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Theoretical basis for data communication

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Transmission of data Data must be transformed to electromagnetic signals to be transmitted.

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Data : Analog or Digital Analog data : human voice, chirping of birds etc, converted to –Analog or digital signals Digital : data stored in computer memory, converted to –Analog or digital signals

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Examples Analog data as analog signal : Human voice from our houses to the telephone exchange. Analog data as digital signal : most of the systems today : Say Human voice, images sent on digital lines.. New telephone system (digital exchanges) Digital data as analog signal : computer data sent over internet using analog line.. Say telephone line ( say our house to the exchange) Digital data as digital signal : say from one digital exchange to another

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Signals : Analog or digital Analog signal has infinitely many levels of intensity (infinitely many values, continuous values) over a period of time. Digital signal has only a limited number of defined values(discrete values) say, 0,1.

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Figure 3.1 Comparison of analog and digital signals

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Figure 3.2 A sine wave

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Figure 3.3 Amplitude

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Figure 3.4 Period and frequency

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If a signal does not change at all, its frequency is zero. If it changes instantaneously, its frequency is infinite.

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An analog signal is best represented in the frequency domain.

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Figure 3.7 Time and frequency domains

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Figure 3.7 Time and frequency domains (continued)

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Single-frequency sine wave isnot useful for data communication A single sine wave can carry electric energy from one place to another. For eg., the power company sends a single sine wave with a frequency of say 60Hz to distribute electric energy to our houses.

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Contd.. If a single sine wave was used to convey conversation over the phone, we would always hear just a buzz. If we sent one sine wave to transfer data, we would always be sending alternating 0’s and 1’s, which does not have any communication value.

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Composite Signals If we want to use sine wave for communication, we need to change one or more of its characteristics. For eg., to send 1 bit, we send a maximum amplitude, and to send 0, the minimum amplitude. When we change one or more characteristics of a single-frequency signal, it becomes a composite signal made up of many frequenies.

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Figure 3.9 Three harmonics

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Figure 3.10 Adding first three harmonics

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Fourier Analysis In early 1900s, French Mathematician Jean- Baptiste Fourier showed that any composite signal can be represented as a combination of simple sine waves with different frequencies, phases and amplitudes. More is the number of components included better is the approximation For eg., let us consider the square wave …

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Time-Voltage graph Time on x-axis in msec, Voltage on y-axis

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The first trace in the above figure is the sum of 2 sine waves with amplitudes chosen to approximate a 3 Hz square wave (time base is msec). One sine wave has a frequency of 3 Hz and the other has a frequency of 9 Hz. The second trace starts with the first but adds a 15 Hz sine wave and a 21 Hz sine wave. It is clearly a better approximation.

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Figure 3.8 Square wave

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It can be shown (ref Kreyzsig) that this signal consists of a series of sine waves with frequencies f, 3f, 5f, 7f, … and amplitudes 4A/pi, 4A/3Pi, 4A/5Pi, 4A/7Pi,… where f is the fundamental frequency(1/T, T the time period) and A the maximum amplitude. The term with frequency f, 3f.. are called the first harmonic, 3 rd harmonic,… respectively.

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Frequency spectrum of a signal The description of a signal using the frequency domain and containing all its components is called the frequency spectrum of the signal.

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Figure 3.11 Frequency spectrum comparison

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Composite Signal and Transmission Medium A signal needs to pass thru a transmission medium. A transmission medium may pass some frequencies, may block few and weaken others. This means when a composite signal, containing many frequencies, is passed thru a transmission medium, we may not receive the same signal at the other end.

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Figure 3.12 Signal corruption

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Bandwidth of a channel The range of frequencies that a medium can pass without loosing one-half of the power contained in that signal is called its bandwidth.

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Figure 3.13 Bandwidth

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Representing data as Digital Signals 1 can be encoded as a positive voltage say 5 volts, 0 as zero voltage (or negative voltage say –5 volts) Most digital signals are aperiodic. Thus we use Bit interval (instead of period) : time required to send one bit = 1/ bit rate. Bit rate (instead of frequency) :number of bits per second.

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Figure 3.17 Bit rate and bit interval

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Digital signal as Composite Signal Digital signal is nothing but a composite analog signal with an infinite bandwidth. A digital signal theoretically needs a bandwidth between 0 and infinity. The lower limit 0 is fixed. The upper limit may be compromised.

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Relationship b/w bit rate and reqd. channel b/w (informal) Imagine that our computer creates 6bps In 1 second, the data created may be , no change in the value, best case In another, , maximum change in the values, worst case In another, , change in between the above two cases We have already shown.. More the changes higher are the frequency components

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Figure 3.18 Digital versus analog

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Using single harmonic – just to get the intuition The signal (or ) can be simulated by sending a single-frequency signal with frequency 0. The signal (010101) can be simulated by sending a single-frequency signal with frequency 3 Hz. (3 signals or sine waves per second)

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All other cases are between the best and the worst cases. We can simulate other cases with a single frequency of 1 0r 2 Hz (using appropriate phase). I.e. to simulate the digital signal at data rate 6bps, sometimes we need to send a signal of frequency 0, sometimes 1,sometimes 2 and sometimes 3. We need that our medium should be able to pass frequencies of 0-3 Hz.

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Generalizing the example above Bit rate = n bps Best case ---- frequency 0 Hz Worst case frequency n/2 Hz Hence B (bandwidth) = n/2

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Using more harmonics However, as said earlier, one harmonic does not approximate the digital signal nicely and more harmonics are required to approximate the digital signal. As shown earlier, such a signal consists of odd harmonics When we add 3 rd harmonic to the worst case, we need B = n/2 + 3n/2 = 4n/2 When we add 5 th harmonic to the worst case, we need B = n/2 + 3n/2 + 5n/2= 9n/2 and so on. In other words, B >= n/2 or n <= 2B

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Relationship b/w bit rate and reqd. channel b/w (informal) Hence we conclude that bit rate and the bandwidth of a channel are proportional to each other.

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Analog vs Digital Low-pass channel : has a bandwidth with frequencies between 0 and f (f could be anything including infinity). Band-pass channel : has a bandwidth with frequencies between f1 (>=0) and f2 A band-pass channel is more easily available than a low-pass channel.

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Figure 3.19 Low-pass and band-pass

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Digital Rate limits Data rate depends on 3 factors: –The bandwidth available –Number of levels of signals –Quality of the channel (noise level)

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Figure 3.18 Digital versus analog

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Noiseless Channel: Nyquist Bit rate b = 2 B log L (log is to base 2) b : bit rate B : Bandwidth L : number of levels

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Noisy channel : Shannon Capacity C = B log (1 + SNR) C = capacity of the channel in bps B = Bandwidth SNR = signal to noise ratio

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Digital vs Analog contd… Digital signal needs a low-pass channel Analog signal can use a band-pass channel. Moreover, bandwidth of a signal can always be shifted ( a property required for FDM – The bandwidth of a medium can be divided into several band-pass channels to carry several analog transmissions at the same time.)

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Example 9 Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity is calculated as C = B log 2 (1 + SNR) = B log 2 (1 + 0) = B log 2 (1) = B 0 = 0

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Example 10 We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal- to-noise ratio is usually For this channel the capacity is calculated as C = B log 2 (1 + SNR) = 3000 log 2 ( ) = 3000 log 2 (3163) C = 3000 = 34,860 bps

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Using both the limits In practice we use both the limits to determine, given the channel bandwidth, what should be the number of levels a signal should have.

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Example 11 We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level? Solution C = B log 2 (1 + SNR) = 10 6 log 2 (1 + 63) = 10 6 log 2 (64) = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. 4 Mbps = 2 1 MHz log 2 L L = 4 First, we use the Shannon formula to find our upper limit.

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I Acknowledge Help from the following site In preparing this lecture.

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