#5: Transmission Lines & Fibre Optic

Question 1

Why do different coaxial terminations have specific impedances?

Answer 1

Source and load impedances must be matched to prevent reflections.

Question 2

What does UTP stand for? Give an example of a UTP cable.

Answer 2

Unshielded Twisted Pair - e.g. cat 5e.

Question 3

Why are there two pairs in UTP? Why are they twisted?

Answer 3

UTP uses differential encoding (i.e. difference in voltage used to represent change in state). Therefore, one core for +1V, and another for -1V. Twisted to reduce inteference.

Question 4

Maximum power is transferred when the load impedance _____ the source impedance. However, in this case, the efficiency is __%.

Answer 4

equals; 50%

Question 5

What causes reflection in a line?

Answer 5

Unmatched source/load/cable impedances.

Question 6

How might one prevent reflection?

Answer 6

Z(load) = Z(cable) = Z(source)

Question 7

What is Time Domain Reflectonomy?

Answer 7

Uses pulse reflections to detect location of a break in a cable.

Question 8

What is the formula for the voltage of a reflected waveform?

Answer 8

V(reflected) = [(load impedance - cable impedance) / (load impedance + cable impedance)] * V(incident)

Question 9

A reflected wave is _____ of _____ with the incident wave.

Answer 9

Out of phase

Question 10

In an open-circuit, the reflected wave is (in / out) of phase with the incident wave.

Answer 10

In phase

Question 11

What is a standing wave, and how can it happen?

Answer 11

If two waves of identical frequency move in an opposite direction to each other, they can add on to each other and create a wave that doesn’t really propagate; it just oscillates from up and down.

Question 12

Why are standing waves dangerous in a telecommunications setting?

Answer 12

The high peaks (where both waves are added together) generates dangerously high voltages/currents that can damage equipment and cable.

Question 13

What is CSMA/CA, and how does it work?

Answer 13

Carrier Sense, Multiple Access, Collision Avoidance. Used in wireless networks; devices must request a timeslot to ‘talk’; other devices expected to honour it.

Question 14

What are the standard equations for an incident + reflected waveform travelling along a transmission line?

Answer 14

Question 15

Why is the ‘wait-for-ACK’ concept inefficient?

Answer 15

Throughput is limited by network delay (how long it takes for a bit to travel to the router) rather than bandwidth.

Question 16

Why can the impedance of a line (at high frequency) be said to equal 0? Explain; include a diagram + equation of characteristic impedance.

Answer 16

If we take a small section of line, its impedance is made of an inductor and capacitor in parallel. Through some nifty maths, the line impedance is found to be independant of length and frequency, and dependant on a ratio of its inductance/capacitance:

Z = sqrt(L / C)

Question 17

What is the ISM band in the context of wireless transmission? What are its advantages/disadvantages?

Answer 17

Industrial, Scientific & Medical band.

• no license required
• lots of other users + interference (i.e. microwaves)
• anyone can listen (security concern)
• not standardized internationally
• power restrictions in some countries.

Question 18

Why is ADSL’s downlink bigger than its uplink?

Answer 18

People use more downlink (e.g. watching videos, looking at websites, etc) than uplink.

Question 19

What is the ‘VSWR’? Explain; include its equation.

Answer 19

Voltage Standing Wave Ratio. Max/min voltage levels of reflected waveform/

Question 20

What does this symbol mean: ‘𝞀’. Include both equations (load/impedance resistance + incident wave).

Answer 20

Reflection co-efficient. Ratio of original (incident) wave to reflected wave.

Rowe = V(incident) / V(reflected)

Question 21

What would a 𝞀 = 0 scenario mean? What would the VSWR mean, and what would it say about the load/source impedance ratio?

Answer 21

No reflection - all power used by load.

VSWR = 1

Z(load) = Z(characteristic impedance of line)

Question 23

What is the ‘Return Loss’? Include its equation. What value do we want it to be?

Answer 23

ratio of reflected power to incident power.

We want it as high as possible (i.e. rowe = 0, no reflection)

Question 24

What is ‘DCT’ in the context of image compression?

Answer 24

Discrete Cosine Transform. A type of discrete fourier transform that seperate the high and low frequency parts of image blocks (8x8 pixels). It then discards unecessary high frequency elements.

Question 25

What are some advantages and disadvantages of using fibre optic transmission lines?

Answer 25

Good:

• enormous bandwidth
• very fast, and low attenuation with distance
• not susceptible to corrosion or electrical noise/crosstalk

Bad:

• harder to make
• can’t just ‘tee off’; difficult to join (must fusion splice)
• minimum bend radius
• need lasers to transmit; hard to make repeaters

Question 26

What are the two common wavelengtsh used in fibre optic transmission, and why are these chosen?

Answer 26

1300nm and 1550nm. These wavelengths have minimal loss and can most easily be generated by low-powered lasers.

Question 27

What’s a laser diode, and how does it work?

Answer 27

Uses mirrors to create standing waves and reflected the combination standing wave out through the transmission line.

Question 29

What is the characteristic impedance? What is its formula?

Answer 29

It’s the impedance of a line at frequencies other than zero (theoretically any frequency). It is a characteristic of that line. Typically 50 Ω, 75 Ω, or 300 Ω. Symbol Zo. Can relate to reflection coefficient

Question 30

What’s the difference between the reflection co-efficient and the VSWR? Explain, with formulae.

Answer 30

• Reflection Coefficient:* ratio of the amount of reflected voltage to the amount incident (rowe = v(Ref) / v(inc) ).
• VSWR:* Voltage Standing Wave Ratio. Measure voltages along a line where reflection is present, the ratio of max to min amplitude is the VSWR.

Question 31

A fibre optic cable and its cladding have refractive indexes of 1.535 and 1.490 respectively. Determine the numerical aperture.

Answer 31

θ_e = sqrt(n₁²-n₂²) = 0.3690 rad.

Question 32

With use of a diagram, derive the first ‘step’ of the numerical aperture equation with use of Snell’s law.

Answer 32

At A, from Snell’s Law, n_air:

sin θ_e = n₁ sin θ₁

Since n_air ≈ 1, sin θ_e = n₁ sin θ₁

At B, n₁ sin θ₂ = n₂ sin θ₃

Question 33

An optical fibre has a loss of 0.1 dB/km. The transmit power is 10 mW. Over 200 km, what is the received signal strength?

Answer 33

Question 34

Convert a 0.1dB/km loss to the total loss in ‘gain’ unit (e.g. ‘G’) for a 100km line.

Answer 34

-0.1dB x 100km = -10dB (loss)

Therefore: -10dB = 10 log G

find G:

G = 10^-1 = 0.1