(a) For the circuit shown below, if the power loss in 1 Ω resistor is 16 W, find the value of k in the dependent source using…

(a) Nodal analysis with a dependent source (10 marks)

The ideal 20 V source is connected directly across node B, so V_B = 20 V and the 5 Ω resistor plays no part in the nodal equation. Let I be the current through the 1 Ω resistor from A towards the dependent source, and i₀ = V_A/2 the current in the 2 Ω resistor.

The power loss in the 1 Ω resistor is I² × 1 = 16 W, so |I| = 4 A. Taking I = +4 A (from A to B):

Node A (KCL): 10 = V_A/2 + I, so V_A/2 = 6 and V_A = 12 V, i₀ = 6 A.

The dependent source is drawn with its negative terminal on the 1 Ω side and its positive terminal towards B. The voltage at its left terminal is V_A − 1 × I = 12 − 4 = 8 V, so

V_B = (V_A − I) + k i₀, that is 20 = 8 + 6k, giving k = 2.

In general form, with the source voltage k i₀ = k V_A/2, the node-A equation is V_A/2 + (V_A − V_x) = 10 with V_x = 20 − kV_A/2, that is V_A (3 + k)/2 = 30. For k = 2 this gives V_A = 12 V, and the loss in 1 Ω is (12 − 8)² = 16 W, as required.

The reversed direction I = −4 A (current from B to A) also gives 16 W and leads to V_A = 28 V, i₀ = 14 A and k = −6/7. The intended, physically natural solution, with the 10 A source feeding current towards B, is k = 2.

(b) Cascaded moving-window integrators (10 marks)

System 1 is w(t) = ∫ x(τ) dτ over [t − 3, t] and System 2 is y(t) = ∫ w(τ) dτ over [t − 6, t].

(i) Linearity and time-invariance. Each system is an integral of the input over a window that moves with t, so each obeys superposition and each shifts with the input: replacing x(τ) by x(τ − t₀) shifts the output by t₀. Both stages are therefore linear and time-invariant, and so is their cascade. A cascade of LTI systems is LTI, and the order of the two stages does not matter.

(ii) Impulse response. Put x(t) = δ(t). For System 1, h₁(t) = ∫ δ(τ) dτ over [t − 3, t] = 1 for 0 ≤ t ≤ 3, so h₁(t) = u(t) − u(t − 3), a rectangular pulse of height 1 and width 3. Likewise h₂(t) = u(t) − u(t − 6), a rectangle of width 6. The overall impulse response is the convolution

h(t) = h₁(t) * h₂(t) = ∫ h₁(τ) h₂(t − τ) dτ.

Convolving the two rectangles gives a trapezoid:

• h(t) = 0 for t < 0 and t > 9;
• h(t) = t for 0 ≤ t ≤ 3 (rising);
• h(t) = 3 for 3 ≤ t ≤ 6 (flat);
• h(t) = 9 − t for 6 ≤ t ≤ 9 (falling).

The base width is 3 + 6 = 9 s, the flat top has height equal to the shorter width, 3, and the area is 3 × 6 = 18, equal to the product of the areas of the two rectangles.

(c) Magnetic field of two equal line currents (10 marks)

Taking both currents I as equal and in the same direction, out of the page (+z), as the dots in the figure show, use Ampere's law for an infinite line current: B = μ₀I/(2πr) in the circumferential direction. A point P at (x, 0) is at distance r = √(x² + d²/4) from each wire, and the line from each wire to P makes an angle θ with the x-axis, where sin θ = (d/2)/r and cos θ = x/r.

Each field is perpendicular to the line joining that wire to P. The two fields have equal magnitude, μ₀I/(2πr). Their y-components add, since the wires are symmetric about y = 0, while their x-components, ±(μ₀I/2πr) sin θ, are equal and opposite and cancel. The resultant is

B = 2 × μ₀I/(2πr) × cos θ = μ₀I x/(π r²) = μ₀ I x/(π (x² + d²/4)) = 4μ₀ I x/(π (4x² + d²)),

directed along +y for x > 0 and along −y for x < 0 (the field lies wholly in the plane, perpendicular to the line joining the wires).

Checks: B = 0 at x = 0, midway between the wires, where the two fields are equal and opposite; and for x ≫ d, B → μ₀(2I)/(2πx), the field of a single wire carrying 2I. The magnitude is greatest at x = ±d/2, where B_max = μ₀I/(πd). If the currents were opposite, the y-components would cancel and the field would be along x.

(d) Combinational logic with two decoders (10 marks)

Take A₁ (= X) as the most significant input and active-high outputs, so the first decoder gives D₀ = X′Y′, D₁ = X′Y, D₂ = XY′ and D₃ = XY.

In the figure only D₀ and D₃ are connected: both go to the two gates, with D₁ and D₂ left open. The upper gate is an XOR and the lower is an XNOR (a bubble on its output). The inputs to the second decoder are

A₂ = D₀ ⊕ D₃ and B₂ = (D₀ ⊕ D₃)′.

D₀ and D₃ can never be 1 together, so D₀ ⊕ D₃ = D₀ + D₃ = X′Y′ + XY, that is A₂ = X ⊙ Y (XNOR) and B₂ = X ⊕ Y (XOR), the complement of A₂.

The second decoder outputs are then Y₀ = A₂′B₂′, Y₁ = A₂′B₂, Y₂ = A₂B₂′ and Y₃ = A₂B₂:

• Y₁ = A₂′B₂ = (X ⊕ Y)(X ⊕ Y) = X ⊕ Y = XY′ + X′Y;
• Y₂ = A₂B₂′ = (X ⊙ Y)(X ⊙ Y) = X ⊙ Y = XY + X′Y′.

Since A₂ and B₂ are always complementary, Y₀ = Y₃ = 0 and only Y₁ or Y₂ can be high: Y₁ = 1 when X ≠ Y and Y₂ = 1 when X = Y.

(e) DC shunt generator with a higher speed and added field resistance (10 marks)

Rated data. Rated output current I_L = 4500/220 = 20.45 A. Field current I_f = P_f/V = 50/220 = 0.2273 A, so the field resistance is R_f = 220/0.2273 = 968 Ω. Armature current I_a = 20.45 + 0.227 = 20.68 A, so R_a = 200/20.68² = 0.4676 Ω and the generated emf is E = 220 + 20.68 × 0.4676 = 229.67 V.

(i) Added resistance. At 1.2 N the emf must remain 229.67 V. With a linear magnetic circuit, E ∝ Nφ ∝ N I_f, so the flux, and hence I_f, must fall to 1/1.2 of its value: I_f′ = 0.2273/1.2 = 0.1894 A. The terminal voltage stays 220 V, so the total field-circuit resistance is 220/0.1894 = 1161.6 Ω, and the added resistance is 1161.6 − 968 = 193.6 Ω.

(ii) Loss in the external resistance and field. Field-circuit loss = V I_f′ = 220 × 0.1894 = 41.7 W, made up of 0.1894² × 968 = 34.7 W in the field winding and 0.1894² × 193.6 = 6.9 W in the external resistance (before the change the field loss was 50 W).

(iii) Input power. The load is unchanged, so the output is 4500 W at 220 V and I_L = 20.45 A. The armature current is I_a′ = 20.45 + 0.189 = 20.64 A, and the armature copper loss is 20.64² × 0.4676 = 199.3 W. The core loss is constant at 250 W, and friction and windage are neglected. Then

P_in = 4500 + 199.3 + 41.7 + 250 = 4991 W ≈ 4.99 kW.

(iv) Armature developed voltage. E′ = V + I_a′ R_a = 220 + 20.64 × 0.4676 = 229.65 V ≈ 229.7 V, the same as before, as the problem requires.