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ENE 428 Microwave Engineering

ENE 428 Microwave Engineering. Lecture 2 Uniform plane waves. Propagation in lossless-charge free media. Attenuation constant  = 0, conductivity  = 0 Propagation constant Propagation velocity for free space u p = 310 8 m/s (speed of light)

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ENE 428 Microwave Engineering

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  1. ENE 428Microwave Engineering Lecture 2 Uniform plane waves RS

  2. Propagation in lossless-charge free media • Attenuation constant  = 0, conductivity  = 0 • Propagation constant • Propagation velocity • for free space up = 3108 m/s (speed of light) • for non-magnetic lossless dielectric (r = 1), RS

  3. Propagation in lossless-charge free media • intrinsic impedance • wavelength RS

  4. Ex1 A 9.375 GHz uniform plane wave is propagating in polyethelene (r = 2.26). If the amplitude of the electric field intensity is 500 V/m and the material is assumed to be lossless, find a) phase constant b) wavelength in the polyethelene RS

  5. c) propagation velocity d) Intrinsic impedance e) Amplitude of the magnetic field intensity RS

  6. Propagation in dielectrics • Cause • finite conductivity • polarization loss ( = ’-j” ) • Assume homogeneous and isotropic medium RS

  7. Propagation in dielectrics Define From and RS

  8. Propagation in dielectrics We can derive and RS

  9. Loss tangent • A standard measure of lossiness, used to classify a material as a good dielectric or a good conductor RS

  10. Low loss material or a good dielectric (tan « 1) • If or < 0.1 , consider the material ‘low loss’, then and RS

  11. Low loss material or a good dielectric (tan « 1) • propagation velocity • wavelength RS

  12. High loss material or a good conductor (tan » 1) • In this case or > 10, we can approximate therefore and RS

  13. High loss material or a good conductor (tan » 1) • depth of penetration or skin depth,  is a distance where the field decreases to e-1or 0.368 times of the initial field • propagation velocity • wavelength RS

  14. Ex2 Given a nonmagnetic material having r= 3.2 and  = 1.510-4 S/m, at f = 3 MHz, find a) loss tangent  b) attenuation constant  RS

  15. c) phase constant  d)intrinsic impedance RS

  16. Ex3 Calculate the followings for the wave with the frequency f = 60 Hz propagating in a copper with the conductivity,  = 5.8107 S/m: a) wavelength b) propagation velocity RS

  17. c) compare these answers with the same wave propagating in a free space RS

  18. Attenuation constant  • Attenuation constant determines the penetration of the wave into a medium • Attenuation constant are different for different applications • The penetration depth or skin depth,  is the distance z that causes to reduce to z = 1  z = 1/  =  RS

  19. Good conductor • At high operation frequency, skin depth decreases • A magnetic material is not suitable for signal carrier • A high conductivity material has low skin depth RS

  20. Currents in conductor • To understand a concept of sheet resistance from Rsheet() sheet resistance At high frequency, it will be adapted to skin effect resistance RS

  21. Currents in conductor Therefore the current that flows through the slab at t   is RS

  22. Currents in conductor From Jxor current density decreases as the slab gets thicker RS

  23. Currents in conductor For distance L in x-direction Ris called skin resistance Rskinis called skin-effect resistance For finite thickness, RS

  24. Currents in conductor Current is confined within a skin depth of the coaxial cable RS

  25. Ex A steel pipe is constructed of a material for which r = 180 and  = 4106 S/m. The two radii are 5 and 7 mm, and the length is 75 m. If the total current I(t) carried by the pipe is 8cost A, where  = 1200 rad/s, find: • The skin depth • The skin resistance RS

  26. c) The dc resistance RS

  27. The Poynting theorem and power transmission Poynting theorem Total power leaving the surface Joule’s law for instantaneous power dissipated per volume (dissi- pated by heat) Rate of change of energy stored In the fields Instantaneous poynting vector RS

  28. Example of Poynting theorem in DC case Rate of change of energy stored In the fields = 0 RS

  29. Example of Poynting theorem in DC case From By using Ohm’s law, RS

  30. Example of Poynting theorem in DC case Verify with From Ampère’s circuital law, RS

  31. Example of Poynting theorem in DC case Total power W RS

  32. Uniform plane wave (UPW) power transmission • Time-averaged power density W/m2 amount of power for lossless case, W RS

  33. Uniform plane wave (UPW) power transmission for lossy medium, we can write intrinsic impedance for lossy medium RS

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