Topic: Wireless Communication
A cell phone transmits at a power level of 800 mW with an antenna gain of 3.0 dB. The cell tower has an antenna gain of 10.0 dB and is at a distance of 5 km away. Transmission frequency is 600 MHz. Noise level at Receiver Input is -95 dBm and required Signal to Noise ratio to close the link is 5 dB. Find the link margin in dB. (assume $\pi^2=10)$
- A. 150 dB
- B. 60 dB
- C. 30 dB
- D. 35 dB
Answer: C. 30 dB
Explanation: Transmit power $P_t = 800$ mW $\approx 29$ dBm. Path loss is calculated as $L_p = 20 \log_{10}(\frac{4\pi d}{\lambda})$. At 600 MHz, $\lambda = 0.5$ m. $L_p = 20 \log_{10}(\frac{4\pi \times 5000}{0.5}) = 20 \log_{10}(40000\pi)$. Given $\pi \approx \sqrt{10}$, $L_p \approx 102$ dB. Received power $P_r = P_t + G_t + G_r - L_p = 29 + 3 + 10 - 102 = -60$ dBm. Received SNR $= P_r - \text{Noise} = -60 - (-95) = 35$ dB. The link margin is Received SNR - Required SNR $= 35 - 5 = 30$ dB.