Radio Propagation Channels Exercise 2 with solutions. Polarization / Wave Vector
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1 /8 Polarization / Wave Vector Assume the following three magnetic fields of homogeneous, plane waves H (t) H A cos (ωt kz) e x H A sin (ωt kz) e y () H 2 (t) H A cos (ωt kz) e x + H A sin (ωt kz) e y (2) H 3 (t) H B cos (ωt kz) e x + H B sin (ωt kz) e y, (3) where H A > 0, H B > 0, H A < H B. Problem. Determine the phasor of the magnetic field H..2 What is the polarization type of the magnetic fields H (t) and H 2 (t)?.3 What is the polarization type of the superposition of H (t) and H 2 (t)?.4 What is the polarization type of the superposition of H (t) and H 3 (t)? Given is the following electric and magnetic field of a homogeneous, plane wave that propagates in a dielectric fluid with ε r 3 and µ r ( ) E E 2 e x e y + 2e z e jk r (4) Problem 2 H H ( 2e x + e y + e z ) e jk r. (5) 2. In which direction does the wave propagate? Determine the normalized vector n, where k k n (k is the wave number in the dielectric fluid). 2.2 Determine the wave vector k. 2.3 Determine the phase velocity v ph of a wave propagating in this dielectric fluid.
2 2/8 Solution of Problem (. Using sin (ωt) cos ωt π ) and H A > 0, H (t) can be rewritten as 2 ( H (t) H A cos (ωt kz) e x H A cos ωt kz π ) e y 2 } Re {H A e j(ωt kz) e x H A e j(ωt kz π 2 ) ey (HA ) } Re{ e jkz e x H A e j π 2 }{{} e jkz e y e jωt j { (HA ) Re e jkz e x + jh A e jkz e y e jωt} { } Re H A (e x + je y ) e jkz e jωt. }{{} H Therefore, the phasor of the magnetic field H is H H A (e x + je y ) e jkz..2 The period of H (t) H A cos (ωt kz) e x H A sin (ωt kz) e y is T 2π ω. Now consider the constant phase plane z 0 and the different time instants t 0, T 4, T 2, 3T 4, we can get with H A > 0 H (t 0, z 0) H A cos 0 e x H A sin 0 e y H A e x point at the positive x-axis, H (t T 4, z 0) H A cos π 2 e x H A sin π 2 e y H A e y point at the negative y-axis, H (t T 2, z 0) H A cos π e x H A sin π e y H A e x point at the negative x-axis, H (t 3T 4, z 0) H A cos 3π 2 e x H A sin 3π 2 e y H A e y point at the positive y-axis. Figure illustrates the polarization type of H (t). Observe the figure and note the clockwise rotation direction with the same length H A we can draw the following conclusion that H (t) is left circularly polarized.
3 3/8 y H (t 3T 4, z 0) H Ae y H (t T 2, z 0) H Ae x z 0 clockwise rotation x H (t 0, z 0) H A e x H (t T 4, z 0) H Ae y Figure : Polarization type of H (t). For magnetic field H 2 (t) H A cos (ωt kz) e x + H A sin (ωt kz) e y, analogously at the constant phase plane z 0 and t 0, T 4, T 2, 3T 4, we can get with H A > 0 H 2 (t 0, z 0) H A cos 0 e x + H A sin 0 e y H A e x point at the positive x-axis, H 2 (t T 4, z 0) H A cos π 2 e x + H A sin π 2 e y H A e y point at the positive y-axis, H 2 (t T 2, z 0) H A cos π e x + H A sin π e y H A e x point at the negative x-axis, H 2 (t 3T 4, z 0) H A cos 3π 2 e x + H A sin 3π 2 e y H A e y point at the negative y-axis. Note the anticlockwise rotation direction with the same length H A we can draw the following conclusion that H 2 (t) is right circularly polarized..3 H (t) + H 2 (t) 2H A cos (ωt kz) e x is also a solution of the wave equation. Consider the equiphase plane z 0 and t 0, T 4, T 2, 3T 4, we can get with H A > 0 H (t 0, z 0) + H 2 (t 0, z 0) 2H A cos 0 e x 2H A e x H (t T 4, z 0) + H 2(t T 4, z 0) 2H A cos π 2 e x 0 H (t T 2, z 0) + H 2(t T 2, z 0) 2H A cos π e x 2H A e x
4 4/8 H (t 3T 2, z 0) + H 2(t 3T 2, z 0) 2H A cos 3π 2 e x 0. Vectors always lie on the same line (x-axis) linearly polarized..4 H (t)+h 3 (t) (H A + H B ) cos (ωt kz) e x +(H B H A ) sin (ωt kz) e y. Consider the equiphase plane z 0 and time instants t 0, T 4, T 2, 3T, we can get with 4 H A > 0, H B > 0, H A < H B H B H A > 0 H (t 0, z 0) + H 3 (t 0, z 0) (H A + H B ) e x H (t T 4, z 0) + H 3(t T 4, z 0) (H B H A ) e y H (t T 2, z 0) + H 3(t T 2, z 0) (H A + H B ) e x H (t 3T 4, z 0) + H 3(t 3T 4, z 0) (H B H A ) e y. y (H B H A ) at (t T 4, z 0) (H A + H B ) at (t T 2, z 0) z 0 x (H A + H B ) at (t 0, z 0) (H B H A ) at (t 3T 4, z 0) Figure 2: Polarization type of H (t) + H 3 (t). Figure 2 illustrates the polarization type of H (t) + H 3 (t). Observe the figure and note the anticlockwise rotation direction with the different length we can draw the following conclusion that H (t) + H 3 (t) is right elliptically polarized.
5 5/8 Some Remarks on Planar Wave in an Arbitrary Direction From the lecture we know that a generalized planar harmonic wave propagating in an arbitrary direction, which is specified by its electric field, can be represented as E (E e + E 2 e 2 ) e jk r (6) and a time-dependent electric wave propagating along the direction of k can be expressed as E(t) Re { E e jωt} ] [E (t)e + E 2 (t)e 2 e jk r [Ê cos (ωt + ϕ ) e + Ê2 cos (ωt + ϕ 2 ) e 2 ] e jk r (7) with e k 0, e 2 k 0 and e e 2 0, i.e., they follow the right hand orthogonal rule, see Figure 3. e 2 e k Figure 3: Right hand orthogonal rule of e, e 2 and k. In the cartesian coordinate system, a location vector r can be represented as k is vector wave number and defined as r xe x + ye y + ze z. k k x e x + k y e y + k z e z with relationship to scalar wave number k k k k 2 k k 2 x + k 2 y + k 2 z. We have found that the fields of the electromagnetic wave are perpendicular to each other, and that they are also perpendicular (or transverse) to the direction of propagation k.
6 6/8 Electromagnetic power flows with the wave along the direction of propagation and it is also constant on the equiphase planes. The power density is described by the time dependent Poynting vector P(t) E(t) H(t). The Poynting vector is perpendicular to both field components, and is parallel to the direction of wave propagation. It means that the following relationships hold true. k E, k H, E H and k (E H P) Solution of Problem 2 2. A homogeneous planar wave one holds n E, n H, E H n (E H) as illustrated in Figure 4. H E k k n Figure 4: Right hand orthogonal rule of E, H and k k n. Since E and H are constants and e jk r denotes a certain phase, we can obtain (E H) ( ) 2 e x e y + 2e z ( 2e x + e y + e z ) (E H) e x e y e z ( 2 3e x e y + 3 ) 2 e z. 2 Hence, n ( 3e x e y + 3 ) 2 e z.
7 7/8 Normalization n yields (both propagation directions are possible!) n ± ( ) 2 ( ) (3e x e y + 3 ) 2 e z n ± 4 (2e x + 3e y + e z ). 2.2 k ω ε µ ω ε r ε 0 µ r µ 0 ε r µ r ω ε 0 µ 0 }{{}}{{} 3 k 0 where k 0 is the scalar wave number of vacuum. Therefore, 3k0, k k n 3 3k 0 n ± 4 k 0 (2e x + 3e y + e z ). 2.3 A time dependent electric field E (t) can be described as E (t) Re { E e jωt} Re { E e jk r e jωt} Re { E e j(ωt k r) } E cos (ωt k r), where E E ( 2 e x e y + 2e z ) and the planar electromagnetic wave propagates along the direction of k. Figure 5: Example of a homogeneous planar wave E + x (z, t), which is propagating along the positive z-axis in vacuum. Here, β 0 is the wave number of vacuum.
8 8/8 For reasons of simplicity, we assume the wave with the electric field E (t) E cos (ωt k z) E cos (ωt k z), which propagates along the positive z-axis. For the isotropic medium one can assume that the wave does not change behavior with its direction. See Figure 5. Looking at equiphase planes, one obtains Thus, we can get ωt k z ωt k z constant. v ph dz dt ω k ω ω ε µ ε0 µ 0 εr µ r c c. εr µ r 3
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