(a) (i) Show that the Smith chart constructed for a lossless transmission line gives a family of r-circles, having a radius of 1/((1+r)) for each circle which is centred at Γᵣ = r/((1+r)) and Γᵢ = 0. Here, r = normalized resistance of the load impeda

Question

(a) (i) Show that the Smith chart constructed for a lossless transmission line gives a family of r-circles, having a radius of 1/((1+r)) for each circle which is centred at Γᵣ = r/((1+r)) and Γᵢ = 0. Here, r = normalized resistance of the load impedance, Γᵣ and Γᵢ = real and imaginary parts of voltage reflection coefficient of the load impedance, respectively. (10 marks)

(ii) In the figure given below, determine the amplitudes of the reflected and transmitted E⃗ and H⃗ at the interface, if E₀ⁱ = 1.2 × 10⁻³ V/m in region 1, where εᵣ₁ = 7.5, μᵣ₁ = 1 and σ₁ = 0. Given : Region 2 is a free space and assume normal incidence. Also, μ₀ = 4π × 10⁻⁷ H/m and ε₀ = 1/(36π) × 10⁻⁹ F/m. (10 marks)

(b)

A sinusoidal modulating signal m(t) of frequency fₘ produces an AM signal : u(t) = A_c [1 + β cos (2π fₘ t)] cos (2π f_c t), where f_c is carrier frequency. Here, f_c >> fₘ and β = 2. This u(t) is applied to an ideal envelope detector which produces an output x(t).

(i)

Determine the Fourier series representation of x(t).

(ii)

Also determine the ratio of second harmonic amplitude to fundamental amplitude in x(t). 20 marks

(i) Determine the Fourier series representation of x(t).

(ii) Also determine the ratio of second harmonic amplitude to fundamental amplitude in x(t). (20 marks)

(c) Discuss in brief various methods of voltage control within 3-phase inverters. (10 marks)

UPSC Answer Check · Accepted Answer

(a)(i)
Let the normalized load impedance be z = r + jx and Γ = Γᵣ + jΓᵢ. For a lossless line,
Γ = (z − 1)/(z + 1), so z = (1 + Γ)/(1 − Γ).

Thus
r + jx = (1 + Γᵣ + jΓᵢ)/(1 − Γᵣ − jΓᵢ).

Equating real parts,
r[(1 − Γᵣ)² + Γᵢ²] = 1 − Γᵣ² − Γᵢ².

Expanding and collecting terms:
(1 + r)(Γᵣ² + Γᵢ²) − 2rΓᵣ + (r − 1) = 0.

Divide by (1 + r):
Γᵣ² + Γᵢ² − [2r/(1 + r)]Γᵣ + (r − 1)/(1 + r) = 0.

Completing the square in Γᵣ:
(Γᵣ − r/(1 + r))² + Γᵢ² = 1/(1 + r)².

Hence each constant-r locus is a circle with centre
Γᵣ = r/(1 + r), Γᵢ = 0
and radius
1/(1 + r).
This proves the Smith-chart r-circle family.

(a)(ii)
For normal incidence between two lossless dielectrics, use the intrinsic-impedance method.

η₀ = √(μ₀/ε₀) = √[(4π × 10⁻⁷)/((1/(36π)) × 10⁻⁹)] = 120π Ω.

Region 1:
η₁ = η₀/√εᵣ₁ = 120π/√7.5 Ω = 137.66 Ω.

Region 2 is free space:
η₂ = η₀ = 120π Ω = 376.99 Ω.

Reflection coefficient for the electric field:
Γ = (η₂ − η₁)/(η₂ + η₁) = (√7.5 − 1)/(√7.5 + 1)
= (17 − 2√30)/13 ≈ 0.4650.

Transmission coefficient:
τ = 1 + Γ = 2η₂/(η₁ + η₂)
= (30 − 2√30)/13 ≈ 1.4650.

Given E₀ⁱ = 1.2 × 10⁻³ V/m:
E₀ʳ = ΓE₀ⁱ = 0.4650 × 1.2 × 10⁻³
= 5.5805 × 10⁻⁴ V/m.

E₀ᵗ = τE₀ⁱ = 1.4650 × 1.2 × 10⁻³
= 1.7581 × 10⁻³ V/m.

Magnetic-field amplitudes:
H₀ⁱ = E₀ⁱ/η₁ = (1.2 × 10⁻³)/(137.66)
= 8.717 × 10⁻⁶ A/m.

H₀ʳ = E₀ʳ/η₁ = (5.5805 × 10⁻⁴)/(137.66)
= 4.054 × 10⁻⁶ A/m.

H₀ᵗ = E₀ᵗ/η₂ = (1.7581 × 10⁻³)/(376.99)
= 4.663 × 10⁻⁶ A/m.

These are valid for σ₁ = 0 and normal incidence.

(b)(i)
An ideal envelope detector removes the carrier and follows the positive envelope. Hence
x(t) = A_c|1 + 2 cos(2πfₘt)|.

Let θ = 2πfₘt and f(θ) = |1 + 2 cos θ|. Since f is even, only cosine terms appear:
x(t) = a₀ + Σ aₙ cos(nθ).

The DC term is
a₀ = (A_c/2π)∫₀²π |1 + 2 cos θ| dθ
= A_c(1/3 + 2√3/π).

For n ≥ 1,
aₙ = (A_c/π)∫₀²π |1 + 2 cos θ| cos nθ dθ.

Splitting where 1 + 2 cos θ is positive and negative:
positive intervals: [0, 2π/3] ∪ [4π/3, 2π];
negative interval: [2π/3, 4π/3].

This gives
a₁ = A_c(2/3 + √3/π),
a₂ = A_c√3/π.

For n ≥ 2:
if n ≡ 0 mod 3, aₙ = −4A_c√3/[π(n² − 1)];
if n ≡ 1 mod 3, aₙ = 2A_c√3/[πn(n + 1)];
if n ≡ 2 mod 3, aₙ = 2A_c√3/[πn(n − 1)].

Therefore,
x(t) = A_c(1/3 + 2√3/π)
+ A_c(2/3 + √3/π) cos(2πfₘt)
+ (A_c√3/π) cos(4πfₘt)
− (A_c√3/2π) cos(6πfₘt)
+ (A_c√3/10π) cos(8πfₘt)
+ (A_c√3/10π) cos(10πfₘt) + ⋯.

(b)(ii)
Second harmonic amplitude = A_c√3/π.
Fundamental amplitude = A_c(2/3 + √3/π).

Ratio = (√3/π)/(2/3 + √3/π)
= 3√3/(2π + 3√3)
≈ 0.45265.

So the second harmonic is approximately 45.27 % of the fundamental amplitude.

(c)
Voltage control in three-phase inverters is achieved mainly by:
Variable DC-link voltage: a controlled rectifier, chopper, or PWM rectifier changes V_dc; the AC output varies in proportion. It is simple but slow and needs a controlled front end.
PWM control: vary the modulation index in sinusoidal PWM. The fundamental output voltage is approximately proportional to the modulation index.
Third-harmonic injection PWM: adds a third-harmonic component to the reference, extending the linear modulation range and improving DC utilisation.
Space Vector PWM: controls the inverter voltage vector by dwell-time selection; it gives better DC utilisation and lower harmonic distortion.
Selective harmonic elimination PWM: switching angles are chosen to eliminate selected harmonics while setting the desired fundamental.
Phase-shift control: two or more inverter bridges feed a common transformer; varying their phase displacement changes the resultant output voltage.
Multilevel or transformer-tap/star-delta switching: connection changes or module count changes give stepped voltage control.
AC-side voltage control: thyristor phase control or AC chopper between inverter and load can vary voltage, but harmonics and power factor are poorer.

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