Plane Waves GATE Problems (Part I). A plane electromagnetic wave traveling along the + z direction, has its electric field given by E x = cos(ωt) and E y = cos(ω + 90 0 ) the wave is (a) linearly polarized (c) left circularly polarized (b) right circularly polarized (d) elliptically polarized [GATE 994: Mark] Soln. E x (t) = cos ωt E y (t) = cos(ωt + 90 0 ) = sin ωt E x (t) + E y (t) = It represents a circle in the E x E y plane with radius as shown in figure. Hence the wave is circularly polarized. E y (t) - - - - E x (t) ωt 0 to π ωt = 0 π π 3π π E x = 0 0 E y = 0 0 0 When the fingers of the left hand follows the clock wise direction (direction of rotation of the E vector), the thumb is pointing in the given direction of propagation (+ z direction). The wave is left circularly polarized.
. The intrinsic impedance of a lossy dielectric medium is given by (a) jωμ σ (c) jωμ (b) jω (σ+jωε) μ (d) μ [GATE 995: Mark] Soln. Conductivity = σ mhos/m Permittivity = farad / m Permeability = μ henry / m E / H for a lossy dielectric medium = η = jωμ σ+jωε 3. Copper behaves as a (a) Conductor always. (b) Conductor or dielectric depending on the applied electric field strength (c) Conductor or dielectric depending on the frequency (d) Conductor or dielectric depending on the electric current density [GATE 995: Mark] Soln. For a conductor σ ωε For copper with σ = 5. 8 0 7 mhos/m = 0 = 36π 0 9 farad/m at relatively large frequency f = 3 0 5 Hz σ ωε = 5. 8 07 36π 0 9 π 3 0 5 = 348 Copper is good conductor for the frequencies used in practice Option (a)
4. The intrinsic impedance of copper at high frequency is (a) Purely resistive (b) Purely inductive (c) Complex with a capacitive component (d) Complex with a inductive component [GATE 998: Mark] Soln. The intrinsic impedance η = jωμ σ+jωε For a good conductor σ ω η = jωμ σ = ωμ σ e j450 = η R + jη x η R = η x = ωμ σ η is complex with inductive component Option (d) 5. The wavelength of wave with propagation constant (0. π + j 0.π)m is (a) m (c) 0 m 0.05 (d) 30 m (b) 0 m [GATE 998: Mark] Soln. Propagation constant = α + jβ = (0. π + j 0. π)m α = attenuation constant = 0. π β = phase constant = 0. π wavelength λ = π β = π 0.π = 0m Option (b)
6. The depth of penetration of a wave in a lossy dielectric increases with increasing (a) Conductivity (c) Wavelength (b) Permeability (d) Permittivity [GATE 998: Mark] Soln. Depth of penetration δ = α α = attenuation constant γ = Propagation constant = α + jβ = jωμ (σ + jωε) α = ω με [ + σ ] ω ε β = ω με [ + σ + ] ω ε For a lossy dielectric with σ 0 α increases with increasing µ and σ good conductor with σ ωε α = β = ω με σ ωε = ω μσ ω = ωμσ α f or δ f or δ λ
7. The polarization of a wave with electric field vector E = E 0 e j(ωt βz) (a x + a y ) (a) Linear (b) Elliptical Soln. E = E 0 e j(ωt βz) (a x + a y ) (c) Left hand circular (d) Right hand circular [GATE 998: Mark] It is a wave propagating in Z direction with electric field components in x and y direction at z = 0 E x = E 0 cos ωt E y = E 0 cos ωt E y = E x at any time t E y E 0 A ωt = 0, π ωt = π/, 3π/ E 0 0 45 E 0 E x ωt = π B E 0 As ωt varies from 0 to π the tip of E vector moves along the straight line AB from A to B and as ωt varices from π to π the tip of E vector moves back from B to A and the cycle repeats. The polarization of the wave is linear Option (a)
8. A TEAM wave is incident normally upon a perfect conductor. The E and H fields at the boundary will be respectively (a) Minimum and minimum (c) Minimum and maximum (b) Maximum and maximum (d) Maximum and minimum [GATE 000: Mark] Soln. IN the case of a plane wave incident normally upon the surface of a perfect conductor, the wave is entirely reflected, neither E nor H can exist within a perfect conductor. According to the boundary condition E tangential = E inc + E ref = 0 E inc is reflected with phase reversal E is minimum equal to zero at the surface of a perfect conductor The magnetic field H must be reflected without reversal of phase. It both E and H are reversed with phase reversal there would be no reversal of direction of propagation. The phase of the reflected magnetic field strength H r is the same. H tan = H i + H r = H i = J S Where J S is the linear current density in A/m on the surface. H is maximum equal to twice the incident value at the boundary E T H T Perfect Conductor
9. If a plane electromagnetic wave satisfies the equation E x = z c E x, t the wave propagates in the (a) x direction (b) z direction (c) y direction (d) x z plane at an angle of 45 0 between the x and z directions [GATE 00: Mark] Soln. The wave equations for free space (In a perfect dielectric containing no charges) E = με E H = με H The wave equation reduces to a simple form where E and H are considered to be independent of two dimensions (x and y) E z = με E t For uniform plane propagating in the Z direction, E may have components E x and E y Where v 0 = E x z με Option (b) = με E x t is the velocity of propagation 0. The depth of penetration of electromagnetic wave in a medium having conductivity σ at a frequency KHz is 5 cm. The depth of penetration at a frequency of 4 KHz will be (a) 6.5 cm (b).50 cm (c) 50.00 cm (d) 00. 00 cm [GATE 003: Mark]
Soln. Depth of penetration δ = 5cm at f = KHz conductivity σ For a medium to be good conductor σ ωε δ = where = attenuation constant γ = α + jβ = jωμ + (σ + jωε) For a lossy dielectric, considered as a good conductor σ ωε α = β = ωμσ = πfμσ δ = α = πfμσ or δα f, δ be the depth of penetration at f = 4KHz δ δ = f f δ = δ Option (b) δ δ = 4 = =. 5cm. The magnetic field intensity vector of a plane wave is given by H (x, y, z, t) = 0 sin(50000t + 0.004x + 30)a y where a y denotes the unit vector in y direction. The wave is propagating with a phase velocity (a) 5 0 4 m/s (c).5 0 7 m/s (b) 3 0 8 m/s (d) 3 0 8 m/s [GATE 005: Mark] Soln. H(x, y, z, t) = 0 sin[50000t + 0. 004x + 30] a y = H y a y H y = 0 sin[ωt + βx + 30] ω = 50, 000 radians / sec β = 0. 004 radians / m
Phase velocity V p = ω β = 50 03 4 0 3 =. 5 0 6 m/sec =. 5 0 7 m/sec Represents a wave traveling in the negative x direction. The electric field of an electromagnetic wave propagating in the positive z direction is given by E = a x sin(ωt βz) + a y sin(ωt βz + π/) The wave is (a) linearly polarized in the z direction (b) elliptically polarized (c) left hand circularly polarized (d) right hand circularly polarized [GATE 006: Mark] Soln. E = a x sin(ωt βz) + a y sin (ωt βz + π ) = a x sin(ωt βz) + a y cos(ωt βz) E x (z, t) = sin(ωt βz) E y (z, t) = cos(ωt βz) E x + E y = It represents a circle in the E x E y plane with radius The tip of the E vector is tracing the circle in the clock wise direction over a cycle from ωt = 0 to π. The fingers of the left hand follows the clockwise direction, thumb is pointing in the direction of propagation (+ z)
ωt = π E y (t) - E x (t) ωt = 3π ωt = π - ωt = π 3. The electric field of a uniform plane electromagnetic wave in free space, along the positive X direction is given by E = 0(a y + ja z)e j5x. The frequency and polarization of the wave respectively are (a). GHz and left circular (c). GHz and right circular (b) 4 Hz and left circular (d) 4 Hz and right circular [GATE 0: Mark] Soln. E = 0(a y + ja z )e j5x in free space E = (E y a y + E z a z )e jβx The wave is propagating in free space in x direction with components in y and z direction. β = 5, v p = ω β = 3 08 m/s f = 75 08 π ω = 3 0 8 5 = 75 0 8 rad / sec = 7.5 09 π. GHz The field in circular polarization is found to be E s = E o = (a y ja z )e jβx Propagating in +ve X direction where plus sign is used for left circular polarization and minus for right circular polarization. Option (a)
4. A plane wave propagating in air with E = (8a x + 6a y + 5a z)e j(ωt+3x 4y) V/M is incident on a perfectly conducting slab positioned at x 0. The E field of the reflected wave is (a) ( 8a x 6a y 5a z) e j(ωt+3x+4y) V/M (b) ( 8a x + 6a y 5a z) e j(ωt+3x+4y) V/M (c) ( 8a x 6a y 5a z) e j(ωt 3x 4y) V/M (d) ( 8a x + 6a y 5a z) e j(ωt 3x 4y) V/M [GATE 0: Mark] Soln. E = (8a x + 6a y + 5a z )e j(ωt+3x 4y) v/m E i σ = E r x = 0 Electric field inside a perfect conductor is zero E transmitted = 0 E i + E r = 0 E r = E i = 8a x 6a y 8a z The x component of Eincident which is normal to slab gets reflected with 80 0 phase change
5. A two port network has scattering parameters given by [S] = [ S S ]. If the port of the two port is short circuited, the S S S parameter for the resultant one port network is (a) (b) S S S +S S +S S +S S S S +S Soln. In put reflection coefficient = T in (c) (d) S +S S +S S S S S S +S S S [GATE 04: Mark] a b [S] a b Z L Z l = 0 b = S a + S a (I) b = S a + S a (II) T in = b a = S + S a a From (II) b a = S a a + S b a = T L or or where T L is load reflection coefficient T L =, Z L = 0 T L = S a a + S S T L T L = S a a or a a = S T L S T L T in = S + S S T L S T L T L =
so, T in = S S S +S Option (b) = S +S S S S +S 6. Which one of the following filed patterns represents a TEM wave traveling in the positive x direction? (a) E = +8y, H = 4z (c) E = +z, H = +y (b) E = y, H = 3z (d) E = 3y, H = +4z [GATE 04: Mark] Soln. The possible combinations are a y a z = a x a z a y = a x a y a z = a x Y X Z Option (b)