### Citations

510 |
Time-Harmonic Electromagnetic Fields
- Harrington
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Citation Context ...ic Maxwell equations have a unique solution if and only if the region is filled with lossy medium, while the time-domain Maxwell equations always have a unique solution even if the medium is lossless =-=[9, 10]-=-. Therefore introducing losses in the metal cavity is required by the time-harmonic electromagnetic theory, which guarantees that the solution is unique and has no singularities. Therefore the time-do... |

466 |
Integral Equation Methods in Scattering Theory
- Colton, Kress
- 1983
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Citation Context ...ic Maxwell equations have a unique solution if and only if the region is filled with lossy medium, while the time-domain Maxwell equations always have a unique solution even if the medium is lossless =-=[9, 10]-=-. Therefore introducing losses in the metal cavity is required by the time-harmonic electromagnetic theory, which guarantees that the solution is unique and has no singularities. Therefore the time-do... |

41 |
Foundations for Microwave Engineering. 2nd edition,
- Collin
- 1992
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Citation Context ...l vector functions. When these expansions are introduced into the timedomain Maxwell equations one may find that the expansion coefficients satisfy the ordinary differential equations of second order =-=[3]-=-, which can be easily solved once the initial conditions and the excitations are known. Recently this approach has been used to investigate the responses of the metal cavity to digital signals [4]. Al... |

13 |
An Introduction to the Theory of Microwave Circuits
- Kurokawa
- 1969
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Citation Context ...a perfectly conducting metal cavity filled with lossy medium. A fundamental problem in metal cavity theory is to prove the completeness of its vector modal functions, which has been tried by Kurokawa =-=[1]-=-. But there is a loophole in Kurokawa’s approach in which he fails to show the existence of the vector modal functions. The main purpose of Section 2 is to provide a rigorous proof of the completeness... |

8 | A time-domain theory of waveguides
- Geyi
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Citation Context ...d to a closed metal cavity. Section 4 is dedicated to a metal cavity formed by a section of uniform waveguide, i.e., the waveguide cavity. The time-domain theory of the waveguide developed previously =-=[6]-=- is applicable to this case, and the fields inside the waveguide cavity can be expanded in terms of the transverse vector modal functions of the corresponding waveguide. The expansion coefficients of ... |

7 | Analysis of Transmission of a Signal Through a Complex Cylindrical/Coaxial Cavity by Transmission Line Methods
- Bopp, Butler
- 2006
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Citation Context ...to category II in the two sets of modal functions {en} and {hn} are related to each other. Actually let en belong to category II. Then ke,n ̸= 0 and one can define a function hn through ∇×en = ke,nhn =-=(13)-=- Therefore hn belongs to category II. Furthermore and ∇×∇×h − k 2 e,nh = k −1 e,n∇× ( ∇×∇×e − k 2 e,ne ) =0, r ∈ V un ×∇×hn = k −1 e,nun ×∇×∇×en = k −1 e,nun × k 2 e,nen =0, r ∈ S Consider the integra... |

5 |
Radiation from multiple annular slots on a circular cylindrical cavity
- Kim, Eom
- 2007
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Citation Context ...(t)hτ (r) n ∇×E(r,t) = ∑ τ ∫ hn(r) n V ∫ + ∑ hτ (r) τ τ v ∇×E(r,t) · hn(r)dv V ∇×E(r,t) · hτ (r)dv ∇×H(r,t) = ∑ ∫ en(r) ∇×H(r,t) · en(r)dv n V + ∑ ∫ ev(r) ∇×H(r,t) · ev(r)dv v V V V H(r,t) · hτ (r)dv =-=(16)-=- (17) where the subscript n denotes the modes belonging to category II, and the Greek subscript v and τ for the modes belonging to category I or228 Geyi III, and V n(v)(t) = I n(τ)(t) = ∫ V∫ V E(r,t)... |

4 |
The evolution equations in study of the cavity oscillations excited by a digital signal,”
- Aksoy, Tretyakov
- 2004
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Citation Context ...der [3], which can be easily solved once the initial conditions and the excitations are known. Recently this approach has been used to investigate the responses of the metal cavity to digital signals =-=[4]-=-. Although the study of metal cavity resonator has a long history, there are still some open questions which need to be investigated. This paper attempts to answer these questions, and presents a thor... |

3 |
Numerical solution on coupling of UWB pulse into a rectangular cavity through slots
- Yang, Liao, et al.
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Citation Context ...he cavity contains an impressed electric current source J and a magnetic current source J m, the fields excited by these sources satisfy ∇×H(r,t)=ε ∂E(r,t) + σE + J(r,t), r ∈ V ∂t ∇×E(r,t)=−µ ∂H(r,t) =-=(15)-=- − J m(r,t), r ∈ V ∂t and can be expanded in terms of the vector modal functions as E(r,t)= ∑ ∫ en(r) E(r,t) · en(r)dv + ∑ ∫ ev(r) E(r,t) · ev(r)dv n n V = ∑ Vn(t)en(r)+ n ∑ Vv(t)ev(r) v H(r,t)= ∑ ∫ h... |

2 |
Applied Functional Analysis-Applications to
- Zeidler
- 1995
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Citation Context ...0, r ∈ S { ∇×∇×h − k 2 h h =0, ∇·h =0, r ∈ V un · h =0, un ×∇×h =0, r ∈ S (4) (5) The functions u(t) and v(t) satisfy 1 c2 ∂2u η ∂u + σ ∂t2 c ∂t + k2 eu = 0 (6) 1 c2 ∂2v η ∂v + σ ∂t2 c ∂t + k2 hv = 0 =-=(7)-=- where k 2 e and k 2 h are separation constants, η = √ µ/ε and c = 1/ √ µε. Both (4) and (5) form an eigenvalue problem. However theirProgress In Electromagnetics Research, PIER 78, 2008 223 eigenfun... |

2 |
Abo El-Hadeed, “Analysis of microwave cavities using LTL-FD method
- Mohsen, Elkaramany, et al.
- 2005
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Citation Context ...+(ke,n/ke,m) V em · endv Therefore the eigenfunctions hn in category II can be derived from the eigenfunction en in category II. Conversely if hn is in category II, one can define en by ∇×hn = kh,nen =-=(14)-=-Progress In Electromagnetics Research, PIER 78, 2008 227 and a similar discussion shows that en is an eigenfunction of (4) with kh,n being the eigenvalue. So the completeness of the two sets are stil... |

2 |
A field theoretical method for analyzing microwave cavity with arbitrary crosssection
- Xiao, Ji, et al.
- 2006
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Citation Context ... (r) n ∇×E(r,t) = ∑ τ ∫ hn(r) n V ∫ + ∑ hτ (r) τ τ v ∇×E(r,t) · hn(r)dv V ∇×E(r,t) · hτ (r)dv ∇×H(r,t) = ∑ ∫ en(r) ∇×H(r,t) · en(r)dv n V + ∑ ∫ ev(r) ∇×H(r,t) · ev(r)dv v V V V H(r,t) · hτ (r)dv (16) =-=(17)-=- where the subscript n denotes the modes belonging to category II, and the Greek subscript v and τ for the modes belonging to category I or228 Geyi III, and V n(v)(t) = I n(τ)(t) = ∫ V∫ V E(r,t) · e ... |

2 |
Frequency domain analysis of waveguides and resonators with FIT on non-orthogonal triangular grids,”
- Rienen
- 2001
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Citation Context ... Vn − kn ε In = − 1 ε ∂Vv ∂t ∂In ∂t σ + ε Vv = − 1 ε kn + µ Vn = − 1 µ ∂Iτ ∂t 1 = − µ n knInen (18) ∑ + σ enVn + σ ∑ evVv + J n ∂Iτ ∂t − J m ∫ V ∫ V ∫ V ∫ V J · endv J · evdv J m · hndv J m · hτ dv v =-=(19)-=-Progress In Electromagnetics Research, PIER 78, 2008 229 From the above equations one may obtain ∂ 2 In ∂t 2 ∂ 2 Vn ∂t 2 where ωn = knc, γ = σ/2ε and S I ∫ n = c S V n ∂In +2γ ∂t + ω2 nIn = ωnSI n ∂V... |

2 |
3-dimensional implementation of the field iterative method for cavity modeling
- Wang, Xu, et al.
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Citation Context ...c, γ = σ/2ε and S I ∫ n = c S V n ∂In +2γ ∂t + ω2 nIn = ωnSI n ∂Vn +2γ ∂t + ω2 nVn = ωnSV n J · endv − V 1 ∫ ∂ J m · hndv − knη ∂t V σc ∫ J m · hndv kn V = − η ∫ ∫ ∂ J · endv − c J m · hndv ∂t kn V V =-=(20)-=- To find In and Vn, one may use the retarded Green’s function defined by ⎧ ⎪⎨ ∂ ⎪⎩ 2Gn(t, t ′) ∂t2 +2γ ∂Gn(t, t ′) + ω ∂t 2 nGn(t, t ′ )=−δ(t − t ′ ) Gn(t, t ′ ) ∣ t<t ′ =0 (21) It is easy to show tha... |

1 |
Electromagnetic Fields, 1st edition, Hemisphere Publishing Corporation
- Bladel
- 1985
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Citation Context ...lso be seen from (31) that In(t) becomes singular when ω approaches ωn, which implies that the fields are infinite everywhere232 Geyi inside the cavity. This phenomenon is discussed in Bladel’s book =-=[5]-=- and is compared to a lossless resonant LC network. In Collin’s book [3], these singularities do not occur because of the introduction of losses in the metal cavity. However, the time-domain solution ... |

1 |
Mode matching analysis of the induced electric field in a material sample placed within an energized cylindrical cavity
- Zhang, Chen
(Show Context)
Citation Context ...∑ knInen = ε n ∑ n ∑ n Thus knVnhn = −µ ∑ ∂Vn ∂t ∂Vn en ∂t n ∂In hn ∂t + ε ∑ v − µ ∑ ∂Vv ev ∂t τ hτ σ + ε Vn − kn ε In = − 1 ε ∂Vv ∂t ∂In ∂t σ + ε Vv = − 1 ε kn + µ Vn = − 1 µ ∂Iτ ∂t 1 = − µ n knInen =-=(18)-=- ∑ + σ enVn + σ ∑ evVv + J n ∂Iτ ∂t − J m ∫ V ∫ V ∫ V ∫ V J · endv J · evdv J m · hndv J m · hτ dv v (19)Progress In Electromagnetics Research, PIER 78, 2008 229 From the above equations one may obta... |