What is the tradeoff between bandwidth and power in a communication link budget?

The fundamental tradeoff between bandwidth and power in a communication link is described by the Shannon capacity theorem: C = BW × log2(1 + SNR), where C is the channel capacity (bits/second), BW is the bandwidth (Hz), and SNR is the signal-to-noise ratio (linear). This equation reveals: (1) Increasing bandwidth (BW) at constant transmit power: reduces the SNR (because noise power = kTBW increases). But the capacity still increases (logarithmically) because the wider bandwidth captures more signal dimensions. In the limit: as BW → infinity with fixed C and power P: SNR = P/(kT×BW) → 0, but the capacity approaches C = P/(kT×ln2) = P/N0 × 1.44 bits/second. This is the ultimate power-limited capacity (bandwidth is free). Required Eb/No → -1.59 dB (Shannon limit). (2) Increasing power at constant bandwidth: increases SNR, which increases capacity logarithmically. Doubling power (3 dB) increases capacity by log2(2×SNR) - log2(SNR) = log2(2) = 1 bit/s/Hz (only when SNR >> 1). At low SNR: doubling power nearly doubles capacity (linear regime). At high SNR: doubling power provides diminishing returns (logarithmic regime). (3) Practical tradeoff: power-limited systems (satellite, deep space, IoT): use wide bandwidth with low spectral efficiency (spread spectrum, low-order modulation with strong coding). The link budget is dominated by Eb/No, and bandwidth is traded for power. Bandwidth-limited systems (cellular, microwave backhaul): use high-order modulation (64-QAM, 256-QAM) to maximize bits/Hz. More transmit power or antenna gain is needed to support the higher SNR requirements. (4) Design example: a satellite link with 1 W transmit power and 1 MHz bandwidth: SNR = 1/(1.38e-23 × 290 × 1e6) = 250 (24 dB). C = 1e6 × log2(1+250) = 8 Mbps. Same power, 10 MHz bandwidth: SNR = 25 (14 dB). C = 10e6 × log2(1+25) = 47 Mbps. 10× more bandwidth yields 5.9× more capacity (not 10×, due to the SNR reduction).