(a) In the figure given below, region 1 is the side of the plane y+z=1 containing the origin and in this region, μᵣ₁ = 5. In…

(a) B₂ and H₂ across the plane y + z = 1

Normal to the interface. The plane y + z = 1 has normal direction (0, 1, 1). Region 1 contains the origin (y + z < 1), so the unit normal pointing from region 1 into region 2 is n̂ = (a_y + a_z)/√2

Boundary conditions (no surface current on the interface): the normal component of B is continuous, B₁ₙ = B₂ₙ, and the tangential component of H is continuous, H₁ₜ = H₂ₜ. The second condition gives B₂ₜ/μᵣ₂ = B₁ₜ/μᵣ₁, so B₂ₜ = (μᵣ₂/μᵣ₁)B₁ₜ.

Normal component of B₁. B₁ · n̂ = (3a_x + a_y) · (a_y + a_z)/√2 = 1/√2 = 0.707 T B₁ₙ = 0.707 n̂ = 0.5a_y + 0.5a_z T

Tangential component of B₁. B₁ₜ = B₁ − B₁ₙ = 3a_x + 0.5a_y − 0.5a_z T (check: B₁ₜ · n̂ = (0.5 − 0.5)/√2 = 0)

Region 2. B₂ₙ = B₁ₙ = 0.5a_y + 0.5a_z T B₂ₜ = (7/5)B₁ₜ = 4.2a_x + 0.7a_y − 0.7a_z T B₂ = B₂ₜ + B₂ₙ = 4.2a_x + 1.2a_y − 0.2a_z T (|B₂| = 4.37 T)

H₂. μ₂ = 7μ₀ = 7 × 4π×10⁻⁷ = 8.796×10⁻⁶ H/m H₂ = B₂/μ₂ = (4.2a_x + 1.2a_y − 0.2a_z)/(8.796×10⁻⁶) H₂ = 4.77×10⁵ a_x + 1.36×10⁵ a_y − 0.23×10⁵ a_z A/m (≈ 477.5a_x + 136.4a_y − 22.7a_z kA/m)

Checks. H₁ = B₁/(5μ₀) = 4.77×10⁵ a_x + 1.59×10⁵ a_y A/m. Its tangential part, H₁ₜ = B₁ₜ/(5μ₀) = 4.77×10⁵ a_x + 0.80×10⁵ a_y − 0.80×10⁵ a_z A/m, equals H₂ₜ = B₂ₜ/(7μ₀), as it must. Measured from the normal, tan θ₁ = |B₁ₜ|/B₁ₙ = 3.082/0.707 = 4.36 (θ₁ = 77.1°) and tan θ₂ = 4.315/0.707 = 6.10 (θ₂ = 80.7°). The ratio tan θ₁/tan θ₂ = 0.714 = μᵣ₁/μᵣ₂, so the field bends away from the normal on entering the more permeable region.

(b) Transmitter power and channel bandwidth for DSB-SC, SSB and AM

Data. Message bandwidth W = 20 kHz; message power m²‾ = 20 W; peak amplitude m_p = 8; channel loss 80 dB (a factor of 10⁸); N₀/2 = 0.5×10⁻¹² W/Hz, so N₀ = 10⁻¹² W/Hz. The 50 dB requirement is read as the SNR at the demodulator (receiver) output: S_o/N_o ≥ 10⁵.

Reference SNR. γ = S_i/(N₀W), where S_i is the received signal power. N₀W = 10⁻¹² × 2×10⁴ = 2×10⁻⁸ W Transmitted power P_T = S_i × 10⁸

(i) DSB-SC AM. With coherent detection, S_o/N_o = γ. The detector's 3 dB gain exactly offsets the doubled noise bandwidth (2W). S_i = 10⁵ × 2×10⁻⁸ = 2×10⁻³ W P_T = 2×10⁻³ × 10⁸ = 2×10⁵ W = 200 kW; channel bandwidth B_T = 2W = 40 kHz

(ii) SSB AM. Also S_o/N_o = γ, with no detection gain and noise bandwidth W. P_T = 200 kW; B_T = W = 20 kHz (the same power as DSB-SC, in half the bandwidth)

(iii) Conventional AM with μ = 0.6. The transmitted signal is [A + m(t)] cos ω_c t with μ = m_p/A. A = 8/0.6 = 13.33, so A² = 177.8 With envelope detection above threshold (or with coherent detection), S_o/N_o = [m²‾/(A² + m²‾)]γ = ηγ η = 20/(177.8 + 20) = 0.101 (only 10.1% of the power carries the message) γ = 10⁵/0.101 = 9.89×10⁵, so S_i = 9.89×10⁵ × 2×10⁻⁸ = 1.98×10⁻² W P_T = 1.98×10⁶ W ≈ 1.98 MW; B_T = 2W = 40 kHz

Comparison. DSB-SC and SSB need the same power, and SSB uses half the bandwidth. AM with μ = 0.6 needs about 9.9 times the power (about 10 dB more), because about 90% of the transmitted power is in the carrier. At this received SNR the envelope detector is well above its threshold, so the high-SNR formula applies.

(c) DC-DC converter: inductor ripple and peak switch current

Identifying the converter. The switch S is in series with the source, and the inductor L connects the switch node to the common return. The diode points from the output towards the switch node: its anode is at the capacitor and its cathode at the switch node. This is the buck-boost (inverting) converter, not a boost converter. When S is on, V_s appears across L and the diode is reverse-biased. When S is off, the inductor current keeps flowing down through L and is drawn through the diode from the capacitor. The top plate of C therefore charges negative with respect to the common line.

Output voltage (volt-second balance on L): V_sD = |V_o|(1 − D)

|V_o| = DV_s/(1 − D) = 0.25 × 20/0.75 = 6.67 V. The polarity is inverted, so V_o = −6.67 V with respect to the terminal marked +.

Average inductor current. The diode carries the inductor current only during the off time, so I_D = I_L(1 − D). I_L = 1.2/0.75 = 1.6 A. The load current is I_o = I_D = 1.2 A, which gives R = 6.67/1.2 = 5.56 Ω.

(i) Peak-to-peak inductor ripple. During the on time, L di/dt = V_s: Δi_L = V_sDT/L = V_sD/(Lf) = (20 × 0.25)/(150×10⁻⁶ × 20×10³) = 1.67 A Check over the off interval: |V_o|(1 − D)/(Lf) = 6.67 × 0.75/3 = 1.67 A.

(ii) Peak current through the switch. The switch carries the inductor current during the on time, so its peak equals the peak inductor current: I_S,pk = I_L + Δi_L/2 = 1.6 + 0.833 = 2.43 A The minimum inductor current is 1.6 − 0.833 = 0.77 A, which is above zero. The converter is in continuous conduction, so the formulas above apply.

Output ripple (for completeness). During the on time the capacitor alone supplies the 1.2 A load: ΔV_o = I_oD/(fC) = 1.2 × 0.25/(20×10³ × 240×10⁻⁶) = 62.5 mV (about 0.94% of 6.67 V)

A boost reading of the circuit happens to give the same two requested currents, because I_L = I_D/(1 − D) and Δi_L = V_sDT/L hold for both converters. It would give a wrong output voltage of 26.7 V, however.