Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

HFSS boundary condition

Status
Not open for further replies.

guo-hai

Junior Member level 1
Joined
May 21, 2011
Messages
19
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,412
Hello,
Is there anyone know what is the difference of perfect E, perfect H, and finite conductivity? And when we must use it? or say which is more accurate ?
THanks
 

Check the ansoft Help_antenna pdf... it has what you're looking for..

here's a copy paste from the text contained inside:

Boundary Conditions
Boundaries specify the behavior of magnetic and electric fields at various
surfaces. They can also be used to identify special surfaces —such as
resistors— whose characteristics differ from the default.
The following four types of boundary conditions will be used for this
antenna problem:
Radiation This type of boundary simulates an open problem that allows
waves to radiate infinitely far into space, such as antenna
designs. HFSS absorbs the wave at the radiation boundary,
essentially ballooning the boundary infinitely far away from the
structure. In this antenna model, the air volume object is
defined as a radiation boundary.
Perfect E This type of boundary models a perfectly conducting surface in
a structure, which forces the electric filed to be normal to the
surface. In this antenna model, the bottom face of the air
volume object is defined as a perfect E boundary.
Perfect H This type of boundary forces the tangential component of the
H-field to be the same on both sides of the boundary. In this
antenna model, the annular feed ring is the aperture that is
assigned this boundary. Because the aperture is defined as a
perfect H boundary, the E-fields will radiate through it. If it
was not defined as a perfect H boundary, the E-field would not
radiate through and the signal would terminate at the
aperture.
 

    JeongWon

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top