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ECE 480 Wireless Systems Lecture 3 Propagation and Modulation of RF Waves 1 Circular Polarization Magnitudes of the x – and y – components of E z are equal Phase difference is 2 2 2 Left – Hand Circular (LHC) Polarization Right – Hand Circular (RHC) Polarization 2 Left – Hand Circular (LHC) Polarization ax ay a 2 j E z xˆ a yˆ a e 2 e j k z jkz ˆ ˆ ax j y e e j 2 j 3 jt E z,t Re E z e xˆ a cos t k z yˆ a cos t k z 2 xˆ a cos t k z yˆ a sin t k z Convert to polar form E z,t E x2 z,t E y2 z,t 1 2 1 2 a 2 cos 2 t k z a 2 sin 2 t k z a E y z,t 1 z,t tan E x z,t tan 1 a sin t k z t k z a cos t k z 4 Linear polarization E z,t a x cos t k z 2 a y cos t k z 2 = f (z, t) ay tan ax 1 constant f (z, t) Circular polarization E z,t a f (z, t) z,t t k z = f (z, t) 5 1 2 Back to LHC Polarization Consider the LHC wave at z = 0 t Inclination angle decreases with time 6 Right – Hand Circular (RHC) Polarization ax ay a E z,t a 2 z,t t k z 7 The direction of polarization is defined in terms of the rotation of E as a function of time in a fixed plane orthogonal to the direction of propagation 8 Example RHC Polarized Wave An RHC polarized plane wave with electric field modulus of 3 mV/m is traveling in the + y direction in a dielectric medium with 4 0 0 0 f = 100 MHz Obtain expressions for E y,t and H y,t 9 Solution The wave is traveling in the + y direction. Therefore, the field components are in the x and z directions. zˆ xˆ yˆ EH direction of propagation 10 Assign a phase angle of 0 o to the z component of E y (arbitrary) The x component of E y will have a phase shift 2 Both components have a magnitude of a = 3 E y xˆ E x zˆ E z j e j k y zˆ a e j k y mV jky xˆ j zˆ 3 e m xˆ a e 2 11 H y 1 1 3 yˆ E y y xˆ j zˆ 3 e j k y zˆ j xˆ e 2 f 2 10 k r 8 rad s 2 10 8 4 4 8 3 10 3 c mA m jky 0 120 60 4 rad m 12 Converting back to the time domain E Re E y e j t Re xˆ j zˆ 3 e j k y e j t 3 xˆ sin t k y zˆ cos t k y mV m H Re H y e j t 3 jky jt Re zˆ j xˆ e e 1 mA xˆ cos t k y zˆ sin t k y 20 m 13 Elliptical Polarization Most general case ax 0 The tip of ay 0 0 E traces an ellipse in the x – y plane Can be left – handed or right - handed Major axis: Minor axis: Rotation Angle Ellipticity Angle 14 Rotation angle cos 2 1 ax a 2 The shape and rotation are defined by the ellipticity angle 1 tan a R 4 4 a R = 1 Circular R axial ratio a R = Linear a 15 tan 0 ay ax 0 0 90 tan 2 tan 2 0 cos tan 2 sin 2 0 sin 45 0 o o o 2 2 4 4 Circular Linear 16 Polarization States for Various Combinations of and 17 How is the type of polarization determined? Positive values of corresponding to sin > 0 define left – handed rotation Negative values of corresponding to sin < 0 define right – handed rotation a x and a y are, by definition, > 0 ay ax 0 0 90 will vary from 0 Two possible values of 0 if cos 0 0 if cos 0 and o in this range 0 if sin 0 0 if sin 0 18 Example: Polarization State Determine the polarization of a plane wave with an electric field given by E z, t xˆ 3 cos t k z 30 o yˆ 3 sin t k z 45 mV m o Solution Convert the sin term to a cos term by subtracting 90 o E z, t xˆ 3 cos t k z 30 o yˆ 3 cos t k z 45 o mV m Convert the – cos term to a + cos term by adding 180 o E z, t xˆ 3 cos t k z 30 o yˆ 3 cos t k z 135 o mV m 19 Convert to phasor form E z xˆ 3 e jkz e j 30 o yˆ 4 e jkz e j 135 o y x 135 o 30 o 105 o 0 tan 1 ay ax 1 tan 4 o 53 . 1 3 tan 2 tan 2 0 cos tan 106.2 o cos 105 o 3.442 0.2588 0.8900 20 There are two possible solutions for since the tan function is positive in both the first and third quadrants 2 tan 1 0.89 41.7 o , 221.7 o 138.3 o 41.7 o 1 20.85 o 2 138.3 o 2 69.15 o 2 Which is correct? cos 0 69.15 o 21 sin 2 sin 2 0 sin sin 106.2 o sin 105 o 0.9603 0.9659 0.9276 By a similar analysis, 34.0 o The wave is elliptically polarized and the rotation is left - handed 22 Plane – Wave Propagation in Lossy Media 2 E 2 E 0 2 2 C 2 ' j " " ' can be written as j = attenuation constant = phase constant 23 2 2 j j 2 2 2 ' j 2 " Equate the real and imaginary parts 2 2 2 ' 2 " 2 ' 2 Solve for and ' 2 1 1 " 1 ' 2 1 2 " 1 ' 2 1 2 24 For a uniform plane wave traveling in the + z direction with an electric field E xˆ E x z the wave equation becomes d 2 E x z dz 2 2 E x z 0 The solution is E z xˆ E x z xˆ E x 0 e z xˆ E x 0 e z e j z H kˆ E c yˆ E x0 c e z " c 1 j c ' ' e 1 2 j 25 The magnitude of E x z is E x z E x0 e z e jz E x0 e z Decreases exponentially with e - z Hy Ex c also decreases exponentially with e - z Define: Skin Depth, s s 1 Distance that a wave must travel before it is attenuated by 1 0.3679 e 26 In a perfect dielectric 0 , 0 , s In a perfect conductor , , s 0 27 Expressions are valid for any linear, isotropic, homogeneous medium " 10 2 ' " 2 2 10 10 ' " 2 10 ' Low – Loss Dielectric Quasi – Conductor (Semiconductor) Good Conductor 28 Low – Loss Dielectric Consider For 1 x x 1 , 1 2 1 x 1 2 x 1 2 1 2 " j ' 1 j j ' ' " 1 j 2 ' Divide into real and imaginary parts " 2 ' 2 ' - Same as for lossless medium Np m 29 " c 1 j c ' ' x 1 , c ' 1 x " 1 j 2 ' c 1 2 1 2 x 1 2 " 1 j 2 ' Same as for the lossless case 30 Good Conductor " 2 f 2 f c f j 1 j 1 j " 1 j j 2 31 Semiconductors – Must use exact solution 32 Example – Plane Wave in Seawater A uniform plane wave is traveling downward in the + z direction in seawater, with the x – y plane denoting the sea surface and z = 0 denoting a point just below the surface. The constitutive parameters of seawater are: r 1, S 4 m r 80 , The magnetic field intensity at z = 0 is given by H 0,t yˆ 100 cos 2 10 t 15 a. Determine expressions for 3 o mA m E z,t and H z,t b. The depth at which the amplitude of E is 1% of its value at z = 0 33 Solution a. The general expressions for the phasor fields are E z xˆ E x 0 e z e j z H z yˆ E x0 c e z e j z " 4 ' r 0 2 10 3 80 10 9 36 9 10 5 Seawater is a good conductor at 1 KHz 34 f 10 4 10 3 7 0.126 Np 4 0.126 m j 0.126 j c 1 j 2 e 4 0.044 e 4 4 The general expression for E x0 is E x0 E x0 e j 0 j E z,t Re xˆ E x 0 e 0 e z e j z e j t x̂ E x 0 e 0.126 z cos 2 10 t 0.126 z 0 3 V m 35 E e j0 x0 z jz jt H z, t Re yˆ e e e j 4 0.044 e ŷ 22.5 E x 0 e 0.126 z cos 2 10 3 t 0.126 z 0 45 o A m at z = 0: H 0, t yˆ 22.5 E x 0 cos 2 10 3 t 0 45 o A m Compare with original expression H 0,t yˆ 100 cos 2 10 t 15 22.5 E x 0 100 10 E x0 mV 4.44 m 3 o mA m 3 0 45 o 15 o 0 60 o 36 E z,t xˆ 4.44 e 0.126 z cos 2 10 3 t 0.126 z 60 o H z,t yˆ 100 e 0.126 z cos 2 10 3 t 0.126 z 15 o mV m mA m Note that they are no longer in phase. The electric field always leads the magnetic field by 45 o. b. Set the amplitude to 0.01 0.01 e z 0.126 z ln 0.01 0.126 36 m 37 Electromagnetic Power Density Define: Poynting Vector S E H W m2 Direction of S is in the direction of propagation, k n̂ unit vector normal to the surface Power through a surface, A P S A ˆ ndA S A cos 38 Plane Wave in a Lossless Medium Consider a plane wave traveling in the + z direction E z xˆ E x 0 e j k z H z yˆ E x0 ejkz Want to find the power density vector, S 39 Time – Domain Approach E z ,t xˆ E x 0 cos t k z H z ,t yˆ E x0 cos t k z S z ,t E z ,t H z ,t ẑ E x0 2 cos 2 t k z 40 Time average of T S av 1 S z ,t d t T 0 2 2 z ẑ 0 E x0 E x0 S 1 2 T f 2 cos 2 t k z d t 2 2 41 Phasor – Domain Approach S av Re E H * Ex 0 * 1 jkz Re xˆ E x 0 e yˆ e 2 ẑ E x0 jkz 2 2 S av Re E H * is valid for any media 42 Plane Wave in a Lossy Medium E z xˆ E x z yˆ E y z xˆ E x 0 yˆ E y 0 e z e j z H z 1 c xˆ E x 0 yˆ E y 0 e z e j z S av Re E H * ẑ E x 0 2 E y0 2 c c e 2 e j 2 z 1 Re c * 43 S av z zˆ E0 2 2c e c c e S av z zˆ E0 cos j 2 2c E 0 E x0 2 z 2 e 2 z cos E y0 2 1 2 2 z e Note that the average power decays with 44 Homework The electric field of a plane wave is given by E z ,t xˆ a x cos t k z yˆ a y cos t k z Identify the polarization state, determine the polarization angles (, ), and sketch the locus of E (0, t) for each of the following cases V V ax 3 , ay 4 , 0o m m V V ax 3 , ay 4 , 180 o m m V V ax 3 , ay 3 , 45 o m m V V ax 3 , ay 4 , 135 o 45 m m Homework In a medium characterized by S r 9 , r 1 , 0.1 m Determine the phase angle by which the magnetic field leads the electric field 46 Radiation and Antennas • An antenna may be considered as a transducer that converts a guided EM wave to a transmitted wave or an incident wave to a guided EM wave • Antenna dimensions are generally referred to in wavelength units 47 48 Reciprocity • Antenna radiation pattern: The directional function that characterizes the distribution pattern radiated by an antenna • Isotropic antenna: A hypothetical antenna that radiates equally in all directions • Used as a reference radiator to compare antennas • Reciprocal antennas: Antennas that have the same radiation patterns for transmission as for reception 49 Two aspects of antenna performance 1. Radiation Properties • Direction of the radiation pattern • Polarization state of the radiated wave in the TX mode (Antenna Polarization) • In the RX mode, the antenna can extract only that component of the wave whose E – field is parallel to that of the antennas polarization direction 2. Antenna Impedance • Pertains to the impedance match between the antenna and the generator 50 Radiation Sources Two basic types 1. Current sources • Dipole and loop antennas 2. Aperture fields • Horn antennas 51 Far – Field Region The far – field region is at a distance R where the wave may be considered to be a plane wave R ff 2D 2 D = Maximum effective size of the antenna = Wavelength of the signal 52 Example: Far – Field Distance of an Antenna A parabolic reflector antenna is 18" in diameter operates at 12.4 GHz. Find the operating wavelength and the far – field distance of this antenna. Solution c 3 10 0.0242 m 9 f 12.4 10 1m 18" x 0.457 m 39.37" 8 R ff 2D 2 2 0.457 0.0242 2 17.3 m 53 Antenna Arrays • Can control the phase and magnitude of each antenna individually • Can steer the direction of the beam electronically 54 Retarded Potentials Consider a charge distribution as shown V R 1 4 v R i ' R' d ' The electric potential V (R) at a point in space specified by the position vector R is given by Ri = position vector of an elemental volume ' v = elemental volume R' R R i = charge density inside the volume 55 = distance between the volume and the point If the charge density is time – varying, the obvious solution is V R ,t 1 4 v R i , t ' R' d ' Problem: Does not account for reaction time Any change in the charge distribution will require a finite amount of time to change the potential 56 R' t' up Delay Time Retarded Scalar Potential V R ,t 1 4 v R i , t t ' R' ' d ' Retarded Vector Potential A R ,t 4 J R i ,t t ' ' R' d ' Valid under both static and dynamic conditions 57 Time – Harmonic Potentials In a linear system, the parameters all have the same functional dependence on time Consider a sinusoidal time – varying charge distribution v jt R i ,t Re v R i e v R i = phasor representation of v R i ,t 58 V R 1 4 A R 4 v R i e j k R' R' ' k J R i e j k R' R' ' d ' V d ' up 59 In terms of A H 1 A H j E 1 E H j E j H 1 H E j 60 The Short (Hertzian) Dipole Approach: Develop the radiation properties of a differential antenna and use that model to predict other configurations Characteristics of a Short Dipole • Current is uniform over the length l 50 jt i t I 0 cos t Re I 0 e I I0 A 61 At the point Q 0 A R 4 k c ' J e j k R' d ' R' 2 J zˆ I0 s s = cross – sectional area of dipole d ' s d z limits of integration Assume l l z 2 2 R' ~ R 62 0 e jkR A 4 R l 2 zˆ I 0 dz l 2 0 e jkR ẑ I0 l 4 R e jkR Spherical propagation factor R Considers both the magnitude and phase change wrt R Change to spherical coordinates 63 R Range Zenith angle Azimuth angle 64 ẑ Rˆ cos ˆ sin I 0 l e jkR 0 A Rˆ cos ˆ sin 4 R Rˆ A R ˆ A ˆ A 0 I0 l e jkR AR cos 4 R 0 I0 l e jkR A sin 4 R A 0 65 H E 1 A 1 j H j 1 H e sin 2 4 k R k R 1 2 I0 lk 2 j jkR ER 0 e cos 2 3 4 k R k R j I0 lk 2 1 j jkR E 0 e sin 2 3 4 k R k R k R I0 lk 2 jkR H R H E 0 66 Electric Field lines 67 Far – Field Approximation R 2 R kR 1 1 k R 2 , 1 k R 3 negligible j I 0 l k e jkR E sin 4 R ER 0 H E 0 Independent of Proportional to sin 68 Power Density S av Re E H * For the short dipole: S av Rˆ S R , 0 k 2 I 02 l 2 2 S R , sin 2 2 32 R W 2 S 0 sin m2 69 Define: Normalized Radiation Intensity F , S R , , S max Radiation is maximum when S max 2 (azimuth plane) 0 k 2 I 02 l 2 2 S0 sin 2 2 32 R 2 2 15 I 0 l R2 70 F , F sin 2 No energy is radiated by the dipole along the direction of the dipole axis and maximum radiation (F = 1) occurs in the broadside direction. 71