Protocol layers and the physical layer

This week we build the skeleton of the network. A network is a complex system in which dozens of protocols work together. The way to manage this complexity is layering: each layer provides a service, uses only the service of the layer below it, and “talks” only to the same layer on the other side. The exam has three types of questions: layer matching (which job belongs to which layer), encapsulation overhead, and physical-layer capacity calculation (Nyquist, Shannon).

Golden rules:

1. Know the five layers and the PDU names by heart: message, segment, datagram, frame, bit.
2. Encapsulation: each layer adds a header on the way down and strips it on the way up.
3. Bandwidth is in Hz, bit rate is in bps. They are not the same thing.
4. Noiseless channel: Nyquist. Noisy channel: Shannon. In Shannon the SNR must be a ratio; if dB is given, convert it.

Why layers?

Think of an airline trip: buy a ticket, check a bag, go through the gate, take off, fly, land, leave through the gate, collect the bag. Each step corresponds to the reverse step at the destination airport. The baggage handler doesn’t need to know how the pilot flies. The same holds in a network: a programmer writing HTTP doesn’t know whether the data goes over fiber or Wi-Fi.

Benefits of layering:

• Modularity: changing one layer (5G instead of Wi-Fi) doesn’t break the others.
• Clarity: each layer has a single responsibility.

The downside: some work is repeated at two layers (error checking at both the link and the transport layer), and one layer may need information from another.

The Internet (TCP/IP) model: 5 layers

The OSI model: 7 layers

ISO’s OSI (Open Systems Interconnection) model has two more layers:

TIP · memorizing the order. From bottom to top: “Please Do Not Throw Sausage Pizza Away” (Physical, Data Link, Network, Transport, Session, Presentation, Application).

Encapsulation

At the sending host, data moves from top to bottom. Each layer takes the packet from above as a payload and puts its own header in front of it:

Application : [ M ]                                   message
Transport   : [ Ht | M ]                              segment
Network     : [ Hn | Ht | M ]                         datagram
Link        : [ Hl | Hn | Ht | M | trailer ]          frame
Physical    : 0110101011...                           bit

At the receiver the reverse happens (decapsulation): each layer reads its own header, strips it, and hands the contents to the layer above. A router on the path looks only at the link and network headers; a switch only at the link header. End hosts process all five layers.

Worked example: overhead

1000 bytes of application data are sent with TCP (20 bytes), IPv4 (20 bytes) and Ethernet (14-byte header + 4-byte FCS = 18 bytes).

Frame      = 1000 + 20 + 20 + 18 = 1058 bytes
Efficiency = 1000 / 1058 = 0.9452  -> 94.52%
Overhead   = 58 / 1058   = 5.48%

For small messages the efficiency collapses: a 1-byte keystroke (SSH) becomes a 59-byte frame. This is why protocols try to combine small packets.

PITFALL · number of headers. Each layer adds one header; the physical layer adds no header (the preamble is a separate topic). Ethernet also puts a trailer (FCS, CRC) at the end. You will compute this as the CRC in week 3.

Physical layer basics

The physical layer turns bits into signals: voltage on copper, light in fiber, radio waves in the air.

Basic concepts

If a symbol can take $V$ different values, it carries $\log_2 V$ bits:

$\text{bit rate} = \text{baud rate} \times \log_2 V$

Example: at 9600 bps and 2400 baud, each symbol carries $9600/2400 = 4$ bits, so $2^4 = 16$ levels.

PITFALL · “bandwidth” in everyday language. In marketing, people say “100 Mbps bandwidth”. In a physical-layer question in this course, bandwidth means Hz. If a question says “bandwidth of 3 kHz”, that is a frequency range, not a bit rate.

Nyquist: the noiseless channel

On a noiseless channel with bandwidth $B$ Hz, at most $2B$ symbols per second can be sent. With $V$ levels:

$C_{Nyquist} = 2B\log_2 V$

B = 3000 Hz, V = 2 :  2 x 3000 x 1 = 6,000 bps
B = 3000 Hz, V = 4 :  2 x 3000 x 2 = 12,000 bps

According to Nyquist, you can raise the rate without limit by increasing the number of levels. In reality, noise does not allow this.

Shannon: the noisy channel

Noise makes levels that are close to each other indistinguishable. The Shannon capacity is an absolute upper limit, independent of the number of levels:

$C_{Shannon} = B\log_2(1 + \text{SNR})$

SNR is usually given in decibels:

$\text{SNR}_{dB} = 10\log_{10}(\text{SNR}) \quad\Longleftrightarrow\quad \text{SNR} = 10^{\text{SNR}_{dB}/10}$

Worked example: a telephone line

$B = 3000$ Hz, SNR = 30 dB.

SNR ratio = 10^(30/10) = 1000
C = 3000 x log2(1 + 1000) = 3000 x 9.967 = 29,902 bps (approximately)

This limit is why classic dial-up modems got stuck at around 30 kbps.

Using the two formulas together

A common exam question: “How many levels are needed to reach the Shannon capacity?” First find $C$ from Shannon, then solve for $V$ in Nyquist.

B = 1 MHz, SNR = 255
Shannon: C = 10^6 x log2(256) = 8 Mbps
Nyquist: 8 x 10^6 = 2 x 10^6 x log2 V  ->  log2 V = 4  ->  V = 16

Line coding

In long runs of 0s or 1s, the receiver’s clock drifts with NRZ. Because Manchester makes a transition in every bit, the clock comes from the signal itself (self-clocking), but it needs twice the signal rate: 10 Mbps Ethernet is 20 Mbaud.

Transmission media

Security note: a wireless medium is a broadcast medium. Anyone within range can receive the signal. This is why encryption (WPA2/WPA3) is a must on Wi-Fi. Tapping fiber requires physical access and is much harder.

Common mistakes