COMPREHENSIVE DIGITAL COMMUNICATION & IOT
NETWORK CALCULATIONS EXAM SOLUTION LATEST
UPDATE 2024-2025
Questions 1 & 2 Solutions
March 5, 2025
Question 1 (30 marks)
(a) Frequency and Period Calculations
Given: Period T = 540 µs
1. Frequency calculation:
The frequency f of a wave is the reciprocal of its period:
1
f=
T
Convert microseconds to seconds:
T = 540 µs = 540 × 10−6 s
Calculate frequency:
1
f= = 1851.85 Hz = 1.85 kHz (3 s.f.)
540 × 10−6
2. New period calculation:
The frequency is reduced by a factor of 2.5:
1.85185 kHz
fnew = = 0.74074 kHz
2.5
The new period is the reciprocal of the reduced frequency:
1 1
Tnew = = = 1.35 × 10−3 s = 1.35 ms (3 s.f.)
fnew 0.74074 × 103
(b) Optical Fiber Attenuation
Given: Initial power P0 = 0.325 W, Final power Pfinal = 3.25 µW
1
, The power drops by a factor of 10 every 27 km. The number of 27 km intervals
is determined by:
P0
n = log10
Pfinal
Substitute values:
0.325
n = log10 = log (105) = 5
10
3.25 × 10−6
Total distance:
Distance = 5 × 27 km = 135 km
Distance (km) Power (W)
0 0.325
27 0.325 × 10−1 = 0.0325
54 0.0325 × 10−1 = 0.00325
81 0.00325 × 10−1 = 0.000325
108 0.000325 × 10−1 = 0.0000325
135 0.0000325 × 10−1 = 0.00000325
Coaxial Cable Comparison: Coaxial cables have higher attenuation than
optical fibers. This means the signal power drops faster, resulting in a shorter
maximum transmission distance.
(c) MAC Address Conversion
Binary Input: 00100100 11001101 10111001 11100000 10001001 10101101
Each 8-bit segment is converted to hexadecimal:
• Split into 8-bit groups: 00100100, 11001101, etc.
• Group the binary into 4-bit segments:
0010 0100 1100 1101 1011 1001 1110 0000 1000 1001 1010 1101
• Convert each 4-bit segment to its hexadecimal equivalent:
• Regroup the hexadecimal values into MAC address format:
24 : CD : B9 : E0 : 89 : AD
Final MAC Address: 24 : CD : B9 : E0 : 89 : AD
2
NETWORK CALCULATIONS EXAM SOLUTION LATEST
UPDATE 2024-2025
Questions 1 & 2 Solutions
March 5, 2025
Question 1 (30 marks)
(a) Frequency and Period Calculations
Given: Period T = 540 µs
1. Frequency calculation:
The frequency f of a wave is the reciprocal of its period:
1
f=
T
Convert microseconds to seconds:
T = 540 µs = 540 × 10−6 s
Calculate frequency:
1
f= = 1851.85 Hz = 1.85 kHz (3 s.f.)
540 × 10−6
2. New period calculation:
The frequency is reduced by a factor of 2.5:
1.85185 kHz
fnew = = 0.74074 kHz
2.5
The new period is the reciprocal of the reduced frequency:
1 1
Tnew = = = 1.35 × 10−3 s = 1.35 ms (3 s.f.)
fnew 0.74074 × 103
(b) Optical Fiber Attenuation
Given: Initial power P0 = 0.325 W, Final power Pfinal = 3.25 µW
1
, The power drops by a factor of 10 every 27 km. The number of 27 km intervals
is determined by:
P0
n = log10
Pfinal
Substitute values:
0.325
n = log10 = log (105) = 5
10
3.25 × 10−6
Total distance:
Distance = 5 × 27 km = 135 km
Distance (km) Power (W)
0 0.325
27 0.325 × 10−1 = 0.0325
54 0.0325 × 10−1 = 0.00325
81 0.00325 × 10−1 = 0.000325
108 0.000325 × 10−1 = 0.0000325
135 0.0000325 × 10−1 = 0.00000325
Coaxial Cable Comparison: Coaxial cables have higher attenuation than
optical fibers. This means the signal power drops faster, resulting in a shorter
maximum transmission distance.
(c) MAC Address Conversion
Binary Input: 00100100 11001101 10111001 11100000 10001001 10101101
Each 8-bit segment is converted to hexadecimal:
• Split into 8-bit groups: 00100100, 11001101, etc.
• Group the binary into 4-bit segments:
0010 0100 1100 1101 1011 1001 1110 0000 1000 1001 1010 1101
• Convert each 4-bit segment to its hexadecimal equivalent:
• Regroup the hexadecimal values into MAC address format:
24 : CD : B9 : E0 : 89 : AD
Final MAC Address: 24 : CD : B9 : E0 : 89 : AD
2
