EE468G NOTES (7B) Contents: Power density and Poynting Vector Reflection and transmissionPoyinting theorem (Power conservation in a closed system): The net power flowing into a closed, source-free system (V) is equal to the sum of the dissipated power and the increasing in energy storage. This theorem can be expressed in time domain as: VVSdVHEtdVJESdHE222121 This equation states that the total power passing through S into V is equal to two parts: one part is the total power dissipation in V, and the other part is the increase in energy storage. In the above, JE is the dissipated power density [3W/m] 221E is the electric field stored energy density 3[J/m ] 221H is the magnetic field stored energy density HE is the power flow (in) density. ,, S JComplex Poynting vector: ][W/m2*HES ( * stands for complex conjugate). Time average power flow through a surface:  SdHEPS*aveRe21 [W] The integrand: *ave1Re2p E H is the power density (2W/m) For UPW wave in lossless media, the above expression can be simplified. To this end, we note that zjzzeEaEaH000ˆ1ˆ1 Hence zzjzzjaEeEaeEHEˆ1ˆ1200*000* If the wave propagates in k-direction, we will have kaEHEˆ1200*, SdaEPSkˆ21200ave Here, kaˆis the direction of wave propagation. It is clear that if the power density is in the normal direction of S, then the integral is S, the area of the surface. S pExample: A plane wave in free-space is specified by zjxeaE4ˆ3 Calculate the time average power density and the time average power flow through a rectangular area that is specified by: 31,42  yx. Solution: The magnitude of the plane wave is 3, the power density is ]W/m[ˆ0119.0ˆ312021ˆ2122200ave zzzaaaEp  ][095.00119.0ˆˆ0119.031423142aveaveWdxdydxdyaaSdpPzzS  Example (12-7): A linearly polarized plane wave propagates through free-space at an angle  with the z-axis (as shown). If the peak amplitude of the wave is 10 [mV/m], calculate the average power that passes through the 1 [2m] surface on the z=0 plane. Solution:  ][W/mˆ106.132ˆ101012021ˆ212923200avekkkaaaEp In the above, kaˆ is the direction of propagation. The time average power that passes the specified area is ][cos106.132ˆˆ106.132ˆˆ106.132999aveaveWdxdyaadxdyaaSdpPSzkSzkS (It can be seen that the power flow through a surface is maximum when it propagates perpendicular to the surface). x y z  kaˆReview examples: Example 1: For a conductor with 007,],/[104 mS Calculate the intrinsic impedance , attenuation constant , the skin depth , and the wavelength, at the frequency of 100 Hz. Solution: At 100 Hz, 715124 10tan 71.9 10 12 100 8.854 10        ]/1[4010410410077mf  m][05.04022  m][008.01     ][10)77.177.1(10421041002121677jjjPlane wave Reflection and Transmission What will happen to a wave when it encounters a discontinuity? Answer: two waves will be induced, one is the reflected wave, and one is the transmitted wave. What is the transmitted wave and what is the reflected wave? (How do we determine the two waves?) Answer: Use boundary condition What you need to know in this topics: (1) What are the forms of the reflected wave and transmitted wave (2) How the reflection coefficient and transmission coefficient are derived? (3) Able to calculate reflection and transmission coefficients for single interface case. (4) Properties of reflection and transmission.Two regions, interface at z=0, incident wave from region 1. Boundary condition: 0zat ,,2121ttttHHEE At z=0 interface: TEREtt21,1 At z=0 interface:  122111,1THRHtt  12212121211211TRTRTR Summary: z 111,, :1Region  222,, :2Region  0z iiHE, rrHE, ttHE, x Incident wave: zjyizjxieaHeaE11ˆ,ˆ1 Reflected wave: zjyrzjxraHeRaE11Reˆ,ˆ11 Transmitted wave: zjytzjxtTeaHeTaE22ˆ,ˆ12R is the (electric field) reflection coefficient: irEER11 T is the (electric field) transmission coefficient: itEET12 Wave impedance for region 1: 1111jj Wave impedance for region 2: 2222jj 12212122, :incident Normal TR Example: A plane wave normal incident from air to glass. Assume the glass has 0,,420202. Calculate R and T. Solution: ][605.04],[ 12000020001 67.012060602,33.01206012060 TR We can also get T by: T=1+R=1-0.33=0.67Reflection of conducting surfaces: If the second region is perfect conductor (2), we have 02, hence Normal incident on conductor: 1, 0RT   In this case, an electric current will be induced on the surface of the conductor. The magnetic field in region 1: 11111 1 1垐Rej z j ziryyH H H a e a    Magnetic field in region 2: 0 Boundary condition for magnetic field intensity at z=0:    SyzSnJaaJHHa0ˆ2ˆˆ1121 In general, the induced electric current density on a conducting surface is given by: inSHaJˆ2 Note: This is for an infinitely large conducting interface. In practical applications, when the conducting surface is flat and is much large than a wavelength, this formula can be used to approximate the induced current. z 111,, :1Region  2 :2Region  0z iiHE, rrHE, 0,0 ttHEOblique incident Plane of incidence: the x-z is the