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

3-2008UP-Copyrights reserved2 Communication Techniques

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

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)

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

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

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

3-2008UP-Copyrights reserved10 Digital signals

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

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 = = 41.6 ms b. F = 8 MHz T= 1/f x T= 1/(8 x 10 6) = 0.125µs c. F = 140kHz T= 1/f x T= 1/(140 x 1000)= 7.14 µs

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

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 = = 180 Hz

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

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

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 = = 1.01s

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 = s = 83 ms Figure (b) : T = 1/F F = 6 T = 1/6 = s = 166 ms

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.

3-2008UP-Copyrights reserved22 20 V

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