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ECET 2026 ECE

ECET 2026 ECE – Multiplexing Explained

Why This Topic Is Important for ECET 2026

Multiplexing is one of the most repeated topics in ECET ECE Communication Systems.
It is used in TV broadcasting, telephone networks, fiber systems, 5G, satellite links, OFDM, Wi-Fi, and IoT communication.

Questions come from:

  • Types of multiplexing
  • Bandwidth usage
  • Guard bands
  • Time slot calculations
  • Numerical problems on sampling & channel capacity

Mastering multiplexing helps you score full marks from this chapter.


📘 Concept Notes (Deep Explanation)

1. What is Multiplexing?

Multiplexing is a technique where multiple signals share the same communication channel without interfering with each other.

Goal:

  • Increase efficiency
  • Reduce cost of communication
  • Utilize bandwidth properly

Two broad categories:

  1. Analog Multiplexing
  2. Digital Multiplexing

2. Frequency Division Multiplexing (FDM)

  • Multiple signals are transmitted simultaneously in a channel
  • Each signal is assigned a unique frequency band
  • Used in: Radio, TV broadcasting, Cable TV, Satellite

Each channel has a guard band to avoid overlap.
Guard band example formula:

BW_{total} = \sum BW_i + \sum GB_i

Example:

Three FDM channels each with 4 kHz, guard band of 1 kHz
Total bandwidth:

BW_{total} = 3\times 4 + 2\times 1 = 14\text{ kHz}


3. Time Division Multiplexing (TDM)

  • Signals occupy the same channel at different time slots
  • Used in: Digital telephony, PCM, 4G/5G, Optical systems

Frame time formula:

T_f = n \times T_s

Example:

4 users, each requires 1 ms sampling time
Frame time =

T_f = 4\times 1\text{ ms} = 4\text{ ms}


4. Synchronous TDM vs. Statistical TDM

Synchronous TDM

  • Each user gets fixed time slot, even if no data
  • Wastage occurs

Statistical TDM

  • Slots assigned only when needed
  • Higher efficiency

5. WDM – Wavelength Division Multiplexing

Used in fiber optic communication.

  • Each signal is a different wavelength (λ)
  • Similar to FDM but in optical domain

Channel spacing:

\Delta \lambda = \lambda_2 - \lambda_1


6. Code Division Multiplexing (CDMA)

  • All users transmit simultaneously
  • Differentiated by unique codes
  • Used in 3G, GPS

Orthogonality condition:

\sum_{i=1}^{n} C_{1i}C_{2i} = 0


⚙️ Formulas List (QuickLaTeX Only)

BW_{total} = \sum BW_i + \sum GB_i
T_f = nT_s
R_b = \frac{n}{T_f}
\Delta\lambda = \lambda_2 - \lambda_1
\sum_{i=1}^{n} C_{1i}C_{2i} = 0
R = \frac{1}{T_s}
BW_{PCM} = n \times f_s \times \text{bits/sample}

C = B\log_2(1+\text{SNR})


🔟 10 MCQs (Conceptual + Numerical)

Q1. Multiplexing is used to:
a) Increase power
b) Reduce bandwidth
c) Share a channel among users
d) Improve noise immunity

Q2. FDM requires ______ for separating channels.
a) Time slots
b) Guard bands
c) Chips
d) Codes

Q3. 4 FDM channels each require 5 kHz and guard band = 1 kHz. Total BW?

Q4. TDM frame consists of 8 users, each needs a sampling period of 0.5 ms. Find frame time.

Q5. In TDM, if slot time = 200 μs and 6 slots exist, what is frame time?

Q6. WDM is used in:
a) Coaxial cable
b) Fiber optics
c) Wireless
d) Coax + Wireless

Q7. CDMA uses ______ to separate users.
a) Different frequencies
b) Time slots
c) Unique codes
d) Guard bands

Q8. If sampling frequency is 8 kHz and 8 channels exist in TDM PCM, what is bit rate?
(bits/sample = 8)

Q9. Orthogonal codes satisfy:
a) Sum = 1
b) Sum = -1
c) Sum = 0
d) Sum = 10

Q10. In FDM, if guard band is reduced to zero, what occurs?
a) No distortion
b) Overlapping of channels
c) Higher SNR
d) Fixed time slots


Answer Key (WordPress-friendly table)

Q NoAnswer
1c
2b
323 kHz
44 ms
51.2 ms
6b
7c
8512 kbps
9c
10b

🧠 Explanations (Step-by-Step)

Q1 → c
Multiplexing shares one channel among multiple users. Others are unrelated.

Q2 → b
FDM assigns different frequency bands and uses guard bands to prevent overlap.

Q3
4 channels × 5 kHz = 20 kHz
Guard bands = 3 × 1 kHz = 3 kHz
Total = 23 kHz

Q4

T_f = nT_s = 8 \times 0.5 = 4\text{ ms}

Q5
6 slots × 200 μs
= 1200 μs = 1.2 ms

Q6 → b
WDM works ONLY in fiber optics.

Q7 → c
CDMA separates users by unique spreading codes.

Q8
BW_{PCM} = n f_s \times \text{bits/sample}
= 8 × 8000 × 8
= 512,000 bps = 512 kbps

Q9 → c
Orthogonal codes produce zero cross-correlation.

Q10 → b
Zero guard band → channels overlap → severe interference.


🎯 Motivation / Why Practice Matters

Multiplexing questions in ECET are fast-solve, formula-based, and perfect for boosting your score. Practicing daily helps you:

  • Quickly calculate bandwidth, frame time, bit rate
  • Understand differences between FDM, TDM, WDM, CDMA
  • Gain accuracy in numerical MCQs
  • Build strong communication fundamentals for GATE-level questions

If you master multiplexing, you secure easy marks in the exam.


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