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3-2008UP-Copyrights reserved1 ITGD4103 Data Communications and Networks Lecture-9: Communication Techniques,Spectrum and bandwidth week 10- q-2/ 2008 Dr. Anwar Mousa University of Palestine Faculty of Information Technology

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3-2008UP-Copyrights reserved2 Communication Techniques

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3-2008UP-Copyrights reserved3 Major topics This chapter explains the differences between analog and digital transmission discusses Transmission impairments and channel capacity describes how digital data can be encoded by means of a modem so that they can be transmitted over analog telephone lines

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3-2008UP-Copyrights reserved4 Analog and digital Analog data vs. digital data Analog: Continuous values on some interval (sound, light, temperature, pressure) Digital: Discrete values (text, integers, binary) Analog signal vs. digital signal Analog: Continuously varying electromagnetic wave Digital: Series of voltage pulses (square wave)

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3-2008UP-Copyrights reserved5 Analog Data-->Signal Options Analog data to analog signal Inexpensive, easy conversion (e.g. telephone) Data may be shifted to a different part of the available spectrum (multiplexing) Used in traditional analog telephony Analog data to digital signal (PCM) Requires a codec (encoder/decoder) sender converts the voice data by a bit stream, receiver reconstruct the bit stream to the analog data Allows use of digital telephony

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3-2008UP-Copyrights reserved6 Digital Data-->Signal Options Digital data to analog signal Requires modem (modulator/demodulator) Allows use of PSTN (public-switched Telephone Network) to send data Necessary when analog transmission is used Digital data to digital signal (LINE CODING) More reliable because no conversion is involved Less expensive when large amounts of data are involved

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3-2008UP-Copyrights reserved8 Advantages of digital transmission (1/2) Cost the LSI and VLSI technologies has caused a continuing drop in the cost and size of digital circuitry the maintenance costs for digital circuitry are a fraction of those for analog circuitry Data integrity with the use of digital repeaters, the effects of noise and other signal impairments are not cumulative it is possible to transmit data longer distances and over lesser-quality lines while maintaining the integrity of the data

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3-2008UP-Copyrights reserved9 Advantages of digital transmission (2/2) Capacity utilization it has become economical to build transmission links of very high bandwidth including satellite channels and optical fiber it is more easily and cheaply achieved with digital transmission for a high degree of multiplexing to effectively utilize the capacity Security and privacy encryption techniques can be readily applied to digital data and to analog data that have been digitized

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3-2008UP-Copyrights reserved10 Digital signals

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3-2008UP-Copyrights reserved11 Digital signals Digital signals usually refers to the transmission of electromagnetic pulses that represent two binary digits, 1 and 0 binary information is generated and then converted into digital voltage pulses for transmission

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3-2008UP-Copyrights reserved13 EXAMPLES

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3-2008UP-Copyrights reserved14 1. Given the frequencies listed below, calculate the corresponding periods. a. 24 Hz b. 8 MHz c.140 KHz Solution a.F = 24 Hz T= 1/f T= 1/24 = 0.0416= 41.6 ms b. F = 8 MHz T= 1/f x T= 1/(8 x 10 6) = 0.125s c. F = 140kHz T= 1/f x T= 1/(140 x 1000)= 7.14 s

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3-2008UP-Copyrights reserved15 2. Given the following periods, calculate the corresponding frequencies. a. 5 s b. 12 s c. 220 nsa. 5 s b. 12 s c. 220 ns Solution a.f=1/T = 1/5= 0.2 Hz xx b.f=1/T = 1/(12 x 10 -6) = 0.083 x 10 6 Hz x c. f=1/T = 1/(220 x 10 -9) = 4.5 MHz

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3-2008UP-Copyrights reserved16 4. What is the bandwidth of a signal that can be decomposed into five sine waves with frequencies at 20,50,100, and 200 Hz? All peak amplitudes are the same. Draw the bandwidth. Solution B = fh - fl = 200 - 20 = 180 Hz 2050100200 20 180

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3-2008UP-Copyrights reserved17 5. What is the transmission time of a message sent by a station if the length of the message one million bytes (each byte consists of 8 bits) and the data transmission rate is 200 kbps? Solution x 10 6 x 8/ 200 x 10 3 Transmission time = Message size / data transmission rate = 1 x 10 6 x 8/ 200 x 10 3 = 40s

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3-2008UP-Copyrights reserved18 6. What is the total delay (latency) for a message of size 5 million bits that is being sent on a link? The length of the link is 2000Km. The speed of electromagnetic wave inside the link is 2 10 8 m/s and the data transmission rate is 5Mbps? Solution Message size = 5 million bits = 5 x 10 6 Distance = 2000Km = 2 x 10 6 m Speed = 2 10 8 m/s data transmission rate = 5 10 6 b/s Latency = Propagation time + Transmission time

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3-2008UP-Copyrights reserved19 Propagation time = Distance / Propagation speed = 2 x 10 6 /2 10 8 =1/100 = 0.01s Transmission time = Message size / data transmission rate = 5 x 10 6 / 5 x 10 6 = 1 s Latency = Propagation time + Transmission time = 0.01 +1 = 1.01s

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3-2008UP-Copyrights reserved20 8. Shown in the next figure, two signals with the same amplitude and phase, but different frequencies. Compare between the calculated periods for each signal. Solution Figure (a) : T = 1/F F = 12 T = 1/12 = 0.083 s = 83 ms Figure (b) : T = 1/F F = 6 T = 1/6 = 0.166 s = 166 ms

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3-2008UP-Copyrights reserved21 9. A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have amplitude of 0. Draw the frequency domain of the signal. Solution Bandwidth = 200 kHz = F h - F l Where f h is the highest frequency, and f l is the lowest frequency. Then F h F l = 200 (F h +F l )/2 = 140 The lowest frequency must be at 40 kHz and the highest at 240 kHz.

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3-2008UP-Copyrights reserved22 20 V

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3-2008UP-Copyrights reserved23 UnitEquivalentUnitEquivalent Seconds (s)1 shertz (Hz)1 Hz Milliseconds (ms)10 3 skilohertz (KHz)10 3 Hz Microseconds (ms)10 6 smegahertz (MHz)10 6 Hz Nanoseconds (ns)10 9 sgigahertz (GHz)10 9 Hz Picoseconds (ps)10 12 sterahertz (THz)10 12 Hz