions inside the metal cavity; therefore, ϵ = ϵ₀ = 8.854×10^-12 F/m and μ = μ₀ = 4π × 10^-7 H/m.

This study establishes an FDTD model using a three-dimensional Cartesian coordinate system. By substituting the components of the electric and magnetic fields into Eq. (1), we obtain the following scalar equation: $｛\begin{array}{l}ϵ\frac{\partial E_x}{\partial t}=\frac{\partial H_z}{\partial y}-\frac{\partial H_y}{\partial z}-J_x\hfill \\ ϵ\frac{\partial E_y}{\partial t}=\frac{\partial H_x}{\partial z}-\frac{\partial H_z}{\partial x}-J_y\hfill \\ ϵ\frac{\partial E_z}{\partial t}=\frac{\partial H_y}{\partial x}-\frac{\partial H_x}{\partial y}-J_z\hfill \\ \mu \frac{\partial H_x}{\partial t}=\frac{\partial E_y}{\partial z}-\frac{\partial E_z}{\partial y}\hfill \\ \mu \frac{\partial H_y}{\partial t}=\frac{\partial E_z}{\partial x}-\frac{\partial E_x}{\partial z}\hfill \\ \mu \frac{\partial H_z}{\partial t}=\frac{\partial E_x}{\partial y}-\frac{\partial E_y}{\partial x}\hfill \end{array}$ (3) Taking the first formula of Eq. (3) as an example, performing the center difference can yield the FDTD formula in a three-dimensional Cartesian coordinate system as follows: $｛\begin{array}{l}\begin{array}{c}{E}_x^{n+1}\left(i+\frac{1}{2}\mathrm{,}j\mathrm{,}k\right)=E_x^n\left(i+\frac{1}{2}\mathrm{,}j\mathrm{,}k\right)+\frac{\text{Δ}t}{ϵ}\cdot [\frac{H_z^{n+1}\left(i+\frac{1}{2}\mathrm{,}j+\frac{1}{2}\mathrm{,}k\right)-H_z^{2+1}\left(i+\frac{1}{2}\mathrm{,}j-\frac{1}{2}\mathrm{,}k\right)}{\text{Δ}y}\\ -\frac{H_y^{n+1}\left(i+\frac{1}{2}\mathrm{,}j\mathrm{,}k+\frac{1}{2}\right)-H_y^{n+1}\left(i+\frac{1}{2}\mathrm{,}j\mathrm{,}k-\frac{1}{2}\right)}{\text{Δ}z}-J_x^{n+\frac{1}{2}}\left(i+\frac{1}{2}\mathrm{,}j\mathrm{,}k\right)]\end{array}\hfill \end{array}$ (4)

Similarly, the other components of the FDTD formulae in Eq. (3) can be obtained.

Owing to the introduction of discrete errors because of the central difference, the time and spatial step sizes must satisfy the Courant stability condition [16]: $c\text{Δ}t\le \frac{1}{\sqrt{\frac{1}{{\left(\text{Δ}x\right)}^2}+\frac{1}{{\left(\text{Δ}y\right)}^2}+\frac{1}{{\left(\text{Δ}z\right)}^2}}}\mathrm{.}$ (5)

Particle in cell Method

After establishing the FDTD formula in Section 2.1, the key to solving for the electromagnetic field lies in the current density $\overrightarrow{J}$. The PIC method was used to calculate the motion parameters of the electrons to solve the current density $\overrightarrow{J}$ [17]. Because of the large number of emitted electrons, this study equates some electrons with similar positions, velocities, and other characteristics to a macroparticle and completes the simulation of actual physical processes containing a large number of electrons by simulating a small number of macroparticles.

The relationship between the position vector $\overrightarrow{r}$ of the macroparticles and time is $\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}\mathrm{.}$ (6) In the three-dimensional Cartesian coordinate system, we perform central difference processing on the above equation and obtain the iterative equation as follows: $｛\begin{array}{l}{x}^{n+1}=x^n+v_x^{n+\frac{1}{2}}\text{Δ}t\hfill \\ y^{n+1}=y^n+v_y^{n+\frac{1}{2}}\text{Δ}t\hfill \\ z^{n+1}=z^n+v_z^{n+\frac{1}{2}}\text{Δ}t\hfill \end{array}$ (7) Electron motion is mainly influenced by electromagnetic forces, and relativistic effects must be considered for high-velocity electrons. Therefore, the momentum equation is $\frac{\text{d}\left(\gamma m_0\overrightarrow{v}\right)}{\text{d}t}=q\left(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B}\right)\mathrm{,}$ (8) where m₀ is the stationary mass of the macroparticle and $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is the Lorentz factor.

The effect of an electromagnetic field on electrons is decomposed into half-acceleration caused by two electric fields and rotation caused by one magnetic field such that $\overrightarrow{u}=\gamma \overrightarrow{v}$. By centrally differencing the momentum equation, we obtain $\begin{array}{c}{\overrightarrow{u}}^{n+\frac{1}{2}}=\left({\overrightarrow{u}}^{n-\frac{1}{2}}+\frac{q{\overrightarrow{E}}^n}{m_0}\frac{\text{Δ}t}{2}\right)+\frac{q\left({\overrightarrow{u}}^{n+\frac{1}{2}}+{\overrightarrow{u}}^{n-\frac{1}{2}}\right)\times {\overrightarrow{B}}^n}{2{\gamma }^n{m}_0}\cdot \text{Δ}t\\ +\frac{q{\overrightarrow{E}}^n}{m_0}\frac{\text{Δ}t}{2}\mathrm{.}\end{array}$ (9) The particle velocity can be iterated via calculation, thereby achieving iterative updates of the particle position.

To determine the current density, the charge-conservation equation is $\nabla \cdot \overrightarrow{J}=-\frac{\rho }{t}$. The differential form in the three-dimensional Cartesian coordinate system is $\frac{\partial J_x}{\partial x}+\frac{\partial J_y}{\partial y}+\frac{\partial J_z}{\partial z}=-\frac{\partial \rho }{\partial t}\mathrm{.}$ (10) Therefore, on grid nodes (i, j, k), the following can be obtained: $\begin{array}{c}\frac{J_x^{n+\frac{1}{2}}\left(x+\frac{1}{2}\mathrm{,}y\mathrm{,}z\right)}{\text{Δ}x}+\frac{J_y^{n+\frac{1}{2}}\left(x\mathrm{,}y+\frac{1}{2}\mathrm{,}z\right)}{\text{Δ}x}+\frac{J_z^{n+\frac{1}{2}}\left(x\mathrm{,}y\mathrm{,}z+\frac{1}{2}\right)}{\text{Δ}x}.\\ =-\frac{{\rho }^{n+1}\left(x\mathrm{,}y\mathrm{,}z\right)-{\rho }^n\left(x\mathrm{,}y\mathrm{,}z\right)}{\text{Δ}t}\end{array}$ (11) Performing the same operation on grid nodes $(i\mathrm{,}j+\mathrm{1,}k)$, $(i\mathrm{,}j\mathrm{,}k+1)$, and $(i\mathrm{,}j+\mathrm{1,}k+1)$, and adding the four equations yields $\begin{array}{c}{J}_x\left(i+\frac{1}{2}\mathrm{,}j\mathrm{,}k\right)+J_x\left(i+\frac{1}{2}\mathrm{,}j+\mathrm{1,}k+1\right)\\ +J_x\left(i+\frac{1}{2}\mathrm{,}j\mathrm{,}k+1\right)+J_x\left(i+\frac{1}{2}\mathrm{,}j+\mathrm{1,}k\right)\\ ={\rho }_0\frac{x^{n+1}\left(p\right)-x^n\left(p\right)}{\text{Δ}t}=J_x.\end{array}$ (12) Similarly, the values of each current-density component can be obtained, and the variation in the cavity SGEMP over time can be solved by combining it with the FDTD method mentioned earlier. Many studies have previously combined the FDTD method with the PIC method. Chen et al. proposed the conformal PIC method to simulate the cavity SGEMP [18, 19] using the UNIPIC code [20, 21]. The most important features of the conformal method are that the code can handle complex geometric structures without staircased approximation and can simulate the electromagnetic fields generated from electrons distributed at different angles.

Transmission-Line Model of Shielded Cable under External Excitation

After obtaining the electromagnetic environment inside the cavity from Section 2, the coupling response of the cable inside the cavity can be solved as follows.

The transmission-line model is often used to analyze the field-line coupling problem under electromagnetic-pulse excitation. Shielded cables can be considered as two transmission-line systems [22]: the shield of the cable and external circuit (external wire or ground) form an external transmission-line system, and the shield and inner conductor form an internal transmission-line system, as shown in Fig. 2. The connection between these two transmission-line systems can be described by the transfer impedance and admittance of the shielded cables.

Fig. 2Full-screen View

Unit-length circuit diagram for cable shield and ground and cable shield and inner conductor

The field–line coupling problem has been studied by many researchers, and commonly used models include the Agrawal, Taylor, and Rachidi models [23-25]. This study used the Agrawal model to study the response of an external transmission-line system under an external-field excitation. The model assumes that the coupling between the external electromagnetic field and transmission line can be considered an electromagnetic-scattering phenomenon, and the coupling effect of the electromagnetic field on the cable is equivalent to the combined voltage source at both ends and the distributed voltage source on the line.

Assuming that the cable is located at a height above the reference ground plane, the equation set for the transmission line of the external system is given as $｛\begin{array}{l}\frac{\partial V_{\text{os}}\left(x\mathrm{,}t\right)}{\partial x}+l_{\text{o}}\frac{\partial I_{\text{o}}\left(x\mathrm{,}t\right)}{\partial t}=E_{\text{L}}\hfill \\ \frac{\partial I_{\text{o}}\left(x\mathrm{,}t\right)}{\partial x}+c_{\text{o}}\frac{\partial V_{\text{os}}\left(x\mathrm{,}t\right)}{\partial t}=0\hfill \end{array}$ (13) where $V_{\text{os}}\left(x\mathrm{,}t\right)$ and $I_{\text{o}}\left(x\mathrm{,}t\right)$ represent the scattering voltage and current of the external system, respectively, lo represents the distributed inductance, co is the distributed capacitance, and the distributed voltage source $E_{\text{L}}=E_x^{\text{ex}}\left(x\mathrm{,}h\mathrm{,}t\right)$ represents the polarization of the electric field in the x direction at position x and height h from the ground at time t. The boundary conditions at both ends of the cable shield are as follows: $｛\begin{array}{l}{V}_{\text{os}}\left(\mathrm{0,}t\right)=-R_{\text{oS}}{I}_{\text{o}}\left(\mathrm{0,}t\right)+V_{\text{S}}\hfill \\ V_{\text{os}}\left(L\mathrm{,}t\right)=R_{\text{oL}}{I}_{\text{o}}\left(L\mathrm{,}t\right)+V_{\text{L}}\hfill \end{array}$ (14) where RoS and RoL represent the grounding load values at the left and right ends of the cable shield, respectively. VS and VL represent the voltage sources formed by the excitation of the electric field at the left and right terminals of the transmission line, respectively, which can be expressed as $｛\begin{array}{l}{V}_{\text{S}}={\displaystyle {\int }_0^h}{E}_z^{\text{ex}}\left(\mathrm{0,}z\mathrm{,}t\right)\text{d}z\hfill \\ V_{\text{L}}={\displaystyle {\int }_0^h}{E}_z^{\text{ex}}\left(L\mathrm{,}z\mathrm{,}t\right)\text{d}z\hfill \end{array}$ (15) where $E_z^{\text{ex}}\left(x\mathrm{,}z\mathrm{,}t\right)$ represents the z-direction electric-field intensity at position x and height h above the ground at time t.

The relationship between the scattering voltage $V_{\text{os}}\left(x\mathrm{,}t\right)$ and total voltage $V_{\text{o}}\left(x\mathrm{,}t\right)$ of the shield is $V_{\text{o}}\left(x\mathrm{,}t\right)=V_{\text{os}}\left(x\mathrm{,}t\right)+V_{\text{oi}}\left(x\mathrm{,}t\right)\mathrm{,}$ (16) where $V_{\text{oi}}\left(x\mathrm{,}t\right)=-{\displaystyle {\int }_0^h}{E}_z^{\text{ex}}\left(x\mathrm{,}z\mathrm{,}t\right)\text{d}z$.

After the current and voltage responses of the external shield are obtained, the transmission-line equation for the internal system of the shielded cable is established as follows: $｛\begin{array}{l}\frac{\partial V_{\text{i}}\left(x\mathrm{,}t\right)}{\partial x}+l_i\frac{\partial I_{\text{i}}\left(x\mathrm{,}t\right)}{\partial t}=V_{\text{si}}\left(x\mathrm{,}t\right)\hfill \\ \frac{\partial I_{\text{i}}\left(x\mathrm{,}t\right)}{\partial x}+c_i\frac{\partial V_{\text{i}}\left(x\mathrm{,}t\right)}{\partial t}=I_{\text{si}}\left(x\mathrm{,}t\right)\hfill \end{array}$ (17) where $V_{\text{i}}\left(x\mathrm{,}t\right)$ and $I_{\text{i}}\left(x\mathrm{,}t\right)$ represent the voltage and current of the internal transmission-line system, respectively, li represents the distributed inductance, ci represents the distributed capacitance, and $V_{\text{si}}\left(x\mathrm{,}t\right)$ and $I_{\text{si}}\left(x\mathrm{,}t\right)$ are the distributed excitation sources of the internal transmission-line system, expressed as $｛\begin{array}{l}{V}_{\text{si}}\left(x\mathrm{,}t\right)=Z_{\text{t}}\left(t\right)I_{\text{o}}\left(x\mathrm{,}t\right)\hfill \\ I_{\text{si}}\left(x\mathrm{,}t\right)=-Y_{\text{t}}\left(t\right)V_{\text{o}}\left(x\mathrm{,}t\right)\hfill \end{array}$ (18) where Z_t(t) and Y_t(t) are the transfer impedance and transfer admittance of the shielded cable, respectively. The commonly used simple model is [18] $｛\begin{array}{l}{Z}_t\left(t\right)=R_{\text{dc}}+j\omega L_{\text{t}}\hfill \\ Y_t\left(t\right)=j\omega C_t\hfill \end{array}$ (19) where Rdc denotes the transfer resistance, ω is the angular frequency, Lt is the transfer inductance, and Ct is the transfer admittance. This model assumes Rdc is constant, independent of the frequency, and larger than the actual value; however, it satisfies the worst-case estimation [26].

Based on the above equations, we conclude that $｛\begin{array}{l}\frac{\partial V_{\text{i}}\left(x\mathrm{,}t\right)}{\partial x}+l_{\text{i}}\frac{\partial I_{\text{i}}\left(x\mathrm{,}t\right)}{\partial t}=R_{\text{dc}}{I}_{\text{o}}\left(x\mathrm{,}t\right)+L_{\text{t}}\frac{\partial I_{\text{o}}\left(x\mathrm{,}t\right)}{\partial x}\hfill \\ \frac{\partial I_{\text{i}}\left(x\mathrm{,}t\right)}{\partial x}+c_{\text{i}}\frac{\partial V_{\text{i}}\left(x\mathrm{,}t\right)}{\partial t}=-C_{\text{t}}\frac{\partial V_{\text{o}}\left(x\mathrm{,}t\right)}{\partial x}\hfill \end{array}$ (20)

Differential approximation of cable transmission-line equations

Using the one-dimensional FDTD method to solve the transmission-line equation, the cable is first discretized and divided into NDZ segments, with each segment having a length of Δ z. Similarly, the entire solution time is divided into NDT segments, with each segment being Δ t [27].

The voltage at point NDZ+1 is interleaved with the current at point NDZ for a solution with a spatial interval Δ z/2 between each voltage point and adjacent current points. Simultaneously, the time points should also be interwoven with a time interval of Δ t/2 between each voltage point and adjacent current points.

For external and internal transmission-line systems, the iterative equation obtained by performing the central difference between the voltage and current is $\begin{array}{c}{I}_{{\text{o}}_k}^{n+\frac{3}{2}}=I_{{\text{o}}_k}^{n+\frac{1}{2}}-\frac{\text{Δ}t}{l_{\text{o}}\text{Δ}z}\left(V_{{\text{os}}_{k+1}}^{n+1}-V_{{\text{os}}_k}^{n+1}\right)\\ +\frac{\text{Δ}t}{2l_o}\left(E_{{\text{L}}_{k+1}}^{n+1}+E_{{\text{L}}_k}^{n+1}\right)\end{array}$ (21) $V_{{\text{os}}_k}^{n+1}=V_{{\text{os}}_k}^n-\frac{\text{Δ}t}{c_{\text{o}}\text{Δ}z}\left(I_{{\text{o}}_k}^{n+\frac{1}{2}}-I_{{\text{o}}_{k-1}}^{n+\frac{1}{2}}\right)$ (22) $\begin{array}{c}{I}_{{\text{i}}_k}^{n+\frac{3}{2}}=I_{{\text{i}}_k}^{n+\frac{1}{2}}-\frac{\text{Δ}t}{l_{\text{i}}\text{Δ}z}\left(V_{{\text{i}}_{k+1}}^{n+1}-V_{{\text{i}}_k}^{n+1}\right)\\ +\left(R_{\text{dc}}\frac{\text{Δ}t}{2l_{\text{i}}}+\frac{L_{\text{t}}}{l_{\text{i}}}\right)I_{{\text{o}}_k}^{n+\frac{3}{2}}+\left(R_{\text{dc}}\frac{\text{Δ}t}{2l_{\text{i}}}-\frac{L_{\text{t}}}{l_{\text{i}}}\right)I_{{\text{o}}_k}^{n+\frac{1}{2}}\end{array}$ (23) $\begin{array}{c}{V}_{{\text{i}}_k}^{n+1}=V_{{\text{i}}_k}^n-\frac{\text{Δ}t}{c_{\text{i}}\text{Δ}z}\left(I_{{\text{i}}_k}^{n+\frac{1}{2}}-I_{{\text{i}}_{k-1}}^{n+\frac{1}{2}}\right)\\ -\frac{C_{\text{t}}}{c_{\text{i}}}\left(V_{{\text{o}}_k}^{n+1}-V_{{\text{o}}_k}^n\right)\mathrm{.}\end{array}$ (24) The boundary conditions satisfy the following equations: $\begin{array}{c}{V}_{{\text{os}}_1}^{n+1}={\left(1+c_{\text{o}}{R}_{\text{oS}}\frac{\text{Δ}z}{\text{Δ}t}\right)}^{-1}[\left(c_{\text{o}}{R}_{\text{oS}}\frac{\text{Δ}z}{\text{Δ}t}-1\right)V_{{\text{os}}_1}^n\\ -2R_{\text{oS}}{I}_{{\text{o}}_1}^{n+\frac{1}{2}}+\left(V_{\text{S}}^{n+1}+V_{\text{S}}^n\right)]\end{array}$ (25) $\begin{array}{c}{V}_{{\text{os}}_{NDZ+1}}^{n+1}={\left(1+c_{\text{o}}{R}_{\text{oL}}\frac{\text{Δ}z}{\text{Δ}t}\right)}^{-1}[\left(c_o{R}_{\text{oL}}\frac{\text{Δ}z}{\text{Δ}t}-1\right)\\ V_{{\text{os}}_{NDZ+1}}^n+2R_{\text{oL}}{I}_{{\text{o}}_{NDZ}}^{n+\frac{1}{2}}+\left(V_{\text{L}}^{n+1}+V_{\text{L}}^n\right)]\end{array}$ (26) $\begin{array}{c}{V}_{{\text{i}}_1}^{n+1}={\left(1+c_{\text{i}}{R}_{\text{oS}}\frac{\text{Δ}z}{\text{Δ}t}\right)}^{-1}[\left(c_{\text{i}}{R}_{\text{iS}}\frac{\text{Δ}z}{\text{Δ}t}-1\right)V_{{\text{i}}_1}^n\\ +2R_{\text{iS}}{I}_{{\text{i}}_1}^{n+\frac{1}{2}}-C_{\text{t}}{R}_{\text{iS}}\frac{\text{Δ}z}{\text{Δ}t}\left(V_{{\text{os}}_1}^{n+1}-V_{{\text{os}}_1}^n\right)]\end{array}$ (27) $\begin{array}{c}{V}_{{\text{i}}_{NDZ+1}}^{n+1}={\left(1+c_{\text{i}}{R}_{\text{iL}}\frac{\text{Δ}z}{\text{Δ}t}\right)}^{-1}[\left(c_{\text{i}}{R}_{\text{iL}}\frac{\text{Δ}z}{\text{Δ}t}-1\right)V_{{\text{i}}_{NDZ+1}}^n\\ +2R_{\text{iL}}{I}_{{\text{i}}_{NDZ}}^{n+\frac{1}{2}}-C_{\text{t}}{R}_{\text{iL}}\frac{\text{Δ}z}{\text{Δ}t}\left(V_{{\text{os}}_{NDZ+1}}^{n+1}-V_{{\text{os}}_{NDZ+1}}^n\right)]\mathrm{.}\end{array}$ (28)

Cable length

The cable lengths are set to 0.1 m, 0.2 m, 0.3 m, 0.4 m, and 0.5 m, keeping the other parameters unchanged. The coupling voltage of the inner conductor terminal load is calculated, as shown in Fig. 6a. As the cable length increases, the amplitude of the coupling voltage increases. This is because the external electromagnetic field remains unchanged, and as the cable length increases, the distribution of the voltage sources increases, leading to an increase in the coupling voltage.

Fig. 6Full-screen View

(a) Time-domain waveform of coupling voltage ViL for cables of different lengths. (b) Relationship between peak amplitude of coupling voltage ViL and length. (c) Time-domain waveform of coupling voltage ViL for cables of different heights. (d) Relationship between peak amplitude of coupling voltage ViL and height

Fig. 6b shows the relationship between the peak amplitude of the coupling voltage ViL and cable length, and the two are approximately linearly related.

Cable height

The distances between the cable and the bottom of the cavity are set to 0.05 m, 0.1 m, 0.15 m, 0.2 m, and 0.25 m, while keeping the other parameters unchanged. The calculated coupling voltage of the inner conductor terminal load is shown in Fig. 6c. As the cable height increases, the amplitude of the coupling voltage increases. This is mainly because, as the height increases, the area between the transmission line and the ground as a reference conductor increases, leading to more electromagnetic energy coupling into the transmission line and an increase in the coupling current.

Fig. 6d shows the relationship between the peak amplitude of coupling voltage ViL and the height of the cable, and the two are also approximately linearly related.

Number of emitted electrons

According to the physical model of the cavity SGEMP described in Section 2, when the cavity size and the X-ray energy are constant, the number of emitted electrons determines the cavity-SGEMP field-strength amplitude. Therefore, we set the number of emitted electrons to 1×10¹⁰, 1×10¹¹, 1×10¹², 4.365×10¹², and 1×10¹³, while keeping other parameters unchanged. The calculated Ez field strength and coupling voltage ViL are shown in Fig. 7. As the number of emitted electrons increases, both the field strength and coupling voltage increase, which is in line with the expected outcomes [3].

Fig. 7Full-screen View

(a) Time-domain waveform of Ez generated by different numbers of emitted electrons. (b) Relationship between peak amplitude of Ez and number of emitted electrons. (c) Time-domain waveform of ViL generated by different numbers of emitted electrons. (d) Relationship between peak amplitude of ViL and number of emitted electrons

According to the calculation results, the key to reducing coupling interference is to reduce the number of emitted electrons. In practical applications, increasing the thickness or atomic number of the metal cavity (e.g., adding lead shielding outside the cavity) can increase the attenuation of X-rays and reduce the number of electrons emitted. However, this method significantly increases the weight of the entire system and is unsuitable for scenarios with weight requirements. Here, we propose an alternative method. In general, the larger the atomic number of a material, the larger is the cross-section of the photoelectric effect, and more electrons are emitted [31]. Therefore, a layer of a low-Z material (such as polyethylene (PE), (C₂H₄)_n) can be attached to the inner side of the metal cavity to reduce the forward electron emission.

Here, we use Monte Carlo calculations to compare the forward electron-emission yields of 1 mm aluminum, 1 mm aluminum with50 μmPE, 1 mm aluminum with 50 μm epoxy resin ((C₁₁H₁₂O₃)_n), and 1 mm aluminum with 50 μm polytetrafluoroethylene (PTFE, (C₂F₄))_n) under different energies of X-ray irradiation, as shown in Table 1. Adding even a 50-μm-thick low-Z material can reduce the electron-emission yield by an order of magnitude. Notably, the electron-emission yield is the lowest after adding PE, whereas the electron-emission yield is the highest after adding PTFE because PTFE contains fluorine elements with higher atomic numbers(Z=9). Therefore, this method can be used to reduce the cavity SGEMP and coupling response of the cable.