Theoretical basis for data communication

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

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

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

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.

Figure 3.1 Comparison of analog and digital signals

Figure 3.2 A sine wave

Figure 3.3 Amplitude

Figure 3.4 Period and frequency

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

An analog signal is best represented in the frequency domain.

Figure 3.7 Time and frequency domains

Figure 3.7 Time and frequency domains (continued)

Figure 3.7 Time and frequency domains (continued)

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.

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.

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.

Figure 3.9 Three harmonics

Figure 3.10 Adding first three harmonics

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 …

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

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.

Figure 3.8 Square wave

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, 3rd harmonic,… respectively.

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.

Figure 3.11 Frequency spectrum comparison

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.

Figure 3.12 Signal corruption

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.

Figure 3.13 Bandwidth

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.

Figure 3.17 Bit rate and bit interval

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.

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 111111, no change in the value, best case In another, 101010, maximum change in the values, worst case In another, 001010, change in between the above two cases We have already shown .. More the changes higher are the frequency components

Figure 3.18 Digital versus analog

Using single harmonic – just to get the intuition The signal 111111 (or 00000 ) can be simulated by sending a single-frequency signal with frequency 0. The signal 101010 (010101) can be simulated by sending a single-frequency signal with frequency 3 Hz. (3 signals or sine waves per second)

All other cases are between the best and the worst cases 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.

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

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 3rd harmonic to the worst case, we need B = n/2 + 3n/2 = 4n/2 When we add 5th 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

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.

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.

Figure 3.19 Low-pass and band-pass

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

Figure 3.18 Digital versus analog

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

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

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.)

C = B log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B  0 = 0 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 log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B  0 = 0

C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163) 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 3162. For this channel the capacity is calculated as C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163) C = 3000  11.62 = 34,860 bps

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.

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 First, we use the Shannon formula to find our upper limit. C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. 4 Mbps = 2  1 MHz  log2 L  L = 4

I Acknowledge Help from the following site http://www.mhhe.com/engcs/compsci/forouzan/ In preparing this lecture.