CAP7001 · Unit 2: Physical layer
Guided and unguided transmission media
11 practice questions · in the MTE and ETE syllabus
Key points
- Guided: twisted pair, coax, fiber; unguided: radio, microwave, infrared
- Fiber uses total internal reflection and ignores EMI
- Microwave needs line of sight
- Infrared cannot penetrate walls
Guided media carry signals along a physical path; unguided media use free space.
| Medium | Notes |
|---|---|
| Twisted pair (UTP, STP) | Twisting makes interference hit both wires equally; shielding reduces crosstalk |
| Coaxial cable | Better shielding and bandwidth than twisted pair; cable TV uses RG-59 |
| Optical fiber | Light in a glass core, guided by total internal reflection; immune to EMI; highest bandwidth |
| Radio waves | Omnidirectional; can penetrate walls |
| Microwaves | Unidirectional; need line-of-sight antennas |
| Infrared | Short range; cannot penetrate walls |
In fiber, a ray hitting the core-cladding boundary at an angle greater than the critical angle is reflected and stays in the core. At a smaller angle it refracts into the cladding and is lost.
Worked example: a 2 km link beside heavy machinery with strong interference and high bandwidth needs fiber, not UTP.
Exam pitfall: radio is omnidirectional and goes through walls, microwave is directional and needs line of sight, and infrared does not pass walls. Students often swap these.
Try two questions
A long-haul link must carry 10 Gbps over 40 km using a fiber with a core only about 9 micrometres wide. Which statement correctly explains why this fiber suits the job?
- Light is reflected along many ray paths at once, so the pulses arrive together at the far end
- The wide cladding lets every ray bounce at its own angle, which spreads pulses out and so aids distance
- Light follows essentially one path, so there is very little modal dispersion over a long distance
- The tiny core makes the light spread into the cladding, which is what lowers the attenuation
Show the answer
C. A core of about 9 micrometres is single-mode fiber, so one ray path exists and modal dispersion is small. Multimode fibers have wider cores and many ray paths, which spreads pulses over distance.
Why not A: Many ray paths are a feature of multimode fiber and they make pulses arrive at different times.
Why not B: Rays at many angles are the cause of modal dispersion and do not help over distance.
Why not D: Light is guided by total internal reflection in the core and does not spread into the cladding.
Optical fiber has an attenuation of 0.4 dB per km. A 2 mW signal is sent along 50 km of this fiber with no amplifier. What power arrives at the far end?
- 0.02 mW
- 0.20 mW
- 0.04 mW
- 0.002 mW
Show the answer
A. The loss is 0.4 x 50 = 20 dB, a factor of 100, so 2 mW falls to 0.02 mW. Dividing by the length instead of using decibels is a common error.
Why not B: That corresponds to a 10 dB loss, half the true loss.
Why not C: That divides 2 mW by the 50 km length and ignores decibels.
Why not D: That corresponds to a 30 dB loss, more than the true loss.