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Satellite Comparison

Satellite Comparison. Why do we need a worldwide fiber optic system if we can reach practically any point on the globe using satellite system? Answer: Many advantages of optical fiber over satellite system

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Satellite Comparison

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  1. Satellite Comparison • Why do we need a worldwide fiber optic system if we can reach practically any point on the globe using satellite system? • Answer: • Many advantages of optical fiber over satellite system • High information carrying capability: optical fiber can carry up to 24-50 Tbps. There is not indication that satellite communication can ever achieve this capacity. Current satellite operate at 300Mhz to 100GHz as carriers • Low data latency: Signal delay in satellite communication is caused by long distance that a signal travels from earth station to satellite and back. This distance from ground station to satellite station is (~40,000km). Therefore, one way traveling signal needs ~0.27 second and bi directional communication can take 0.54 second. Time delay in satellite communication causes echo disturbance. In optical fiber, signal travels through a much shorter and straighter route. • Satellite communication quality is susceptible to adverse atmosphere conditions. • Satellite communication risks the broadcast of sensitive information

  2. Basic Concepts • Bandwidth and Data Transmission Rate • For any periodical signal, Bandwidth is a measure of the spectral contents of the signal, which is determined mathematically through Fourier transformation • Sound signal bandwidth 3.1KHz (0.3 KHz-3.4 KHz), to make a good quality digital conversation it require a sampling rate 8KHz, 8 bits are transmitted by each sample. Thus required BW is at least 64 Kbps

  3. Basic Concepts • Bandwidth is a measure of information carrying capacity of a transmission medium • For example, communication carrier wavelength λ = 1550nm, a single mode fiber carrier is • v=c/ λ=3x108/1.55x10-6=2x1014 Hz • Assuming a usable BW is just 1% of the carrier frequency (1% rule of thumb in estimating BW from carrier frequency) • Δv=BW~2x1012 Hz = 2THz • Δv α (1/Δt), such large BW corresponds to an extremely small temporal extent 5x10-13 sec or 0.5 psec. Small temporal extent allows great number of data to be carried in a single second • If Δv is further extended such as from (1.45-1.65) μm (instead of 1%) then the BW is 2.5x1013 Hz = 25 THz. It can carry 390 million simultaneous voice conversation or 2.5x1013 / 90Mbps TV Channels

  4. Basic Concepts • Bit: Binary digit 1 or 0 • Bit rate addresses characteristics of digital signal while bandwidth is really for analog system. How are they related? • Noiseless channel (Nyquist bit rate), the maximum bit rate given by • Bit Rate = 2 x Bandwidth x log2L bps • Log2L is number of bit per signal level • Noisy channel (Shannon capacity) • Capacity = Bandwidth x log2 (1+SNR) bps • Simplest approach is to assume that the number of bit per second and number of cycles per second are the same i.e. if you have a bandwidth of 1 Hz you can transmit 1 bit each second • In this course, we will concern ourselves with Bandwidth = Bit Rate. This is convenient because fiber optic communication technology uses the terms bandwidth and bit rate interchangeably

  5. Basic Concepts • Actual data transmission rate is limited in fiber by dispersion and ultimately limited by material dispersion over the frequency range within BW • Channel capacity is also noise limited and cannot be increased indefinitely by increasing BW i.e. C=BW x log 2(1+S/N) When S/N = 0, then 2 (C/BW)=1. This means capacity is zero even if channel bandwidth is large

  6. Basic Concepts: Bit Rate Distance Product • Bit Rate Distance Product: a figure of merit for communication systems. B is bit rate [bit/sec], L is the regenerator spacing. • Normally, a re-amplification of signal is necessary when power is dropped by 10-5=50dB. • Copper has attenuation coefficient of 0.2dB/km, and thus need to be regenerated every 2.5km. • Optical fiber, for alpha = 0.2dB/km, • needs to be regenerated every 250km.

  7. Power Definition in Fiber Optics • Launching power is defined in [dBm]= 10 log 10 (PTX/1mW) • Receiver sensitivity is defined as: Sensitivity [dBm]=10log10(PRX/1mW) • Optical power received depends on • Light coupled into the fiber • Loss occurring in fiber • Loss at the connectors and splices • Link power margin to allow for component aging, temperature fluctuations, losses arising from components that might be added • The Link loss budget is derived from the sequential loss contributions of each element in the link • Launching power (dBm) = Total power loss allowed (dB) + Receiver Sensitivity (dBm) • Total allowed loss [dB]= 10log (PTX/PRX) • = 10log (PTX/1mW)/(PRX/1mW) =10log (PTX/1mW)-10log(PRX/1mW) =Launching power – Receiver sensitivity

  8. Example • A data rate 20 Mbps has error rate of 10-9. Silicon PIN photodiode working at 850nm. Receiver input signal -42 dbm (42dB below 1mW); GalAlAs LED couples 50μW (-13dBm) average optical powr into a fiber. Find the allowable power loss for the system. Total power loss allowed = Launching power – Receiver sensitivity= -13+42 = 29dB

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