Loss tangent, α, β, η at 5 MHz

Given: ε_r = 2.25, σ = 10⁻⁴ S/m, μ = μ₀ (non-magnetic), f = 5 MHz.

ω  = 2π × 5×10⁶ = 3.1416×10⁷ rad/s
ε  = 2.25 × 8.854×10⁻¹² = 1.9922×10⁻¹¹ F/m
μ₀ = 4π×10⁻⁷ H/m

(a) Loss tangent

tan δ = σ/(ωε) = 10⁻⁴ / (3.1416×10⁷ × 1.9922×10⁻¹¹)
      = 10⁻⁴ / 6.2587×10⁻⁴
      = 0.160

Answer (a): tan δ ≈ 0.16. Since 0.16 ≪ 1, the medium is a low-loss dielectric.

Common factor: ω√(με/2) = 3.1416×10⁷ × √(4π×10⁻⁷ × 1.9922×10⁻¹¹ / 2) = 0.1111 (used below).

(b) Attenuation constant

α = ω√(με/2) · √(√(1 + tan²δ) − 1)
√(1 + 0.02553) = 1.01269 → 1.01269 − 1 = 0.01269 → √ = 0.11265
α = 0.1111 × 0.11265 = 0.0125 Np/m

Check with the low-loss formula α ≈ (σ/2)√(μ/ε) = 0.5×10⁻⁴×√(4π×10⁻⁷/1.9922×10⁻¹¹) = 0.0126 Np/m.

Answer (b): α ≈ 0.0125 Np/m.

(c) Phase constant

β = ω√(με/2) · √(√(1 + tan²δ) + 1)
  = 0.1111 × √(2.01269) = 0.1111 × 1.41869
  = 0.158 rad/m

Answer (c): β ≈ 0.158 rad/m.

(d) Intrinsic impedance

|η| = √(μ/ε) / (1 + tan²δ)^(1/4)
√(μ/ε) = √(4π×10⁻⁷ / 1.9922×10⁻¹¹) = 251.2 Ω
(1.02553)^(1/4) = 1.00634
|η| = 251.2 / 1.00634 = 249.6 Ω
θ_η = ½ tan⁻¹(tan δ) = ½ × 9.08° = 4.54°

Answer (d): η ≈ 249.6∠4.54° Ω (about 250 Ω).