Physical Layer | Computer Science

flowchart TD
    A[2_Physical Layer] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

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.

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.

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.

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.

• Computer Networks
• Network Layer
• Transport Layer