Answer :
Final answer:
The receiver sensitivity in an IEEE 802.11a system using 64 QAM modulation to achieve a data rate of 54 Mbps with Eb/N0 = 33 dB, an implementation loss of 4 dB, and a noise figure of 8 dB, is -81.7 dBm.
Explanation:
To find the required sensitivity of the receiver in this IEEE 802.11a (WLAN standard) that uses 64 QAM modulation to achieve a data rate of 54 Mbps with Eb/N0 = 33 dB, we must first convert the Eb/N0 from dB into linear scale. The linear scale is 10^(Eb/N0/10) = 1995.26.
Given that the bit energy to noise ratio is known, we can find the noise spectrum density (N0) by dividing the bit energy (Eb) by this ratio. If we take the bit energy as -174 dBm, the noise figure is 8 dB and the implementation loss is 4 dB, we can sum these values together to find N0 = -174 + 8 + 4 = -162 dBm.
Finally, under these circumstances, we can calculate the receiver sensitivity as the sum of N0 and Eb/N0. This results in sensitivity = N0 + 10log10(1995.26) = -162 + 10log10(1995.26) = -81.7 dBm. So, the sensitivity of the receiver is -81.7 dBm.
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