or this purpose, it believe this implies that the noise level of the receiver is obvious that differential modulation carries consider- must be less than (-80-12.5) = -92.5 dBm. However, able benefits and little penalty. recall that 11 QPSK chips at 11 Mbps are "averaged" to- The maximum transmit power of an 802.11 radio is gether (correlated with the Barker sequence) to arrive at limited in the standard to comply with the requirements one 2-bit data symbol. This averaging process provides us of regulatory bodies. In the United States, power must be with some extra gain-the spreading gain described less than 1W, but in practice a typical access point uses above-so that we can tolerate an (S/N) of 12.5 - 10.5 = 2 a transmit power of about 30 to 100 mW. The standard dB and still get the data right. The noise level we can tol- then requires that a receiver achieve a frame error rate of erate in the receiver is thus -80 - 2 = -82 dBm. Recalling less than 8% for 1024-byte frames of 2 Mbps QPSK at that the signal is about 16 MHz wide, the unavoidable a signal power of -80 decibels from a milliwatt (dBm). thermal noise in the receiver is (-174 dBm + 10(log 16) + That sounds like a lot of errors, but because each frame has 60) = -102 dBm. The specification has left room for an 1024 X 8 = 8192 bytes, the bit error rate is a respectable additional 20 dB of excess noise in the receiver. Receivers 0.08/8192 = 10-5. To understand what the transmit and with noise figures of 20 dB at 2.4 GHz are very easy to receive power levels imply, let's do a link budget calcula- build, and in fact 802.11 commercial products do much tion. Let the transmitted power be 100 mW (20 dBm). better than this. The standard has specified a level of Assume the transmitting antenna concentrates its energy performance that can be reached inexpensively, appro- in, for example, the horizontal plane, achieving a 6-dB priate to equipment meant for an LAN. The link budget increase in signal power over that of an ideal isotropic calculation is displayed graph