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ELECTROMAGNETIC WAVES Uniform plane waves (TEM) in lossless media Solutions – time-domain E r,t = aE E cos t - r = aE E cos t - a r E r,t = aE E cos t - x cos x + y cos y + z cosz Solutions – frequency-domain E r,t = Re E e-jt E(r, ) = aEE e-j r The electric and magnetic vectors perpendicular and both are perpendicular to the direction of propagation. aE aH = a 1 H = a E or use Faraday’s law E = H a or use Ampere’s law 2 = = v = f = (general) 1 v= (lossless only) all kinds of applets on transmission lines, waves etc. http://www.educypedia.be/electronics/javatransmissinlines.htm another excellent site http://www.amanogawa.com 2 Example: TEM wave propagation ☼ x 3y An E-field is given by E = a z 50cos 109 t - 5 + 2 V/m . ( = o) Find 2 i) wave number, wave vector, direction of travel ii) velocity iii) wavelength iv) wave (or intrinsic) impedance v) H (time-domain vector), H (frequency-domain vector) EM waves 3 Energy, conduction current, displacement current ☼ From the point form of Ohm’s law, J = E, two points can be appreciated: 1. The conduction current and the electric field vary alike. Changes in the electric field produce like variations in the conduction current. 2. The conduction current and the electric field are in the same direction. The result is that plots of E and J versus time are scaled versions of one another as shown below. From Joule’s law, dP =E J dv Given that E and J are in phase and parallel, the power dissipated per volume in conductive material is simply the product of the magnitudes of E and J. dP =E J dv EM waves 4 Displacement current density is the time rate-of-change of the electric flux density vector. The flux density can be broken into two pieces, the first being oE which is present even in the absence of material being present (it’s there when material is present too) and the second being P which is the polarization vector associated with the separation of bound change. The displacement current is associated with the time rate of change of these two components. D = E = r o E = oE + ( r - 1) oE D = oE + P D E P JD = = o + t t t Note that both components of the displacement current have the same phase and direction (true at least for ―low‖ frequencies—more on this on the next page). This allows us to frame the discussion of power absorbed by a dielectric in terms of the behavior of bound change with respect to changes in electric field. Below, a sinusoidal variation in E is assumed and the corresponding bound charge separation is tracked with respect to time and correlated to the associated displacement current. The power per unit volume absorbed by a dielectric is E JD EM waves 5 If the material under consideration have both displacement and conduction currents present, there will be both energy storage and average power dissipation. The net current will be between 0° and 90° out-of-phase with the electric field. This situation can also occur in dielectrics at high frequencies in which the time constants associated with the dynamics of the bound charge mass can no longer be neglected with respect to the time variation of the field. Displacement current at high-frequencies The result in either case, whether the material has both conduction and displacement current or whether the frequency is sufficiently high so that the displacement current is not in phase with the electric field, the result is that there is an average power lost from the field in the material. Conduction loss and dielectric loss both lead to the same result, which is that the propagation constant becomes complex, = + j . Conduction loss is accounted for by a non-zero conductivity, , in Ampere's law, and dielectric loss is accounted for by including an imaginary component in the permittivity ( - jp). EM waves 6 Waves in lossy material Consider Ampere and Faraday (with no charge or current sources). There are two basic source for loss – conductive loss and dielectric loss. The dielectric loss is readily incorporated by including an imaginary component in the permittivity [( - jp) used here—other notations are used, e.g. (' - j")] . It is important to note that this imaginary component describes how the bound charge separation lags the electric field excitation and leads to loss. Conductive loss is due to a non-zero conductivity . Note: no current sources DOES NOT mean there are no currents. In conductive materials (0), currents exist in response to an electric field in conductive material. E E H = J + = E + - j p t t H E = - t Combining these equations E E = g E - 2E = - H = - E + - j p t t t E 2E 2E - - 2 = 0 t t In the frequency domain, 2E - j E + 2 - j p E = 2E + 2 - j p - j E = 2E + 2 - j p - j E = 0 2E + 2 E = 0 where = - j p - j is the complex permittitivty In the frequency domain, solutions are E = aE E e - r + p where = a = j a = j - j p - j = j 1 + j = + j a Terminology complex propagation vector complex propagation constant attenuation constant (equal to zero for lossless materials) phase constant IMPORTANT NOTE: From this point, p will be neglected for simplicity so that = - j . EM waves 7 Details of solution The solution must satisfy the wave equation, 2E + 2 E = 0. E = aE E e - r E = aE E e j e - + j a r E = aE E e j e - + j x cos x + y cos y + z cosz 2 2 2 - + j x cos x + y cos y + z cosz 2 + + 2 E a E e j e x y z 2 - + j x cos x + y cos y + z cosz + 2 aE E e j e = 0. Each derivative pulls down - + j cos i from the exponential. The derivatives hear are second derivatives. - + j cos x 2 + - + j cos y + - + j cos z 2 2 E + E = 0 2 - + j cos 2 2 x + cos2 y + cos2 z E + 2 E = 0 The sum of the squares of the directional cosines must equal zero. - + j cos 2 2 x + cos2 y + cos2 z E + 2 E = 0 2 E + 2 E = 0 2 = - 2 = j The result is the propagation constant is complex = + j . The propagation vector also indicates direction of travel, = a = + j a . Propagation along a E = aE E e - r E = aE E e - a r cos t - a r + + EM waves 8 Plane waves in lossy materials E = aE E e - r E = a E E e j e - + j x cos x + y cos y + z cos z E = aE E e - a r e- a r e j In the time domain E = aE E e a r cos t - a r + - x cosx + y cosy + z cosz cos t - x cos + y cos + z cos + E = aE E e x y z EM waves in lossy material are exponentially decaying sinusoids ☼ Consider an electromagnetic wave, pointing in ax direction, traveling in a lossy material in the az direction (assume = 0). E = a xEe- z - 2 = 2 ( - j ) = j ( - j ) = + j In the time domain, E = a xEe- zcos( t - z) The intrinsic impedance is complex (E and H are NOT in phase for lossy materials) j = = = = e 1 - j -j 1 H = a E or use Faraday’s law E = H a or use Ampere’s law 2 1 = v = f = (general) v= (lossless only) EM waves 9 Example: Wave propagation in lossy material ☼ - z An electric field is given as E = a x 100e V/m is traveling through material with =0.1/m, r=1, r=4. The frequency is 2.45 GHz. Find and and the dB/m attenuation in the material. EM waves 10 Waves and power flow in lossy material ☼ What is the relation between E and H in the diagram shown? Take E as pointing along ax, traveling in az direction and take phase of E z = 0 = 0° . t=0 S = E H = a xEe- z cos( t - z) a yHe- zcos( t - z - ) cos + cos(2 t - 2 z - ) EH -2 z = az e 2 time-average Poynting vector 1 2 1 S = Re E H = Re a xEe- z e-j z a yH e e- z e j z % % 2 j 1 1 = a z EH cos e-2 z j -2 z = a zRe EHe e 2 2 EM waves 11 Special Cases ☼ High loss (good conductors) ~ ? = j (1 - j ) j -j j2 j = = j = e 2 j 1+j = = 1 + j = + j 2 2 Low loss (good dielectric) ~ = = j 1 - j 1 2 j 1 - j + 2 8 1 2 + j 1 + 2 8 = + j For very low loss, the second order term is often neglected. + j 2 EM waves 12 Good dielectrics (low loss) and good conductors (high loss) Note that, in the expression for complex permittivity, the imaginary term within the parentheses gives the ratio between conduction current and displacement current. = 1 - j Jc = E Jc = E D E Jd = = Jd = jD = j E t t Jc E = = = -j Jd j E j ―good dielectrics‖ are defined by Jd >> Jc ―good conductors‖ are defined by Jc >> Jd Example: Good conductor? Good dielectric? Find the propagation constant for f = 100Hz, = 10/m, r=1, r=10. Is the material a good conductor or a good dielectric? Neither? Example: Copper At what frequency does Cu cease to be a good conductor? =5.8(107)/m, r=1, r=1. EM waves 13 High loss - skin depth ☼ The distance a wave must travel into a good conductor for its amplitude to be reduced by a factor of e-1 is called skin depth, . Recall that, for a good conductor, conduction current is much greater than displacement current ( >> ). Jc = E, Jd = dD/dt = dE/dt - 2 = 2 (1 - j ) + j = +j 2 2 is the attenuation constant and is the reciprocal of skin depth. Consider an electric wave pointing in the x-direction and traveling in the z-direction: This implies that the magnetic field must lag the electric field High loss - wave impedance (E = H ) = = = 45° 1 - j - j EM waves 14 Example: skin effect ☼ The distance in a good conductor that is required to reduce the amplitude of the field vectors by a factor of e-1 is called the skin depth, . Assuming copper: z - E(z) = E(z) = Eo e- z = Eo e = Eo e-1 for z = = -1 2 1 1 0.0661 = 1 = = 2 5.8f f copper, non-magnetic = o = 4(10 ) H/m = 5.8(10 ) S/m -7 7 frequency (Hz) 60 Hz 1 KHz 1 MHz 1 GHz skin depth 8.5 mm 2.1 mm 66.1 m 2.1 m EM waves 15 Skin effect in circuits ☼ Consider the important special case in which the skin depth is much smaller than the cross sectional dimensions of the conductor. In this case, the field are confined to a thin layer around the surface. Even though the fields actually decay exponentially, it is sometimes convenient to suppose the current density is uniformly distributed over a thin layer—the thickness of which turns out to be a skin depth (=). V R = I The AC resistance of conductors is a function of frequency since the skin depth is a function of frequency. i) circular conductors ii) square conductors iii) stranded wire EM waves 16 Example: Skin effect ☼ A voltage, vs(t) = 10 cos (2ft) V is applied to a 100 load and a solid copper conductor as shown. The radius of the conductors is 2 mm and their length is 5 cm. a) What power is dissipated in the copper conductor at 60 Hz? b) What is the resistance of the conductors at 4 GHz? What power is dissipated in the copper conductors at 4 GHz? A voltage, vs(t) = 10 cos (2ft) V is applied to a similar conductor-load, this time the conductors consist of a 10 m copper coating with the interior of the conductors being a good insulator like Teflon. a) What is the resistance of the Cu-coated conductors at 60 Hz? What power is dissipated by the Cu-coated conductors at 60 Hz? b) What is the resistance of the Cu-coated conductors at 4 GHz? What power is dissipated by the Cu-coated conductors at 4 GHz? EM waves 17 Poynting’s theorem ☼ B H E =H - t John Henry Poynting D E H =E J + t Using this and the vector identity E H =H E -E H EM waves 18 Poynting’s theorem ☼ Looking at the integral form of Poynting’s theorem, it is readily seen that the theorem is an expression of conservation of energy IF Poynting’s vector is power flux density in W/m2. Conservation of energy is the justification for the interpretation of S = E H as power flux density. Example: units What are the units of a) E D b) H B c) S ? Example: Calculations of power flow ☼ An electric vector, pointing in the y-direction, travels in the z-direction. At z=0, the electric vector has an amplitude of 2 V/m. f=10 MHz with material properties r=1, r=4, =10-5/m. Calculate the terms in the integral form of Poynting’s theorem for the cube 0 < x < 2 m, 0 < y < 2 m, 0 < z < 2 m. EM waves 19 Instantaneous and time-averaged power density ☼ Calculate Poynting’s vector and the time-averaged Poynting’s vector for the plane wave below, traveling in air. -j y E = a x 2e 2 V/m Repeat, this time for the wave traveling in a lossy material. - + j y E = ax E e V/m = a x E e y e j y V/m EM waves 20 EM waves incident on boundaries (normal incidence) ☼ How are EM waves reflected and transmitted when a plane dielectric boundary is encountered? Consider the situation below: Define a transmission coefficient, T, and reflection coefficient, . Et Er T= = Ei Ei -j z Let Ei = a xEie 1 Boundary conditions: E1tan z=0 = E2tan z=0 H1tan z=0 = H2tan z=0 EM waves 21 Transmission and reflection ☼ Find the reflection and transmission coefficients for a wave traveling in a printed circuit board made from FR-4 (most common PCB material—fiberglass epoxy base) into air. FR-4 r=4.4, r=1 air r=1, r=1 EM waves 22 Polarization ☼ Polarization refers to the path traced by the electric vector in a traveling electromagnetic wave. Assume a wave in traveling in the +az direction and take the point-of-view seen in front of the oncoming EM wave. For a given value of z, what path is traced by the electric vector in the x-y plan as time advances? In general, a wave traveling in the az direction can have an x-component and a y- component. The resultant vector traces an ellipse—referred to as elliptical polarization— in the x-y plan as it travels, the shape of which depends on Ex, Ey, x, and y. E = axExcos t - z + x + ayEycos t - z + y Linear polarization is a special case E = a xEx cos t - z + x E = ayEycos t - z + y E = a xE x cos t - z + a yE y cos t - z When Ex = Ey and x and y differ by ±90°, the result is circular polarization. For circular and elliptical polarization, the sense of the rotation is referred to as left-handed or right-handed, depending on which hand, if the fingers follow the rotation of E, the thumb is in the direction of propagation. Polarization examples ☼ What is the polarization for E = a x 2 cos t - z + 45° - a y 10 cos t - z - 90° ? (LHEP) What is the polarization for E = a x 2 cos t - z + 45° + a y 2 cos t - z - 45° ? (RHCP) What is the polarization for E = a x 3 cos t - z + 45° + a y cos t - z + 45° ? (LP, = 18.43°) EM waves 23 EM waves incident on boundaries (oblique incidence) Now consider an EM wave, traveling in region 1, which encounters a second region at an angle other than normal. In this case, we must consider two types of polarization. 1. Perpendicular polarization: E to plane of incidence (POI). 2. Parallel polarization: E P to POI E POI E P to POI Definitions POI – plane formed by propagation vector, , and the vector normal to the boundary between regions 1 and 2. Polarization: in general, any non-random orientation of an electric and magnetic field. In particular, polarization more usually describes the path the electric vector takes, in planes of constant phase as the wave travels. Note: We’re considering only linear polarization, where the electric vector points in one direction. It reaches its maximum positive, grows smaller, becomes zero then negative, becomes maximum negative, grows smaller, becomes zero, etc. That is the vector remains on a straight line—thus, linear polarization. Any TEM wave can be expressed in terms of linearly polarized waves. EM waves 24 Refection and transmission at dielectric boundaries ( E POI) ☼ 1st, give expressions for the electric and magnetic vectors. Then, to find reflection and transmission coefficients, require tangential H and E to be continuous at the boundary (z = 0). EM waves 25 Refection and transmission at dielectric boundaries ( E P POI) Answers 2cos t - 1cos i 2 2cos i = = 2cos t + 1cos i 2cos t + 1cos i EM waves 26 Example: Reflection and transmission ☼ A uniform plane wave, in air, impinges on a dielectric material at i = 45°. The transmitted wave propagates with t = 30°. The frequency is 300 MHz and the electric vector is perpendicular to the plane-of-incidence. i) Find 2 in terms of 0 (assume 1=2=o) ii) Find the reflection coefficient, . iii) Find the transmission coefficient, T. iv) Give expressions for the incident E and H -fields, the reflected E and H fields, and the transmitted E and H -fields. EM waves 27 EM waves normally incident on good conductors ☼ Consider EM waves encountering a conducting material 2 - 1 22 = T= 2 + 1 2 + 1 EM waves 28 Reflection and transmission for an air/metal interface ☼ A 1-MHz plane wave, in air is incident on a copper sheet. The amplitude of the incident E-field is 100 V/m. Find , T, , and the |E| and |H| fields at a distance of into the copper. EM waves 29 Transmission through a conducting sheet – an air/metal/air barrier ☼ To shield from potential EM interference, a product manufacturer is considering using a copper sheet with a thickness, t, of several skin depths (say 4 or 5). As a measure of the shielding available, find an expression for the ratio of the magnitudes of the electric fields, Eo/Ei, at f = 106 Hz. In developing the expression, incorporate the good-conductor approximations for the copper shield. copper = o, copper = o ) 7 (copper = 5.810 S/m, EM waves 30 Standing Waves ☼ Consider normal incidence. In steady-state, the incident and reflected waves interact to produce, in region 1, a traveling wave and a standing wave. -j E1 = Ei + Er choose reference phase Ei = Eie 2 e-j z -j -j z j -j E1 = Eie 2 e + e Ei e 2 e j z -j j -j -j E1 = Eie 2 e-j z + Eie 2 e j z + Eie 2 e-j z - Eie 2 e-j z EM waves 31 Standing Wave Ratio (SWR) ☼ E 1+ SWR = max = Emin 1- = SWR - 1 SWR + 1 = The SWR ratio can readily be measured—so too = . For no reflection, SWR=1. As more of the incident wave is reflected, the SWR ratio grows. EM waves

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