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

2. RS 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) – for non-magnetic lossless dielectric (  r = 1),

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

4. RS 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  = 295 rad/m = 2.13 cm

5. RS c) propagation velocity d) Intrinsic impedance e) Amplitude of the magnetic field intensity v = 2x10 8 m/s  = 250.77  H = 1.99 A/m

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

7. RS Propagation in dielectrics From and

8. RS Propagation in dielectrics We can derive and

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

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

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

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

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

14. RS Ex2 Given a nonmagnetic material having  r = 3.2 and  = 1.5  10 -4 S/m, at f = 30 MHz, find a) loss tangent  b) attenuation constant  tan  = 0.03  = 0.016 Np/m

15. RS c) phase constant  d) intrinsic impedance  = 1.12 rad/m  = 210.74(1+j0.015) 

16. RS Ex3 Calculate the followings for the wave with the frequency f = 60 Hz propagating in a copper with the conductivity,  = 5.8  10 7 S/m: a) wavelength b) propagation velocity  = 117.21 rad/m = 5.36 cm v = 3.22 m/s

17. RS c) compare these answers with the same wave propagating in a free space  = 1.26x10 -6 rad/m = 5000 km v = 3x10 8 m/s

18. RS 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/  = 

19. RS 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

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

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

22. RS Currents in conductor From J x or current density decreases as the slab gets thicker

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

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

25. RS Ex4 A steel pipe is constructed of a material for which  r = 180 and  = 4  10 6 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  = 7.66x10 -4 m

26. RS c) The dc resistance

27. RS 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

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

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

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

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

32. RS Uniform plane wave (UPW) power transmission Time-averaged power density amount of power for lossless case, W/m 2

33. RS Uniform plane wave (UPW) power transmission for lossless case, we can write power average in terms of H W/m 2

34. Uniform plane wave (UPW) power transmission intrinsic impedance for lossy medium for lossy medium, we can write The phasor is Instantaneous magnetic field: which is seen to be out of phase with E(z,t) by the angle  

35. RS Homework Prob.6.19: In seawater, a propagating electric field is given by V/m Assuming  = 5 S/m,  r ’ = 72,  r ” = 0, find (a)  and β and (b) the instantaneous form of H(z,t) Prob. 6.26: In air,A/m Determine the power passing through a 1.0 m 2 surface that is normal to the direction of propagation.