(a) As shown in the figure, just inside the surface of a dielectric slab, the electric field (E₁) is 15 V/m and it makes an angle…

(a) Take region 1 as the dielectric slab and region 2 as air. With no free surface charge at the boundary, the boundary conditions are: tangential electric field continuous, and normal electric flux density continuous. The given angles are with the surface, so for a field E making angle θ with the surface, the tangential component is E cosθ and the normal component is E sinθ.

• Tangential E: E₁ cos30° = E₂ cos65.5°.
• Normal D: ε₁ E₁ sin30° = ε₂ E₂ sin65.5°.

Here ε₂ = ε₀ and ε₁ = εr ε₀, where εr is the relative permittivity, or dielectric constant, of the slab.

From the tangential condition, E₂ = 15 cos30° / cos65.5° = 15(√3/2)/cos65.5°. Using cos30° = 0.866025 and cos65.5° ≈ 0.414694, E₂ ≈ 15(0.866025)/0.414694 ≈ 31.3 V/m.

Divide the normal condition by the tangential condition: (ε₁/ε₂)(E₁ sin30°)/(E₁ cos30°) = (E₂ sin65.5°)/(E₂ cos65.5°), so (ε₁/ε₂) tan30° = tan65.5°. Since ε₁/ε₂ = εr, εr = tan65.5°/tan30°. With tan65.5° ≈ 2.194 and tan30° = 1/√3 ≈ 0.577350, εr ≈ 2.194/0.577350 ≈ 3.80.

Final: E₂ ≈ 31.3 V/m and εr ≈ 3.80, dimensionless.

(b) Let E₂ be the no-load secondary induced emf in volts, V₂ the full-load terminal voltage in volts, I₂ the full-load current in amperes, and R and X the equivalent resistance and reactance in ohms referred to the secondary. Voltage regulation is VR = (E₂ - V₂)/V₂, often expressed as 100 × (E₂ - V₂)/V₂ percent. Take V₂ as the reference phasor.

For a lagging power factor cosφ, the current phasor is I₂∠-φ. The series drop is I₂(R + jX)∠-φ. Hence E₂ = [V₂ + I₂(R cosφ + X sinφ)] + j[I₂(X cosφ - R sinφ)]. The exact magnitude is E₂ = √((V₂ + I₂(R cosφ + X sinφ))² + (I₂(X cosφ - R sinφ))²). For the usual small transformer voltage drop, the quadrature term is neglected in the magnitude approximation, giving E₂ ≈ V₂ + I₂(R cosφ + X sinφ). Therefore VR_lag ≈ I₂(R cosφ + X sinφ)/V₂.

For a leading power factor, the current phasor is I₂∠+φ. Repeating the resolution gives the in-phase drop as I₂(R cosφ - X sinφ), so VR_lead ≈ I₂(R cosφ - X sinφ)/V₂. These expressions show that regulation depends on both cosφ and sinφ, and the sign of the reactive drop changes with the power-factor angle.

(i) Zero regulation requires the approximate in-phase drop to be zero: R cosφ - X sinφ = 0. Thus tanφ = R/X. This is a leading power factor because the reactive drop opposes the resistive drop. The corresponding power factor is cosφ = 1/√(1 + (R/X)²) = X/√(R² + X²), leading.

(ii) Maximum regulation occurs for a lagging power factor, where the resistive and reactive drops add. Maximize f(φ) = R cosφ + X sinφ. Differentiate: df/dφ = -R sinφ + X cosφ. Set df/dφ = 0: X cosφ = R sinφ, so tanφ = X/R. The second derivative is -R cosφ - X sinφ, which is negative for positive R and X, so this gives a maximum. The power factor is cosφ = R/√(R² + X²), lagging. The maximum approximate regulation is I₂√(R² + X²)/V₂. Validity: small drops, I₂√(R² + X²) << V₂.

(c) Given: Vs = 230 V rms, f = 50 Hz, Ls = 2 mH = 0.002 H, Id = 10 A, α = 30°. The thyristors are ideal, so their voltage drops are neglected; only source inductance causes overlap and output voltage drop.

The peak supply voltage is Vm = 230√2 V ≈ 325.27 V. The angular frequency is ω = 2πf = 100π rad/s ≈ 314.16 rad/s.

For a single-phase full-bridge thyristor converter with source inductance, the commutation relation from volt-second balance is cosα - cos(α + u) = 2ωLsId/Vm. Substitute the data: 2ωLsId/Vm = 2(100π)(0.002)(10)/(230√2) = 4π/(230√2) ≈ 0.03863. Therefore cos(30° + u) = cos30° - 0.03863 = √3/2 - 0.03863 ≈ 0.866025 - 0.038634 = 0.827391. Hence 30° + u = cos⁻¹(0.827391) ≈ 34.17°, and u ≈ 34.17° - 30° = 4.17°.

The average output voltage drop is the average of the voltage lost during the overlap interval. The lost volt-seconds per half cycle are ∫ from α to α+u of Vm sin(ωt) d(ωt) = Vm[cosα - cos(α + u)]. Using the commutation relation, this equals 2ωLsId. Averaging over a half cycle of π radians gives ΔV = 2ωLsId/π. Substitute: ΔV = 2(100π)(0.002)(10)/π = 4 V.

Final: u ≈ 4.17° and ΔV = 4.00 V.

(d) Let X be a binomial random variable with n trials and success probability p. Put q = 1 - p. The probability mass function is P(X = k) = C(n,k)pᵏqⁿ⁻ᵏ, for k = 0,1,...,n, where C(n,k) is the binomial coefficient.

The mean is E[X] = Σ over k = 0 to n of k C(n,k)pᵏqⁿ⁻ᵏ. The k = 0 term is zero. Use the identity k C(n,k) = n C(n-1,k-1). Then E[X] = n p Σ over k = 1 to n of C(n-1,k-1)pᵏ⁻¹qⁿ⁻ᵏ. Let j = k - 1. Then j runs from 0 to n-1, and E[X] = n p Σ over j = 0 to n-1 of C(n-1,j)pʲqⁿ⁻¹⁻ʲ. By the binomial theorem, the sum is (p + q)ⁿ⁻¹ = 1. Therefore E[X] = np.

For the variance, compute E[X(X-1)] first: E[X(X-1)] = Σ over k = 0 to n of k(k-1) C(n,k)pᵏqⁿ⁻ᵏ. Terms k = 0 and k = 1 are zero. Use k(k-1) C(n,k) = n(n-1) C(n-2,k-2). Then E[X(X-1)] = n(n-1)p² Σ over k = 2 to n of C(n-2,k-2)pᵏ⁻²qⁿ⁻ᵏ. Let j = k - 2. Then E[X(X-1)] = n(n-1)p² Σ over j = 0 to n-2 of C(n-2,j)pʲqⁿ⁻²⁻ʲ = n(n-1)p²(p + q)ⁿ⁻² = n(n-1)p².

Now E[X²] = E[X(X-1) + X] = E[X(X-1)] + E[X] = n(n-1)p² + np. Hence Var(X) = E[X²] - (E[X])² = n(n-1)p² + np - (np)² = n(n-1)p² + np - n²p² = np - np² = np(1-p). This holds for n a non-negative integer and 0 ≤ p ≤ 1.

Final: mean = np and variance = np(1-p).

(e) For the standard high-side injection FM superheterodyne arrangement, the local oscillator is above the carrier: fLO = fc + fIF. The mixer produces the difference fLO - fc = fIF. The image frequency f′c is the unwanted RF that also gives the same IF: |f′c - fLO| = fIF. With high-side injection, the image lies above the LO, so f′c = fLO + fIF = fc + 2fIF.

The receiver tuning range is fc = 88 MHz to 108 MHz. The image must fall outside this band. Since f′c = fc + 2fIF increases with fc, the lowest image frequency occurs at fc = 88 MHz. To keep even this lowest image outside the band, it must be at least the upper end of the band: 88 MHz + 2fIF ≥ 108 MHz. Therefore 2fIF ≥ 20 MHz, fIF ≥ 10 MHz. The minimum required IF is fIF(min) = 10 MHz.

For this chosen IF, the local oscillator frequency is fLO = fc + 10 MHz. When fc = 88 MHz, fLO = 98 MHz. When fc = 108 MHz, fLO = 118 MHz. Thus the LO must vary from 98 MHz to 118 MHz. This also satisfies fIF < fLO because 10 MHz is less than every LO frequency in the range.

Final: minimum fIF = 10 MHz; corresponding fLO range = 98 MHz to 118 MHz.