(a) (i) Find the values of R and C in the figure shown below, so that V_b = 3V_a, also V̄_b and V̄_a are in quadrature. Find also…

(a)(i) Series R–C design with Vb = 3Va and quadrature (10 marks)

The reactance of the coil is X_L = 2π × 50 × 0.0255 = 8.01 ≈ 8 Ω, so the first section has impedance Z_b = 6 + j8, with |Z_b| = 10 Ω and an angle of tan⁻¹(8/6) = 53.13° (V_b leads I). The second section is Zₐ = R − jX_C, with Vₐ lagging I by φₐ = tan⁻¹(X_C/R).

Magnitude condition. The same current I flows in both, so V_b = 3Vₐ means |Zₐ| = |Z_b|/3 = 10/3 = 3.333 Ω.

Quadrature condition. V_b leads I by 53.13° and Vₐ lags I by φₐ, so the angle between them is 53.13° + φₐ = 90°, giving φₐ = 36.87°. Then

• R = |Zₐ| cos 36.87° = 3.333 × 0.8 = 2.67 Ω;
• X_C = |Zₐ| sin 36.87° = 3.333 × 0.6 = 2.00 Ω, so C = 1/(2π × 50 × 2) = 1.592 × 10⁻³ F = 1592 μF (≈ 1.59 mF).

Current and phase relations. The total impedance is Z = (6 + 2.667) + j(8 − 2) = 8.667 + j6 Ω, with |Z| = 10.54 Ω and angle tan⁻¹(6/8.667) = 34.70°. Hence I = 240/10.54 = 22.77 A, V_b = 227.7 V and Vₐ = 75.9 V. Check: √(227.7² + 75.9²) = 240 V, as it must be for quadrature phasors.

• V and V_b: V leads I by 34.70° and V_b leads I by 53.13°, so V lags V_b by 18.43° (tan⁻¹(Vₐ/V_b) = tan⁻¹(1/3)).
• Vₐ and I: Vₐ lags I by 36.87°, since the section is R–C. It follows that V leads Vₐ by 71.57°.

(a)(ii) Value of R for equal source powers (10 marks)

Let V₁ be the node voltage across the 10 Ω resistor and V₂ the node voltage across R (both above ground). The current source drives V₂.

Node 1: (V₁ − 100)/5 + V₁/10 + (V₁ − V₂)/20 = 0, that is 4(V₁ − 100) + 2V₁ + (V₁ − V₂) = 0, so 7V₁ − V₂ = 400.

Node 2: (V₂ − V₁)/20 + V₂/R = 10.

Equal powers. The 100 V source delivers Pᵥ = 100 × (100 − V₁)/5 = 20(100 − V₁). The 10 A source delivers Pᵢ = 10 V₂. Setting Pᵥ = Pᵢ gives 2000 − 20V₁ = 10V₂, that is V₂ = 200 − 2V₁.

Substituting into the node-1 equation: 7V₁ − 200 + 2V₁ = 400, so 9V₁ = 600, V₁ = 66.67 V and V₂ = 66.67 V. No current flows in the 20 Ω resistor, and node 2 gives V₂/R = 10, so

R = 66.67/10 = 6.67 Ω (= 20/3 Ω).

Check: Pᵥ = 20 × 33.33 = 666.7 W and Pᵢ = 10 × 66.67 = 666.7 W.

(b)(i) Fourier-transform properties of a triangular pulse (10 marks)

The pulse rises from 0 at t = −1 to 2 at t = 1 and falls to 0 at t = 3, with x(0) = 1. It is symmetric about t = 1.

1. X(0) is the area under x(t): X(0) = ∫x(t) dt = ½ × 4 × 2 = 4.
2. ∫X(ω) dω. From the inverse transform x(t) = (1/2π) ∫X(ω) e^jωt dω, at t = 0: ∫X(ω) dω = 2π x(0) = 2π.
3. ∫|X(ω)|² dω. By Parseval's theorem this is 2π ∫x²(t) dt. The left half gives ∫(t + 1)² dt from −1 to 1 = 8/3, and the right half is the same: total 16/3. Hence ∫|X(ω)|² dω = 2π × 16/3 = 32π/3 ≈ 33.5.
4. Phase of X(ω). Write x(t) = g(t − 1), where g(t) is a real even triangle (peak 2, half-width 2). Then G(ω) is real, and by the time-shift property X(ω) = G(ω) e^−jω. Here G(ω) = 4 sin²ω/ω² ≥ 0, so the phase is ∠X(ω) = −ω (linear, with slope −1 rad per rad/s), with no jumps of π.

(b)(ii) Z-transform of x(n) = |n|(1/2)^|n| (10 marks)

Split the sequence into a right-sided part (n ≥ 0) and a left-sided part (n < 0). For n ≥ 0 the term is n(1/2)ⁿ, and for n < 0, with n = −m, it is m(1/2)^m.

Right-sided part. Using Σ n aⁿ z⁻ⁿ = a z⁻¹/(1 − a z⁻¹)² with a = 1/2:

X₁(z) = (½ z⁻¹)/(1 − ½ z⁻¹)² = ½ z/(z − ½)² = 2z/(2z − 1)², for |z| > 1/2.

Left-sided part. Σₘ≥1 m (z/2)^m = (z/2)/(1 − z/2)² = 2z/(z − 2)², for |z| < 2.

The two-sided transform is the sum:

X(z) = 2z/(2z − 1)² + 2z/(z − 2)² = 2z(5z² − 8z + 5)/[(2z − 1)²(z − 2)²],

with the region of convergence the annulus 1/2 < |z| < 2, the intersection of the two parts. Check: at z = 1 (inside the ROC) X(1) = 2 + 2 = 4, equal to Σ|n|(1/2)^|n| = 2 × 2 = 4. The unit circle is inside the ROC, so the Fourier transform exists.

(c) Standing-wave ratio in front of a lossless dielectric slab (10 marks)

The slab has air on both sides, so it is terminated in η₀ = 377 Ω. Taking the thickness λ/8 as measured in the dielectric, its electrical length is βd = (2π/λ)(λ/8) = π/4 (45°), so tan βd = 1. The input impedance seen at the front face is

Z_in = η (η₀ + jη tan βd)/(η + jη₀ tan βd) = 260 (377 + j260)/(260 + j377) = 243.0 − j92.4 Ω.

The reflection coefficient at the front face, with air (η₀) in front, is

Γ = (Z_in − η₀)/(Z_in + η₀) = (−134.0 − j92.4)/(620.0 − j92.4), with |Γ| = 162.8/626.9 = 0.260.

Alternatively, in closed form |Γ| = (η₀² − η²)/√[4η²η₀² + (η² + η₀²)²] = 74 522/287 077 = 0.260.

The standing-wave ratio in front of the plate is

S = (1 + |Γ|)/(1 − |Γ|) = 1.260/0.740 ≈ 1.70.

At a quarter-wave thickness the slab would reflect more strongly, and at half-wave thickness not at all; at λ/8 the reflection is intermediate, with about 6.7% of the incident power reflected (|Γ|² = 0.067).