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Published byGabriel Goodman Modified over 8 years ago
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This chapter describes how the link- power budget calculations are made. In this text [square] bracket are used to denote decibel quantities using the basic power definition. Boltzmann’s constant is given as -228.6 dB, although strictly speaking, this should be given as -228.6 decilogs relative to 1 J/K.
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To operate as efficiently as possible, a power amplifier should be operated as close as possible to saturation. The saturated output power is designated P o (sat) or simply Pt(dBW).
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E b = P t T b E b = energy of a single bit (joules per bit) Pt = total saturated output power (watts or joules per second) Tb = time of a single bit (seconds). = 1/f b
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For a total transmit power (Pt) of 1000W, determine the energy per bit (Eb) for a transmission rate of 50Mbps. (20uJ or - 47dBW/bps)
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A satellite downlink at 12 GHz operates with a transmit power of 6W and an antenna gain of 48.2 dB. Calculate the EIRP in dBW. Answer: 56 dBW
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Calculate the gain in decibels of a 3-m paraboloidal antenna operating at a frequency of 12 GHz. Assume an aperture efficiency of 0.55. Answer: 48.9 dB
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Free-space transmission Feeder losses Antenna misalignment losses Fixed atmospheric and ionospheric losses
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The range between a ground station and a satellite is 42,000 km. Calculate the free-space loss at a frequency of 6 GHz. Answer: 200.4 dB
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Where: [PR] = received power, dBW [EIRP] = equivalent isotropic radiated power, dBW [FSL] = free-space spreading loss, dB [RFL] = receiver feeder loss, dB [AML] = antenna misalignment loss, dB [AA] = atmospheric absorption loss, dB [PL] = polarization mismatch loss, dB [PR] = [EIRP] + [G R ] – [LOSSES] [LOSSES] = [FSL] + [RFL] + [AML] + [AA] + [PL]
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A satellite link operating at 14 GHz has receiver feeder losses of 1.5 dB and a free-space loss of 207 dB. The atmospheric absorption loss is 0.5 dB, and the antenna pointing loss is 0.5 dB. Depolarization losses may be neglected. Calculate the total link loss for clear sky conditions. Answer: 209.5 dB
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Thermal Noise Antenna Noise Amplifier Noise Temperature Amplifier in Cascade Noise Factor Noise Temperature of absorptive networks Overall system noise temperature
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Where: P N = Noise Power, W k = Boltzmann’s constant = 1.38 x 10 -23 J/K T N = equivalent noise temperature, K B N = equivalent noise bandwidth, Hz N 0 = Noise Power Spectral Density, J
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An antenna has a noise temperature of 35 K and is matched into a receiver which has a noise temperature of 100 K. Calculate (a) the noise power density and (b) the noise power for a bandwidth of 36 MHz. Answer: 1.86 x 10-21 J ; 0.067 pW
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Amplifier Power Gain G N 0,in N 0,out Amplifier Power Gain G 1 T e1 N 0,1 N 0,out T ant N 0,2 Amplifier Power Gain G 2 T e2
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An LNA is connected to a receiver which has a noise figure of 12 dB. The gain of the LNA is 40 dB, and its noise temperature is 120 K. Calculate the overall noise temperature referred to the LNA input. Answer: 120.43 K
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Where N rad = Noise energy radiated L = Power Loss T X = Ambient Temperature T NW,0 = Equivalent Temperature Network Noise Lossy network power loss L:1 N RAD RTRT TXTX Ambient Temperature
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LNA G 1 T e1 T ANT Cable Receiver Noise Factor F loss L : 1
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For the system shown in the previous figure, the receiver noise figure is 12 dB, the cable loss is 5 dB, the LNA gain is 50 dB, and its noise temperature 150 K. The antenna noise temperature is 35K. Calculate the noise temperature referred to the input.
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Repeat the calculation when the system is arranged as shown below. LNA G 1 T e1 T ANT Cable Receiver Noise Factor F loss L : 1 Input
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In a link-budget calculation at 12 GHz, the free-space loss is 206 dB, the antenna pointing loss is 1 dB, and the atmospheric absorption is 2 dB. The receiver [G/T] is 19.5 dB/K, and a receiver feeder losses are 1 dB. The EIRP is 48 dBW. Calculate the carrier-to-noise spectral density ratio.
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QuantityDecilogs Free-Space Loss-206 Atmospheric absorption loss-2 Antenna Pointing Loss Receiver feeder losses Polarization mismatch loss0 Receiver G/T ratio19.5 EIRP48 -[k]228.6 [C/N 0 ]86.1
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Where: E b = Bit Energy, joules/bps P t = Transmit Power, watts f b = bit rate, bit per second
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In satellite communication system, for a total transmit power of 500 watts, determine the energy per bit for a transmission rate of 50 Mbps expressed in dBW. Answer: -50 dBW/bps
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Where: E b = Bit Energy, joules/bps P t = Transmit Power, watts f b = bit rate, bit per second N o = noise density, joules BW = bandwidth, Hz N = Noise, Watts C = Carrier Power, Watts
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A coherent binary phase shift keyed BPSK transmitter operates at a bit rate of 20 Mbps with a carrier-to-noise ratio C/N of 8.8 dB. Find the E b /N 0. Answer: 8.8 dB
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