(a) In Figure 1(a) shown below, the two-port network is characterized in terms of y-parameters with y₁₁ = 3·3 × 10⁻³ S, y₂₂ = 5 × 10⁻³ S and y₁₂ = y₂₁ = 0. Find the voltage across 200 Ω load. (10 marks)

(b) For the signal shown in Figure 1(b), calcu

Question

(a) In Figure 1(a) shown below, the two-port network is characterized in terms of y-parameters with y₁₁ = 3·3 × 10⁻³ S, y₂₂ = 5 × 10⁻³ S and y₁₂ = y₂₁ = 0. Find the voltage across 200 Ω load. (10 marks)

(b) For the signal shown in Figure 1(b), calculate the total energy of the signal X(t). Also sketch y(t) = X(10t – 5). (10 marks)

(c) A 220 V dc shunt motor has armature resistance Rₐ = 0·13 Ω, field resistance Rf = 250 Ω and rotational loss 230 W. On full-load, the line current is 9·5 A with the motor running at 1440 rpm. Determine the following:
(i) The mechanical power developed
(ii) The power output
(iii) The load torque
(iv) The full-load efficiency (10 marks)

(d) For the transistor circuit shown in Figure 1(d), determine the value of reverse saturation current, I_S, that would give a collector current of 1 mA, if β = 80, V_A = ∞ and V_T = 26 mV at T = 300 K. (10 marks)

(e) Consider the four variables logic function defined as follows:
F (A, B, C, D) = ĀC + ĀD + B̄C + B̄D + ABC̄ D̄
Assuming input variables as A, B, C and D, propose a logic circuit using only three logic gates to implement the function. (10 marks)

UPSC Answer Check · Accepted Answer

(a)
Using the y-parameter port equations:
I₁ = y₁₁V₁ + y₁₂V₂
I₂ = y₂₁V₁ + y₂₂V₂
Given y₁₂ = y₂₁ = 0, the two ports are decoupled:
I₁ = 3.3×10⁻³ V₁
I₂ = 5×10⁻³ V₂

For the 200 Ω load at port 2, with I₂ entering the two-port, the load equation is:
I₂ = −V₂/200

Substitute I₂ = 5×10⁻³ V₂:
5×10⁻³ V₂ = −V₂/200
5×10⁻³ V₂ + 5×10⁻³ V₂ = 0
10×10⁻³ V₂ = 0
V₂ = 0 V

The capacitor between the lower terminals does not create a transfer because y₁₂ = y₂₁ = 0; the input loop current returns through port 1 and no common-mode signal reaches port 2.
Final answer: V_load = 0 V.

(b)
The total energy is:
E = ∫₋∞^∞ X²(t) dt

Break the integral into the six intervals of the waveform:
E = ∫₋₃^−2 [2(t+3)]² dt + ∫₋₂^−1 2² dt + ∫₋₁^0 (1−t)² dt + ∫₀^1 (1+t)² dt + ∫₁^2 2² dt + ∫₂^3 [2(3−t)]² dt

Evaluate:
∫₋₃^−2 4(t+3)² dt = 4/3
∫₋₂^−1 4 dt = 4
∫₋₁^0 (1−t)² dt = 7/3
∫₀^1 (1+t)² dt = 7/3
∫₁^2 4 dt = 4
∫₂^3 4(3−t)² dt = 4/3

Therefore:
E = 4/3 + 4 + 7/3 + 7/3 + 4 + 4/3 = 46/3

Final answer: E = 46/3 units²·s (if X is in volts, V²·s; if in amperes, A²·s).

For y(t) = X(10t − 5), let τ = 10t − 5. The nonzero interval is −3 ≤ τ ≤ 3, so:
−3 ≤ 10t − 5 ≤ 3
2 ≤ 10t ≤ 8
0.2 ≤ t ≤ 0.8

The piecewise sketch is:
0.2 ≤ t ≤ 0.3: y = 20t − 4
0.3 ≤ t ≤ 0.4: y = 2
0.4 ≤ t ≤ 0.5: y = 6 − 10t
0.5 ≤ t ≤ 0.6: y = 10t − 4
0.6 ≤ t ≤ 0.7: y = 2
0.7 ≤ t ≤ 0.8: y = 16 − 20t
Outside [0.2, 0.8]: y = 0

At t = 0.2 and 0.8, y = 0; at t = 0.3, 0.4, 0.6, 0.7, y = 2; at t = 0.5, y = 1.

(c)
Using DC shunt-motor power balance:

Field current:
I_f = V/R_f = 220/250 = 0.88 A

Armature current:
I_a = I_L − I_f = 9.5 − 0.88 = 8.62 A

Input power:
P_in = V I_L = 220 × 9.5 = 2090 W

Field copper loss:
P_f = V I_f = 220 × 0.88 = 193.6 W

Armature copper loss:
P_cu = I_a²R_a = (8.62)² × 0.13 = 9.659572 W

(i) Mechanical power developed:
P_dev = P_in − P_f − P_cu
P_dev = 2090 − 193.6 − 9.659572 = 1886.740428 W
P_dev ≈ 1886.74 W

(ii) Power output:
P_out = P_dev − rotational loss
P_out = 1886.740428 − 230 = 1656.740428 W
P_out ≈ 1656.74 W

(iii) Load torque:
ω = 2πN/60 = 2π × 1440/60 = 48π rad/s
T = P_out/ω = 1656.740428/(48π)
T ≈ 10.9866 N·m
T ≈ 10.99 N·m

(iv) Full-load efficiency:
η = P_out/P_in = 1656.740428/2090 = 0.792698
η ≈ 79.27%
η ≈ 79.27%

(d)
For the BJT in active region, use:
I_C = I_S exp(V_BE/V_T)

Given:
I_C = 1 mA = 1×10⁻³ A
β = 80
I_B = I_C/β = 1×10⁻³/80 = 12.5×10⁻⁶ A = 12.5 μA
I_E = I_C + I_B = 1×10⁻³ + 12.5×10⁻⁶ = 1.0125×10⁻³ A

The emitter voltage is:
V_E = I_E × 1.6 kΩ = 1.0125×10⁻³ × 1600 = 1.62 V

The collector is at V_CC = 2.5 V, so the base voltage is:
V_B = V_C − I_B × 20 kΩ = 2.5 − 12.5×10⁻⁶ × 20000
V_B = 2.5 − 0.25 = 2.25 V

Thus:
V_BE = V_B − V_E = 2.25 − 1.62 = 0.63 V

Now:
I_S = I_C exp(−V_BE/V_T)
I_S = 1×10⁻³ exp(−0.63/0.026)
0.63/0.026 = 24.230769
exp(24.230769) ≈ 3.34×10¹⁰

Therefore:
I_S ≈ 1×10⁻³/(3.34×10¹⁰) ≈ 2.99×10⁻¹⁴ A

Final answer: I_S ≈ 3.0×10⁻¹⁴ A.

(e)
Simplify the Boolean function using Boolean algebra.

F = ¬A C + ¬A D + ¬B C + ¬B D + A B ¬C ¬D
F = ¬A(C + D) + ¬B(C + D) + A B ¬C ¬D
F = (¬A + ¬B)(C + D) + A B ¬(C + D)
F = ¬(AB)(C + D) + AB ¬(C + D)

This is the XOR form:
F = AB ⊕ (C + D)

So the three-gate implementation is:
Gate 1: AND gate with inputs A and B, output X = AB.
Gate 2: OR gate with inputs C and D, output S = C + D.
Gate 3: XOR gate with inputs X and S, output F = X ⊕ S.

Final answer: F = (A·B) ⊕ (C + D), using one AND gate, one OR gate and one XOR gate only.

(a) In Figure 1(a) shown below, the two-port network is characterized in terms of y-parameters with y₁₁ = 3·3 × 10⁻³ S, y₂₂ = 5 ×…

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