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Prof. David R. Jackson ECE Dept.

ECE 6341. Spring 2014. Prof. David R. Jackson ECE Dept. Notes 43. Radiated Power from Patch. Patch. Lossless substrate and metal. Radiated Power (cont.). Use Parseval’s Theorem:. Radiated Power (cont.). Hence. From SDI analysis,. where. Radiated Power (cont.). We then have.

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Prof. David R. Jackson ECE Dept.

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  1. ECE 6341 Spring 2014 Prof. David R. Jackson ECE Dept. Notes 43

  2. Radiated Power from Patch Patch Lossless substrate and metal

  3. Radiated Power (cont.) Use Parseval’s Theorem:

  4. Radiated Power (cont.) Hence From SDI analysis, where

  5. Radiated Power (cont.) We then have Using symmetry, Polar coordinates:

  6. Radiated Power (cont.) Note that Hence Note: is analytic but is not. This is a preferable form!

  7. Radiated Power (cont.) TM0 pole Branch point For realkt we have (proof on next page): Hence, we can neglect the region kt>k0, except possibly for the pole.

  8. Radiated Power (cont.) Proof of complex property: always imaginary Consider the following term (De is similar):

  9. Space-Wave Power • The region gives the power radiated into space. • The residue contribution gives the power launched into the TM0 surface wave.

  10. Surface-Wave Power The pole contribution gives the surface-wave power: From the residue theorem, we have

  11. Surface-Wave Power (cont.) The residue of the spectral-domain Green’s function at the TM pole is Note: This can be calculated in closed form, but the result is omitted here. where

  12. Surface-Wave Power (cont.) Since the transform of the current is real (assuming that ktp is real), we have Since the residue is pure imaginary, we then have

  13. Total Power The total power may also be calculated directly: Note: hb=0.05k0 is a good choice for numerical purposes. hb

  14. Surface-Wave Power: Alternative Method The surface-wave power may then be calculated from: This avoids calculating any residues. Total hb Space

  15. Results Results: Conductor and dielectric losses are neglected 2.2 10.8 r = 2.2 or 10.8 W/L = 1.5

  16. Results Results: Accounting for all losses (including conductor and dielectric loss) r = 2.2 or 10.8 W/L = 1.5

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