UNIT 2 Mc GrawHill www bookspar com Website

UNIT - 2 Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Unit 2 Title: Data, Signals & Digital Transmission Syllabus: Analog & Digital Signals, Transmission impairments, Data rate limits, Performance, Digital to analog conversion, Analog to digital conversion, Transmission modes. www. bookspar. com | Website for students | VTU NOTES

Analog & Digital data Data can be in analog or digital. i) Analog Data – refers to information that is continuous and takes continuous values. Example: Human voice. ii) Digital data – refers to information that has discrete states and take discrete values. Example: Data stored in a computer memory. www. bookspar. com | Website for students | VTU NOTES

Analog & Digital signals Signals can be Analog or Digital. Before transmitting the data over a medium, the data must be converted in to electromagnetic signals. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Analog signal An analog signal is a continuous signal. It has an infinite number of values in a range. Example - Human Voice. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Digital signal A digital signal is a discrete signal. It has a limited number of values. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

A sine wave Mathematically, a Sine wave can be represented by S(t) = A sin(2Π f t + Φ) Where, S – instantaneous amplitude, A – peak amplitude, f – frequency and Φ - phase www. bookspar. com | Website for students | VTU NOTES

Amplitude Highest intensity of a signal represent the peak amplitude of the signal. The intensity is proportional to the energy it carries. Amplitude is measured in volts. www. bookspar. com | Website for students | VTU NOTES

Period & Frequency Period refers to the time taken by a signal to complete one cycle & expressed in seconds. It is denoted by ‘ T ‘. Frequency refers to numbers of signals produced in one second & and expressed in hertz ( Hz ). It is denoted by ‘ f ‘. The relation between period and frequency is given by T x f = 1. Frequency and period are inverses of each other. www. bookspar. com | Website for students | VTU NOTES

Period and frequency www. bookspar. com | Website for students | VTU NOTES

Units of periods and frequencies Unit Seconds (s) Equivalent 1 s Unit hertz (Hz) Equivalent 1 Hz Milliseconds (ms) 10– 3 s kilohertz (KHz) 103 Hz Microseconds (ms) 10– 6 s megahertz (MHz) 106 Hz Nanoseconds (ns) 10– 9 s gigahertz (GHz) 109 Hz Picoseconds (ps) 10– 12 s terahertz (THz) 1012 Hz www. bookspar. com | Website for students | VTU NOTES

Example 1 Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz. Solution We Know that, 1 ms = 10 3 µs. 100 ms = 100 103 µs = 105 ms Now we use the inverse relationship to find the frequency, changing hertz to kilohertz 100 ms = 100 10 -3 s = 10 -1 s = T f = 1/10 -1 Hz = 10 10 -3 KHz = 10 -2 KHz www. bookspar. com | Website for students | VTU NOTES

Note: • Frequency is the rate of change of the signal with respect to time. • Change in a short span of time means high frequency. • Change over a long span of time means low frequency. www. bookspar. com | Website for students | VTU NOTES

Note: • If a signal does not change at all, its frequency is zero. • If a signal changes instantaneously (it jumps from one level to another in no time), its frequency is infinite, because its period is zero. www. bookspar. com | Website for students | VTU NOTES

Phase describes the position of the waveform relative to time zero. It is measured in degrees or radians. www. bookspar. com | Website for students | VTU NOTES

Signals with different phases www. bookspar. com | Website for students | VTU NOTES

Example A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians? Solution We know that one complete cycle is 360 degrees. Therefore, 1/6 cycle is (1/6) 360 = 60 degrees = 60 x 2 p /360 rad = 1. 046 rad www. bookspar. com | Website for students | VTU NOTES

Sine wave examples www. bookspar. com | Website for students | VTU NOTES

Sine wave examples (continued) www. bookspar. com | Website for students | VTU NOTES

Sine wave examples (continued) www. bookspar. com | Website for students | VTU NOTES

Time and frequency domains Time-domain plot shows changes in signal amplitude with respect to time. Frequency-domain plot show a relations between amplitude and frequency. A signal with peak amplitude= 5 volts, and frequency =0 www. bookspar. com | Website for students | VTU NOTES

Time and frequency domains (continued) A signal with peak amplitude= 5 volts, and frequency = 8 www. bookspar. com | Website for students | VTU NOTES

Time and frequency domains (continued) A signal with peak amplitude= 5 volts, and frequency = 16 www. bookspar. com | Website for students | VTU NOTES

Note: A single-frequency sine wave is not useful in data communications; we need to change one or more of its characteristics to make it useful. www. bookspar. com | Website for students | VTU NOTES

Note: When we change one or more characteristics of a singlefrequency signal, it becomes a composite signal made of many frequencies. www. bookspar. com | Website for students | VTU NOTES

Note: According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes. Any composite signal is a sum of set sine waves of different frequencies, phases and amplitudes. Mathematically it is represented by S(t) = A 1 sin(2Π f 1 t + Φ 1) + A 2 sin(2Π f 2 t + Φ 2) + A 3 sin(2Π f 3 t + Φ 3 ) +…. . www. bookspar. com | Website for students | VTU NOTES

Three harmonics A graph with three harmonic waves www. bookspar. com | Website for students | VTU NOTES

Adding first three harmonics Fig. Adding first three harmonics To create a complete square wave sum up all the odd harmonics up to infinity. www. bookspar. com | Website for students | VTU NOTES

Signal corruption & Bandwidth A signal has to pass through a medium. One of the characteristics of the medium is frequency. The medium needs to pass every frequency and also preserve the amplitude and phase. No medium is perfect. A medium passes some frequencies and blocks some others. www. bookspar. com | Website for students | VTU NOTES

Bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal. For example, Voice normally has a spectrum of 300 – 3300 Hz. Thus, requires a bandwidth of 3000 Hz. The bandwidth is a property of a medium. Bandwidth is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass. www. bookspar. com | Website for students | VTU NOTES

Bandwidth (continued) The medium can pass some frequencies above 5000 and below 1000, but the amplitude of those frequencies are less than those in the middle. www. bookspar. com | Website for students | VTU NOTES

Example: If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V. Solution B = fh - fl = 900 - 100 = 800 Hz The spectrum has only five spikes, at 100, 300, 500, 700, and 900 www. bookspar. com | Website for students | VTU NOTES

Example A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all integral frequencies of the same amplitude. Solution B = f h - fl → 20 = 60 - fl → fl = 60 - 20 = 40 Hz www. bookspar. com | Website for students | VTU NOTES

Example A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium? Solution The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz. The signal with frequency 1000 & 2000 Hz is totally lost. www. bookspar. com | Website for students | VTU NOTES

Digital Signals Data can be represented by a digital signal. Bit 1 can be encoded by positive voltage and bit 0 can be encoded by zero voltage. www. bookspar. com | Website for students | VTU NOTES

Bit rate & Bit interval - is the time required to send one bit. Bit rate - is the number of bits sent in 1 second. It is expressed in bps. www. bookspar. com | Website for students | VTU NOTES

Example A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval) Solution The bit interval is the inverse of the bit rate. Bit interval = 1/ 2000 s = 0. 000500 x 106 ms = 500 ms www. bookspar. com | Website for students | VTU NOTES

Example Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel? . Answer A page is an average of 24 lines with 80 characters in each line. If we assume that one character require 8 bits. The bit rate of the channel = 100 x 24 x 80 x 8 = 1, 636, 000 bps = 1. 636 Mbps Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Bit Length It is the distance one bit occupies on the transmission medium. Bit Length = Propagation speed x Bit interval. Transmission of Digital Signals A digital signal can transmit by using either baseband transmission or Broadband transmission. Baseband transmission -means sending a digital signal with out changing it to an analog signal. Baseband transmission requires a low-pass channel. Digital transmission use a low-pass channel. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Broadband transmission (modulation)- means sending a digital signal after changing it to an analog signal. Broadband transmission requires a bandpass channel. Analog transmission use a band-pass channel. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Low-pass & Band-pass A channel is either low pass or band pass. A low – pass channel has a B/W With frequencies between 0 & f. A band pass has B/W with frequencies between f 1 & f 2. www. bookspar. com | Website for students | VTU NOTES

Data Rate Limit The data rate over a channel depends on 3 factors. i) The Band width available. ii) The levels of signals that can use for data transmission. iii) The quality of the channel. Two theoretical formulas were developed to calculate the data rate. 1) Nyquist Bit Rate - Noiseless Channel 2) Shannon Capacity - Noisy Channel Nyquist Bit Rate Maximum Bit rate = 2 x Bandwidth x log 2 L where L is number of levels used to represent data. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Example Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. Calculate the bit rate. Answer: Bit Rate = 2 3000 log 2 2 = 6000 bps www. bookspar. com | Website for students | VTU NOTES

Example Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). Calculate the bit rate. Answer: Bit Rate = 2 x 3000 x log 2 4 = 12, 000 bps www. bookspar. com | Website for students | VTU NOTES

Shannon capacity of Bit rate Maximum data rate of a noisy channel, C=Band width x log 2 (1+SNR) Where SNR is the Signal to Noise Ratio. www. bookspar. com | Website for students | VTU NOTES

Signal – to – Noise Ratio (SNR) SNR is defined as SNR = Average signal power / Average noise power. It is described in decibel units, SNR d. B = 10 log 10 SNR Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Example: 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. Calculate the channel capacity. Answer: C = B log 2 (1 + SNR) = B log 2 (1 + 0) = B log 2 (1) = B 0 = 0 www. bookspar. com | Website for students | VTU NOTES

Example: A telephone line has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is 3162. Find the channel capacity. Answer: C = B log 2 (1 + SNR) = 3000 log 2 (1 + 3162) = 3000 log 2 (3163) = 3000 11. 62 = 34, 860 bps www. bookspar. com | Website for students | VTU NOTES

Example 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 log 2 (1 + SNR) = 106 log 2 (1 + 63) = 106 log 2 (64) = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. 6 Mbps = 2 1 MHz log 2 L L = 8 www. bookspar. com | Website for students | VTU NOTES

Transmission Impairments www. bookspar. com | Website for students | VTU NOTES

Attenuation means loss of energy. When a signal (simple /composite) travels through a medium, it loses some of its energy. To compensate the loss, amplifiers are used to amplify the signal. It is measured in decibels (d. B) www. bookspar. com | Website for students | VTU NOTES

Decibel (d. B) To show that a signal has lost or gained strength, we use the unit called decibel. The decibel measures the relative strengths of two signals or one signal at two different points. The decibel is negative if a signal is attenuated and positive if a signal is amplified. db = 10 log 10 p 2 / p 1 where p 1 & p 2 are the powers of a signal at point 1 & 2 respectively. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Example Imagine a signal travels through a transmission medium and its power is reduced to half. This means that P 2 = 1/2 P 1. Find the attenuation (loss of power). Solution Attenuation Loss =10 x log 10 (P 2/P 1) = 10 x log 10 (0. 5 P 1/P 1) = 10 x log 10 (0. 5) = 10 x(– 0. 3) = – 3 d. B www. bookspar. com | Website for students | VTU NOTES

Example Imagine a signal travels through an amplifier and its power is increased 10 times. This means that P 2 = 10 x P 1. find the amplification (gain of power). Answer: Amplification= 10 log 10 (P 2/P 1) = 10 log 10 (10 P 1/P 1) = 10 log 10 (10) = 10 (1) = 10 d. B www. bookspar. com | Website for students | VTU NOTES

Example The loss in a cable is measured in decibels per kilometer ( d. B/km). If the signal at the beginning of a cable with -0. 3 db/km has a power of 2 m. W. What is the power of the signal at 5 km? . Solution The loss of energy in the cable in decibels is = 5 x (-0. 3) = -1. 5 d. B. Power of the signal at 5 kms= d. B = 10 x log p 2/p 1 = - 1. 5 or p 2/p 1 = 10 -0. 15 = 0. 71 P 2 = 0. 71 P 1 = 0. 7 x 2 = 1. 4 m. W Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Distortion means the signal changes its shape. Distortion occurs in a composite signal because a it is made up of signals of different frequencies. Each signal has its own propagation speed. So, its own delay in arriving at the destination. www. bookspar. com | Website for students | VTU NOTES

Noise Several types of noise such as thermal noise, induced noise, crass talk and impales noise may corrupt the signals. www. bookspar. com | Website for students | VTU NOTES

Example The power of a signal is 10 m. W and the power of the noise is 1 µW. What are the values of SNR & SNR d. B? . Solution SNR = Average signal power / Average noise power. = 10 x 10 -6 W / 10 -9 W = 10, 000 SNR d. B = 10 x log 10 SNR = 10 x log 10 10000 = 40 Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Example What is the channel capacity for a teleprinter channel with a 300 Hz bandwidth and a signal to noise ratio of 3 d. B. Solution Using Shannon's equation: C = B log 2 We have B = 300 Hz, (SNR) (1 + SNR) d. B C = 300 log 2 (1 +3) = 300 log 2 (4) = 600 bps www. bookspar. com | Website for students | VTU NOTES =3

Example A digital signaling system is required to operate at 9600 bps. a) If a signal element encodes a 4 bit word, what is the minimum required bandwidth of the channel? b) Repeat part (a) for the case of 8 – bit words. Solution By Nyquist's Theorem: C = 2 x B x log 2 M We have C = 9600 bps a) log 2 M = 4, because a signal element encodes a 4 -bit word Therefore, C = 2 x B x 4 9600 = 2 x. Bx 4 Hence, B = 1200 Hz b) 9600 = 2 x B x 8 and B = 600 Hz www. bookspar. com | Website for students | VTU NOTES

Example Given a channel with an intended capacity of 20 Mbps, the bandwidth of the channel is 3 MHz. Assuming thermal noise, what signal to noise ratio is required to achieve this capacity? Solution C = B log 2 (1 + SNR) 20 x 106 = 3 x 106 x log 2 (1 + SNR) = 6. 67 1 + SNR = 2 6. 67 - 1 www. bookspar. com | Website for students | VTU NOTES

Performance Bandwidth is a potential measure of a link. Bandwidth in Hz – The range of frequencies that a channel can pass. Bandwidth in bps – Number of bits per second that a channel can transmit. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Throughput is an actual measurement of data received at the receiver. Example A network with bandwidth 10 Mbps can pass only an average of 12, 000 frames per minute with each frame carrying an average of 10, 000 bits. What is the throughput of this network? Solution Throughput = ( 12000 x 10000) /60 = 2 Mbps www. bookspar. com | Website for students | VTU NOTES

Latency ( Delay ) The latency the time taken for an entire message to arrive completely at the destination taken from the time the first bit is sent out from the source. Latency = propagation time + transmission time + queuing time + Processing delay. Mc. Graw-Hill www. bookspar. com | Website for students | VTU NOTES ©The Mc. Graw-Hill Companies, Inc. , 2004

Propagation time is the time required for a bit to travel from source to the destination. Propagation time = Distance / Propagation speed. www. bookspar. com | Website for students | VTU NOTES

Wavelength Wave length is the distance a simple signal can travel in one period. www. bookspar. com | Website for students | VTU NOTES

Examples Question paper: January / February 2005 Using Shannon’s theorem, compute the maximum bit rate for a channel having a bandwidth of 3100 Hz and signal to noise ratio is 20 d. B. www. bookspar. com | Website for students | VTU NOTES

Digital Transmission www. bookspar. com | Website for students | VTU NOTES

Line coding Line Coding is the process of converting binary data into digital data. Characteristics of line coding techniques are 1) Signal level Vs Data level 2) DC ( Direct Current ) – component & self synchronization 3) Pulse rate Vs Bit rate www. bookspar. com | Website for students | VTU NOTES

Signal level versus data level The number of values allowed in a signal are called Signal Levels. The number of values used to represent data are called Data Levels. www. bookspar. com | Website for students | VTU NOTES

DC component Some line coding schemes leave have residual dc component. It is undesirable because, i) it causes distortion and may create errors in the output. ii) This component is the extra energy residing on the line and useless. a) + voltage is not cancelled by – voltage b) + voltage are cancelled by - voltage. www. bookspar. com | Website for students | VTU NOTES

Pulse Rate & Bit Rate A pulse is the minimum amount of time required to transmit a symbol. The number of pulses per second is called pulse rate. Bit Rate- The number of bits transmitted per second is called bit rate. If a pulse carries a bit, then the pulse rate = bit rate. If pulse carries more than one bit, then bit rate > pulse rate. Bit Rate = Pulse Rate x log 2 L www. bookspar. com | Website for students | VTU NOTES

Example A signal has two data levels with a pulse duration of 1 ms. Calculate the pulse rate and bit rate. Answer: Pulse Rate = 1/ 10 -3= 1000 pulses/s Bit Rate = Pulse Rate x log 2 L = 1000 x log 2 2 = 1000 bps www. bookspar. com | Website for students | VTU NOTES

Example A signal has four data levels with a pulse duration of 1 ms. Calculate the pulse rate and bit rate. Answer: Pulse Rate = = 1000 pulses/s Bit Rate = Pulse. Rate x log 2 L = 1000 x log 2 4 = 2000 bps Example Can the bit rate be less than pulse rate. Solution The bit rate is always greater than or equal to the pulse rate because a pulse contains one or more bits. www. bookspar. com | Website for students | VTU NOTES

Synchronization At the receiver, to interpret the signal correctly , the receiver bit interval must match exactly to the senders bit interval. If the receiver clock is faster or slower, the bit intervals are not matched. So the output will be wrong. www. bookspar. com | Website for students | VTU NOTES

Line coding schemes There are three classifications of line coding 1) Unipolar 2) Polar & 3) Bipolar www. bookspar. com | Website for students | VTU NOTES

Unipolar encoding The signal levels are on one side of the time axis either above or below. The 1’s are encoded as positive voltage value and 0’s are encoded as zero voltage value. Disadvantages The average amplitude of a unipolar encoded signal is nonzero. This method lacks synchronization. www. bookspar. com | Website for students | VTU NOTES

Polar Encoding Note: Polar encoding uses two voltage levels (positive and negative). The average voltage level on the line is reduced. The D. C component problem is alleviated. www. bookspar. com | Website for students | VTU NOTES

polar encoding There are 4 types of polar encoding schemes i) Non. Return to Zero ( NRZ ) ii) Returned to Zero ( RZ ) iii) Manchester iv) Differential Manchester www. bookspar. com | Website for students | VTU NOTES

NRZ – L ( Non. Return Zero – Level ) Encoding A positive voltage represent the bit 0, a negative voltage represent the bit 1. Disadvantage: When the data contains a long stream of 0’s or 1’s cases the problem of synchronization with the sender clock. www. bookspar. com | Website for students | VTU NOTES

NRZ – I ( Non. Return Zero – Invert ) Encoding The inversion is the transition between a positive and a negative voltage. A 1 bit is represented when there is a transition. A 0 bit is represented when no change. www. bookspar. com | Website for students | VTU NOTES

NRZ-L and NRZ-I encoding www. bookspar. com | Website for students | VTU NOTES

Example How does NRZ – L differs from NRZ – I ? Solution In NRZ-L the signal depends on the state of the bit: a positive voltage is a 0, and the negative a 1. In NRZ-I the signal is inverted when a 1 is encountered. www. bookspar. com | Website for students | VTU NOTES

RZ ( Return to Zero ) encoding RZ encoding uses three values- positive, negative & zero. Signal changes for each bit. At the mid of each bit interval, the signal returns to zero. A transition from positive to zero represents bit 1. A transition from negative to zero represents bit 0. www. bookspar. com | Website for students | VTU NOTES

Manchester encoding uses two level of amplitude. Signal changes for each bit. Inversion takes place at the middle of each bit interval. A transition from negative to positive represents bit 1. A transition from positive to negative represents bit 0. www. bookspar. com | Website for students | VTU NOTES

Differential Manchester encoding The differential Manchester encoding uses two level of amplitude. Signal changes for each bit. Inversion takes place at the middle of each bit interval is used for synchronization. The additional transition at the beginning of the interval represents bit 0. The absence of additional transition at the beginning of the interval represents bit 1. www. bookspar. com | Website for students | VTU NOTES

Bipolar Encoding Note: In bipolar encoding, we use three levels: positive, zero, and negative. www. bookspar. com | Website for students | VTU NOTES

Bipolar AMI (Alternate Mark Inversion) encoding The AMI encoding uses three voltage levels, positive, negative & zero. The zero voltage level represents bit 0. The alternate positive and negative voltage levels represent bit 1. www. bookspar. com | Website for students | VTU NOTES

2 B 1 Q ( two binary, one quaternary) 2 BIQ encoding uses four voltage levels. Each pulse represent 2 bits. - www. bookspar. com | Website for students | VTU NOTES

MLT-3(Multi Line Transmission three level) signal MLT-3 encoding uses three level signals, +1, 0 & -1. The signal transitions from one level to the next at the beginning of a bit interval represent bit 1 , there is no transition at the beginning of a bit 0 www. bookspar. com | Website for students | VTU NOTES

Example Assume a data stream is made of ten 0’s. Encode this stream, using the following encoding schemes. How many changes (vertical line) can you find for each scheme? a) Unipolar b) NRZ- L c) NRZ-I d) RZ e) Manchester f) Differential Manchester g) AMI www. bookspar. com | Website for students | VTU NOTES

Example Assume a data stream is made of ten alternate 0’s & 1’s. Encode this stream, using the following encoding schemes. How many changes (vertical line) can you find for each scheme? a) Unipolar b) NRZ- L c) NRZ-I d) RZ e) Manchester f) Differential Manchester g) AMI Solution www. bookspar. com | Website for students | VTU NOTES

EXAMPLE Jan/Feb 2005 Sketch the signal waveforms when 00110101 is transmitted in the following signal codes. i) NRZ – L ii) Manchester Code www. bookspar. com | Website for students | VTU NOTES

Block Coding www. bookspar. com | Website for students | VTU NOTES

Block coding is used to improve the performance of line coding. The method has three steps- Division, Substitution & Line coding. Division – The sequence of bits is divided into group of m bits. www. bookspar. com | Website for students | VTU NOTES

Substitution in block coding Substitute an m-bit code for an n-bit group. For example in the figure, 4 B/5 B , a 4 -bit group is substituted to a 5 -bit code. While substituting choose a policy such that it helps in synchronization and error detection. www. bookspar. com | Website for students | VTU NOTES

Table : 4 B/5 B encoding Data Code 0000 11110 10010 0001 010011 0010 1010 10110 0011 10101 10111 0100 01010 11010 01011 11011 0110 011100 01111 11101 www. bookspar. com | Website for students | VTU NOTES

Analog to Digital conversion SAMPLING • Pulse Amplitude Modulation • Pulse Code Modulation • Sampling Rate: Nyquist Theorem www. bookspar. com | Website for students | VTU NOTES

PCM Technique www. bookspar. com | Website for students | VTU NOTES

PAM ( Pulse Amplitude Modulation) PAM is a method used to convert analog signal level in to discrete digital signal value at constant interval of time called sample. PAM uses a method called sample and hold. At a given interval, the signal level is read, then held briefly. www. bookspar. com | Website for students | VTU NOTES

Quantized PAM signal Quantization is a method of assigning integral values in a specific range to the sampled instances. www. bookspar. com | Website for students | VTU NOTES

Binary Encoding The quantized samples are assigned by sign and magnitude. Each value is translated into 8 bit representation. In eight bits first bit is used to indicate the sign and the other seven bits to represent the quantized value. www. bookspar. com | Website for students | VTU NOTES

PCM The binary digits are transformed to a digital signal by using a line coding method called Pulse Coding Method. www. bookspar. com | Website for students | VTU NOTES

Sampling Rate : Nyquist Theorem Note: According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency. Example: We want to sample telephone voice with a maximum frequency of 3000 Hz, we need a sampling rate of 6000 samples. www. bookspar. com | Website for students | VTU NOTES

Nyquist theorem For voice-over –phone –lines, take a sample for every 1/6000 s. www. bookspar. com | Website for students | VTU NOTES

Example What sampling rate is needed for a signal with a bandwidth of 10, 000 Hz (1000 to 11, 000 Hz)? Solution The sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11, 000) = 22, 000 samples/s www. bookspar. com | Website for students | VTU NOTES

Bit Rate – Bit Rate = Sampling rate x Number of bits per sample Example We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/s Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64, 000 bps = 64 Kbps www. bookspar. com | Website for students | VTU NOTES

Transmission Mode Parallel Transmission Serial Transmission www. bookspar. com | Website for students | VTU NOTES

Data transmission The transmission of binary data across a medium can be made in either parallel or serial mode. www. bookspar. com | Website for students | VTU NOTES

Parallel transmission In parallel mode, a group of n bits are sent with each clock tick. It needs n lines Advantage - Speed. Disadvantage – Cost. So, parallel transmission restricted to short distance. www. bookspar. com | Website for students | VTU NOTES

Serial transmission In serial mode, 1 bit is sent with each clock tick. There are two classes of serial transmission – i) Synchronous & ii) Asynchronous Advantage – Less cost. www. bookspar. com | Website for students | VTU NOTES

Asynchronous transmission The start bits are 0’s and the stop bits are 1’s. The gap is represented by an idle line. www. bookspar. com | Website for students | VTU NOTES

Synchronous transmission In Synchronous transmission, the bit stream is a combination of many bytes. Each byte is introduced with out a gap. The receiver reconstruct the information. The advantage of synchronous transmission is speed. Synchronous transmission is more faster than Asynchronous transmission. So, Synchronous transmission is useful for high-speed applications. www. bookspar. com | Website for students | VTU NOTES