Low-Frequency Magnetic Field Shielding Physics and Discovery for Fabric Enclosures Using Numerical Modeling

Transcription

1 Low-Frequency Magnetic Field Shielding Physics and Discovery for Fabric Enclosures Using Numerical Modeling Bruce Archambeault, Andrey Radchenko, Tamar Makharashvili and James Drewniak Missouri University of Science and Technology 214 IEEE Symposium on Electromagnetic Compatibility Session ID: MO-PM-5 1

2 Outline The infinite planar shield as a canonical geometry for the design of magnetic shields Superiority of various materials such as Copper, Steel and Permalloy for different thicknesses Comparison of simulated results with the several design approximations for infinite planar shields Cylindrical and Spherical magnetic conducting shield A simulation tool for LF magnetic field discovery and design Canonical loop problems from the literature MIL-STD enclosure with loop antenna 2

3 Introduction The infinite planar shield has been studied as a canonical geometry for the design of EM shields. The shield consists of an infinite planar sheet with thickness Δ, large value of the conductivity σ, and/or of the relative magnetic permeability μ r. [1] Following figures show geometry of the problem for excitation source current loop placed parallel or perpendicular to the shielding plane. Benchmark problem 1 Circular current loop parallel to an infinite plane Benchmark problem 2 Circular current loop perpendicular to an infinite plane 3

4 Parallel Loop Excitation - Geometry and Model h = 3.5 cm Finite Size Shielding Plane Field Probe R = cm 61cm Current Loop =.5mm Number of triangles: ~ m Top View Side View The loop center is at the XYZ origin and the loop is in XY plane. Magnetic field is monitored at symmetric location behind the shielding plane at: Z = 2h = 61 cm In simulation model the infinite plate was replaced by a sufficiently large plate with length and width: a = 7h = 3.5 m. Modeling is performed in EMCoS EMC Studio [4] using Low Frequency Magnetic Field solver [5, 6]. 4

5 Perpendicular Loop Excitation - Geometry and Model h = 3.5 cm Finite Size Shielding Plane R = cm Field Probe 61cm Current Loop =.5mm Number of triangles: ~ m Top View Side View The loop center is at the XYZ origin and the loop is in XY plane. Magnetic field is monitored at symmetric location behind the shielding plane at: Z = 2h = 61 cm In simulation model the infinite plate was replaced by a sufficiently large plate with length and width: a = 7h = 3.5 m. Modeling is performed in EMCoS EMC Studio [4] using Low Frequency Magnetic Field solver [5, 6]. 5

6 LF Modeling and Measurements: Loop over Al Plate [6] Modeling is performed in EMCoS EMC Studio. Number of coil turns: 1 Coil Radius: 51.3 mm Height of the coil: 53 mm Wire radius: 3.53 mm Very good agreement between modeling and measurement for input impedance. 6

7 LF Modeling vs. Measurements: Printed Loops over Al [6] Modeling is performed in EMCoS EMC Studio. 7

8 Calculation of Shielding Effectiveness Step 1: Calculation of H i (Incident Magnetic Field) without shield. H i Step 2: Calculation of H t (Transmitted Magnetic Field) with shield. Step 1 Step 3: Shielding Effectiveness is a ratio of the magnitude of the incident magnetic (electric) field without shield, with the magnitude of the transmitted magnetic (electric) field through the shield [2]. In terms of magnetic field, the shielding effectiveness could be defined as: H i H t SE db = 2log 1 H i H t In terms of electric field, the shielding effectiveness could be defined as: H r SE db = 2log 1 E i E t Step 2 8

9 Parallel Loop Simulation Results Validation 8 7 COPPER μ =1 σ = 54 x 1 6 [S/m] Δ =.5 mm Copper, = 54x1 6 [S/m] LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] Δ =.5 mm Low Carbon Steel, = 9x1 6 [S/m] Exact Small Dipole Simulation 1 5 Exact Small Dipole Simulation Copper, Frequency = 54x1 [Hz] 6 [S/m] Low Carbon Frequency Steel, [Hz] = 9x1 6 [S/m] Exact Small Dipole Simulation X: 9833 Y: X: 1e+4 Y: % Exact and Small Dipole curves are provided from [1], Fig. B.12, pg Exact Small Dipole Simulation X: 4.847e+4 Y: 17.5 X: 5e+4 Y: % 9

10 Parallel Loop Simulation Results Validation 8 7 COPPER μ =1 σ = 54 x 1 6 [S/m] Δ =.5 mm LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] Δ =.5 mm Exact Small Dipole Simulation Exact Small Dipole Simulation % X: 113 Y: X: 1 Y: Exact Small Dipole Simulation Exact and Small Dipole curves are provided from [1], Fig. B.14, pg Exact Small Dipole Simulation 9.37 % X: 2.445e+4 Y: X: 2.5e+4 Y:

11 Simulation - Thickness Variation Test COPPER μ =1 σ = 54 x 1 6 [S/m] Δ =.5 mm At 1 KHz frequency: 11 LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] Δ =.5 mm At 1 KHz frequency: Thickness SE Thickness SE 15 um ~ 17.1 db 15 um ~ 5 db um ~27.4 db 1 um ~33.3 db 3 um 1 um 5 um 25 um 1 um 5 um 3 um 15 um 1 um 5 um 1 um Copper Shielding plate thickness, was swept in range from 1 micron to 3 mm um 1 um 5 um 25 um 1 um 5 um 3 um 15 um 1 um 5 um 1 um 5 um ~12.7 db 1 um ~18.7 db Low Carbon Steel

12 Copper and Low Carbon Steel Comparison of SE for Copper and Low Carbon Steel for different thicknesses: Copper um - LC Steel 3 um - Copper 1 um - LC Steel 1 um - Copper 5 um - LC Steel 5 um - Copper 25 um - LC Steel 25 um - Copper 1 um - LC Steel 1 um - Copper 5 um - LC Steel 5 um - Copper Low Carbon Steel um - LC Steel 3 um - Copper 15 um - LC Steel 15 um - Copper 1 um - LC Steel 1 um - Copper 5 um - LC Steel 5 um - Copper 1 um - LC Steel 1 um - Copper COPPER μ =1 5 Frequency σ = 54 [Hz] x 1 6 [S/m] LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m]

13 SE vs. 1Hz, 6Hz, 1Hz, 1KHz, 1KHz COPPER μ =1 σ = 54 x 1 6 [S/m] 11 LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] PERMALLOY μ =5 σ = 1.7 x 1 6 [S/m] Copper Steel Permalloy 1 Hz Copper 1 Hz 6 Hz Thickness [micron] Steel Permalloy 6 Hz Thickness [micron] Copper Steel Permalloy 1 Hz Copper Steel Permalloy 1 KHz Copper Steel Permalloy 1 KHz 2 1 Hz 1 KHz KHz Thickness [micron] Thickness [micron] Thickness [micron] 13

14 SE vs. 1Hz, 6Hz, 1Hz, 1KHz, 1KHz 8 6 Copper Steel 1 Hz 2 15 Copper Steel 6 Hz COPPER μ =1 σ = 54 x 1 6 [S/m] 11 LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] Hz 6 Hz Thickness [micron] Thickness [micron] Hz 1 KHz 8 Copper Copper Steel Steel 6 1 Hz 1 KHz KHz Copper Steel 1 KHz Thickness [micron] Thickness [micron] Thickness [micron] 14

15 SE vs. Thickness Hz Copper 1Hz LC Steel 6Hz Copper 6Hz LC Steel 1Hz Copper 1Hz LC Steel SE vs. 1KHz, 1KHz 1KHz Copper 1KHz LC Steel 1KHz Copper 1KHz LC Steel Thickness [micron] SE vs. 1Hz, 6Hz, 1Hz COPPER μ =1 σ = 54 x 1 6 [S/m] 11 LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] Thickness [micron] 15

16 Permeability Variation - Low Carbon Steel - 1 KHz 11 LOW CARBON STEEL σ = 9 x 1 6 [S/m] um 15 um 2 um Low Carbon Steel, = 9x1 6 [S/m] 2 With fixed steel conductivity, permeability was tested in the range of values from μ = 2 up to μ = 8 to find out required value to achieve 2 db Shielding Effectiveness Spec. 2 db 1 KHz requires: Thickness Permeability μ 1 um ~12, 15 um ~7, 2 um ~4, 2, 4, 6, 8, Permeability 16

17 Skin Depth [um] Skin Depth vs. Frequency δ = 6685μm δ = 2898μm δ = 11863μm δ = 669μm Copper LC Steel Permalloy skin depth 2 2 f δ = 1726μm δ = 3751μm δ = 29μm 1 3 δ = 1186μm δ = 66.9μm 11 LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] δ = 545.9μm δ = 172.6μm δ = 375.1μm δ = 54.59μm δ = 29μm δ = 118.6μm δ = 66.9μm δ = 37.51μm δ = 17.26μm δ = 11.86μm COPPER μ =1 σ = 54 x 1 6 [S/m] PERMALLOY μ =5 σ = 1.7 x 1 6 [S/m] δ = 5.46μm δ = 1.73μm 17

18 Parallel Loop - Bannister Approximation Two quasi-near approximations are introduced: 1. When the measurement distance is much smaller than the operating wavelength (L λ ), the propagation constant in air can be neglected 2. When the measurement distance is much greater than the skin depth in the shield (L δ) and the shield is thicker than twice the skin depth, the integration variable λ can be neglected Bannister Approximation in the low-frequency case [1]: =.5mm Shield Thickness δ = With the following assumptions: L λ L δ /δ >.5 L/(δμ r ) > 1 2 2πfμσ Skin Depth SE db = δ + 2log 1 Absorption A term in the TL theory of planar shield z = 1m Distance from the loop center to field probe R = 2cm Loop radius L = R 2 + z 2 Measurement distance L 8.485μ r δ SE db = A + R L z L R 2+ z 2 Reflection coefficient R term in the TL theory of planar shield R reflection-loss term, due to the mismatch between the two impedances at both interfaces. A absorption-loss term, a function of shield characteristics. 3 (1) 18

19 A and R terms, [db] SE, [db] Bannister Approximation - >1Hz COPPER μ =1 σ = 54 x 1 6 [S/m] 11 LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] PERMALLOY μ =5 σ = 1.7 x 1 6 [S/m] 15 1 Copper A term Copper R term LCSteel A term LCSteel R term Permalloy A term Permalloy R term 15 1 Copper - eq. 1 LCSteel - eq. 1 Permalloy - eq. 1 Copper - Simulation LCSteel - Simulation Permalloy - Simulation A and R terms according to the eq. (1) If the shield is thinner compare to the skin depth, multiple reflections occurs between boundaries, because of the small absorption loss [**] For Copper restriction /δ >.5 is not fulfilled. For Permalloy restriction L/(δμ r ) > 1 is not fulfilled. Bannister Approximation is good as long as the quasi-near approximation restrictions are fulfilled, for frequencies >1Hz. [*] S. Celozzi, R. Araneo, G. Lovat, Electromagnetic Shielding, John Wiley & Sons, Inc., 28, Fig. B.12, page 36 [**] H. W. Ott, Electromagnetic Compatibility Engineering, John Wiley & Sons, Inc., 29, Chapter 6, page

20 A and R terms, [db] SE, [db] TL theory Approximation [7] Steel, Permalloy With the additional assumptions: K > 1 and μ r 1 SE db Δ δ + 2log 1.354K K +.48 Absorption A term SE db = A + R For materials with μ r 1, expression (2) is not valid [7]. Reflection coefficient R term (2) K = z δμ r COPPER μ =1 σ = 54 x 16 [S/m] 11 LOW CARBON STEEL μ =2 σ = 9 x 16 [S/m] PERMALLOY μ =5 σ = 1.7 x 16 [S/m] A - Copper R - Copper A - Steel R - Steel A - Permaloy R - Permaloy Copper - eq.2 Steel - eq.2 Permalloy - eq.2 Copper - Simulation Steel - Simulation Permalloy - Simulation

21 A and R terms, [db] SE, [db] TL theory Approximation [7] Copper, Steel With the following assumptions: L λ, L > 1δ, > 2δ, z SE db Δ δ + 2log 1 1 R 2 + z δμ r z (3) Absorption A term SE db = A + R Reflection loss R term In case of a negative value of reflection loss, R= is manually defined [3]. 11 LOW CARBON STEEL μ =2 σ = 9 x 1 6 [S/m] COPPER μ =1 σ = 54 x 1 6 [S/m] PERMALLOY μ =5 σ = 1.7 x 1 6 [S/m] A - Copper R - Copper A - Steel R - Steel A - Permaloy R - Permaloy Copper - eq.3 Steel - eq.3 Permalloy - eq.3 Copper - Simulation Steel - Simulation Permalloy - Simulation

22 Attenuation, [db] Attenuation, [db] Attenuation, [db] Measurement vs. Simulation Graph represents the measured low-frequency magnetic field shielding effectiveness of various type of metallic sheets [3]. The measurements were made in the near field with the source and receptor.1in apart COPPER STEEL ALUMINUM FREQUENCY = 1 KHz MUMETAL MUMETAL MUMETAL FREQUENCY = 1 KHz FREQUENCY = 1 KHz STEEL COPPER STEEL COPPER ALUMINUM ALUMINUM Measured data At 1KHz, 1KHz, 1KHz Aluminum 3 Copper Steel Mumetal 25 2 Material Relative Conductivity σ r Dash lines Simulation results Copper - 1KHz Copper - 1KHz Steel - 1KHz Relative permeability μ r Copper Steel.2 5 Aluminum Mumetal With 5.6 respect.8 to Copper Thickness [mm] Data is acquired from [3] Thickness [mm] H. W. Ott, Electromagnetic Compatibility Engineering, John Wiley.2 & Sons,.4 Inc.,.6 29,.8 ISBN , Chapter 6. Thicknes 22

23 Absorption Loss, [db] Absorption Loss, [db] Absorption Loss Term Skin depth of copper, [m]: δ c = A = Δ δ 1 πfμ μ c σ Cu Skin depth of arbitrary material, [m]: 1 δ m = σ πfμ μ m r σ σcu Cu δ m = μ = 4π 1 7 Permeability of free space μ c = 1 Relative permeability of copper σ Cu = Conductivity of copper 1 πfμ μ r σ rel,cu = A m = 128 μ r Relative permeability of material with respect to copper σ m = σ σ rel,cu Relative conductivity of material 18 with respect to copper Cu A c = πμ σ Cu fμ r σ rel,cu = fμ r σ rel, Cu Provided figure shows the advantage of steel over copper in providing absorption loss. Also thin sheet of copper has no significant loss below 1KHz. f fμ r σ rel,cu COPPER μ r =1 σ Cu = 54 x 1 6 [S/m] 11 LOW CARBON STEEL μ r =2 σ rel,cu = Copper - = 3mm Steel - = 3mm Copper - =.5mm Steel - =.5mm 23

24 A and R terms, [db] SE, [db] Magnetic Field Reflection Loss The reflection loss, R term, for magnetic field (according to eq. 3 copper and steel): 1 R 2 + z 2 R = 2log δμ r z δ m = 1 πfμ μ r σ rel,cu σ Cu COPPER μ r =1 σ Cu = 54 x 1 6 [S/m] 11 LOW CARBON STEEL μ r =2 σ rel,cu =.17 R m = 2log 1 πμ σ Cu fμ r σ rel,cu R 2 + z 2 + 2log 1 μ r z = log 1 f σ rel,cu R 2 + z 2 μ r z PERMALLOY μ r =5 σ rel,cu =.3 SE = A m + R m = 132 fμ r σ rel,cu log 1 f σ rel,cu R 2 +z 2 μ r z A - Copper R - Copper A - Steel R - Steel A - Permaloy R - Permaloy Copper - eq.3 Steel - eq.3 Permalloy - eq.3 Copper - Simulation Steel - Simulation Permalloy - Simulation

25 A and R terms, [db] SE, [db] Magnetic Field Reflection Loss The reflection loss, R term, for magnetic field (according to eq. 2 steel, permalloy): R = 2log δμ r z z δμ r +.48 δ m = 1 πfμ μ r σ rel,cu σ Cu COPPER μ r =1 σ Cu = 54 x 1 6 [S/m] 11 LOW CARBON STEEL μ r =2 σ rel,cu =.17 R m = 2log z f μ r σ rel,cu z f σ rel,cu μ r +.48 PERMALLOY μ r =5 σ rel,cu =.3 SE = A m + R m = 132 fμ r σ rel,cu +2log z f μ r σ rel,cu z f σ rel,cu μ r A - Copper R - Copper A - Steel R - Steel A - Permaloy R - Permaloy Copper - eq.2 Steel - eq.2 Permalloy - eq.2 Copper - Simulation Steel - Simulation Permalloy - Simulation

26 Introduction Consider an infinitely long cylindrical shell with inner radius α, outer radius b and wall thickness (i.e., = b - α ). The shell is placed in uniform ac magnetic field of amplitude H. The infinitely long cylindrical magnetic conducting shield has been studied as a canonical geometry for the design of EM shields. The shield consists of an infinitely long cylindrical shell with radius ρ = 3cm and thickness =.15mm, with large value of the conductivity σ, and/or of the relative magnetic permeability μ r [*]. Following figures show geometry of the problem for cylindrical shell placed in an uniform external transverse or parallel magnetic field. Benchmark problem 1 Cylindrical shell placed in an uniform external parallel magnetic field Benchmark problem 2 Cylindrical shell placed in an uniform external transverse magnetic field [*] S. Celozzi, R. Araneo, G. Lovat, Electromagnetic Shielding, John Wiley & Sons, Inc., 28, ISBN , pages

27 Geometry and Materials Benchmark problem 1 Cylindrical shell placed in uniform external parallel magnetic field IRON-NICKEL ALLOY IN Alloy μ =75 σ = 2 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 1.5mm Benchmark problem 2 Cylindrical shell placed in uniform external transverse magnetic field DURANICKEL STAINLESS STEEL DS Steel μ =1.58 σ = 2.35 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 2mm COPPER CASTING ALLOY CC Alloy μ =1.9 σ = 1.18 x 1 7 [S/m] Radius - ρ = 3cm Thickness - = 2mm 27

28 Model of Parallel Magnetic Field Parameters of cylinder: Length: L = 2 ρ =6cm Radius: ρ = 3cm Parameters of coil: Length: L = 2 ρ =6cm Radius: R = 7ρ = 21cm Number of turns: 2 Number of triangles: ~14 Y X Z Validation of Coil Approach Field probes across the cylinder for monitoring H field. Observation point in the center at (,,) Total H Field (Magnitude), [A/m] at 1KHz (Linear Scale) 28

29 Parallel Magnetic Field IRON-NICKEL ALLOY IN Alloy μ =75 σ = 2 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 1.5mm 2 15 Simulation Exact Simulation Exact 1 X: 8 Y: X: 8.36 Y:

30 Parallel Magnetic Field Benchmark problem 2 Cylindrical shell placed in an uniform external transverse magnetic field IRON-NICKEL ALLOY IN Alloy μ =75 σ = 2 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 1.5mm DURANICKEL STAINLESS STEEL DS Steel μ =1.58 σ = 2.35 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 2mm COPPER CASTING ALLOY CC Alloy μ =1.9 σ = 1.18 x 1 7 [S/m] Radius - ρ = 3cm Thickness - = 2mm Spherical shell placed in an uniform external transverse magnetic field 3

31 Validation of Spherical and Cylindrical Shells Equivalently Data acquired from [*]. According to the graph we can conclude that even as shield geometry changes, the shielding mechanisms remain always the same. So we can place Spherical shell instead of the Cylindrical in an uniform external transverse magnetic field. IRON-NICKEL ALLOY IN Alloy μ =75 σ = 2 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 1.5mm Cylindrical Spherical Number of triangles: ~ Z Y X Parameters of sphere: Radius: ρ = 3cm Parameters of coil: Length: L = 2 ρ =6cm Radius: R = 7ρ = 21cm Number of turns: 2 [*] S. Celozzi, R. Araneo, G. Lovat, Electromagnetic Shielding, John Wiley & Sons, Inc., 28, ISBN , pages 294,

32 Transverse Magnetic Field Exact Simulation Iron-Nickel Alloy COPPER CASTING ALLOY CC Alloy μ =1.9 σ = 1.18 x 1 7 [S/m] Radius - ρ = 3cm Thickness - = 2mm Exact Simulation Duranickel Stainless Steel Exact Simulation Copper Casting Alloy IRON-NICKEL ALLOY IN Alloy μ =75 σ = 2 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 1.5mm DURANICKEL STAINLESS STEEL DS Steel μ =1.58 σ = 2.35 x 1 6 [S/m] Radius - ρ = 3cm Thickness - = 2mm 32

33 MIL STD Test Setup Antenna diameter is 3 cm (12 inches) Antenna position is 1.5 m from the exterior wall, and 1. m inside the interior wall Antenna locations are shown as #1, 2, 33

34 MIL STD : Model View 2.5 m 12.5 m Copper wall thickness: 2 m Current Loop Diameter: 3cm Value: 1A Field Probes Step: 1cm Side View 8 m S S Loop parallel to the wall Loop perpendicular to the wall 3D View 34

35 Shielding Effectiveness at Various Distances to Wall 45 Shielding 1 KHz - PARALLEL 45 Shielding 1 KHz - PERPENDICULAR cm cm 25 1 cm 15 cm cm 125 cm 1 cm 5 cm cm 35 cm 3 cm 5 cm cm cm 3 cm 2 cm 25 cm 1 15 cm cm Location [m] Location [m] Parallel Loop F = 1 KHz Perpendicular Loop COPPER μ =1 σ = 58 x 1 6 [S/m] = 2um 35

36 Simulation Shielding 1 KHz MIL STD distance Antenna distance from the wall (cm) CENTER FIELD PROBE F = 1 KHz COPPER μ =1 σ = 58 x 1 6 [S/m] = 2um 36

37 Simulation Magnetic 1 KHz Side View Room Field Probes Excitation Loop Top View 3D View 37

38 Magnetic Field [dba/m] Simulation Magnetic 1 KHz Parallel Loop Field Probes along the loop central axis with 1 cm step 1A Current Loop 3 cm diameter 1.5 m 6 m 4 2 Parallel Loop :: 1 KHz Free Space With the Shielded Room Distance from the loop [m] 38

39 Magnetic Field [dba/m] Simulation Magnetic 1 KHz Parallel Loop Field Probes in plane of the loop with 1 cm step 1A Current Loop 3 cm diameter 1.5 m 6 m 4 2 Perpendicular Loop :: 1 KHz Free Space With the Shielded Room H 1 r Distance from the loop [m] 39

40 References 1. S. Celozzi, R. Araneo, G. Lovat, Electromagnetic Shielding, John Wiley & Sons, Inc., 28, ISBN , Chapter 4 and Appendix B 2. C. R. Paul, Introduction to Electromagnetic Compatibility, John Wiley & Sons, Inc., 26, Second Edition, ISBN , pages H. W. Ott, Electromagnetic Compatibility Engineering, John Wiley & Sons, Inc., 29, ISBN , Chapter 6 4. EMCoS EMC Studio 7., Low Frequency Magnetic Field solver 5. R. Jobava, A. Gheonjian, D. Karkashadze, J. Hippeli, Interaction of Low Frequency Magnetic Fields with Thin 3D Sheets of Combined Resistive and Magnetic Properties, Proceedings of the 4th European Microwave Conference, EuMW 21, Paris, France 6. F. Bogdanov, R. Jobava, A. Gheonjian, K. Khasaia, Application of Loop-Star and Loop-Tree Basis Functions to MoM Solution of Radiation and Scattering Problems on Complicated Surface and Wire Geometries From Low to Microwave Frequencies, 6th European Conference on Antennas and Propagation, EUCAP 211, Rome, Italy 7. P. Bannister. New theoretical expressions for predicting shielding effectiveness for the plane shield case, IEEE Trans. Electromagnetic Compatibility, vol. 1, no. 1, pp. 2-7, Mar