Physical Layer

Intuition

The physical layer deals with transmitting raw bits over a physical medium. Bandwidth determines how many signal changes per second a channel can carry, while noise limits how many distinct signal levels can be reliably distinguished. The Nyquist theorem bounds the maximum bit rate for noiseless channels, and Shannon’s theorem extends this to noisy channels by incorporating signal-to-noise ratio. These fundamental limits apply regardless of the encoding scheme or technology used.

2.1 Transmission Media

Guided media: Twisted pair (UTP, STP), coaxial cable, fibre optic.

Twisted pair: Category 5e/6/6a for Ethernet. Bandwidth up to 10 Gbps (Cat 6a, 100 m).
Fibre optic: Single-mode (long distance, laser) and multi-mode (shorter distance, LED). Bandwidth up to 100+ Gbps.

Unguided media: Radio waves, microwaves, infrared. Subject to attenuation, interference, and Line-of-sight constraints.

2.2 Signaling

Analog vs. Digital. Analog signals vary continuously; digital signals are discrete.

Bandwidth: Range of frequencies a channel can carry, measured in Hz.
Bit rate: Number of bits transmitted per second (bps).
Nyquist theorem: For a noiseless channel of bandwidth $H$ Hz with $V$ discrete signal levels:

C = 2H \log_2 V ;\mathrm{bps}

Theorem 2.1 (Nyquist—Shannon Sampling Theorem). A bandlimited signal of bandwidth $H$ Hz can Be perfectly reconstructed from samples taken at a rate of at least $2H$ samples per second.

Proof. Let $x(t)$ be a signal with Fourier transform $X(f)$ such that $X(f) = 0$ for $\lvert f \rvert \gt H$ . Sampling at rate $f_s$ produces $x_s(t) = x(t) \cdot \sum_{n=-\infty}^{\infty} \delta(t - nT_s)$ Where $T_s = 1/f_s$ . In the frequency domain, $X_s(f) = f_s \sum_{k=-\infty}^{\infty} X(f - kf_s)$ . When $f_s \geq 2H$ The spectral copies do not overlap, and $x(t)$ can be recovered by an ideal Lowpass filter with cutoff $H$ . When $f_s \lt 2H$ Aliasing occurs and perfect recovery is Impossible. $\blacksquare$

Shannon capacity: For a noisy channel with signal-to-noise ratio $\mathrm{SNR}$ :

C = H \log_2(1 + \mathrm{SNR}) ;\mathrm{bps}

Theorem 2.2 (Shannon—Hartley Theorem). The channel capacity $C$ is the maximum error-free data Rate achievable on a channel of bandwidth $H$ with signal-to-noise ratio $\mathrm{SNR}$ .

Proof. For a bandlimited AWGN channel, the number of distinguishable signal levels is constrained by the noise power. Let $\mathrm{SNR} = S/N$ where $S$ is signal power and $N = N_0 H$ is noise Power. The number of distinguishable amplitude levels is proportional to $\sqrt{1 + \mathrm{SNR}}$ . With $\log_2$ levels per signal element and $2H$ signal elements per second (Nyquist), the maximum Error-free rate is $C = 2H \cdot \tfrac{1}{2}\log_2(1 + \mathrm{SNR}) = H \log_2(1 + \mathrm{SNR})$ . $\blacksquare$

Example. A telephone line has $H = 3100$ Hz and $\mathrm{SNR} = 3162$ (35 dB). Shannon limit: $C = 3100 \times \log_2(3163) \approx 34860$ bps.

Worked Example: Nyquist Bit Rate

A noiseless channel has a bandwidth of 4000 Hz. How many signal levels are needed to achieve a data Rate of 56000 bps?

Using Nyquist’s formula: $C = 2H \log_2 V$ $56000 = 2 \times 4000 \times \log_2 V$ \log_2 V = \frac{56000}{8000} = 7V = 2^7 = 128$$

Answer: 128 signal levels are required.

Worked Example: Shannon Channel Capacity

A satellite channel has a bandwidth of 36 MHz and an SNR of 30 dB. Find the maximum data rate.

First convert SNR from dB to linear: $$\mathrm{SNR_}{\mathrm{linear} = 10^{30/10} = 1000}

Apply Shannon’s formula: $$C = H \log_2(1 + \mathrm{SNR}) = 36 \times 10^6 \times \log_2(1001) \log_2(1001) = \frac{\ln(1001)}{\ln(2)} \approx 9.967 C = 36 \times 10^6 \times 9.967 \approx 358.8 \times 10^6 ;\mathrm{bps} \approx 358.8;\mathrm{Mbps}

Answer: The maximum achievable data rate is approximately 358.8 Mbps. Any attempt to exceed This rate will result in an unacceptable error rate regardless of the modulation scheme used.

Worked Example: Comparing Nyquist and Shannon Limits

A channel has $H = 6000$ Hz and $\mathrm{SNR} = 1023$ (30 dB).

Shannon limit: $$C = 6000 \times \log_2(1024) = 6000 \times 10 = 60000;\mathrm{bps}

Nyquist limit with $V = 8$ : C = 2 \times 6000 \times \log_2(8) = 12000 \times 3 = 36000;\mathrm{bps}

The Nyquist limit (36 kbps) is below the Shannon limit (60 kbps), so 8 signal levels are Achievable. With $V = 64$ : $$C = 12000 \times 6 = 72000;\mathrm{bps}

This exceeds Shannon’s limit of 60 kbps, meaning 64 levels would produce errors. The maximum Number of levels consistent with Shannon: C_{\mathrm{Shannon} = 2H \log_2 V \implies 60000 = 12000 \times \log_2 V \implies V = 32}

Answer: At most 32 signal levels can be used reliably on this channel.

the range of frequencies The channel can carry; bit rate is the number of bits transmitted per second. Shannon’s theorem Relates the maximum bit rate to bandwidth and SNR, but they are not interchangeable.

2.3 Multiplexing

Frequency-Division Multiplexing (FDM). Divide bandwidth into non-overlapping frequency bands. Each user gets a dedicated band. Used in radio and cable TV.

Time-Division Multiplexing (TDM). Divide time into fixed slots; each user gets a slot per cycle. Synchronous TDM assigns slots statically; statistical TDM assigns dynamically based on demand.

Wavelength-Division Multiplexing (WDM). FDM for fibre optics. Multiple wavelengths share a Single fibre. Dense WDM (DWDM) supports 80+ channels.

Code-Division Multiple Access (CDMA). Each user is assigned a unique code. All users transmit Simultaneously on the same frequency; codes are mathematically orthogonal so receivers can isolate their signal.

2.4 Modulation

Digital-to-analog modulation: ASK (Amplitude Shift Keying), FSK (Frequency Shift Keying), PSK (Phase Shift Keying). QAM combines ASK and PSK for higher data rates.

16-QAM encodes 4 bits per symbol (16 combinations of amplitude and phase).
256-QAM encodes 8 bits per symbol.
1024-QAM encodes 10 bits per symbol (used in Wi-Fi 6).

Theorem 2.3 (QAM spectral efficiency). An $M$ -ary QAM scheme where $M = 2^{2k}$ has a spectral Efficiency of $2k$ bits/symbol, i.e., the bit rate equals $2k \times B$ where $B$ is the bandwidth in Hz.

Proof. QAM modulates both amplitude and phase of a carrier. With $M$ symbols, each symbol carries $\log_2 M = 2k$ bits. The symbol rate equals the bandwidth $B$ (Nyquist: 2 symbols/Hz for Baseband, 1 symbol/Hz for passband). Therefore bit rate = $2k \times B$ . $\blacksquare$

Worked Example: QAM Data Rate Calculation

A 256-QAM modem operates over a 20 MHz channel. What is the maximum data rate?

M = 256, \quad \log_2 256 = 8 ;\mathrm{bits}/symbol

\mathrm{Bit};rate = 8 \times 20 \times 10^6 = 160;\mathrm{Mbps}

If the channel has SNR = 24 dB, verify against Shannon:

\mathrm{SNR_}{\mathrm{linear} = 10^{24/10} = 251.2}

C = 20 \times 10^6 \times \log_2(252.2) \approx 20 \times 10^6 \times 7.98 \approx 159.6;\mathrm{Mbps}

The Nyquist-based rate (160 Mbps) is very close to the Shannon limit (159.6 Mbps), meaning 256-QAM is near-optimal for this channel but has almost no margin for noise or interference. </details> ### 2.5 Line Coding Line codes map binary data to signals suitable for the physical medium. <div class="table-wrap"> | Encoding | Description | Example | | ----------------------- | -------------------------------------- | ------------------ | | NRZ (L) | High = 1, Low = 0 | USB | | NRZI | Transition = 1, no transition = 0 | USB | | Manchester | Transition at mid-bit; low-to-high = 0 | 802.3 (10 Mbps) | | Differential Manchester | Transition at start of 0 bit only | 802.5 (Token Ring) | | 4B/5B | 4 data bits encoded as 5-bit codes | 100BASE-TX | | 8B/10B | 8 data bits encoded as 10-bit codes | Gigabit Ethernet | | 64B/66B | 64 data bits encoded as 66-bit codes | 10GBASE-R | </div> **Spectral efficiency.** Manchester encoding doubles the bandwidth requirement (two signal levels per Bit). 4B/5B adds 25% overhead. 8B/10B adds 25%. 64B/66B adds only 3% overhead. **Scrambling.** High-density bipolar 3 (HDB3) and other scrambling techniques ensure sufficient Transitions for clock recovery, preventing long runs of identical bits.

Common Mistakes

Confusing bandwidth with bit rate: Bandwidth is the range of frequencies a channel can carry. Bit rate is the number of bits transmitted per second. They’re related but different concepts.

Forgetting that noise limits channel capacity: Shannon’s theorem shows that even with perfect encoding, noise limits the maximum reliable bit rate. Don’t assume infinite capacity is possible.

Mixing up guided with unguided media: Guided media use physical cables (twisted pair, fibre). Unguided media use wireless (radio, microwave). Don’t confuse the two categories.