1 / 11

Chapter 9 Waveguides and Resonant Cavities optional content

2. In a conductor, we have If s?8, we must have E=0, else , which us unphysical. Therefore, E=0 inside perfect conductor. Also, the bulk charge density (free charge density) ?=0, because if there were any charge, there would be E?0, and hence J ? 0 implies that free charge i

jabari
Download Presentation

Chapter 9 Waveguides and Resonant Cavities optional content

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

    1. 1 Chapter 9 Waveguides and Resonant Cavities (optional content)

    2. 2

    3. 3

    4. 4

    5. 5

    6. 6 Modes in a Rectangular Waveguides (TE modes)

    7. 7 TE modes

    8. 8 Energy Flow in Waveguides Time-averaged energy density is Energy per unit length is therefore U=abu. The Poynting vector The z-component of flux is Power in z-direction is thus If we take the fields as given above and compute all the above quantities, we find that for the ?-th mode. Thus energy travels with the group velocity.

    9. 9 These are regions enclosed on all sides by walls. Then instead of propagating waves, we have discrete modes (standing waves). E.g. cylindrical cavity of length c. TE modes: But at the end of cavity we must have Bz=0 and therefore Hz=0. Therefore, must have a linear combination which vanishes at ends. This would be with freq. TM modes: For TM mode, we have to be more careful and explicitly calculate all the fields then Resonant Cavities

    10. 10 Similarly, if , we get Therefore, if , we get At z=0 and z=c, we need Ex=0 or k=pp/c and hence just as in TE case.

    11. 11 Cylindrical Cavity For TE modes and , since Hz=0 at ends of cavity. For TM modes where For cylinder, we use cylindrical co-ords: Let ?=Hz or Ez, This is Bessels equation. Solutions are Nm blows up at origin so allowed solutions are for Hz in TE modes and Ez in TM modes.

    12. 12 TM mode solution at ?=R, must have and ?mn denotes to n-th zero of Jm(x). The corresponding freq. is and TE mode solution Can also show that Therefore, we must have Its solution ?mn denotes to nth zero of Jm(x). . .

More Related