Channel Capacity

Channel capacity represents the theoretical maximum rate at which information can be reliably transmitted over a communication channel. Shannon's groundbreaking 1948 theorem established that error-free communication is possible at any rate below capacity, fundamentally changing how we think about communication system design.

Nyquist Rate (Noiseless Channel)

For an ideal, noiseless channel with bandwidth B Hz:

Binary Signaling (2 levels):

C = 2B bits/second

M-ary Signaling (M levels):

C = 2B × log₂(M) bits/second

Example: A noiseless channel with B = 3 kHz using 4 levels: C = 2 × 3000 × log₂(4) = 2 × 3000 × 2 = 12,000 bps

Limitation: This assumes perfect channel — in reality, noise limits how many levels can be reliably distinguished.

Shannon-Hartley Theorem (Noisy Channel)

The maximum rate of error-free transmission over a noisy channel:

C = B × log₂(1 + S/N) bits/second

Where:

• C = channel capacity (bps)
• B = channel bandwidth (Hz)
• S/N = signal-to-noise ratio (linear, not dB)

Equivalent forms:

C = B × log₂(1 + P/(N₀B)) where P = signal power, N₀ = noise PSD

C/B = log₂(1 + SNR) (spectral efficiency limit in bits/s/Hz)

Key Implications of Shannon's Theorem

1. Capacity is finite: Even with infinite SNR, bandwidth limits capacity
2. Error-free communication is possible: Below capacity, codes exist that achieve arbitrarily low error rates
3. Above capacity, errors are unavoidable: No coding scheme can achieve reliable communication above C
4. Bandwidth-power trade-off: Capacity depends on both — they are partially interchangeable

Channel Capacity Examples

Voice-Band Channel (PSTN)

• B = 3.1 kHz (300-3400 Hz)
• SNR = 30 dB (linear = 1000)
• C = 3100 × log₂(1001) ≈ 3100 × 9.97 = 30,900 bps
• Practical modem: V.34 achieves 33.6 kbps (near Shannon limit!)

Satellite Channel

• B = 36 MHz
• SNR = 10 dB (linear = 10)
• C = 36×10⁶ × log₂(11) = 36×10⁶ × 3.46 = 124.5 Mbps

Wi-Fi Channel (80 MHz)

• B = 80 MHz
• SNR = 25 dB (linear = 316)
• C = 80×10⁶ × log₂(317) = 80×10⁶ × 8.31 = 664.6 Mbps

Bandwidth-Limited vs. Power-Limited Regimes

Bandwidth-Limited (High SNR)

When SNR >> 1: C ≈ B × log₂(SNR)

Doubling bandwidth approximately doubles capacity. Adding more power gives diminishing returns (logarithmic). Solution: Use higher-order modulation (64-QAM, 256-QAM).

Power-Limited (Low SNR)

When SNR << 1: C ≈ (S/N₀) × log₂(e) = 1.44 × S/N₀

Capacity depends only on S/N₀, not bandwidth. Increasing bandwidth doesn't help. Solution: Use wider bandwidth with low-order modulation and strong coding.

Ultimate bandwidth limit: As B → ∞: C → (S/N₀) × 1.44 bits/s

This shows even with infinite bandwidth, capacity is finite (limited by power).

Capacity of Parallel Channels

For a frequency-selective channel divided into N sub-channels (like OFDM):

C = Σ Bᵢ × log₂(1 + Sᵢ/Nᵢ) for i = 1 to N

Water-filling power allocation: Allocate more power to sub-channels with better SNR:

Sᵢ = max(0, μ - N₀/|Hᵢ|²)

Where μ is chosen so that ΣSᵢ = P_total

Shannon Capacity vs. Practical Systems

*V.34 exceeds simple Shannon by using line coding techniques beyond binary

Solved Example

Problem: A channel has bandwidth 4 MHz. The required channel capacity is 20 Mbps. Determine: (a) required SNR, (b) if SNR is limited to 20 dB, what minimum bandwidth is needed?

Solution:

(a) Required SNR for C = 20 Mbps in B = 4 MHz:

C = B × log₂(1 + SNR) 20×10⁶ = 4×10⁶ × log₂(1 + SNR) log₂(1 + SNR) = 5 1 + SNR = 2⁵ = 32 SNR = 31 (linear) SNR = 10×log₁₀(31) = 14.9 dB

(b) Minimum bandwidth for C = 20 Mbps with SNR = 20 dB:

SNR(linear) = 10^(20/10) = 100 C = B × log₂(1 + 100) 20×10⁶ = B × log₂(101) 20×10⁶ = B × 6.66 B = 3.003 MHz ≈ 3 MHz

Interview Questions

Q1: Can Shannon capacity be exceeded in practice?

No, Shannon capacity is a theoretical upper bound. However, practical systems can sometimes appear to exceed it through techniques that implicitly increase resources: MIMO creates parallel spatial channels (effectively multiplying bandwidth), successive interference cancellation treats interference as known signal, and some modems exploit non-white noise characteristics. These don't violate Shannon — they use additional degrees of freedom.

Q2: What is the significance of the Shannon limit Eb/N₀ = -1.6 dB?

As bandwidth approaches infinity (with fixed power), the minimum Eb/N₀ for reliable communication approaches ln(2) = 0.693, which is -1.59 dB. This is the ultimate energy efficiency limit — no communication system, regardless of complexity, can reliably transmit a bit with less than -1.6 dB of Eb/N₀.

Q3: How do modern systems approach Shannon capacity?

Modern capacity-approaching techniques include: (1) Turbo codes and LDPC codes with iterative decoding (within 0.1 dB of capacity), (2) Adaptive modulation and coding (match modulation to channel conditions), (3) OFDM with water-filling (optimal for frequency-selective channels), (4) MIMO spatial multiplexing (multiply capacity by min antenna count).

Q4: Explain the bandwidth-power trade-off using Shannon's formula.

From C = B×log₂(1+P/(N₀B)): For fixed capacity C, you can achieve the same rate with more bandwidth and less power, or less bandwidth and more power. Wideband systems (spread spectrum, UWB) use large B with low P/B, while narrowband systems use small B with high P/B. The optimal trade-off depends on whether bandwidth or power is the scarcer resource in the application.