- 2068 Chaitra · 2+2+2+2 marks
Phasor E in free space, find f η Hs
Short answer
Frequency f = βc/(2π) = 2.39 GHz, wave impedance η = 377 Ω. Phasor H_s = 0.265∠45° e^{−j50z} a_y A/m. At z=10 mm and t=20 ps, instantaneous electric field magnitude is 83.3 V/m.
The phasor component of electric field intensity in free space is given by E_s = (100∠45°) e^(−j50z) a_x V/m. Determine the frequency of the wave, wave impedance, H_s, and magnitude of E at z = 10 mm, t = 20 ps.
Given: free space, E_s = (100∠45°) e^{−j50z} a_x V/m. So β = 50 rad/m, E₀ = 100 V/m, phase 45°, and the wave travels in +z.
(a) Frequency
ω = β c = 50 × 3×10⁸ = 1.5×10¹⁰ rad/s
f = ω/2π = 1.5×10¹⁰/6.2832 = 2.387×10⁹ Hz ≈ 2.39 GHz
(b) Wave impedance
η₀ = 120π = 377 Ω
(c) H_s. For travel along +z with E along a_x, H points along a_y (a_x × a_y = a_z):
H_s = (E₀/η₀)∠45° e^{−j50z} a_y
= (100/377) ∠45° e^{−j50z} a_y
= 0.2653 ∠45° e^{−j50z} a_y A/m
(d) |E| at z = 10 mm, t = 20 ps.
E(z,t) = 100 cos(ωt − βz + 45°)
ωt = 1.5×10¹⁰ × 20×10⁻¹² = 0.3 rad
βz = 50 × 0.01 = 0.5 rad
45° = 0.7854 rad
phase = 0.3 − 0.5 + 0.7854 = 0.5854 rad
E = 100 cos(0.5854) = 100 × 0.8335 = 83.3 V/m (a_x direction)
Answer: f = 2.39 GHz; η = 377 Ω; H_s = 0.265∠45° e^{−j50z} a_y A/m; E(10 mm, 20 ps) = 83.3 a_x V/m.
✓ Answer checked by Claude Opus 5.5 (high effort)
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