Channel Capacity

Channel capacity is the maximum rate at which information can be transmitted over a communication channel with arbitrarily small error probability. It is determined by the Shannon-Hartley theorem and depends on bandwidth and signal-to-noise ratio. For GATE aspirants, channel capacity is one of the highest-weightage formulas in digital communications and information theory.

Core Concept Explanation

Channel capacity C is defined as the maximum mutual information I(X;Y) optimized over all possible input distributions P(x). Shannon proved that for an additive white Gaussian noise (AWGN) channel, this maximum is achieved by a Gaussian input distribution, and the resulting capacity is the Shannon-Hartley formula C = B log2(1 + SNR).

The fundamental meaning of capacity is operational: if you transmit at any rate R less than C, there exists a coding scheme that achieves an arbitrarily small probability of error. If you attempt to transmit at any rate R greater than C, no coding scheme can reduce the error probability to zero. This is one of the deepest results in all of engineering.

Capacity is measured in bits per second. It is determined by two factors: bandwidth B in hertz, and the signal-to-noise ratio SNR = S/N where S is average signal power and N is noise power. The relationship is logarithmic in SNR, which means diminishing returns: doubling SNR adds only one extra bit per second per hertz, while doubling bandwidth directly doubles capacity.

Mathematical Expression

The Shannon-Hartley theorem for a continuous AWGN channel gives:

C = B log2(1 + S/N) bits per second

Here B is the channel bandwidth in Hz, S is average received signal power in watts, and N is total noise power N = N0 * B where N0 is the one-sided noise power spectral density in W/Hz. When SNR is expressed in linear scale (not dB), substitute directly. If SNR is given in dB, convert first: SNR_linear = 10^(SNR_dB / 10).

The spectral efficiency eta = C/B = log2(1 + SNR) is measured in bits per second per hertz (bps/Hz). This normalized form allows comparison of different systems independent of bandwidth. Real systems like LTE target specific spectral efficiency values governed by this bound.

As SNR approaches infinity, capacity grows without bound but very slowly. As bandwidth approaches infinity with fixed total noise power N0*B, the capacity approaches a finite limit of C_max = S / (N0 * ln 2) bits per second, showing that infinite bandwidth does not give infinite capacity when noise power scales with bandwidth.

Practical Understanding

In 4G LTE and 5G NR systems, link budget calculations use the Shannon formula to estimate the theoretical maximum throughput of a cell. Real systems operate at efficiencies typically 60 to 80 percent of the Shannon limit due to practical constraints like finite modulation orders, pilot overhead, feedback delay, and hardware imperfections.

The formula explains why increasing transmit power in a congested wireless system gives diminishing returns: capacity grows only as log2(1 + SNR). Allocating more bandwidth or reducing noise floor (through better receivers or shorter distances) is often more effective than simply increasing transmit power.

Solved Numerical Example

To apply the Shannon formula, we need bandwidth and SNR. In exam problems, SNR is often given in dB and must be converted before substitution.

Given:
Bandwidth B = 4 kHz (standard voice-grade telephone channel)
SNR = 30 dB

Why this formula applies:
This is an AWGN channel; Shannon-Hartley theorem gives the maximum achievable rate.

Formula:
C = B * log2(1 + SNR_linear)

Substitution:
SNR_linear = 10^(30/10) = 10^3 = 1000
C = 4000 * log2(1 + 1000)
C = 4000 * log2(1001)

Calculation:
log2(1001) = log10(1001) / log10(2) = 3.0004 / 0.3010 ≈ 9.967 bits
C = 4000 * 9.967

Final Answer: C ≈ 39,868 bits/s ≈ 39.87 kbps

Exam Tip: When SNR is given in dB, always convert to linear before substituting: SNR_linear = 10^(SNR_dB/10). A common GATE trap is substituting the dB value directly into log2(1 + SNR), which gives a dramatically wrong answer. Also remember that log2(x) = log10(x) / log10(2) = log10(x) / 0.301.

• C = B log2(1 + SNR) gives theoretical maximum; no practical system can exceed this limit at any given SNR and bandwidth.
• Doubling bandwidth doubles capacity directly. Doubling SNR adds only log2(2) = 1 extra bit per second per hertz.
• When total noise power scales with bandwidth (N = N0*B), infinite bandwidth gives finite capacity C = S/(N0 ln2) = 1.44 S/N0.
• Spectral efficiency eta = log2(1 + SNR) determines how many bits per hertz the channel can support, independent of absolute bandwidth.
• The Gaussian input distribution achieves capacity in the AWGN channel; for discrete channels, capacity is achieved by specific input distributions found through the Blahut-Arimoto algorithm.

Quick Revision

• C = B log2(1 + SNR) bits/s. Bandwidth in Hz, SNR must be linear (not dB).
• SNR_linear = 10^(SNR_dB / 10). Never substitute dB value directly into the formula.
• Spectral efficiency eta = C/B = log2(1 + SNR) bits/s/Hz. A key figure of merit for modulation comparison.
• Infinite bandwidth limit: C = 1.44 * S/N0 bits/s. This is finite because noise power grows with bandwidth.
• C is a maximum; transmitting at R < C allows error-free communication with proper coding. R > C cannot be achieved reliably.
• Exam trap: Students often forget to include the +1 inside the logarithm, especially when SNR is very large and the +1 seems negligible (it still matters for small SNR).
• Relationship: C = max_{P(x)} I(X;Y). For AWGN, this maximum is the Shannon-Hartley formula.