(a) (i) The continuous-time signal x(t) = cos(20πt + π/4) is ideally sampled at a rate of 30 samples per second. The resulting…

(a)(i) Sampled cosine passed through a band-pass filter (15 marks)

Spectrum of the sampled signal. x(t) = cos(20πt + π/4) has frequency f₀ = 10 Hz. Ideal sampling at fₛ = 30 Hz multiplies x(t) by an impulse train, whose Fourier series is fₛ Σₖ e^j2πk fₛ t, so the sampled spectrum repeats X(f) at every multiple of fₛ:

Xₛ(f) = fₛ Σₖ X(f − k fₛ), with X(f) = ½[e^jπ/4 δ(f − 10) + e^−jπ/4 δ(f + 10)].

The spectral lines therefore lie at f = ±10 + 30k Hz, that is 10, 20, 40, 50, 70, 80 … Hz and their negatives; the one at +40 Hz comes from k = 1 (10 + 30) and has weight fₛ × ½ e^jπ/4 = 15 e^jπ/4.

The filter. Since sin(πt)/(πt) has transform rect(f) (unit height, |f| < 0.5 Hz), and cos(80πt − π/2) = ½[e^−jπ/2 e^j2π40t + e^jπ/2 e^−j2π40t], modulation shifts the rectangle to ±40 Hz:

H(f) = ½ e^−jπ/2 rect(f − 40) + ½ e^jπ/2 rect(f + 40).

It is a narrow band-pass filter with pass band 39.5–40.5 Hz (and its mirror image), gain 1/2 and phase −90° at +40 Hz (H(40) = ½ e^−jπ/2 = −j/2), and the opposite phase at −40 Hz.

Output. Only the ±40 Hz lines fall in the pass band, and no other line (20, 50 Hz …) does. The component at +40 Hz, 15 e^j(80πt + π/4), is multiplied by H(40), and the −40 Hz component by H(−40) = H(40)*. Their sum is

y(t) = 2 Re15 e^j(80πt + π/4) × ½ e^−jπ/2 = 15 cos(80πt − π/4).

The 10 Hz signal has thus been aliased to 40 Hz (= 30 + 10) by the sampling and picked out by the filter, with amplitude 15 = 30 × ½ (line weight) × ½ (filter gain) × 2 (conjugate pair) and phase π/4 − π/2 = −π/4.

(a)(ii) Circular shift by the DFT property (5 marks)

N = 9, and e^j4πk/3 = e^j2πk·6/9. By the circular time-shift property, x((n + m))_N ↔ e^j2πkm/N X(k), so here m = 6 and

y(n) = x((n + 6))₉ = x((n − 3))₉,

a circular right shift of x(n) by 3 samples (the same as a left shift by 6). Shifting 3, 5, 1, 4, −3, 5, −2, −2, 4 right by 3, the last three samples move to the front:

y(n) = −2, −2, 4, 3, 5, 1, 4, −3, 5.

(b)(i) Bias of a Darlington emitter follower (10 marks)

Derivation of the relations. For the pair, the emitter current of Q₁ is the base current of Q₂: I_B2 = I_E1 = (1 + β₁) I_B1. The output current is I_E2 = (1 + β₂) I_B2, so

I_E2 = (1 + β₁)(1 + β₂) I_B1.

The base of Q₁ is one V_BE for Q₁ and one for Q₂ above the output: V_B1 = Vₒ + 2V_BE.

Numbers.

• I_E2 = Vₒ/Rₒ = 5.0/1 kΩ = 5.0 mA.
• I_B1 = I_E2/[(1 + β₁)(1 + β₂)] = 5 mA/(36 × 31) = 4.48 μA.
• V_B1 = 5.0 + 2 × 0.7 = 6.4 V.
• Current in R₁: I_R1 = (V_CC − V_B1)/R₁ = (12 − 6.4)/100 kΩ = 56 μA.
• Current in R₂: I_R2 = I_R1 − I_B1 = 56 − 4.48 = 51.52 μA.

Then

R₂ = V_B1/I_R2 = 6.4/51.52 μA ≈ 124 kΩ.

In general form, R₂ = (Vₒ + 2V_BE)/[V_CC − Vₒ − 2V_BE]/R₁ − Vₒ/[Rₒ(1 + β₁)(1 + β₂)].

(b)(ii) Bridge rectifier with a faulty diode (10 marks)

The bridge is the standard one, with D₁ and D₃ tied to the positive rail and D₂ and D₄ to the negative rail. D₂ behaves as a 1 kΩ resistor in both directions. The supply is vₛ = Vₘ sin ωt with Vₘ = 240√2 = 339.4 V.

Positive half-cycle (upper terminal positive). The path is D₁ → R_L → the faulty D₂ (1 kΩ), so the series resistance is 100 + 1000 = 1100 Ω. i_L = vₛ/1100 (peak 0.309 A) and v_L = vₛ/11 (peak 30.9 V); the source current equals i_L.

Negative half-cycle. The path is D₃ → R_L → D₄, both ideal, so v_L = |vₛ| (peak 339.4 V) and i_L = |vₛ|/100 (peak 3.39 A). In addition the 1 kΩ resistance of D₂ connects the negative rail directly to the lower supply terminal, so it carries |vₛ|/1000, and the source supplies iₛ = −|vₛ| (1/100 + 1/1000) = −|vₛ|/90.9 (peak −3.73 A).

Waveforms. The output is strongly unbalanced: a small hump (30.9 V) in one half-cycle and a full 339.4 V hump in the other. The source current is asymmetric, small and positive (0.31 A peak) in one half-cycle and large and negative (3.73 A peak) in the other, so the supply current has a large dc component.

Average load voltage.

V_L(avg) = (1/2π)[∫₀^π (Vₘ/11) sin θ dθ + ∫_π²π Vₘ |sin θ| dθ] = (Vₘ/2π)(2/11 + 2) = (Vₘ/π)(12/11) = (339.4/π)(12/11) ≈ 117.9 V.

RMS source current. The positive half-cycle has peak 0.3086 A and the negative half-cycle has peak 3.7335 A. Each half-cycle is a half-sine whose mean square over the full period is (peak)²/4, so

Iₛ(rms) = √¼ [0.3086² + 3.7335²] = √¼ (0.0952 + 13.939) = √3.509 ≈ 1.87 A.

(c) Average power delivered to the load on a shorted line (10 marks)

Line constants. Z₀ = √(L/C) = √(0.2 × 10⁻⁶/80 × 10⁻¹²) = 50 Ω and v = 1/√(LC) = 2.5 × 10⁸ m/s, so λ = v/f = 2.5 m and β = 2π/λ = 2.513 rad/m.

Circuit. The load Z_L = 50 + j20 Ω is shunted across the line at z = −20 cm. The section from z = −20 cm to the short at z = 0 is a shorted stub of length 0.2 m = 0.08λ (βℓ = 28.8°):

Z_stub = jZ₀ tan βℓ = j50 tan 28.8° = j27.5 Ω.

At the junction the stub is in parallel with Z_L:

Zⱼ = Z_L Z_stub/(Z_L + Z_stub) = (50 + j20)(j27.5)/(50 + j47.5) = 7.94 + j19.9 Ω.

The section from the junction to the input, z = −50 cm, is 0.3 m = 0.12λ (βℓ = 43.2°, tan = 0.94), so

Z_in = Z₀ (Zⱼ + jZ₀ tan βℓ)/(Z₀ + jZⱼ tan βℓ) = 50 (7.94 + j66.9)/(31.3 + j7.5) ≈ 36.2 + j98.3 Ω.

Power. The line is lossless and the short and the stub absorb no power, so all the real power at the input, taking 100 V as an r.m.s. value, reaches Z_L:

P = |V_in|² Re(Z_in)/|Z_in|² = 100² × 36.2/(36.2² + 98.3²) = 362 000/10 974 ≈ 32.9 W.

Check by the load voltage: V_L = V_in/[cos βℓ + j(Z₀/Zⱼ) sin βℓ] gives |V_L| = 43.7 V, and P = |V_L|² Re(1/Z_L) = 43.7² × 50/2900 = 32.9 W. If 100 V were a peak value, the power would be half, 16.5 W.