CS331 · Annotated SlidesChapter 2 · Part 1 — Physical Layer: Signals, Impairments and Delay
Chapter 2 · Part 1 · Annotated slides

Physical Layer: Signals, Impairments and Delay

Analog vs. digital, frequency and period, bit rate and bit interval, the three impairments (attenuation, distortion, noise), and the four delays: propagation, transmission, queuing and processing. This is the calculation-heavy deck: every formula here appears in Tutorial 3 (the solved Physical Layer Activity), and the worked answers are linked on each slide.

Slide text Study notes Highlight: key words Tips, tricks & traps Seen again elsewhere How it can come up

Physical Layer

CS331: Computer Networks · Chapter 2, Part 1
1
Exam weight Low
  1. Signals → analog vs. digital → frequency, period, bit rate → impairments → delays → bandwidth-delay product.
  2. Calculation deck: practise every example with units (ms, µs, Mbps vs. MB).
TipFormula sheet: f = 1/T · bit interval = 1/bit rate · propagation = distance / speed · transmission = size in bits / bandwidth · BDP = bandwidth × delay · wavelength = speed / f.

Physical Layer

  • Deals with the actual communication media
    • Defines the physical characteristics of interfaces and transmission media
      • the way devices are connected to each other and to the link (topology)
      • Shape, size and number of pins of connectors
      • What voltages and currents are used
      • The type of transmission media
      • Transmission mode (duplex type)①
    • Representation of bits – encoding into Electrical or Optical signals
    • Data Rate = transmission rate = bandwidth (number of bits sent per second)②
    • Synchronization of bits (sender and receiver clocks must be synchronized)
    • Devices that operates at this level: Network card, hub, repeater③
3
Exam weight Medium
  1. Same list as Ch 1 Part 2: interfaces, media, connectors, voltages, duplex mode.
  2. Plus: encoding bits into signals, the data rate, and keeping sender and receiver clocks in sync.
  3. Layer-1 devices: network card, hub, repeater.

Seen again

↻ Ch 1 Part 2, slide 26 · First look at the physical layer

Signals: data must become electromagnetic signals

Left: electrical signal over copper cable. Right: light pulses over fiber-optic cable
4
Exam weight Medium
  1. Data can’t travel as-is: it must be turned into an electromagnetic signal (voltage on copper, light in fiber, radio in air).
🎯 How this can come up · 1 question
True / False

To be transmitted, data must be transformed to electromagnetic signals.

A Transmission System

Transmission: communication of data by propagation and processing of signals through a communication channel

Signal: electromagnetic energy that moves through the transmission medium and carries some information.①

Transmitter

  • Converts information into a signal suitable for transmission
  • Injects energy into the communications medium or channel
    • Telephone converts voice into electric current
    • NIC converts bits into electric current or pulses of light②

Receiver

  • Receives energy from the medium
  • Converts the received signal into a form suitable for delivery to the user
    • Telephone converts current into voice
    • NIC converts electric current or pulses of light into bits③
Transmitter, communication channel, receiver
5
Exam weight Low
  1. Signal = electromagnetic energy carrying information.
  2. The NIC is both: transmitter (bits → current/light) and receiver (current/light → bits).

Analog and Digital

  • Analog Data : Continuous value data (sound, light, temperature)①
  • Digital Data: Discrete value (text, integers, symbols)
  • Analog Signals : Continuously varying electromagnetic wave
    • have an infinite number of values over a period of time
  • Digital Signals : Series of voltage pulses (square wave)
    • have only a limited number of values over a period of time.②
    • Maintains a constant level then changes to another constant level
  • Signals are classified into:
    • Periodic Signals : Consists of repeated patterns over identical period of times (cycles)
    • Aperiodic Signals : change without a pattern that repeats over time③
6
Exam weight High
  1. Analog = infinitely many values (continuous). Digital = a limited set of levels (discrete, square wave).
  2. Periodic = repeats in identical cycles. Aperiodic = no repeating pattern.
Exam trapAnalog/digital is about the VALUES; periodic/aperiodic is about the PATTERN over time. They are separate classifications.
🎯 How this can come up · 2 questions
MCQ

A signal that can have an infinite number of values in a range is:

True / False

A periodic signal repeats its pattern over identical periods called cycles.

Comparison of analog and digital signals (Figure 2.2)

a. Analog signal: a smooth wave. b. Digital signal: square steps
7
Exam weight High
  1. Rule from the slide: data communication uses PERIODIC ANALOG and APERIODIC DIGITAL signals.
🎯 How this can come up · 1 question
MCQ

In data communication we commonly use:

Note: periodic analog, aperiodic digital

In data communication, we commonly

use periodic analog signals and

aperiodic digital signals.

8
Exam weight Low
  1. Repeated on purpose: periodic analog + aperiodic digital is a classic MCQ.

Note: frequency

Frequency is the rate of change with respect to time.

Change in a short span of time means high frequency.①

Change over a long span of time means low frequency.②

9
Exam weight Low
  1. Fast changes = high frequency. Slow changes = low frequency.
  2. A signal that never changes has frequency 0.
★ Tutorial

Note: frequency and period

Frequency and period are the inverse of each other: f = 1 / T and T = 1 / f①

10
Exam weight High
  1. f in hertz (cycles per second), T in seconds.
Memory trickUnits pair up: s ↔ Hz, ms ↔ kHz, µs ↔ MHz, ns ↔ GHz.

Asked in a course tutorial

★ Tutorial 3 (solved) · Q5 · period of a 5 kHz signal
🎯 How this can come up · 2 questions
MCQ

An analog signal has a frequency of 5 kHz. Its period is:

Written

An analog signal has a frequency of 5 kHz. What is its period, and how are frequency and period related?

Example

The power we use at home has a frequency of 60 Hz. The period of this sine wave is:

T = 1/f = 1/60 = 0.0166 s = 0.0166 × 10³ ms = 16.6 ms

1 second = 1000 ms①

11
Exam weight Low
  1. Multiply seconds by 1000 to get ms.

Example

The period of a signal is 100 ms. What is its frequency in kilohertz?

Step 1: convert to seconds: 100 ms = 100/1000 = 0.1 s

Step 2: f = 1/T = 1/0.1 = 10 Hz

Step 3: convert to kHz: 10/1000 = 0.01 kHz①

12
Exam weight Medium
  1. Always convert to base units (s, Hz) first, then to the unit asked.
🎯 How this can come up · 1 question
MCQ

A signal has a period of 4 ms. Its frequency is:

★ Tutorial

Simple and composite signals

Signals that have just one frequency are called simple signals: for example, the sine signal (each cycle takes the same time)①

Signals that are composed of multiple frequencies are called composite signals (each cycle takes different time)②

An analog and a digital signal made of several frequencies
13
Exam weight Medium
  1. Simple = one sine wave = one frequency. Composite = many sine waves added together.

Asked in a course tutorial

★ Tutorial 3 (solved) · Q8 · simple vs. composite signal
🎯 How this can come up · 1 question
Written

What is the key difference between a simple and a composite signal?

Digital Signals

1 = positive voltage > 0

0 = zero voltage

Most digital signals are aperiodic → frequency is not appropriate①

Square wave carrying 1 0 1 1 0 0 0 1
14
Exam weight Low
  1. For digital signals we use bit rate and bit interval instead of frequency and period.

Two digital signals, one with two and one with four bit-levels

a. Two levels, 8 bits in 1 s = 8 bps. b. Four levels, 16 bits in 1 s = 16 bps
15
Exam weight Medium
  1. With L levels, each level carries log₂L bits: 2 levels → 1 bit, 4 levels → 2 bits, so twice the bit rate.
🎯 How this can come up · 1 question
MCQ

A digital signal has 8 levels. How many bits does each level carry?

★ Tutorial

Bit rate and bit interval

Bit interval (digital signal period): the time required to send (represent) one single bit (time units)①

Bit rate (digital bandwidth) = transmission speed = link capacity: number of bits sent per second (bps)②

Eight bit intervals in 1 s give a bit rate of 8 bps
16
Exam weight High
  1. Bit interval = time for ONE bit. Bit rate = bits per second. Bit rate = 1 / bit interval.

Asked in a course tutorial

★ Tutorial 3 (solved) · Q1 · bit rate from a 50 µs bit interval
🎯 How this can come up · 1 question
MCQ

A digital signal has a bit interval of 50 µs. Its bit rate is:

Example: bit interval

A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)?

The bit interval is the inverse of the bit rate.

Bit interval = 1/2000 s = 0.000500 s = 0.000500 × 10⁶ µs = 500 µs① (microseconds)

17
Exam weight Medium
  1. Seconds × 10⁶ = µs.
🎯 How this can come up · 1 question
MCQ

A signal has a bit rate of 4000 bps. Each bit lasts:

Advantages of Digital Transmission

Table: low cost, longer distances, security and privacy, encryption, noise removal, digital multiplexing, analog plus digital processing
18
Exam weight Medium
  1. Digital wins on: cost, distance (repeaters regenerate), security/encryption, noise removal, multiplexing, processing.

Related tutorial questions

Tutorial 3 (solved) · Q2 · what a repeater does

Signal corruption

  • No transmission medium can pass all the signal frequencies safely
  • A medium may
    • Pass some frequencies
    • Block others
    • Weaken others
A square input signal comes out of the medium rounded
19
Exam weight Low
  1. A medium passes some frequencies, blocks others and weakens others, so the output never equals the input exactly.

Transmission Impairments

Impairments: Factors that make the received signal different from the transmitted one①

Impairment splits into attenuation, distortion and noise
20
Exam weight High
  1. Three impairments: attenuation, distortion, noise.
Memory trick“ADN”: Attenuation (weaker), Distortion (reshaped), Noise (extra signal).
🎯 How this can come up · 1 question
MCQ

Which is NOT a transmission impairment?

★ Tutorial

Attenuation

  • Attenuation: Loss of energy due to resisting the

medium (Signal strength falls off with distance)①

  • Increases with signal frequency
  • Ex. A wire carrying electrical signal becomes warm after some time②
  • Amplifiers (analog signals) and repeaters (digital signals) are used to handle attenuation③
Signal weakens along the medium; an amplifier restores it. Repeaters regenerate a corrupted digital signal
21
Exam weight High
  1. Attenuation = LOSS OF ENERGY (the wire heats up). Worse with distance and with higher frequency.
  2. Fix: amplifiers for analog, repeaters for digital (they regenerate a clean copy).
Exam trapAmplifiers boost noise too; repeaters regenerate the original bits.

Asked in a course tutorial

★ Tutorial 3 (solved) · Q2 · repeater regenerates the weakened signal
🎯 How this can come up · 2 questions
MCQ

Loss of signal energy due to the resistance of the medium is called:

Written

What does a repeater do and why is it needed in digital transmission?

Distortion

  • Distortion: Signal changes in shape
  • Example: In a video call, distorted images or audio delays may result from signal distortion during transmission.①
A composite signal changes shape between sender and receiver
22
Exam weight Medium
  1. Distortion = the signal changes SHAPE (its frequency components arrive with different delays). Typical for composite signals.
🎯 How this can come up · 1 question
MCQ

A composite signal arrives with a changed shape. This impairment is:

★ Tutorial

Noise

  • Thermal Noise due to random motion of electrons in a wire which will create an extra signal①
  • Induced Noise: caused by motors and electrical equipments.②
  • crosstalk noise: Two wires beside each others (hearing another conversation in the background while talking with the phone)③
  • Impulse noise: irregular pulses or noise spikes of short duration and high amplitude
    • caused by power lines or lightning
    • Very critical in case of digital signals (primary source of error in digital data communication)④
Noise adds to the transmitted signal, so the received one differs
23
Exam weight High
  1. Thermal: random motion of electrons. Induced: motors and electrical equipment.
  2. Crosstalk: one wire picks up a neighbour’s signal (hearing another call).
  3. Impulse: short high spikes from power lines or lightning; the main error source for digital data.
Memory trickCrosstalk = “cross talk”: someone else’s conversation crossing into yours.

Asked in a course tutorial

★ Tutorial 3 (solved) · Q6 · background conversation = crosstalk★ Tutorial 3 (solved) · Q9 · nearby electrical equipment = induced noise
🎯 How this can come up · 3 questions
Scenario

On a phone call you hear a faint conversation from another call. Which impairment is this, and under which category?

Scenario

A long-distance link has many errors because of nearby electrical equipment. Which impairment, and how would you reduce it?

MCQ

Which noise is the primary source of error in digital data communication?

★ Tutorial

Performance: propagation time and speed

  • Propagation time = time required for a bit to move between two nodes.①
    • Propagation time = distance of the link (between the two nodes) / propagation speed
  • Propagation speed: speed of light in the medium.
  • Depends on the medium
    • Light travels at 3 × 10⁸ m/s in vacuum (free space), lower in the air and much lower in a cable (2/3 of vacuum)②
24
Exam weight High
  1. Propagation time = distance / speed. It does NOT depend on the message size.
  2. Speed: 3 × 10⁸ m/s in vacuum, about 2 × 10⁸ m/s in cable.
Exam trapConvert km to m before dividing by m/s.

Asked in a course tutorial

★ Tutorial 3 (solved) · Q4 · 2,000 km at 2 × 10⁸ m/s
🎯 How this can come up · 1 question
MCQ

Sender and receiver are 2,000 km apart and the signal travels at 2 × 10⁸ m/s. Propagation time?

Propagation Time

A bit leaves at time t1 and arrives distance d later at time t2; propagation time = t2 - t1 = d / propagation speed
25
Exam weight Low
    ★ Tutorial

    Other Physical layer definitions – Transmission Time

    • Transmission time = the time it takes the sender to put all the bits of a frame onto the link (from NIC to medium)①
      • Transmission time (s) = length of frame (message) in bits / bandwidth (bit rate) (bps)
    • Queuing time = time needed for each intermediate device to hold the message before it can be processed.②
    • Latency (delay): total message delivery time = transmission time + propagation time + processing time + queuing time③
    26
    Exam weight High
    1. Transmission time = size / bandwidth: depends on message size, not distance.
    2. Latency = propagation + transmission + queuing + processing.
    Memory trickPropagation = how FAR. Transmission = how BIG. Queuing = how BUSY.
    Exam trap1 MB = 8 Mb: multiply bytes by 8 first.

    Asked in a course tutorial

    ★ Tutorial 3 (solved) · Q7 · 1 MB over 1 Mbps
    🎯 How this can come up · 3 questions
    MCQ

    A 1 MB file is sent over a 1 Mbps link (ignore propagation). Transmission time?

    MCQ

    Which delay depends on the size of the message?

    Written

    List the four components of latency.

    Example

    What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 10⁸ m/s.

    27
    Exam weight Low
    1. Short message + fast link: guess which delay wins before you calculate.

    Example (continued)

    Propagation time = distance / propagation speed = (12,000 × 1000) / (2.4 × 10⁸) = 50 ms①

    Transmission time = message size in bits / bandwidth = (2500 × 8) / 10⁹ = 0.020 ms②

    Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, not the transmission time. The transmission time can be ignored.

    28
    Exam weight Medium
    1. Small message on a fast link: propagation dominates.
    🎯 How this can come up · 1 question
    True / False

    For a short e-mail on a 1 Gbps link across 12,000 km, the transmission time dominates.

    Example

    What are the propagation time and the transmission time for a 5-Mbyte message (an image) if the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 10⁸ m/s.

    29
    Exam weight Low

      Example (continued)

      Propagation time = (12,000 × 1000) / (2.4 × 10⁸) = 50 ms①

      Transmission time = (5,000,000 × 8) / 10⁶ = 40 s②

      Note that in this case, because the message is very long and the bandwidth is not very high, the dominant factor is the transmission time, not the propagation time. The propagation time can be ignored.

      30
      Exam weight Medium
      1. Same distance, same 50 ms propagation. Big message on a slow link: transmission dominates.
      🎯 How this can come up · 1 question
      MCQ

      A 5 MB image is sent on a 1 Mbps link over 12,000 km (2.4 × 10⁸ m/s). Which delay dominates?

      Example: a satellite link

      Two computers A and B are connected by a satellite link. The bandwidth of the link is 1 Gbps and the frame length is 2000 bytes. The distance between the satellite and each ground station is 36,000 km and the speed of light is 3 × 10⁸ m/s.

      Propagation time = distance / propagation speed

      Transmission time = message size in bits / bandwidth

      A sends up 36,000 km to a satellite and down to B
      31
      Exam weight Medium
      1. A → satellite → B is 2 × 36,000 = 72,000 km: propagation = 72 × 10⁶ / 3 × 10⁸ = 0.24 s (0.12 s for one leg).
      2. Transmission = 2000 × 8 / 10⁹ = 16 µs. Propagation dominates by far.
      🎯 How this can come up · 1 question
      MCQ

      In the satellite example (36,000 km up, 36,000 km down, 3 × 10⁸ m/s), the propagation time from A to B is:

      ★ Tutorial

      Bandwidth-delay product (Figure 2.7)

      We can think about the link between two points as a pipe. The cross section of the pipe represents the bandwidth, and the length of the pipe represents the delay. We can say the volume of the pipe defines the bandwidth-delay product.①

      bandwidth-delay product = length of the link in bits

      • Bandwidth = how wide the pipe is → how many bits can enter per second.
      • Delay = how long the pipe is → how long a bit takes to travel from one end to the other.
      • Bandwidth-delay product = how many bits fit inside the pipe.②
      A pipe: cross section = bandwidth, length = delay, volume = bandwidth × delay
      32
      Exam weight High
      1. BDP = bandwidth × delay = how many bits are “in the pipe” at once.

      Asked in a course tutorial

      ★ Tutorial 3 (solved) · Q10 · 5 bps × 5 s
      🎯 How this can come up · 2 questions
      MCQ

      A link has a bandwidth of 5 bps and a delay of 5 s. Its bandwidth-delay product is:

      MCQ

      A 10 Mbps link has a 2 ms delay. How many bits fill the link?

      Filling the link with bits for case 1

      Bandwidth 1 bps, delay 5 s: after 5 s the link holds 5 bits
      33
      Exam weight Low
      1. 1 bps × 5 s = 5 bits in the pipe.

      Other Physical layer definitions - Wavelength

      • Wavelength: the distance a simple signal can travel in one period (or the distance occupied by one cycle)①
      • Depends on both the signal frequency and the medium
      • Wavelength = propagation speed × period = propagation speed / frequency②
      Wavelength is the distance one cycle occupies in the medium
      35
      Exam weight Medium
      1. Wavelength = speed × T = speed / f. It depends on the frequency AND the medium.
      🎯 How this can come up · 1 question
      MCQ

      A 1 MHz signal travels at 2 × 10⁸ m/s. Its wavelength is: