Communication is the transfer of information over distance between a source and a user. A basic communication system consists of a transmitter, a receiver, and a channel
Download 57.7 Kb.
|
- Navigate this page:
- Basic System Considerations
- Message Type Used bandwidth(B)
- Digital transmission rates of US telephone system
- Signaling Designation Data Rate
- Spectral-band classification scheme
- Decibel (dB) for power levels
- Note
 ENE 423
Lecture I Overview: Optical Communication Systems Communication is the transfer of information over distance between a source and a user. A basic communication system consists of a transmitter, a receiver, and a channel. Transmitter Information Channel Receiver
1. Light Emitting Diode (LED) 2. Laser Diode (LD)
y(t)
TX RX
Signal Processing – For analog transmission, the signal processor includes amplification and filtering of the signal. Some random fluctuations in the received signal are called “noise”. For digital system, the signal processor includes decision circuits in addition to amplifiers and filters. This circuit decides if “1” or “0” was received. A figure of merit for characterizing communication systems is the product of information capacity × maximum length of channel without repeaters. Ex. optical system (16 channels at 2.4 Gbps/channel) x (60 - 70 km) = (2 – 3) × 1012 bps.km Basic System Considerations Maximum frequency in a modulating message signal is called “baseband”. Baseband that is adapted for voice message is 4 kHz. Bandwidth requirements of several analog systems
Digital transmission – The sampling theorem says that an analog signal can be accurately transmitted if sampling rate is twice the highest frequency contained in that signal. Let R be the required transmission rate. R can be expressed by 
where m = number of bits/sample fs = sampling frequency = 2(f) Ex. A telephone system has m = 8 bits/sample. Find R. R = m(2Df) = 8x(2x4x103) = 64 kbps Several messages can be combined (multiplexed) onto a single information channel. Most fundamental multiplexing in telephone network is incorporating 24 voice channels in one line. Required rate = 24 x 64 kbps = 1.536 x 106 bps T-1
system 1 2 3 . . 24 Actually, more data bits of 1.536 x 106 are being sent than ones are required (1.536 Mb/s). This actual rate of 1.544 Mb/s includes the bits to identify input frames (synchronization and signaling pulses). 
Digital transmission rates of US telephone system
Data transmission hierarchy used in North America. By putting information in an ATM (asynchronous transfer mode) format, it is possible to transmit simultaneously narrowband and broadband communication data such as telephone, videoconferencing, or imaging, on a single subscriber. A new transmission standard has been developed and it is called SONET (Synchronous Optical NETwork) in North America and called SDH (Synchronous Digital Hierarchy) in other parts of the world. The first level of the SONET is called STS-1 (synchronous transport signal - level 1) where is then scrambled and converted to an OC-1 (optical carrier – level 1) signal. Whereas, for SDH systems, the fundamental building block starts from 155.52 Mb/s or STM-1(synchronous transport module – level 1)
Ex. What level of ATM would be adequate for a commercial TV broadcast if 8 bits/sample is used. Soln 1st approach: An analog signal has a bandwidth of 6 MHz. R = m(2Df) = 8(2×6×10-6) = 96 Mbps From table: T4 at 274.17 Mbps …too much
Audio at Df = 15kHz Consider TV signal Video at Df = 4.5MHz Encoding by using 8 bits/sample for video and keeping 8 bits/sample for audio. R = (8×2×15×103)+(9×2×4.5×106) = 81.24 Mbps Therefore, this signal can be transmitted over the T3-C level.
Soln (a) Minimum bandwidth = (10000×106)×4×103 = 4 ×1013 Hz. (b) f = c/l = 3x108/1×10-6 = 3×1014 Hz. 3×1014 Hz > 4×1013 Hz …O.K. (c) Digital case: R = (10000×106)×64000 = 6.4×1014bps An optical carrier could not handle this rate. It could not be turned on and off fast enough. Spectral-band classification scheme
Optical fiber installations: on poles, in ducts, and undersea Ex. A cable contains 144 single-mode fibers, each operating at 2.3 Gbps. How many digitized voice messages can be transmitted simultaneously along this cable? Soln For 1 message,  kb/s (for voice message)
144 single-mode fibers contain The number of digitized voice message = = 5.175 million messages Decibel (dB) for power levels Loss in fibers can be expressed by attenuation () in units of dB/km P1 P2 < P1 l If the power is P1 W at one point and P2 W at another point further along the link, then P2/P1 is the fraction of the power transmitted between the 2 locations or it is called the efficiency of transmission between 2 points. 
Since P2 < P1 without any amplifier placed between the two, dB < 0. 
For example, P2 = 0.5 P1 and l = 1 km, then = 10log10(0.5) = -3 dB/km. Sometimes, it is convenient to express power at any point relative to 1 mW, this is denoted by the term dBm. 
Ex. A receiver of sensitivity 1 μW. If 10 mW is transmitted at 100 MHz, what would be the maximum length of the link when
Note: Sensitivity describes the minimum power required at the receiver (user). Soln Allowed loss in dB = (a) Max. length = 40 dB/ 22.6 dB/km = 1.8 km (b) Max. length = 40 dB/ (5 dB/km) = 8 km 
History of fiber attenuation. Ex. An LED radiates 2 mW. Compute the dBm value of this radiated power. This power travels through a group of components having a combined loss of 23 dB. Compute the output power. Soln (a) dBm = 10log10 (2mW/1mW) = 3 dBm (b) Output power = 3 dBm – 23 dB = -20 dBm 10log10 P0(mW) = -20 dBm P0 = 10-2 = 0.01 mW As the preceding example, the level of power at the receiver dBmr are related by 
where dBmt = transmitted power (dBm) dBs = system power (“-” for loss and “+” for gain) Download 57.7 Kb. Share with your friends:
|
b/s
messages