(a) A single-phase AC voltage controller is feeding a resistive load of 26·45 Ω from an AC source of 230 V, 50 Hz. Compute the…

(a) The single-phase AC voltage controller uses anti-parallel thyristors (or a TRIAC) in series with the resistive load R = 26.45 Ω. The 230 V, 50 Hz supply is connected as:

AC source --- anti-parallel thyristors T1/T2 --- R = 26.45 Ω --- back to source

T1 is fired at α in the positive half-cycle and T2 at π+α in the negative half-cycle. For R-load, load current has the same waveform as load voltage.

Let V_s = 230 V (rms), V_m = √2 V_s. For phase control, V_o² = (1/π) ∫_α^π V_m² sin²θ dθ = V_s²[(π−α)/π + sin 2α/(2π)].

Thus P = V_o²/R = (V_s²/R)[(π−α)/π + sin 2α/(2π)].

At α = 0, full power is P_full = 230²/26.45 = 52900/26.45 = 2000 W.

For P = 1000 W, the bracket must be 1/2: (π−α)/π + sin 2α/(2π) = 1/2.

Multiply by 2π: 2(π−α) + sin 2α = π => 2α − sin 2α = π.

The solution in 0 < α < π is α = π/2 = 90°.

At this firing angle, V_o² = V_s²/2, so V_o = 230/√2 = 162.63 V. Load current rms = V_o/R = 162.63/26.45 = 6.149 A.

Input power factor: p.f. = P/(V_s I_s) = V_o/V_s = √(P R)/V_s = √(1000 × 26.45)/230 = √26450/230 = 162.63/230 = 1/√2.

Final: α = π/2 = 90°; p.f. = 1/√2 = 0.707 lagging.

Waveforms: v_s is sinusoidal. For α = 90°, v_o = v_s during 90° to 180° and 270° to 360°, and v_o = 0 during 0° to 90° and 180° to 270°. The current i_o = v_o/R has the same shape.

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(b) Take the PD controller as K(s) = K(1 + Td s). Then L(s) = K(1 + Td s)/[s²(τs + 1)].

Unity feedback characteristic equation: 1 + L(s) = 0 => s²(τs + 1) + K(1 + Td s) = 0 => τs³ + s² + K Td s + K = 0.

For oscillation at ω = √2 rad/s, put s = j√2: τ(j√2)³ + (j√2)² + K Td(j√2) + K = 0.

Since (j√2)² = −2 and (j√2)³ = −j2√2, (−j2√2 τ − 2) + j√2 K Td + K = 0.

Separate real and imaginary parts:

Real: K − 2 = 0 => K = 2.

Imaginary: −2√2 τ + √2 K Td = 0 => K Td = 2τ.

With K = 2: 2 Td = 2τ => Td = τ.

Routh array for τs³ + s² + K Td s + K = 0: s³: τ, K Td s²: 1, K s¹: K(Td − τ) s⁰: K.

Marginal oscillation occurs when Td = τ, and the auxiliary equation from the s² row is s² + K = 0 => s = ±j√K.

Given oscillation frequency √2 rad/s: √K = √2 => K = 2.

Therefore Td = τ, with τ > 0.

Final: K = 2; Td = τ; τ > 0. At the specified oscillation, Td = τ. For a stable non-oscillatory closed loop, the Routh condition requires Td > τ and K > 0.

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(c) Let J = moment of inertia of rotor in kg-m², f = windage and friction coefficient in N-m/rad/s, and ω_m = angular speed in rad/s.

The mechanical torque balance is J dω_m/dt = T_e − T_L − f ω_m.

If electrical torque and load torque are removed, the rotor decelerates under windage and friction: J dω_m/dt + f ω_m = 0 => dω_m/dt + (f/J)ω_m = 0.

The solution is ω_m(t) = ω_m(0) exp(−(f/J)t).

Hence the mechanical time constant is τ_m = J/f seconds.

Equivalently, from the transfer function ω_m(s)/T(s) = 1/(Js + f) = (1/f)/[(J/f)s + 1], so τ_m = J/f.

Experimental determination: use the retardation or free deceleration test.

• Run the machine up to a suitable no-load speed, with normal field if core losses are to be included.
• Switch off the driving supply and allow the rotor to decelerate freely.
• Record speed ω_m versus time t using a tachometer and data logger or recorder.
• If f is viscous, plot ln ω_m against t. The graph is a straight line with slope −f/J = −1/τ_m.
• Therefore τ_m = −1/slope.
• Alternatively, τ_m is the time in which the speed falls to 36.8% of its initial value.
• If Coulomb friction is present, fit the deceleration curve to J dω_m/dt = −(fω_m + C), or determine J separately by a known-flywheel retardation test.

Final: τ_m = J/f seconds.

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(d) Let x be the distance of the train from station A in km. The go-and-return resistance per km is r = 0.04 Ω/km. Let I_A and I_B be the currents supplied from A and B respectively: I_A + I_B = 600 A.

Voltage at the train: V_x = 550 − I_A(0.04x) and also V_x = 500 − I_B[0.04(10 − x)].

Substitute I_B = 600 − I_A: 550 − 0.04x I_A = 500 − 0.04(10 − x)(600 − I_A).

Expanding the right side: 500 − 24(10 − x) + 0.04(10 − x)I_A = 500 − 240 + 24x + 0.04(10 − x)I_A = 260 + 24x + 0.04(10 − x)I_A.

Thus, 550 − 0.04x I_A = 260 + 24x + 0.04(10 − x)I_A => 290 − 24x = 0.04[ x + (10 − x) ] I_A => 290 − 24x = 0.4 I_A => I_A = 725 − 60x A.

Then I_B = 600 − I_A = 60x − 125 A.

The train voltage is V_x = 550 − 0.04x(725 − 60x) = 550 − 29x + 2.4x² V.

For minimum potential, dV_x/dx = −29 + 4.8x = 0 => x = 29/4.8 = 145/24 = 6.0417 km from A.

At this point, I_A = 725 − 60(145/24) = 725 − 362.5 = 362.5 A, I_B = 600 − 362.5 = 237.5 A.

Voltage: V_min = 550 − 29(145/24) + 2.4(145/24)² = 22195/48 = 462.3958 V.

Final: point of minimum potential is 145/24 km = 6.0417 km from A (3.9583 km from B). V_min = 22195/48 = 462.4 V. Currents: from A = 362.5 A, from B = 237.5 A.

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(e) Given: x(t) = 1 for −0.5 ≤ t ≤ 0.5, and X(ω) = sin(ω/2)/(ω/2).

For y(t): y(t) = 2 for 0 ≤ t < 1, and −2 for 1 ≤ t ≤ 2.

This can be written as y(t) = 2x(t − 0.5) − 2x(t − 1.5).

Using linearity and time-shift properties: Y(ω) = 2 exp(−jω/2)X(ω) − 2 exp(−j3ω/2)X(ω) = 2X(ω)[exp(−jω/2) − exp(−j3ω/2)] = 2X(ω) exp(−jω)[exp(jω/2) − exp(−jω/2)] = 4j exp(−jω)X(ω) sin(ω/2).

Substitute X(ω): Y(ω) = 4j exp(−jω) [sin(ω/2)/(ω/2)] sin(ω/2) = 8j exp(−jω) sin²(ω/2)/ω.

At ω = 0, use the limit: Y(0) = 0.

For z(t), the triangular pulse has peak 2 at t = 1 and base from t = 0 to t = 2: z(t) = 2(1 − |t − 1|), 0 ≤ t ≤ 2, else 0.

It is the convolution of two identical rectangular pulses of width 1 and height 1, shifted by 0.5: z(t) = 2[x(t − 0.5) * x(t − 0.5)].

By the convolution property and time-shift property: Z(ω) = 2[exp(−jω/2)X(ω)]² = 2 exp(−jω)X²(ω) = 2 exp(−jω)[sin²(ω/2)/(ω/2)²] = 8 exp(−jω) sin²(ω/2)/ω².

At ω = 0, Z(0) = 2.

Also, since y(t) = dz/dt, the derivative property gives Y(ω) = jωZ(ω), which agrees with the results above.

Final: Y(ω) = 8j exp(−jω) sin²(ω/2)/ω, with Y(0) = 0. Z(ω) = 8 exp(−jω) sin²(ω/2)/ω², with Z(0) = 2.