Physical Layer

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.

:::caution Common Pitfall Bandwidth (Hz) and bit rate (bps) are different quantities. Bandwidth is 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 = $2k \times B$. $\blacksquare$

2.5 Line Coding

Line codes map binary data to signals suitable for the physical medium.

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.

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