Copyright 1999, S.D. Personick. All Rights Reserved. Telecommunications Networking I Lectures 14 & 15 Wireless Transmission Systems

Copyright 1999, S.D. Personick. All Rights Reserved. Wireless Point-to-Point Link Radio Transmitter Radio Receiver Feed Line (e.g., coaxial cable) Antenna

Copyright 1999, S.D. Personick. All Rights Reserved. The Electromagnetic Spectrum Hz: SLF 3GHz-30GHz: SHF (DBS) 300Hz - 3kHz ULF 30GHz-300GHz: EHF 3kHz - 30kHz: VLF 300,000GHz: 1um light 30kHz-300kHz: LF 300kHz-3MHz: MF (AM Radio) 3MHz-30MHz: HF (Short Wave Radio) 30MHz-300MHz: VHF (FM, TV) 300MHz-3GHz: UHF (TV, Digital Cordless, …)

Copyright 1999, S.D. Personick. All Rights Reserved. Wireless Transmitter ModulatorMixer RF Oscillator Antenna RF Amp Information IF Oscillator

Copyright 1999, S.D. Personick. All Rights Reserved. Transmitter Subsystems 1 milliwatt -100 milliwatts: low r.f. exposure with hand held appliances, low battery drain (some hand held appliances radiate ~5 watts…but I wouldn’t hold one of these near my head!) 10 watts watts: okay for consumer and small business applications, with 115 volt or 12 volt automobile power. Stay feet from the antenna watts-50,000 watts : broadcast applications

Copyright 1999, S.D. Personick. All Rights Reserved. Transmitter Subsystems R.F. Amplifier Coax Antenna FET R.F. Power = (V**2)/Z V~ 1 volt Z~50 ohms R.F. Power (launched into the coaxial cable) = 1/50 Watt = 20 mW R.F. Power radiated = ?

Copyright 1999, S.D. Personick. All Rights Reserved. The Antenna Forward Power Reflected Power Radiated Power

Copyright 1999, S.D. Personick. All Rights Reserved. The Antenna If the antenna is much smaller in length than the wavelength of the radiation (c/f), then the antenna will be a very inefficient radiator (most of the forward power is reflected) If the antenna is 1/4 wavelength or larger in size, it can be an efficient radiator If the antenna is significantly larger in size than 1 wavelength, it can be an efficient and directional radiator

Copyright 1999, S.D. Personick. All Rights Reserved. A Directional Radiator 1/4 wavelength spacing between two dipoles. 3/4 wavelength delay in crossover cable Forward: fields add Reverse: fields cancel It’s actually not quite that simple: dipole interactions

Copyright 1999, S.D. Personick. All Rights Reserved. The Parabolic Dish Antenna Divergence angle ~Lambda/D

Copyright 1999, S.D. Personick. All Rights Reserved. Wireless Receiver DemodulatorMixer RF Oscillator Antenna RF Amp Information IF Oscillator

Copyright 1999, S.D. Personick. All Rights Reserved. The Receiver Subsystem Low Noise Amplifier Signal + Background Noise + Interference Cable attenuation + thermal noise + Amplifier noise

Copyright 1999, S.D. Personick. All Rights Reserved. Sources of Thermal Noise Hot object (333K) Background (27K) Signal Field of View Mirror Sun

Copyright 1999, S.D. Personick. All Rights Reserved. The Receiver Subsystem Example: Equivalent background temperature: T = 100K (Kelvins) Equivalent background noise = kTB (watts) Coupling loss of antenna into cable ~ 0dB Cable loss = 3 dB Cable temperature = 293K Preamplifier Noise Temperature = 30K Equivalent amplifier total input noise = ???

Copyright 1999, S.D. Personick. All Rights Reserved. The Receiver Subsystem Example (continued): Equivalent input noise = kB [100/ /2 + 30]=226.5kB I.e., half of the background noise + half of the cable noise + the preamplifier noise

Copyright 1999, S.D. Personick. All Rights Reserved. The End-To-End System Example (continued from prior example): Assume that the transmitter power amplifier produces 20 milliwatts. The coupling loss from the transmitter into its coaxial antenna feed cable is 0 dB. The transmitter antenna feed cable has 1.5 dB of loss. The antenna radiates 90% of the power that arrives from the transmitter antenna feed cable, and reflects the rest back into the cable. How much power is radiated? Assume that the bandwidth, B = 6 MHz. How much total propagation loss can we allow if the required signal-to-noise ratio is 40 dB?

Copyright 1999, S.D. Personick. All Rights Reserved. The End-to-End System Example: (continued) The preamplifier produces 20 mW = +13 dBm The loss of the cable, plus the impact of a 10% reflection (90% radiation) is: 1.5 dB (cable loss) - 10 log (0.9) = 1.95 dB The total radiated power = dBm = 12.7 mW The required power at the preamplifier input = x k x 6 x 10**6 x 10**4 (watts)... I.e., 40 dB larger than the noise

Copyright 1999, S.D. Personick. All Rights Reserved. The End-to-End System The radiated power = 12.7 mW The required power at the receiver preamplifier input = 1.85 x 10**-7 mW The required power at the receiver antenna is 2 x 1.85 x 10**-7 mW = 3.7 x 10**-7 mW The allowed propagation loss is the ratio of these ~3.4 x 10**7 ~ 75.3 dB

Copyright 1999, S.D. Personick. All Rights Reserved. Calculating the Propagation Loss Area of surface = 4  r**2 Antenna equivalent area = A

Copyright 1999, S.D. Personick. All Rights Reserved. Calculating the Propagation Loss Suppose the frequency is 100 MHz, and the wavelength = 3 meters. Assume that the equivalent area of the antenna is 2.25 square meters. If the allowable loss is 75.3 dB, then the distance from the transmitter to the receiver, r, can be derived from: 2.25/[4  r**2) > 1/[3.4 x 10**7]; r < 2.5 km (Line-Of-Sight)

Copyright 1999, S.D. Personick. All Rights Reserved. Why Do Broadcast Transmitters Have Such High Power? In the previous example, we could cover 2.5 km with 20 mW. If we want to cover a 100 km radius, we would need 40**2 or 1600 times the power. That would correspond to 32 watts << 10 kW But, what about: high antenna-feeder-cable losses, splitters, noisy preamplifiers, inadequate antennas (low effective area, and poor matching to the feeder cable), attenuation through buildings, and fading