(a) The diagram of Master-Slave S-R flip-flop and the waveform applied to the Master flip-flop is shown in Figure 5(a). Draw the…

(a) Output of the Slave Flip-Flop In a master-slave S-R flip-flop, the master is enabled when CLK = 1, while the slave is enabled when CLK = 0 because of the inverter. Therefore, the slave output Q₂ changes only at the falling edge of CLK, and it copies the master output Q₁ that existed just before the falling edge.

Reading the given input waveform:

• During CLK pulse 1, S = 0, R = 1. Hence master is reset, Q₁ = 0. At the falling edge of pulse 1, slave copies Q₁ = 0, so Q₂ becomes 0.
• During CLK pulse 2, S = 1, R = 0. Hence master is set, Q₁ = 1. At the falling edge of pulse 2, slave copies Q₁ = 1, so Q₂ becomes 1.
• During CLK pulse 3, S = 0, R = 1. Hence master is reset, Q₁ = 0. At the falling edge of pulse 3, slave copies Q₁ = 0, so Q₂ becomes 0.
• During CLK pulse 4, S = 0, R = 0. Master holds Q₁ = 0. At the falling edge of pulse 4, Q₂ remains 0.

The output waveform is therefore:

If the initial state before the first falling edge is not specified, it is indeterminate; from the first falling edge onward, Q₂ = 0, then 1, then 0, then 0.

Final answer: Q₂ changes only at falling edges: 0 after edge 1, 1 from edge 2 to edge 3, and 0 from edge 3 onward.

(b) Input Impedance of the λ/4 Line For a lossless transmission line of length l,

Z_in = Z₀ (Z_L + jZ₀ tan βl) / (Z₀ + jZ_L tan βl).

Here l = λ/4, so

βl = (2π/λ)(λ/4) = π/2.

Thus tan βl → ∞. Taking the limiting value for a λ/4 line,

Z_in = Z₀² / Z_L.

Given Z₀ = 60 Ω and Z_L = 60 Ω,

Z_in = 60² / 60 = 3600 / 60 = 60 Ω.

Final answer: Z_in = 60 Ω.

(c) Induction Motor Power Flow The air-gap power transferred to the rotor is

P_ag = input power − stator losses = 40 − 1.5 = 38.5 kW.

Rotor Cu loss at slip s = 0.04:

P_cu,rotor = s P_ag = 0.04 × 38.5 = 1.54 kW.

Gross mechanical power developed:

P_m = (1 − s) P_ag = 0.96 × 38.5 = 36.96 kW.

Net shaft mechanical power = gross mechanical power − friction and windage loss:

P_shaft = 36.96 − 0.8 = 36.16 kW.

For the second case, speed is reduced to 40% of synchronous speed. Therefore new slip is

s′ = 1 − 0.40 = 0.60 pu.

The torque is assumed unchanged. At original slip, gross mechanical power was 36.96 kW at speed 0.96 N_s. Hence torque is proportional to

T ∝ 36.96 / 0.96 = 38.5 per unit synchronous speed.

At new speed 0.40 N_s,

P_m′ = (38.5)(0.40) = 15.4 kW.

Net shaft power at new speed:

P_shaft′ = 15.4 − 0.8 = 14.6 kW.

Final answers: Rotor Cu loss = 1.54 kW; net shaft power at original slip = 36.16 kW; net shaft power at 40% synchronous speed = 14.6 kW. The corresponding gross mechanical power in the second case is 15.4 kW, and the rotor circuit loss (rotor copper plus external resistance) rises to 0.6 × 38.5 = 23.1 kW.

(d) Amplitude Modulation The modulating signal is

m(t) = 10 sin(2π × 500t), so A_m = 10 V and f_m = 500 Hz.

The carrier is

c(t) = 50 sin(2π × 10⁵t), so A_c = 50 V and f_c = 100 kHz.

(i) Modulation index:

μ = A_m / A_c = 10 / 50 = 0.2.

Final answer: μ = 0.2.

(ii) The amplitude of each sideband is

A_side = μ A_c / 2 = (0.2 × 50) / 2 = 5 V.

The sideband frequencies are

f_c − f_m = 100000 − 500 = 99.5 kHz,

f_c + f_m = 100000 + 500 = 100.5 kHz.

Final answer: Each sideband amplitude = 5 V, at 99.5 kHz and 100.5 kHz.

(iii) Bandwidth required:

BW = 2 f_m = 2 × 500 = 1000 Hz = 1 kHz.

Final answer: Bandwidth = 1 kHz.

(iv) Carrier power in load R = 500 Ω is

P_c = A_c² / (2R) = 50² / (2 × 500) = 2500 / 1000 = 2.5 W.

Total AM power is

P_total = P_c (1 + μ²/2)

= 2.5 (1 + 0.2²/2)

= 2.5 (1 + 0.04/2)

= 2.5 × 1.02 = 2.55 W.

Final answer: Total power delivered to the 500 Ω load = 2.55 W.

(v) The frequency spectrum has three spectral lines:

• Lower sideband at 99.5 kHz with amplitude 5 V.
• Carrier at 100 kHz with amplitude 50 V.
• Upper sideband at 100.5 kHz with amplitude 5 V.

Final answer: Spectrum lines at 99.5 kHz, 5 V, 100 kHz, 50 V, and 100.5 kHz, 5 V.

(e) Transmission and Hybrid Parameters The given equations are

I₁ = 2V₁ − V₂,

I₂ = −V₁ + 2V₂.

This is in admittance-parameter form. Writing the matrix,

[ I₁ ; I₂ ] = [ 2 −1 ; −1 2 ] [ V₁ ; V₂ ].

The determinant is

Δ = (2)(2) − (−1)(−1) = 4 − 1 = 3.

Solving for V₁ and V₂:

V₁ = (2I₁ + I₂) / 3,

V₂ = (I₁ + 2I₂) / 3.

Hybrid parameters

Hybrid parameters are defined by

V₁ = h₁₁ I₁ + h₁₂ V₂,

I₂ = h₂₁ I₁ + h₂₂ V₂.

From I₁ = 2V₁ − V₂,

2V₁ = I₁ + V₂,

V₁ = 0.5 I₁ + 0.5 V₂.

So

h₁₁ = 0.5 Ω,

h₁₂ = 0.5.

Now substitute V₁ into the second equation:

I₂ = −V₁ + 2V₂

= −(0.5 I₁ + 0.5 V₂) + 2V₂

= −0.5 I₁ + 1.5 V₂.

Hence

h₂₁ = −0.5,

h₂₂ = 1.5 S.

Final hybrid parameters:

h₁₁ = 0.5 Ω, h₁₂ = 0.5, h₂₁ = −0.5, h₂₂ = 1.5 S.

Transmission parameters

Using the standard form with I₂ entering port 2,

V₁ = A V₂ − B I₂,

I₁ = C V₂ − D I₂.

From V₂ = (I₁ + 2I₂)/3,

I₁ = 3V₂ − 2I₂.

Thus

C = 3 S,

D = 2.

Now substitute I₁ into V₁:

V₁ = (2I₁ + I₂)/3

= [2(3V₂ − 2I₂) + I₂]/3

= (6V₂ − 4I₂ + I₂)/3

= (6V₂ − 3I₂)/3

= 2V₂ − I₂.

Therefore

A = 2,

B = 1 Ω.

Final transmission parameters:

A = 2, B = 1 Ω, C = 3 S, D = 2.