About this previous year paper
OPTICAL COMMUNICATION (KEC-058) – EXAM ANSWERS
SECTION-A (2 Marks Each)
a) Acceptance Angle & Numerical Aperture
Acceptance Angle: Maximum angle at which light can enter the fiber and still propagate.
Numerical Aperture (NA): Light-gathering ability of fiber.
- NA=n12−n22NA = \sqrt{n_1^2 - n_2^2}NA=n12−n22
b) Normalized Frequency (V-Number)
V=2πaλ×NAV = \frac{2\pi a}{\lambda} \times NAV=λ2πa×NA
Multimode fiber: V > 2.405 Single mode fiber: V < 2.405
c) Non-linear Scattering Types
Stimulated Brillouin Scattering (SBS) Stimulated Raman Scattering (SRS)
d) Electrical vs Optical Bandwidth
Electrical bandwidth: Limited by electronic components.
Optical bandwidth: Wider, determined by fiber dispersion and light source.
e) Importance of Double Hetero-junction
Improves efficiency Better carrier confinement Higher optical output power
f) Temperature Effect on Avalanche Gain
As temperature increases → avalanche gain decreases
Due to increased phonon collisions
g) Receiver Sensitivity & Quantum Limit
Receiver Sensitivity: Minimum power required for acceptable BER.
Quantum Limit: Theoretical minimum optical power for detection.
h) Intrinsic & Extrinsic Absorption
Intrinsic: Due to material itself (UV & IR absorption)
Extrinsic: Due to impurities (OH⁻ ions, metal ions)
i) Minimum Gain Condition (Fabry-Perot)
gth=12Lln(1R1R2)g_{th} = \frac{1}{2L} \ln\left(\frac{1}{R_1 R_2}\right)gth=2L1ln(R1R21)
j) Stimulated Emission
Emission of photons when excited atoms are forced to emit identical photons by incident radiation.
SECTION-B (10 Marks Each – Attempt Any Three)
a) Acceptance Angle & NA Derivation
NA=n12−n22NA = \sqrt{n_1^2 - n_2^2}NA=n12−n22
Given:
n1=1.56n_1 = 1.56n1=1.56
n2=1.40n_2 = 1.40n2=1.40
NA=(1.56)2−(1.40)2=0.69NA = \sqrt{(1.56)^2 - (1.40)^2} = 0.69NA=(1.56)2−(1.40)2=0.69
b) Attenuation & Power Calculation
Pout=Pin×10−αL/10P_{out} = P_{in} \times 10^{-\alpha L /10}Pout=Pin×10−αL/10
Given:
Pin=200µWP_{in} = 200µWPin=200µW
Attenuation = 0.4 dB/km
Length = 30 km
Pout≈12.6µWP_{out} \approx 12.6µWPout≈12.6µW
c) Population Inversion & Threshold Condition
Population inversion occurs when N2>N1N_2 > N_1N2>N1
Threshold gain:
gth=αi+αmΓg_{th} = \frac{\alpha_i + \alpha_m}{\Gamma}gth=Γαi+αm
d) Noise in Photodiode
Shot noise Thermal noise
Dark current noise Quantum noise: Due to random photon arrival
e) Free Space Optics (FSO)
Wireless optical communication through air High data rate Affected by fog, rain, dust
SECTION-C (10 Marks)
a) Classification of Optical Fibers
By modes: Single-mode, Multi-mode By index profile: Step index, Graded index
b) Numerical Problem
Given:
Core diameter = 70µm → a = 35µm
Δ = 1.7% = 0.017 n1=1.48n_1 = 1.48n1=1.48 λ = 0.85µm
NA=n12ΔNA = n_1 \sqrt{2\Delta}NA=n12Δ V≈39V \approx 39V≈39
Number of modes: M=V22≈760M = \frac{V^2}{2} \approx 760M=2V2≈760
SECTION-D (10 Marks)
a) Intermodal Dispersion
σs=n1ΔLc\sigma_s = \frac{n_1 \Delta L}{c}σs=cn1ΔL
Delay difference:
ΔT=n1ΔLc\Delta T = \frac{n_1 \Delta L}{c}ΔT=cn1ΔL
b) Bending Losses Macrobending Microbending
Critical radius:
Rc=3n12λ4π(n12−n22)3/2R_c = \frac{3n_1^2\lambda}{4\pi(n_1^2-n_2^2)^{3/2}}Rc=4π(n12−n22)3/23n12λ
SECTION-E (10 Marks)
a) Fabry-Perot Cavity
Given:
Length = 5 cm n = 1.67 λ = 0.65µm
Mode spacing: Δf=c2nL\Delta f = \frac{c}{2nL}Δf=2nLc
b) S-LED vs E-LED
S-LED: Surface emitting, low power E-LED: Edge emitting, higher efficiency
SECTION-F (10 Marks)
a) PIN Photodiode Fast response
Low noise Speed limited by:
Transit time RC time constant
b) Avalanche Photodiode Internal gain
High sensitivity Requires high bias voltage
SECTION-G (10 Marks)
a) Eye Diagram
Shows ISI, noise, jitter Wide eye opening → good system
b) Power Penalty
Extra power required to maintain BER.