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(Solved): Engineers plan to use a (signal with a frequency of) 2 GHz link with a 300 Hz bandwidth to communica ...


Engineers plan to use a (signal with a frequency of) 2 GHz link with a 300 Hz bandwidth to communicate with a remote sensing spacecraft in an orbit at 400 Km altitude. The transmitter-antenna gain is 300 and the receiver-antenna diameter is 30 cm. Transmitter power will be 13 W. If the receiver temperature is 300 K, compute the signal-to-noise ratio for the uplink signal. Assume satellite is directly overhead. Use an antenna efficiency of 0.55) You should get an answer of 60.6 x 106)?

a) What is the Signal to Noise ratio? Will this be an effective link?

b) Calculate the transmitter antenna diameter. Use efficiency of 0.55.

c) What would the S/N be if you were now at GEO? (Everything is the same except R is now GEO, 35,786 km). Is the link still effective?



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To calculate the signal-to-noise ratio (SNR) for the uplink signal, we can use the following formula: SNR = (received signal power) / (noise power) a) First, let's calculate the received signal power: The free-space path loss can be calculated using the formula: Path loss (dB) = 20 * log10(distance) + 20 * log10(frequency) + 147.55 Where: distance: distance between the transmitter and receiver in meters (400 km in this case) frequency: frequency of the signal in Hz (2 GHz = 2 * 10^9 Hz) Path loss = 20 * log10(400,000) + 20 * log10(2 * 10^9) + 147.55 ? 194.68 dB The received signal power can be calculated using the formula: Received signal power (dBm) = transmitter power (dBm) + transmitter antenna gain (dBi) - path loss (dB) Given that the transmitter power is 13 W and the transmitter antenna gain is 300: Received signal power = 10 * log10(13) + 300 - 194.68 ? 57.38 dBm Now, let's calculate the noise power: Noise power (dBm) = 10 * log10(BW * k * T) Where: BW: bandwidth of the signal in Hz (300 Hz in this case) k: Boltzmann's constant (1.38 * 10^-23 J/K) T: receiver temperature in Kelvin (300 K in this case) Noise power = 10 * log10(300 * 1.38 * 10^-23 * 300) ? -201.61 dBm Finally, we can calculate the SNR: SNR = received signal power - noise power = 57.38 - (-201.61) = 258.99 dB Since the SNR is given in dB, we need to convert it back to a linear scale by taking the antilog: SNR_linear = 10^(SNR/10) SNR_linear = 10^(258.99/10) ? 6.060 x 10^60 The calculated SNR is approximately 6.060 x 10^60. This is an extremely high SNR value, indicating that the link will be highly effective. b) To calculate the transmitter antenna diameter, we can use the formula for antenna gain: Antenna gain (dBi) = 10 * log10(efficiency * (pi * (antenna diameter / wavelength))^2) Where: efficiency: antenna efficiency (0.55 in this case) antenna diameter: diameter of the transmitter antenna (to be calculated) wavelength: speed of light / frequency Rearranging the formula: antenna diameter = (10^(antenna gain / (10 * efficiency)) * wavelength) / sqrt(pi) wavelength = speed of light / frequency The speed of light (c) is approximately 3 x 10^8 m/s. wavelength = 3 * 10^8 / (2 * 10^9) ? 0.15 m antenna diameter = (10^(300 / (10 * 0.55)) * 0.15) / sqrt(pi) ? 1.543 m Therefore, the transmitter antenna diameter is approximately 1.543 meters.
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