 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 dB, although strictly speaking, this should be given as decilogs relative to 1 J/K.

 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).

 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

 For a total transmit power (Pt) of 1000W, determine the energy per bit (Eb) for a transmission rate of 50Mbps. (20uJ or - 47dBW/bps)

 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

 Calculate the gain in decibels of a 3-m paraboloidal antenna operating at a frequency of 12 GHz. Assume an aperture efficiency of  Answer: 48.9 dB

 Free-space transmission  Feeder losses  Antenna misalignment losses  Fixed atmospheric and ionospheric losses

 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: dB

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]

 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: dB

 Thermal Noise  Antenna Noise  Amplifier Noise Temperature  Amplifier in Cascade  Noise Factor  Noise Temperature of absorptive networks  Overall system noise temperature

Where:  P N = Noise Power, W  k = Boltzmann’s constant = 1.38 x J/K  T N = equivalent noise temperature, K  B N = equivalent noise bandwidth, Hz  N 0 = Noise Power Spectral Density, J

 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 J ; pW

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

 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: K

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

LNA G 1 T e1 T ANT Cable Receiver Noise Factor F loss L : 1

 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.

 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

 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.

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

Where:  E b = Bit Energy, joules/bps  P t = Transmit Power, watts  f b = bit rate, bit per second

 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

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

 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