Telcommunication laboratory ybu engineering solution homework quizz exam (7)

Sri Sai Aditya Institute of Science & Technology

ECE Department

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY KAKINADA III Year B. Tech. Electronics and Communication Engineering – I Sem. DIGITAL COMMUNICATIONS UNIT I-PULSE DIGITAL MODULATION: Elements of digital communication systems, advantages of digital communication systems, Elements of PCM: Sampling, Quantization & Coding, Quantization error, Compading in PCM systems. Differential PCM systems (DPCM).

UNIT II-DELTA MODULATION : Delta modulation, its draw backs, adaptive delta modulation, comparison of PCM and DM systems, noise in PCM and DM systems.

UNIT III-DIGITAL MODULATION TECHNIQUES : Introduction, ASK, FSK, PSK, DPSK, DEPSK, QPSK, M-ary PSK, ASK, FSK, similarity of BFSK and BPSK. UNIT IV-DATA TRANSMISSION : Base band signal receiver, probability of error, the optimum filter, matched filter, probability of error using matched filter, coherent reception, non-coherent detection of FSK, calculation of error probability of ASK, BPSK, BFSK,QPSK. UNIT V-INFORMATION THEORY : Discrete messages, concept of amount of information and its properties. Average information, Entropy and its properties. Information rate, Mutual information and its properties, UNIT VI-SOURCE CODING : Introductions, Advantages, Shannon’s theorem, Shanon-Fano coding, Huffman coding, efficiency calculations, channel capacity of discrete and analog Channels, capacity of a Gaussian channel, bandwidth –S/N trade off.

UNIT VII-LINEAR BLOCK CODES : Introduction, Matrix description of Linear Block codes, Error detection and error correction capabilities of Linear block codes, Hamming codes, Binary cyclic codes, Algebraic structure, encoding, syndrome calculation, BCH Codes.

UNIT VIII-CONVOLUTION CODES : Introduction, encoding of convolution codes, time domain approach, transform domain approach. Graphical approach: state, tree and trellis diagram decoding using Viterbi algorithm.

TEXT BOOKS : 1. Digital communications - Simon Haykin, John Wiley, 2005 2. Principles of Communication Systems – H. Taub and D. Schilling, TMH, 2003 REFERENCES : 1. Digital and Analog Communication Systems - Sam Shanmugam, John Wiley, 2005.

2. Digital Communications – John Proakis, TMH, 1983. Communication Systems Analog & Digital – Singh & Sapre, TMH, 2004.

3. Modern Analog and Digital Communication – B.P.Lathi, Oxford reprint, 3rd edition,2004.

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ECE Department

UNIT I-PULSE DIGITAL MODULATION ELEMENTS OF A DIGITAL COMMUNICATION SYSTEM The analysis and design of digital communication systems Involves the transmission of information in digital form from a source that generates the information to one or more destinations.

The source output may be either an analog signal, such as an audio or video signal, or a discrete signal, such as the output of a teletype machine, that is discrete in time and has a finite number of output characters.

In a digital communication system, the messages produced by the source are converted into a sequence of binary digits. The process of efficiently converting the output of either an analog or discrete source into a sequence of binary digits is called source encoding or data compression.

The sequence of binary digits from the source encoder, which we call the information sequence, is passed to the channel encoder

FIGURE 1. Basic elements of a digital communication system. . The purpose of the channel encoder is to introduce, in a controlled manner, some redundancy in the binary information sequence that can be used at the receiver to Prepared by Venkata Satish N

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overcome the effects of noise and interference encountered in the transmission of the signal through the channel. This increases the reliability of the received data and improves the fidelity of the received signal.

The binary sequence at the output of the channel encoder is passed to the digital modulator, which serves as the interface to the communication channel. Since nearly all the communication channels encountered in practice are capable of transmitting electrical signals (waveforms), the primary purpose of the digital modulator is to map the binary information sequence into signal waveforms.

To elaborate on this point, let us suppose that the coded information sequence is to be transmitted one bit at a time at some uniform rate R bits per second (bits/s). The digital modulator may simply map the binary digit 0 into a waveform So(t) and the binary digit 1 into a waveform S1(t). In this manner, each bit from the channel encoder is transmitted separately. We call this binary modulation.

The communication channel is the physical medium that is used to send the signal from the transmitter to the receiver. In wireless transmission, the channel may be the atmosphere (free space). On the other hand, telephone channels usually employ a variety of physical media, including wire lines, optical fiber cables, and wireless (microwave radio).

Whatever the physical medium used for transmission of' the information, the essential feature is that the transmitted signal is corrupted in a random manner by a variety of possible mechanisms, such as additive thermal noise generated by electronic devices; man-made noise, e.g., automobile ignition noise; and atmospheric noise, e.g., electrical lightning discharges during thunderstorms.

At the receiving end of a digital communication system, the digital demodulator processes the channel-corrupted transmitted waveform and reduces the waveforms to a sequence of numbers that represent estimates of the transmitted data symbols. This sequence of numbers is passed to the channel decoder, which attempts to reconstruct the original information sequence from knowledge of the code used by the channel encoder and the redundancy contained in the received data.

A measure of' how well the demodulator and decoder perform is the frequency with which errors occur in the decoded sequence. More precisely, the average probability of a Prepared by Venkata Satish N

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bit-error at the output of the decoder is a measure of the performance of the demodulator decoder combination.

In general, the probability of error is a function of the code characteristics, the types of waveforms used to transmit the information over the channel, the transmitter power, the characteristics of the channel, and the method of' demodulation and decoding.

The source decoder accepts the output sequence from the channel decoder and, from knowledge of the source encoding method used attempts to reconstruct the original signal.

Because of channel decoding errors and possible distortion introduced by the source encoder, and perhaps, the source decoder, the signal at the output of the source decoder is an approximation to the original source output. The difference or some function of the difference between the original signal and the reconstructed signal is a measure of the distortion introduced by the digital communication system.

The points worth noting are: The source coding algorithm plays important role in higher code rate The channel encoder introduced redundancy in data The modulation scheme plays important role in deciding the data rate and immunity of signal towards the errors introduced by the channel Channel introduced many types of errors like multi path, errors due to thermal noise etc.

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ECE Department

ADVANTAGES OF DIGITAL COMMUNICATION OVER ANALOG MODULATION: There are many advantages of using Digital Communication over analog communication. Some of them are listed as below: 1. The Digital communication has mostly common structure of encoding a signal so devices used are mostly similar.

2. The Digital Communication's main advantage is that it provides us added security to our information signal.

3. The Digital Communication system has more immunity to noise and external interference.

4. Digital information can be saved and retrieved when necessary while it is not possible in analog.

5. Digital Communication system is cheaper than Analog Communication.

6. The configuring process of digital communication system is simple as compared to analog communication system.

7. In Digital Communication System, the error correction and detection techniques can be implemented easily.

8. Digital hardware implementation is flexible & permits the use of microprocessors, digital switching elements & layer scale.

9. Digital systems are relatively less expensive than analog systems 10. Transmission rate can be changed easily.

11. Easy for processing and applying multiplexing techniques.

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ECE Department

ANALOG-TO-DIGITAL CONVERSION A digital signal is superior to an analog signal because it is more robust to noise and can easily be recovered, corrected and amplified. For this reason, the tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation and delta modulation.

PULSE CODE MODULATION (PCM)

Definition: Pulse code modulation (PCM) is essentially analog-to-digital conversion of a special type where the information contained in the instantaneous samples of an analog signal is represented by digital words in a serial bit stream.

PCM consists of three steps to digitize an analog signal: 1. Sampling 2. Quantization 3. Binary encoding Before we sample, we have to filter the signal to limit the maximum frequency of the signal as it affects the sampling rate.

Filtering should ensure that we do not distort the signal, ie remove high frequency components that affect the signal shape.

PCM Transmitter Prepared by Venkata Satish N

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SAMPLING

Analog signal is sampled every TS secs. Ts is referred to as the sampling interval. fs = 1/Ts is called the sampling rate or sampling frequency. According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal. There are 3 sampling methods: Ideal - an impulse at each sampling instant Natural - a pulse of short width with varying amplitude Flat top - a pulse of short width with constant amplitude Usually Flat top sampled signal is generated by sampler.

Three different sampling methods for PCM

Nyquist sampling rate for low-pass and bandpass signals

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QUANTIZATION

Sampling results in a series of pulses of varying amplitude values ranging between two limits: a min and a max. The amplitude values are infinite (or many) between the two limits. We need to map the infinite amplitude values onto a finite set of known values.

This is achieved by dividing the distance between min and max into q zones, each of height 



rangeofinputsignal noofQuantiztionlevels

vmax  vmin  q The midpoint of each zone is assigned a value from 0 to q-1 (resulting in q values). Each sample falling in a zone is then approximated to the value of the midpoint. That is quantization is a process of rounding-off each sampled value to the nearest value.

The reason for approximating to the mid point is that minimizes the maximum quantization error.

Example: Assume we have a voltage signal with amplitudes Vmin= -20V and Vmax=+20V. We want to use q=8 quantization levels. Then zone width  = (20 - -20)/8 = 5 The 8 zones are: -20 to -15, -15 to -10, -10 to -5, -5 to 0, 0 to +5, +5 to +10, +10 to +15, +15 to +20 The midpoints are: -17.5, -12.5, -7.5, -2.5, 2.5, 7.5, 12.5, 17.5 Each zone is then assigned a binary code.

The number of bits required to encode the zones, or the number of bits per sample as it is commonly referred to, is obtained as follows: v = log2 q Hence no of bits required to represent each sample are v = 3 The 8 zone (or level) codes are therefore: 000, 001, 010, 011, 100, 101, 110, and 111 Assigning codes to zones: 000 will refer to zone -20 to -15; 001 to zone -15 to -10, etc

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ECE Department

Note: If suppose quantization levels are 16 (2v), no of bits required to represent each sample are 4 (v bits).

If no of quantization levels are not in the power of 2, for example to distinguish 10 > 23 ( > 2v ) quantization levels, 4 bits (v+1) are required. Possible no of 4 bit code words are 16, use any 10 code words out of 16 for representing samples.

Quantization error is defined as the difference between actual sample and quantized sample.

i.e   x(nTs )  xq (nTs )

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TYPES OF QUANTIZERS

1. Uniform Quantizers Types: a) Symmetrical type of mid rise quantizer b) Symmetrical type of mid tread quantizer 2. Non uniform Quantizers Uniform Quantization •

Most ADC‟s use uniform quantizers.

•

The quantization levels of a uniform quantizer are equally spaced apart.

•

Uniform quantizers are optimal when the input distribution is uniform ie when all values within the Dynamic Range of the quantizer are equally likely.

Symmetrical type of mid rise quantizer

a) Symmetrical type of mid rise Quantizer b) Quantization error

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ECE Department

Origin lies in the middle of a rising part of the staircase graph like. Note that in mid rise type, any input value in between 0 to Δ is mapped to an output value of Δ/2, any input value between Δ to 2Δ is mapped to an output value of 3Δ/2 and so on.

Mid rise characteristic is desirable because of symmetry and because it uses the 2 v levels of a v bit coder efficiently. A disadvantage of this mid rise characteristic is that it cannot represent a zero output level.

Symmetrical type of mid tread quantizer

a) Symmetrical type of mid tread Quantizer b) Quantization error Origin lies in the middle of a tread of a staircase like graph. Note that in mid tread type, any input value in between -Δ/2 to + Δ/2 is mapped to an output value of zero, any input value between + Δ/2 to 3 Δ/2 is mapped to an output value of Δ and so on.

Unfortunately, this characteristic has an odd number of levels (if it is symmetric) or it must be non symmetric about zero. Therefore it does not use the 2 v possible levels of a v bit coder efficiently.

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Illustration of Quantization process for an analog signal & discrete time signal and error signal in the approximations

Fig.: (a) An analog signal and its quantized version (b) The error signal

Fig: (c) Equispaced samples of m(t ) (d) Quantized sample sequence Prepared by Venkata Satish N

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NOISE IN PCM SYSTEMS The performance of a PCM system is influenced by two major sources of noise.

1. TRANSMISSION NOISE: It is introduced anywhere between the transmitter output and the receiver input. The effect of transmission noise is to introduce bit errors into the received PCM wave, with the result that, in case of a binary system, a symbol 1 occasionally is mistaken for a symbol 0, or vice versa. Clearly, the more frequently such errors occur, the more dissimilar the receiver output becomes compared with the original message signal.

2. DISTORTION DUE TO QUANTIZING

There are two types of distortions associated with a quantizer: 1. Overload or clipping distortion: Overload distortion occurs when the input signal exceeds the quantizer's input range, then output will remain at its maximum (or minimum) value until the input falls within the quantizer's input range. Overload distortion results in a clipped output signal. To avoid clipping, a quantizer is matched to the input signal.

2. Quantization distortion: Figure below shows the error signal introduced by the quantizer. From this figure, it can be seen that quantization error occurs when the input signal is within the input range of the quantizer. It arises because of the difference between the input amplitude and the quantized sampled amplitude and because of the limited sampling rate. The quantization error signal produces quantization noise or distortion in the reconstructed message signal. Its frequency spectrum covers a large bandwidth. Low-pass filtering which is used to smooth the waveform will remove most of the quantization error above its cutoff frequency.

However, some of the quantization error is in the signal band, and that cannot be removed by the low-pass filter. This will produce a gritty sound at the output of a PCM system called quantization noise.

Figure: Characteristic of quantization and overload errors.

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Quantization noise is the result of the quantization process. Since the quantization process adjusts the height of each sample, the original waveform cannot be exactly reconstructed using a low-pass filter as is the case with PAM signals and the classical sampling theorem. The sampling rate will also affect the quantization noise since the quantization error will become larger as the sampling rate decreases.

Figure below shows an analog input signal and its quantized waveform. Shown below this is the resulting quantization error signal. The maximum amplitude of this error signal is half a quantization interval. The overall amplitude variation is from half a quantization interval to minus half a quantization interval. During a period of small intervals, the error signal appears to be a sawtooth wave.

Figure: Analog input signal, quantized waveform, and quantization error waveform.

Quantization error is another reason for using compressed encoding for digitizing a voice signal. Compressed encoding allows a higher signal-to-quantization-noise ratio (SNQR) than linear encoding. This ratio defined as where S is the voice signal level and NQ is noise due to the quantization error. Clearly, keeping the quantization error small is key to keeping a high SNQR. As signal amplitude gets smaller, NQ must get smaller to keep SNQR from dropping. Compression accomplishes this by forcing quantization error magnitude to decrease with lower amplitudes.

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ECE Department

Illustration of how quantization error is reduced by increasing quantization levels When a signal is quantized, we introduce an error since the coded signal is an approximation of the actual amplitude value. The difference between actual and coded value (midpoint) is referred to as the quantization error.

The more zones, the smaller ď „ which results in smaller errors. But, the more zones, the more bits required to encode the samples which leads higher bit rate.

Example:

In the above example, increasing the no of quantization levels from 5 to 10, decreases the step size by 2, there by decreases the quantization error.

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NON UNIFORM QUANTIZING

In this step size of the quantizer is not fixed over entire input range and it varies according to the input signal. i.e step size of the quantizer is reduced at low levels and increased at high levels.

Non uniform Quantizer of 8 levels

Importance of Non uniform Quantization: Voice signals are more likely to have amplitudes near zero than at extreme peaks.. Signals with lower amplitude values will suffer more from quantization error as the error range: /2, is fixed for all signal levels.

Non linear quantization is used to alleviate this problem. The Goal is to keep SNQR fixed for all sample values.

Two approaches for obtaining Non uniform Quantization: Direct approach: The quantization levels follow a logarithmic curve. Smaller ‟s at lower amplitudes and larger ‟s at higher amplitudes. But this process of varying  directly is very difficult.

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Indirect approach: An Effect of non linear quantizing can be can be obtained by first passing the sample values through a compressor at the sender, then through a uniform quantizer. This technique increase amplitudes near zero. To compensate the effects happened at the sender, pass the sample values through an expander at the receiver. The process of compression, uniform quantization and expansion is called Companding.

A-law and µ -law Companding •

These two are standard companding methods.

•

µ -Law is used in North America and Japan

•

A-Law is used elsewhere to compress digital telephone signals

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Two types according to compression filter –

-law : used in US

y –

ln(1   x ) ln(1   )

sgn( x)

A-law : used in Europe  Ax sgn( x), 0 x  1  A 1  ln A y 1  ln( A x ) sgn( x), 1  x  1  1  ln A A

A-law & µ-law compression curve

Similarities between A−law and µ −law  Both are linear approximations of logarithmic input/output relationship.  Both are implemented using eight−bit code words (256 levels, one for each quantization interval).  Eight−bit code words allow for a bit rate of 64 kilobits per second (kbps). This is calculated by multiplying the sampling rate (twice the input frequency) by the size of the code word (2 x 4 kHz x 8bits = 64 kbps).  Both break a dynamic range into a total of 16 segments

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ECE Department

Differences Between A−law and µ −law  Different linear approximations lead to different lengths and slopes.·  A−law provides a greater dynamic range than u−law. 

u−law provides better signal/distortion performance for low level signals than A−law.

 A−law requires 13−bits for a uniform PCM equivalent. u−law requires 14−bits for a uniform PCM equivalent SNR of Compander

Example: µ-law Companding

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ENCODING

The output of the quantizer is one of “q”possible signal levels. If we want to use a binary transmission system, then we need to map each quantized sample into a “v” bit binary word.

Encoding is the process of representing each quantized sample by an  bit code word. The mapping is one-to-one so there is no distortion introduced by encoding. Some mappings are better than others. 

A Gray code gives the best end-to-end performance.



With gray codes adjacent samples differ only in one bit position.



The weakness of Gray codes is poor performance when the sign bit (MSB) is received in error.

Example (3 bit quantization): With this gray code, a single bit error will result in an amplitude error of only 2.

Unless the MSB is in error.

There are several ways by which binary symbols 1 and 0 can be represented by electrical signals: Unipolar NRZ (on-off signaling): Symbol 1 is represented by transmitting a pulse of constant amplitude for the duration of symbol, and symbol 0 is represented by switching off the pulse. This type of signal is referred to as an on-off signaling or Unipolar non return to zero.

Polar NRZ: Symbols 1 and 0 are represented by pulses of equal positive and negative amplitudes. This type of signal is referred to as a polar Non Return to Zero signal.

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Unipolar RZ: A rectangular pulse (half symbol wide) is used for a 1 and no pulse for a 0. This type of signal is called Unipolar Return to zero.

Bipolar RZ: Positive and negative pulses are used alternatively for symbol 1, and no pulse for symbol 0. This type of signal is called a bipolar signal.

Manchester or Split-phase code: Symbol 1 is represented by a positive pulse followed by a negative pulse, with both pulses being of equal amplitude and half-symbol wide; for symbol 0, the polarities of these pulses are reversed. This type of signal is called a split phase or Manchester code.

Electrical representations of binary data Reasons and advantages of different encodings will be discussed in UNIT 4

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Bit rate of PCM The bit rate of a PCM signal can be calculated form the number of bits per sample x the sampling rate. i.e. Bit rate = v x fs Bandwidth requirements of PCM The bandwidth of (serial) binary PCM waveforms depends on the bit rate R and the waveform pulse shape used to represent the data.

For no aliasing case (fs≼ 2B), the MINIMUM Bandwidth of PCM is: Bpcm(Min) = R/2 = vfs//2.

The Minimum Bandwidth of vfs//2 is obtained only when sin(x)/x pulse is used to generate the PCM waveform.

For PCM waveform generated by rectangular pulses, the First-null Bandwidth is: Bpcm = R = nfs A digitized signal will always need more bandwidth than the original analog signal. Price we pay for robustness and other features of digital transmission.

EXAMPLE: DESIGN OF A PCM SIGNAL FOR TELEPHONE SYSTEMS

Assume that an analog audio voice-frequency (VF) telephone signal occupies a band from 300 to 3,400Hz. The signal is to be converted to a PCM signal for transmission over a digital telephone system. The minimum sampling frequency is 2x3.4 = 6.8 ksample/sec.

To be able to use of a low-cost low-pass anti aliasing filter, the VF signal is oversampled with a sampling frequency of 8ksamples/sec. This is the standard adopted by the Unites States telephone industry. Assume that each sample values is represented by 8 bits; then the bit rate of the binary PCM signal is Bit rate = v x fs = 8 x 8k = 64k bit/sec This 64-kbit/s signal is called a DS-0 signal (digital signal, type zero).

The minimum absolute bandwidth of the binary PCM signal when sin(x)/x pulse is used to generate is Bpcm(Min) = R/2 = vfs//2 = 32k bit/sec If we use a rectangular pulse for sampling the first null bandwidth is given by Bpcm(Min) = R = vfs = 64k bit/sec We require a bandwidth of 64 kHz to transmit this digital voice PCM signal, whereas the bandwidth of the original analog voice signal was, at most, 4 kHz.

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APPLICATIONS OF PCM



With the advent of fibre optic cables, PCM is used in telephony.



In space communication, space craft transmits signal to earth. Here the ransmitted power is quite small and the distances are very large.Hence due to high noise immunity, only pcm systems can be used in such applications.

ADVANTAGES OF PCM

•

Relatively inexpensive digital circuitry may be used extensively.

•

PCM signals derived from all types of analog sources may be merged with data signals and transmitted over a common high-speed digital communication system.

•

In long-distance digital telephone systems requiring repeaters, a clean PCM waveform can be regenerated at the output of each repeater, where the input consists of a noisy PCM waveform.

•

The noise performance of a digital system can be superior to that of an analog system.

•

The probability of error for the system output can be reduced even further by the use of appropriate coding techniques.

DRAW BACKS OF PCM

 Encoding, Decoding and quantizing circuitry of PCM is complex  PCM requires a large bandwidth as compared to other systems

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QUANTIZATION ERROR/NOISE IN PCM Quantization error is defined as the difference between actual sample and quantized i.e   x(nTs )  xq (nTs )

sample.

If the step size of the quantizer (mid rise or mid tread) is  , then maximum quantization error  max is

    and the range of quantization error is   ,  .

2  2 2

   As the error is equally likely in the range   ,  , it is better to assume error as uniform  2 2

random variable. The probability density function of this error is given by f ( ) 

1 1        2  2

Mean Square value of this quantization error (Noise power) is given by  2

 2

1 2 E[ ]    f ( )d     d   12   2

2



2

2



2

SIGNAL TO QUANTIZATION NOISE RATIO IN PCM Case 1: input signal is sinusoidal signal x(t )  Am sin mt

Am2 SNR = Signal Power (rms) / Quantization noise power = 22  12 Where  

2A 2A rangeofinputsignal  m  vm noofQuantiztionlevels q 2

SNR = Signal Power / Quantization noise power

Am2 = 22 =  12

Am2 3 2 BWpcm fm 3 2v 2 2 or  2 2 2 2  2 Am   v   2  12

3 SNR in decibels = 10log10 ( 22v )  1.76  6v 2 Prepared by Venkata Satish N

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Case 2: Input signal is a DC signal ranging between –Am to +Am SNR= Signal Power / Quantization noise power =

Where  

SNR =

Am2  2     12 

2A 2A rangeofinputsignal  m  vm noofQuantiztionlevels q 2

Am2 Am2 =  3* 22 v 2 2     2 Am     v   12   2  12

SNR in decibels = 10log10 (3*22v )  4.76  6v .

Note: Signal to noise ratio of PCM system improved by 6db for every one bit increase.

PCM TRANSMISSION PATH & REGENERATION The path between the PCM transmitter and PCM receiver over which the PCM signal travel, is called as PCM transmission path. The most important feature of PCM system lies in its ability to control the effects of distortion and noise when the PCM signal travels on the channel. This capability is accomplished by reconstructing the PCM wave by means of a chain of regenerative repeaters located at sufficiently close spacing along the transmission route.

There are three basic functions are performed by a regenerative repeater, namely 1. Equalization 2. Timing 3. Decision Making The equalizer shapes the received pulses so as to compensate for the effects of amplitude and phase distortions produced by the transmission characteristics of the channel.

The timing circuitry provides a periodic pulse train, derived from the received pulses, for sampling the equalized pulses at the instants of time where the signal to noise ratio is a maximum.

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The decision making device makes a decision in the favor of 1, if the equalized pulse plus noise is above the threshold level and it makes a decision in the favor of 0 if the equalized pulse plus noise is below the threshold level.

PCM RECEIVER The first operation in the receiver is to regenerate the received pulses. These clean pulses are then regrouped into code words and decoded into a quantized PAM signal. The decoding process involves generating a pulse the amplitude of which is the linear sum of all the pulses in the code word, with each pulse weighted by its place value in the code.

The final operation in the receiver is to recover the signal wave by passing the decoder output through a low-pass reconstruction filter whose cutoff frequency is equal to the message bandwidth W. Assuming that the transmission path is error free, the recovered signal includes no noise with the exception of the initial distortion introduced by the quantization process

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TIME DIVISION MULTIPLEXING PAM SYSTEM Normally, in PAM system, the duration of the pulse đ?œ? is much less than the time period of pulses Ts. Thus no information is being transmitted through the system for most of the time. The time space Ts – đ?œ? can be utilized to transmit information from other signals. The signal numbers 2, 3 and 4 are transmitting information with the help of samples numbered 2, 3 and 4 respectively. This is along with the samples numbered 1 of the signal number 1.

The time period Ts is equally divided between the four signals, thus allocating a time slot of

�� 4

to each signal. Thus the duration of time slot is such that

there is a guard time

�� 4

�� 4

> đ?œ?. Thus,

- đ?œ? between all successive sampling pulses, ensuring that there is

less cross talk between signals. The arrangement by which the information from more than one signal is transmitted in this manner is known as time division multiplexing.

A TDM PAM system is shown in figure below, which transmits information from n signals. The switch 1 and switch 2 respectively known as commutator and decommutator are synchronized electronic switches which rotate at the same speed of 2f M rotations per second. The commutator samples and combines the samples, while the decommutator seperates the samples belonging to individual signals.

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Synchronization is the most crucial in TDM system. Thus, for example, if the commutator is at position 2, the decommutator must also be in position 2. To provide synchronization, a synchronizing puse is transmitted in every frame (time interval between two successive samples of the same signal, i.e Ts).

Thus to multiplex n channels, n+1 time slots are provided in a frame; n for channels and 1 for the synchronizing pulse. The synchronizing pulse is chosen in such a way that it is easily distinguishable. For this purpose, one of its properties is adjusted in such a way that it is never attained by the other pulses. For example, in case of PAM, its amplitude is made larger than the amplitudes of all the other pulses.

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DIFFERENTIAL PULSE CODE MODULATION When a voice or video signal is sampled at a rate slightly higher than the nyquist rate, the resulting sampled signal is found to exhibit a high correlation between adjacent samples. The meaning of this high correlation is that, in an average sense, the signal does not change rapidly from one sample to next with the result that the difference between adjacent samples has a variance that is smaller than the variance of the signal itself.

When these highly correlated samples are encoded, as in standard PCM system, the resulting encoded signal contains redundant information. By removing this redundancy before encoding, we obtain a more efficient coded signal.

For example, we can observe that the samples taken at 4T s, 5Ts and 6Ts are encoded to same value of 110. This information can be carried only by one sample. But three samples are carrying the same information means that it is redundant. Cosider another example of samples taken at 9T s and 10Ts. The difference between these samples only due to last bit and first two bits are redundant, since they do nit change.

If this redundancy is reduced, then overall bit rate will decrease and number of bits will decrease and number of bits required to transmit one sample will also be reduced.

Redundant information in PCM Prepared by Venkata Satish N

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DPCM TRANSMITTER DPCM can be treated as a variation of PCM; it also involves the three basic steps of PCM, namely, sampling, quantization and coding. But, in the case of DPCM, what is quantized is the difference between the actual sample and its predicted value, as explained below.

Let x(t) represent the analog signal that is to be DPCM coded, and let it be sampled with a period Ts. The sampling frequency fs = 1/Ts is such that there is no aliasing in the sampling process. Let x(nTs) = m(t) at t= nTs. Quite a few real world signals such as speech signals, biomedical signals (ECG, EEG, etc.), telemetry signals (temperature inside a space craft, atmospheric pressure, etc.) do exhibit sample-to-sample correlation. This implies that x(n) and x(n + 1) (or x((n) and x (n − 1)) do not differ significantly. In fact, given a set of previous M samples, say x (n − 1), x (n − 2),…… x (n − M) , it may be possible for us to predict (or estimate) x (n) to within a small percentage error.

DPCM transmitter Let x (nTs ) denote the predicted value of x(nTs) and let e(nTs )  x(nTs )  x (nTs ) Which is the difference between the unquantized input sample m(nTs) and a prediction of it, denoted by x (nTs ) . This predicted value is produced bu using a prediction filter whose input, as we will see, consists of a quantized version of the input signal x (nTs).

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The difference signal e(nTs ) is called a prediction error, since it is the amount by which the prediction filter fails to predict the input exactly.

In DPCM, error sequence is quantized, coded and obtained a variation of PCM, which is known as differential pulse code modulation.

The quantizer output may be expressed as eq (nTs )  e( nTs )  qe ( nTs ) , where qe (nTs ) is the quantization error.

According to fig, the quantizer output eq (nTs ) is added to the predicted value

x (nTs ) to produce the prediction filter input xq (nTs )  x (nTs )  eq (nTs ) . xq (nTs )  x (nTs )  e(nTs )  qe (nTs ) xq (nTs )  x(nTs )  qe (nTs )

That is irrespective of the properties of the prediction filter, the quantized signal xq (nTs ) at the prediction filter input differs from the original input signal x(nTs ) by the

quantizing error qe (nTs ) . Accordingly if prediction is good, the variance of the prediction error eq (nTs ) will be smaller than the variance of x(nTs ) .

DPCM RECEIVER The receiver for reconstructing the quantized version of the input is shown in figure. It consists of a decoder to reconstruct the quantized error signal. The quantized version of the original input is reconstructed from the decoder output using the same prediction filter as used in the transmitter.

DPCM receiver

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Output Signal to Noise ration of the DPCM system: By definition, (SNR)o =Variance of the input signal/Variance of the quantized noise =

=

 X2  Q2

 X2  E2 *  E2  Q2

Where  E2 is the variance of the prediction error.

 X2  E2 (SNR)o = 2 * 2 = Gp * prediction error to quantization noise ratio.  E Q Where Gp is the predictive gain. This prediction gain must be high as possible. This Gp is maximized by minimizing the variance  E2 of the prediction error.

THE PREDICTION FILTER

The predicted value x (nTs ) is modeled as a linear combination of past values of the quantized input as shown below p

x (nTs )   wk xq (nTs  kTs ) k 1

Where the tapped delay line weights w1, w2, w3………wp define the desired prediction filter coefficients and p is order of the prediction filter.

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p

The prediction error e(nTs )  x(nTs )   wk xq (nTs  kTs ) k 1

The variance of the prediction error is therefore  E2 = E[e2 (nTs )] p

E[e2 (nTs )]  E[{x(nTs )   wk xq (nTs  kTs )}2 ] k 1

In order to choose a set of weights that minimize the variance  E2 , we must differentiate

 E2 with respect to each weight and then put the resulting derivatives equal to zero.

ADVANTAGES OF DPCM

1. As the difference is being encoded and transmitted by the DPCM technique, a small difference voltage is to be quantized and encoded.

2. This will require less number of quantization levels and hence less number of bits to represent them 3. Thus signaling rate and bandwidth of a DPCM system will be less than that of DPCM. COMPARISON BETWEEN PCM AND DPCM Parameter

of Pulse code modulation

Differential

comparison Number of bits

Pulse

Code

Modulation It can use 4, 8 or 16 bits per sample

Bits can be more than one but are less than PCM

Quantization error

Quantization

error

depends

on

Quantization error is present

number of levels used.

Transmission

Highest bandwidth is required since Bandwidth requires is lower

Bandwidth

number of bits are high

Feed back

There is no feedback in transmitter Here, feedback exists

than PCM

and receiver Complexity

of System Complex

Simple

implementation Signal to noise ratio

Good

Fair

Applications

Audio and video telephony

Speech and video

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Important Questions 1. a) Draw the block diagram of PCM scheme. Explain each block? b) The bandwidth of a TV radio plus audio signal is 4.5MHz. If this signal is converted to PCM with 1024 quantizing levels. Determine the bit rate of the resulting PCM signal. Assume that the signal is sampled at a rate 20% above the Nyquist rate.

2 .a) The signal m(t)=6sin(2∏t) volts is transmitted using a 4-bit binary PCM system. The quantizer is of the midrise type, with a step size of 1 volt . Sketch the resulting PCM wave for one complete cycle of the input. Assume a sampling rate of four samples per second, with samples taken at t=±1/8, t=±3/8, t=±5/8…….seconds. b)What is quantization error? How does it depend upon the step size? Suggest some methods to overcome the difficulties encountered when the modulating signal amplitude swing is large.

3. a)A PCM system uses a uniform quantizer followed by a 7-bit binary encoder. The bit rate of the system is equal to 50x106b/sec. (i) What is the maximum message bandwidth for which the system operates satisfactorily? (ii) Determine the output signal to quantization noise ratio when a full load sinusoidal Modulating wave of frequency 1MHz is applied to the input. (b) Explain the importance of prediction in DPCM & draw the structure of Prediction filter?

4. (a) Show that in a PCM system, the output signal power to quantization noise

3 BWpcm fm power ratio can be expressed as S/NQ= 4 . Where BWpcm is the channel bandwidth and 2 fm is the message bandwidth. (b) Draw and explain different ways of representing binary data by electrical signals?

5. (a) Explain µ - law and A - law companding technique? (b)Draw the block diagram of DPCM system and explain each block. or Explain the process of differential quantizing scheme and compare differential quantizing scheme with direct quantizing scheme. (c) Discuss the advantages & Drawbacks of PCM system?

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Communication Systems- Simon haykin 4th edition Exercise Problems

Problem1: A speech signal has a total duration of 10 s. It is sampled at the rate of 8 kHz and the encoded. The signal to Quantization noise ratio is required to be 40db. Calculate the minimum storage capacity needed to accommodate this digitized speech signal.

Solution: The minimum number of bits per sample is 7 for a signal to quantization noise ratio of 40 dB. Hence The number of samples in a duration of 10 s = 8000*10 = 8*104 samples.

The minimum storage is therefore = 7*8*104 = 560 Kbits

Problem 2 : A PCM system uses a uniform quantizer followed by a 7 bit binary encoder.

The bit rate of the system is equal to 50 106 b / s . (a) What is the maximum message bandwidth for which the system operates satisfactorily? (b) Determine the output signal to quantization noise ratio when a full- load sinusoidal modulating wave of frequency 1 MHz is applied to the input.

Solution: (a) Bit rate of the PCM system is given by R  vf s For the system to operate satisfactorily, sampling rate must be atleast equal to the nyquist rate. Hence R  vf s  v  2 f max

50 106 b / s = 7  2 f max  f max = 3.57*106 Hz (c) The output signal to Quantizing noise ratio is given by SNR in dB= 1.8 + 6v= 43.8 dB

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Problem 3: (a) A sinusoidal signal, with an amplitude of 3.25 volts, is applied to a uniform quantizer of the mid rise type whose output takes on the values 0, ±1, ±2,±3 volts. Sketch the waveform of the resulting quantizer output for one complete cycle of the input. (b) Repeat this evaluation for the case when the quantizer is of the midrise type whose output takes on the values 0.5, ±1.5, ±2.5, ±3.5 volts.

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Problem 4: The signal m (t) = 6sin (2πt) volts is transmitted using a 4 bit binary PCM system. The quantizer is of the midrise type, with a step size of 1 volt. Sketch the resulting PCM wave for one complete cycle of the input. Assume a sampling per second, with samples taken at t = ±18, ±3/8, ±5/8,……, seconds.

Solution:

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UNIT2- DELTA MODULATION Delta modulation, like DPCM is a predictive waveform coding technique and can be considered as a special case of DPCM. It uses the simplest possible quantizer, namely a two level (one bit) quantizer. The price paid for achieving the simplicity of the quantizer is the increased sampling rate (much higher than the Nyquist rate) and the possibility of slope-overload distortion in the waveform reconstruction, as explained in greater detail later on in this section. In DM, the analog signal is highly over-sampled in order to increase the adjacent sample correlation. The implication of this is that there is very little change in two adjacent samples, thereby enabling us to use a simple one bit quantizer, which like in DPCM, acts on the difference (prediction error) signals. In its original form, the DM coder approximates an input time function by a series of linear segments of constant slope. Such a coder is therefore referred to as a Linear (or nonadaptive) Delta Modulator (LDM). Subsequent developments have resulted in delta modulators where the slope of the approximating function is a variable. Such coders are generally classified under Adaptive Delta Modulation (ADM) schemes. We use DM to indicate either of the linear or adaptive variety.

LINEAR DELTA MODULATION PRINCIPAL OF WORKING

The principle of operation of an LDM system can be explained with the help of Fig 2.1 below. The signal x (t), band limited to W Hz is sampled at the rate f s  2W .

If x(nT s) denote the sample of x(t) at t= nTs. The staircase approximation to x(t), denote by x (nTs ) is arrived as follows. One notes, at t=nTs, the polarity of the difference between x(nTs) and the latest approximation to it; that is x (nTs ) at t= nTs. The difference between the input and the previous approximation is quantized into only two levels, namely,  , corresponding to positive and negative differences, respectively. Thus, if the approximation falls below the signal at any sampling epoch, it is increased by  . If on the other hand, the approximation lies above the signal, it is diminished by  . Provided that the signal does not change too rapidly from sample to sample, we find that the staircase approximation remains within  of the input signal.

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Figure 2.1 Illustration of Delta Modulation If e(nTs )  0

then eq (nTs )   & b(nTs )  1

If e(nTs )  0

then eq (nTs )   & b(nTs )  0

DELTA MODULATION TRANSMITTER

The principal virtue of delta modulation is its simplicity. It may be generated by applying the sampled version of the incoming baseband signal to a modulator that involves a summer, quantizer and accumulator interconnected as shown in figure 2.2.

Fig.2.2 Delta Modulation Transmitter

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Denoting the input signal as x(t) and the staircase approximation as x q(t), the basic principal of delta modulation may be formalized in the following set of discrete-time relations. e(nTs )  x(nTs )  xq (nTs  Ts )    eq1 eq (nTs )   sgn(e(nTs ))    eq 2

& xq (nTs )  x q (nTs Ts )  eq (nTs )    eq 3 where Ts is the sampling period; e(nT s) is an error signal representing the difference between the present sample value x(nTs) of the input signal and the latest approximation to it. Namely, x (nTs )  xq (nTs  Ts ) ; and eq (nTs ) is the quantized version of e(nTS).

The quantized output eq (nTs ) is finally coded to produce the desired DM wave.

Figure 2.1 illustrates the way in which the staircase approximation xq (t ) follows variations in the input signal x (t) in accordance with above equations and it also displays the corresponding binary sequence at the delta modulator output

Working of Accumulator (Stair case wave form generator) 1. In particular, quantizer consists of a hard limiter with input and output relation defined by eq2 which is depicted in fig 2.2.1. The quantizer output is applied to an n

n

i 1

i 1

accumulator, producing the result xq (nTs )   sgn(e(iTs ))   eq (iTs ) .

Fig 2.2.1: Input –output characteristic of quantizer for DM system 2. Thus at the sampling instant nTs, the accumulator increments the approximation by a step  in a positive or negative direction, depending upon the algebraic sign of error signal e(nTs).

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3. If the input signal x(nTs) is greater than the most recent approximation x (nTs ) , a positive increment +  is applied to the approximation.

4. If on other hand, the input signal is smaller, a negative increment -  is applied to the approximation.

5. In this way the accumulator does the best it can to track the input samples by one step at a time.

DELTA MODULATION RECEIVER

In the receiver, the staircase approximation xq(t) is reconstructed by passing the sequence of positive and negative pulses, produced at the decoder output, through an accumulator in a manner similar to that used in the transmitter. Then pass this staircase waveform through a low pass filter (with a bandwidth equal to Original signal bandwidth) to recover the original signal.

Fig.2.3 Delta Modulation Receiver

In comparing the DPCM and DM networks, we note that they are basically similar, except for two important differences, namely, the use of a one-bit quantizer in delta modulator and the replacement of the prediction filter by a single delay element.

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QUANTIZING NOISE Delta modulation systems are subject to two types of quantizing error.

(1) Slope overload distortion (2) Granular Noise SLOPE OVERLOAD DISTORTION:

This distortion arises because of large dynamic range of the input signal. The rate of rise of input signal x(t) is so high that the staircase signal cannot approximate it. The slope overload is said to occur when the step size „Δ‟ is too small to follow steep segment of the input waveform x(t). To reduce this error, the step size must be increased when slope of the signal x(t) is high. Since the step size of delta modulator remains fixed, its maximum or minimum slopes occur along straight lines. Therefore this modulator is also known as Linear Delta Modulator.

Quantiztion errors in delta modulation for an arbitrary input To reduce this slope overload distortion, the slope of the quantizer must be greater than the maximum slope of the input signal.

Ie.

 dm(t )  Ts dt max imum

GRANULAR NOISE (IDLE NOISE):

Granularity, on other hand refers to a situation where the stair case function x (nTs ) hunts around a relatively flat segment of the input function, with a step size that is too large relative to local slope characteristic of the input. This means that for very small Prepared by Venkata Satish N

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variations in the input signal, the staircase signal is changed by large amount Δ because of large step size. The solution is to this problem is to make step size small.

Quantiztion errors in delta modulation for another input BIT RATE (SIGNALING RATE) OF DELTA MODULATION

Delta Modulation bit rate (R) = Number of bits transmitted/seconds = Number of samples/sec  Number of bits/sample = f s 1 = f s Therefore, the delta modulation bit rate is (1/N) times the bit rate of a PCM system.

Where N is the number of bits per transmitted PCM codeword. Hence, we can say that the channel bandwidth for the delta modulation system is reduced to a great extent as compared to that for the PCM system.

ADVANTAGES OF DELTA MODULATION

1. Since the delta modulation transmits only one bit for one sample, therefore the signaling rate and transmission channel bandwidth is quite small for delta modulation compared to PCM. 2. The transmitter and receiver implementation is very much simple for delta modulation. There is no analog to digital converter required in delta modulation.

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DRAWBACKS OF DELTA MODULATION

The Delta Modulation has two major drawbacks as under; (i)

Slope overload distortion

(ii)

Granular Noise

SIMON HAYKIN Problem1: Given a sine wave of frequency fm and amplitude Am applied to a delta modulator having step size Δ. Show that the slope overload will occur if Am 

 here Ts is the sampling period.

2 f mTs

Solution: Let us consider that the sine wave is represented as x(t )  Am sin(2 f mt ) Maximum slope of delta modulator is given as

 .

Ts

We know that, the slope overload distortion will take place if slope of the sine wave is greater than slope of delta modulator i.e., max

max

dx(t )  > Ts dt

dAm sin(2 f mt )   dt Ts

max 2 f m Am cos(2 f mt )  2 f m Am > Am 

 Ts

 Ts

 2 f mTs

Note: To avoid slope overload distortion, the condition that must be satisfied is Am 

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 2 f mTs

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QUANTIZATION ERROR/NOISE IN DELTA MODULATION

Quantization error is defined as the difference between actual sample and quantized i.e   x(nTs )  xq (nTs )

sample.

If the step size of the quantizer is  , then maximum quantization error  max is  and the range of quantization error is  ,   .

As the error is equally likely in the range  ,   , it is better to assume error as uniform random variable. The probability density function of this error is given by f ( ) 

1 1       2

Mean Square value of this quantization error (Noise power) is given by E[ 2 ] 



2   f ( )d 





2 



1 2 d   3

SIGNAL TO QUANTIZATION NOISE RATIO IN DELTA MODULATION

Case 1: input signal is sinusoidal signal x(t )  Am sin mt Am2 SNR = Signal Power (rms) / Quantization noise power = 22  3

No slope overload distortion occurs, for Am 

 , then substituting into the above 2 f mTs 2

equation gives

 1  Am2   2 2  f T m s   SNR = 2 = 2  2 3 3

This noise power

2 is uniformly distributed over the frequency band upto f s (which is 3

more than f m ). Then the output quantization power within the bandwidth f BWLPF is given

by Nq'    f BWLPF 2

3

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fs

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In the receiver, at the output of Low pass filter of Bandwidth SNR =

 1    2  2 f mTs  f 2  BWLPF 3 fs

f BWLPF

2



8

2

3 f s3 f BWLPF f

2 m

ADAPTIVE DELTA MODULATION To reduce slope overload distortion, a large step size is required to accommodate wide dynamic range of the input signal and small steps are required to reduce granular noise. In fact, adaptive delta modulation is the modification to overcome these errors.

Finally, we should mention that a delta modulator may also be made adaptive, wherein the variable step size increases during a steep segment of the input signal and decreases when the modulator is quantizing an input signal with a slowly varying segment. In this way the step size is adapted to the level of the input signal. The resulting system is called an adaptive delta modulator.

The problem in adaptive delta modulation, of course, is to specify suitable rules for step size variation. Figure below illustrates the operation of ADM.

Waveforms illustrative of ADM operation

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ADAPTIVE DELTA MODULATION TRANSMITTER:

The logic for step size control is added in the diagram. The step size increases or decreases according to a specified rule depending on one bit quantizer output. As an example, if one bit quantizer output is high (i.e. 1), then the step size may be doubled for next sample. If one bit quantizer output is low, then step size may be reduced by one step.

ADAPTIVE DELTA MODULATION RECEIVER:

In the receiver of Adaptive delta modulator shown in figure, there are two portions. The first portion produces the step size from each incoming bit. Exactly the same process is followed as that in transmitter. The previous input and present input decides the step size.

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It is then applied to an accumulator which builds up staircase waveform. The low pass filter then smoothens out the staircase waveform to reconstruct the original signal.

ADVANTAGES OF ADAPTIVE DELTA MODULATION

1. Signal to Noise ratio becomes better than ordinary delta modulation because of the reduction in slope overload distortion and idle noise.

2. Because of the variable step size, the dynamic range of ADm is wider than simple DM. 3. Bandwidth required for the transmission through channel is also less.

Results have been reported in the literature which compares the (SNR)o performance of Îź -law PCM and the ADM scheme discussed above. One such result is shown in Fig. below for the case of band pass filtered (200-3200 Hz) speech. For PCM telephony, the sampling frequency used is 8 kHz. As can be seen from the figure, the SNR comparison between ADM and PCM is dependent on the bit rate. An interesting consequence of this is, below 50 kbps, ADM which was originally conceived for its simplicity, out-performs the logarithmic PCM, which is now well established commercially all over the world. A 60 channel ADM (continuous adaptation) requiring a bandwidth of 2.048 MHz (the same as used by the 30 channel PCM system) was in commercial use in France for sometime.

French authorities have also used DM equipment for airborne radio communication and air traffic control over Atlantic via satellite. However, DM has not found wide-spread commercial usage simply because PCM was already there first!

Performance of PCM and ADM versus bit rate

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Communication Systems SIMON HAYKIN Exercise Problems: Problem2: A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The specifications of the modulator are as follows; Sampling rate = 10 fNyquist, where is the Nyquist rate of the speech signal, Step size Δ = 100 mV. The modulator is tested with a 1-kHz sinusoidal signal. Determine the maximum amplitude of this test signal required to avoid slope overload.

Solution: fs = 10 fNyquist =10 (2*3.4K) =68k Hz

dx(t )  < dt Ts

To avoid slope overload distortion max

max

dAm sin(2 f mt )   dt Ts

max 2 f m Am cos(2 f mt )  2 f m Am <

Am  Therefore Am 

 Ts

 Ts

 2 f mTs

100m  68k  = = 1.08 v 2 1k 2 f mTs

Problem3: Consider a test signal m(t) defined by a hyberbolic tangent function m(t)= Atanh(βt) where A and β are constants. Determine the minimum step size Δ for delta modulation of the signal, which is required to avoid slope overload.

Solution: m(t)= Atanh(βt) To avoid slope overload distortion max

Or

dx(t )  < Ts dt

 dx(t ) > max Ts dt

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 > Aβ sech2(βt) Ts  > AβT s since the maximum value of sech(βt) is 1 at t=0

Problem4: Consider a DM system designed to accommodate analog message signals limited to bandwidth W= 5kHz. A sinusoidal test signal of amplitude A=1 volt and frequency fm = 1 kHz is applied to the system. The sampling rate of the system is 50 kHz (a) Calculate the step size required to minimize slope overload distortion. (b) Calculate signal to Quantization noise ratio of the system for the specified sinusoidal test signal.

Solution: (a) To avoid slope overload distortion  > 2 f m Am Ts

Therefore  = 2 f m AmTs = 2 f m Am = fs

2 1k 1v =0.126 v 2  50k

(c) Signal to Quantization noise ratio SNR 

3 f s3 8 2 f BWLPF f

SNR in db = 10

2 m

= 

3  (50k )3 = 475 8 2  5k  (1k ) 2

475 log10 = 26.8 db

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Problem 5: Consider a low pass signal with a bandwidth of 3 kHz. A linear delta modulation system with step size Δ=0.1 v is used to process this signal at a sampling rate ten times the nyquist rate. (a) For linear delta modulation, the maximum amplitude of a sinusoidal test signal of frequency 1 kHz which can be processed by the system without slope-overload distortion. (b)For the specifications given in part a, evaluate the output signal to noise ratio under (i) prefilterd and (ii) postfiltered conditions

Solution: (a) For linear delta modulation, the maximum amplitude of a sinusoidal test signal that can be used without slope overload distortion is Am 

 f s = = 2 f mTs 2 f m

0.110  2  3k =0.95 v 2 1k

(b) (i) Under the pre-filtered condition, it is reasonable to assume that the granular quantization noise is uniformly distributed between – Δ and +Δ. Hence the variance of the quantization noise is 



1 2 E[ ]    f ( )d     d   3   2

2

2

When input signal is sinusoidal signal x(t)  Am sin mt

Am2 0.952 2 2 SNR =  2 = 0.12 = 135= 21.3 db 3 3 (iii)

The signal to noise ratio under the post filtered condition is

3 f s3 SNR  8 2 f BWLPF f

2 m



3  (60k )3 =1367== 31.3 db 8 2  3k  (1k ) 2

The filtering gain in signal to noise ratio due to the use of a reconstruction filter at the demodulator output is therefore 31.3 db- 21.3 db= 10db.

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Problem 6: A linear delta modulator has a step size of 100 mV and the minimum output amplitude is + 50 mV. A signal s(t) = 0.5 u(t) is applied to the input of the delta modulator. Show how the modulator tracks the input indicating the distortions in the waveform. Sketch the waveform for 12 clock cycles, beginning at least 2 clock cycles before t = 0. Also, sketch the output waveform in NRZ format.

Solution: Figure (a) below shows the sketch of the delta modulator input and the tracking distortions.

The input is a step signal of amplitude 0.5 volts beginning at t = 0 as shown by the heavy line. The input for t < 0 is 0 volts. Initial amplitude of the DM predictor, at clock instant 1, is assumed to be + 50 mV. The clock instants are shown in (b).

At the clock instant 2 the predictor output is higher than the input (0 V) and hence, a negative step (-100mV) is added to the predictor output. At clock instant 3 the predictor output is lower (- 50 mV) than the input (0.5 V) and hence, a positive step is added to the Prepared by Venkata Satish N

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predictor. At clock instant 4 the predictor output (+50 mV) is still lower than the input.

Hence, a 100 mV step is added. At clock instants 4, 5, 6, 7, and 8 the predictor output is lower than input and at each instant a 100 mV step is added to the previous predictor out.

At clock instant 9 the predictor output (550 mV) is found higher than the input. Hence, a 100 mV step is subtracted from the predictor output. At clock instant 10 a 100 mV step is added. The DM output waveform is shown in figure (c)

Problem 7: A segment of a delta modulated data stream is a sequence given below.

01010111111000110001 This sequence is applied to a linear delta demodulator having a step size of 100 mV. Assuming initial output of the demodulator is 0 V, show the output sample voltages at each input bit and sketch the waveform.

Solution: The output of demodulator is obtained by adding the step voltage for input 1 and subtracting the step voltage for input 0. The output is shown in the table below for the input bits given above.

Input

0

1 0

1 0

1 1

1

1

1

1

0

0

0

1

1

Output -0.1 0 -0.1 0 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.4 0.3 0.2 0.3 0.4

0

0

0

1

0.3

0.2

0.1

0.2

The waveform of the output is shown in figure below.

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Problem 8:

An instantaneously companded delta modulator employs the following step size adaptation algorithm.

where Sk and Sk-1 are the current and previous step sizes, Bk and Bk-1 are the current and previous output bits, Bk and Bk ď&#x20AC;­1 have opposite polarity. The minimum step size is 100 mV, so the amplitude of the steps when the input is zero is Âą50 mV. If a step input x(t) = 1.2 V is applied to the modulator at t=0 show how the predictor output tracks the input by sketching the waveform. Sketch the binary output waveform of the delta modulator.

Solution: We can show the step size Sk, predictor output Pk and modulator outputs for some clock cycles in a tabular form as below. The input of 1.2 V is applied to the modulator at t = 0 and we start at t = -2.

The waveforms of the predictor output and the modulator output are plotted in the figure below.

Prepared by Venkata Satish N

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Problem 9: A linear delta modulator is used to digitize speech signal band limited to 3.4 kHz. An output filter with 4 kHz cutoff frequency is used. Find the sampling frequency required to get a performance equivalent to that of a 6-bit linear PCM coder. Compare the information rates for PC and DM outputs.

Solution: The S/N obtained from a linear PCM coder is S    6v  1.8  6  6  1.8  37.8db  N max

3 f s3 The S/N obtained from a linear DM coder is SNR  8 2 f BWLPF f

2 m

3 f s3 In db we can write the above equation as SNR dB = 10 log 8 2 f BWLPF f

2 m

To get DM performance equivalent to PCM technique SNR of DM must be equal to SNR of PCM. Prepared by Venkata Satish N

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 3  37.8 = 10log  2   30log f s  10log f BWLPF  20log f m  8   3  37.8 = 10log  2   30log f s  10log  4 103   20log  3.4 103   8 

fs = 191 kb/s Assuming 8kHz sampling the information rate for PCM data is R PCM=8*6=48kb/s For DM the information rate is same as the sampling rate, hence R DM = 191 kb/s Thus, DM requires approximately four times the data rate compared to 6-bit PCM for similar performance.

Problem 10: A stereo music signal is sampled at 44.1 kHz and digitized with 16 bits for recording on CD. If the CD stores 80 minutes of music find the total capacity of the CD in bytes. What is the quality of the music if it has an RMS value 15 dB below the peak value of the quantizer?

Solution: We have fs = 44.1 kHz, n = 16 and number of channels is 2.

Hence, the bit rate R = 2.n.fs = 2 x 16 x 44.1 x 103 = 1411.2 kb/s The capacity of the CD is C = 80 x 60 x 1411.2 kbits = 6773760 kbits = 846.72 Mbytes The signal to noise ratio is S/Nq = 6.n +1.8 -15 = 6 x16 +1.8 -15 = 82.8 dB

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Important Questions 1. a) Consider a test signal m(t)= A tanh(ßt) defined by a hyperbolic tangent function.

Where A and ß are constants. Determine the minimum step size ∆ for delta modulation of this signal, which is required to avoid slope overload. b) Comparison between PCM and DM.

2. a) A signal to transmitted is of the form S(t)=10COS 1000πt+5COS1500πt. i) Choose an appropriate f and step size for delta modulator. ii) Find the SNR for your design. b) Draw the block diagram for Adaptive delta modulation system and explain each block?

3. (a) Explain the noise effects in delta modulation (b) A DM system is designed to operate at 3 times the Nyquist rate for a signal with a 3 KHz BW. The quantization step size is 250mv (i) Determine the maximum amplitude of a 1 KHz sinusoid for which delta modulator does not show slope overload. (ii) Determine post filtered output SNR for the signal at part (i)

4. (a) Derive the condition for step size of the quantizer in a DM system to avoid slope over load distortion for the message signal x(t) = A cos(wt) (b) Explain the major drawback of the DM system with relevant waveforms.

5. A sinusoidal modulating signal is represented by m(t)= A cos (w mt) where wm=2πfm.

Derive the expression for the maximum output signal to quantization noise ratio in the DM system system with no slope overload distortion and also determine maximum output signal to quantization noise ratio at the post receiver?

Prepared by Venkata Satish N

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UNIT 3- DIGITAL MODULATION TECHNIQUES Modulation is defined as the process by which some characteristics of a carrier is varied in accordance with a modulating signal. In digital communications, the modulating signal consists of binary data or an M-ary encoded version of it. The data is used to modulate a carrier wave (usually sinusoidal) with fixed frequency.

The modulation process involves switching or keying the amplitude, frequency or phase of the carrier in accordance with the input data.

Thus there are three basic modulation techniques for the transmission of digital data. They are known as amplitude-shift keying (ASK), frequency shift keying (FSK) and phase shift keying (PSK).

If the amplitude of the carrier is switched depending on the input digital signal, then it is called amplitude shift keying (ASK). This process is quite similar to analog amplitude modulation.

If the frequency of the sinusoidal carrier is switched depending upon the input digital signal, then it is known as the frequency shift keying. This is very much similar to the analog frequency modulation.

If the phase of the carrier is switched depending upon the input digital signal, then it is called phase shift keying. This is similar to phase modulation.

ASK, PSK & FSK waveforms (with sine as carrier signal) Prepared by Venkata Satish N

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Since the phase and frequency modulation has constant amplitude envelope, therefore FSK and PSK, the effect of non-linearities, noise interference is minimum on signal detection. However, these effects are more pronounced on ASK. Therefore FSK and PSK are preferred over ASK. Figure shows the waveforms for amplitude-shift keying, phase shift keying and frequency shift keying. In these waveforms, a single feature of the carrier (i.e. amplitude, phase or frequency) undergoes modulation.

In digital modulations, instead of transmitting one bit at a time, we transmit two or more bits simultaneously. This is known as M-ary transmission. This type of transmission results in reduced channel bandwidth.

However, sometimes, we use two quadrature carriers for modulation. This process is known as Quadrature modulation.

Thus we see that there are a number of modulation schemes available to the designer of a digital communication system required for data transmission over a bandpass channel.

Every scheme offers system trade-offs of its own. In particular choice is made in favour of a scheme which possesses as many of the following design characteristics as possible: (i)

Maximum data rate

(ii)

Minimum probability of symbol error

(iii)

Minimum transmitted power

(iv)

Minimum channel bandwidth

(v)

Maximum resistance to interfering signals.

(vi)

Minimum circuit complexity.

DEFINITIONS AND TERMINOLOGY

There are basically two types of transmission of digital signals i)

BASEBAND DATA TRANSMISSION:

The digital data is transmitted over the channel directly. There is no carrier or any modulation. This is suitable for transmission over short distances. ď&#x201A;ˇ

A signal whose frequency content (i.e. its spectrum) is in the vicinity of zero (i.e., f = 0 or dc) is said to be a baseband signal.

ď&#x201A;ˇ

Original source signal are sometimes said to be baseband. Baseband systems transmit baseband signals.

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This is usually not an effective means of communication.

ii) PASSBAND (BAND PASS OR NARROW BAND) DATA TRANSMISSION: The digital data modulates high frequency sinusoidal carrier. Hence it is also called digital CW modulation. It is suitable for transmission over long distances.

Types of passband Modulation are ASK, PSK, FSK and etc. 

Bandpass signal spectrum is nonzero in some band of frequency with BW = 2B centered about f = ±fc, where fc >> 0.



Effective transmission of signal usually requires bandpass signal.



Bandpass transmission involves some translation of the baseband signal to some band of frequency centered around fc.

Types of Reception for Passband transmission There are two types of methods for detection of passband signals. i) COHERENT (SYNCHRONOUS) DETECTION: In this method, the local carrier generated at the receiver is phase locked with the carrier at the transmitter. Hence it is also called synchronous detection. ii) NON COHERENT (ENVELOPE) DETECTION: In this method, the receiver carrier no need to be phase locked with transmitter carrier. Hence it is also called envelope detection.

Noncoherent detection is simple but it has higher probability of error.

Prepared by Venkata Satish N

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Types of Digital Modulation techniques (Classification based on envelope)

Types of Digital Modulation techniques (Classification based on coherence) Prepared by Venkata Satish N

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DIGITAL MODULATION TECHNIQUES

As mentioned earlier, the binary (i.e. Digital) modulation has three basic forms amplitude-shift keying(ASK), phase-shift keying(PSK) and frequency shift keying (FSK).

BINARY AMPLITUDE SHIFT KEYING (ON-OFF KEYING)

Definition: Amplitude shift keying (ASK) or ON-OFF keying (OOK) is the simplest digital modulation technique. In this method, there is only one unit energy carrier and it is switched on or off depending upon the input binary sequence.

a) Binary Modulating signal and b) BASK signal

a) Modulating signal b) spectrum of „a‟ c) Spectrum of BASK signal Prepared by Venkata Satish N

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EXPRESSION:

s(t )  2Ps cos(2 f ct ) s(t )  0

To transmit symbol „1‟

To transmit symbol „0‟ i.e. no signal is transmitted.

Signal s(t) contains some complete cycles of carrier frequency „fc‟.

Hence ASK waveform looks like an ON-OFF of the signal. Therefore it is also known as the ON-OFF keying (OOK).

SIGNAL SPACE DIAGRAM (CONSTELLATION DIAGRAM) OF BASK

Study of signal spaces provides us with a geometrical method of conceptualizing the modulation process.

The ASK waveform of equation for symbol 1 can be represented as,

s(t )  PT s b

2 cos(2 f ct )  PT s b 1 (t ) Tb

This means that there is only one carrier function 1 (t ) which is a unit energy signal over (0, Tb). The signal space diagram will have two points on 1 (t ) . One will be at zero and other will be at PT s b . The collection of all possible signal points is called the signal constellation.

Thus, the distance between the two signal points is d= PT s b = Eb

The decision boundary is determined by the threshold value λ. If x lies in the region Z1, then a decision of a “1” is made. If x lies in the region Z2, then a decision of a “0” is made.

One advantage in using the signal space representation is that it is much easier to identify the “distance” between signal points. The distance between two signal points will be increased which makes the received signal point less probable be located in the wrong region.

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GENERATION OF BASK SIGNAL

ASK signal may be generated by simply applying the incoming binary data and the sinusoidal carrier to the two inputs of a product modulator. The resulting output will be the ASK waveform.

Generation of BASK signal

BASK RECEPTION: COHERENT DETECTION OR DEMODULATION OF BINARY ASK SIGNAL

The demodulation of BASK waveform can be achieved with the help of coherent detector as shown in figure.

It consists of a product modulator which is followed by an integrator and Decision making device. The incoming ASK signal is applied to one input of the product modulator. The other input of the product modulator is supplied with a sinusoidal carrier which is generated with the help of a local oscillator.

The output of the product modulator goes to input of the integrator. The integrator operates on the output of the multiplier for successive bit intervals and essentially performs a low-pass filtering action. The output of the integrator goes to the input of a decision making device. .

Coherent detection of Binary ASK signal Prepared by Venkata Satish N

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Now, the decision making device compares the output of the integrator with a preset threshold. It makes a decision in favour of symbol 1 when the threshold is exceeded and in favour of symbol 0 otherwise.

In this method we assumed that the local carrier is in perfect synchronization with the carriers used in the transmitter. This means that the frequency and phase of the locally generated carrier is same as those of the carriers used in the transmitter NON-COHERENT DETECTION OR DEMODULATION OF BINARY ASK SIGNAL

Figure shows the block diagram of noncoherent ASK receiver. In this figure the received ASK signal is given to the band pass filter. This band pass filter passes only carrier frequency, fo or fc.

The envelope detector generates high output voltage when carrier is present.

When carrier is absent, there is only noise at the input of envelope detector. Hence it produces low output.

Non-Coherent Detection or Demodulation of Binary ASK Signal

The decision device is basically a regenerator. It generates the binary sequence b(t). Threshold is provided to the decision device to overcome effects due to noise. When y(t) is greater than threshold, b(t)= 1 and when y(t) is less than threshold, b(t)=0. Non coherent reception of ASK does not need any carrier synchronization.

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Power Spectral Density The ASK signal, which is basically the product of the binary sequence and the carrier signal, has a power spectral density same as that of the baseband on-off signal but shifted in the frequency domain by Âąfc.

Power Spectral density of ASK Signal (One Sided)

Bandwidth of BASK The Spectrum of the ASK signal shows that it has an infinite bandwidth. However for practical purpose, the bandwidth is often defined as the range of frequency over which ASK contains about 95% of the total average power content.

If suppose the center lobe of PSD contains 95% of the total power, then Bandwidth (Null to Null) is given as 2Rb. B= 2Rb =2/Tb Advantages: The advantage of using BASK is its simplicity. It is easy to generate and detect.

Drawback: It is very sensitive to noise. Therefore it finds limited application in data transmission. It is used at low bit rates, upto 100 bits per sec.

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BINARY PHASE SHIFT KEYING (BPSK)

Binary Phase Shift Keying is the most efficient of the three digital modulation schemes. In Binary PSK, phase of the sinusoidal carrier is changed according to the data bit to be transmitted. Also, a polar NRZ signal is used to represent the digital data coming from the digital source.

Mathematically BPSK signal is expressed as s(t )  b(t ) Ac cos(2 f ct ) Where b (t) is represented in bipolar NRZ format.

For Binary 1, b (t) = +1v then s(t )   Ac cos(2 f ct ) For Binary 0, b (t) = -1v

then s(t )   Ac cos(2 f ct ) = Ac cos(2 fct  1800 )

BPSK Waveforms (cosine as carrier) Prepared by Venkata Satish N

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Spectrum of BPSK signals Input b(t) is a NRZ binary waveform. In this waveform, there are rectangular pulses of amplitude ±Ab. If we assume that each pulse is 

Tb around its centre, then it becomes 2

easy to find the fourier transform of such pulse. The fourier transform of this pulse is given as X ( f )  AbTb

sin( fTb )  fTb

a) Modulating signal b) spectrum of „a‟ c) Spectrum of BPSK signal

For a large number of such positive and negative pulses, the power spectral density is expressed as X avg ( f ) 

X(f ) Ts

2

 Ab 2Tb

sin 2 ( fTb ) sin 2 ( fTb ) PT = b ( fTb )2 ( fTb )2

Where T s is symbol duration. In this case T s = Tb. The above equation gives the power spectral density of baseband signal b (t). Due to modulation of the carrier of frequency fc, the spectral components are translated from f to fc+f and fc-f. The magnitude of these components is divided by half.

Therefore the power spectral density of BPSK signal is given by

PTb 2

2

 sin( ( f  f c )Tb )  PTb  sin( ( f  f c )Tb )       2   ( f  f c )Tb    ( f  f c )Tb 

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2

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Generation of BPSK signal BPSK signal may be generated by applying carrier signal to a product modulator.

The binary data signal b (t) is converted into a NRZ bipolar signal by an encoder. Here the bipolar signal b (t) is applied as modulating signal to the Product modulator. The other input to the product modulator is supplied with a carrier signal.

RECEPTION OF BPSK SIGNAL-COHERENT DETECTION

The signal undergoes the phase change depending upon the time delay from transmitter end to receiver end. This phase change is, usually, a fixed phase shift in the transmitted signal.

Let us consider that this phase shift is θ. Because of this, the signal at the input of the receiver can be written as s(t )  b(t ) 2Ps cos(2 f ct   ) . Now, from this received signal, a carrier is separated because this is coherent detection. The received signal is allowed to pass through a square law device followed by a band pass filter and frequency divider. Thus at the output of frequency divider, we get a carrier signal whose frequency is fc i.e. cos(2 fct   ) .

The synchronous demodulator multiplies the input signal and the recovered carrier. Hence at the output of multiplier, we get b(t ) cos2 (2 fct   ) =

b(t ) {1  cos(4 f ct  2 )} 2

The output of multiplier contains a dc term

b(t ) and a high frequency signal of 2

frequency 2fc Prepared by Venkata Satish N

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Block Diagram of BPSK Receiver

The signal is then applied to the bit synchronizer and integrator. The integrator integrates the signal over one bit period and it essentially performs low pass filtering and it produces a voltage proportional to b(t).

Tb

i.e.

 0

b(t ) b(t ) Tb {1  cos(4 f ct  2 )}dt = 2 2

The output of integrator goes through the decision making device. The decision making device compares the output of integrator with the preset threshold. It makes a decision in the favor of symbol 1 when the threshold is exceeded and in favor of symbol 0 otherwise.

The bit synchronizer takes care of starting and ending times of the bit. At the end of the bit duration Tb, the bit synchronizer closes switch S2 temporarily. This connects the output of an integrator to the decision making device. The synchronizer then opens the switch S2 and switch S1 is closed temporarily. This resets the integrator voltage to zero.

The integrator then integrates next bit.

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SIGNAL SPACE DIAGRAM OF BPSK SIGNAL

We know that BPSK signal carries the information about two symbols. These symbols are symbol 1 and symbol 0.

BPSK signal is given by s(t )  b(t ) Ac cos(wct )

 b(t ) 2Ps cos(wct )  b(t ) PT s b

2 cos( wct ) Tb

 b(t ) PT s b 1 (t ) Where 1 (t ) =

2 cos( wc t ) represents a unit energy signal over (0, Tb).

Tb

If b (t) is binary 1 (+1 v), then s(t )   PT s b 1 (t ) If b (t) is binary 0 (-1 v), then s(t )   PT s b 1 (t ) Thus on the single axis of 1 (t ) , there will be two points. One point will be located at

 PT s b and other will be located at  PT s b . Thus it has been shown in figure below.

Geometrical representation of BPSK signal

At the receiver end, the point at  PT s b on 1 (t ) represents symbol 1 and the point at

 PT s b on 1 (t ) represents symbol 0. The separation between these two points represents the isolation in symbols „1‟ and „0‟ in BPSK signal. This separation is generally called

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Euclidean distance d. As the distance increases, the isolation between the symbols in BPSk signal is more. Thus probability of error reduces.

Bandwidth for BPSK signal

BW= Highest frequency – Lowest frequency in the main lobe  f c  fb  ( f c  fb )  2 fb

Hence the minimum bandwidth of BPSK signal is equal to twice the highest frequency contained in baseband signal.

Salient Features of BPSK 1. BPSK has a bandwidth which is lower than that of a BFSK signal.

2. BPSK has the best performance of all the three digital modulation techniques in presence of noise. It yields the minimum value of probability of error.

3. Binary PSK has very good noise immunity.

Drawbacks 1. Providing synchronization in the reception of BPSK is very difficult.

2. Discontinuity of PSK signal creates high frequencies as side effects.

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BINARY FREQUENCY SHIFT KEYING

In BFSK, the frequency of the carrier is shifted according to the binary symbol. In other words, the frequency of a sinusoidal carrier is shifted between two discrete values. This means that we have two different frequency signals according to binary symbols. However the phase of the carrier is unaffected.

If b (t) =1 then sH(t) =

2 Ps cos(2 f H t )  2 Ps cos(2 ( f c 

 )t ) 2

If b (t) =0 then sL(t) =

2 Ps cos(2 f Lt )  2 Ps cos(2 ( f c 

 )t ) 2

These equations combinely may be written as s(t )  2 Ps cos(2 ( f c  d (t)

 ) t) .

2

Where d(t) = + 1 for binary 1 and d(t)=-1 for binary 0.

GENERATION OF BFSK

It may be observed from table that PH(t) is same as b(t) and also PL(t) is inverted version of b(t).

Input b(t)

d(t)

PH(t)

PL(t)

1

+1 v

+1 v

0v

0

-1 v

0 v

+1v

Conversion table used in the generation of BFSK Prepared by Venkata Satish N

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The block diagram of BFSK generation is shown in figure below.

We know that input sequence b(t) is same as PH(t). An inverter is added after b(t) to get PL(t). Then these PH(t) & PL(t) signals are applied to the product modulators separately and carriers of frequencies fH & fL are also applied to product modulators respectively. The outputs of multipliers are then applied to the adder. The adder produces the BFSK signal.

BFSK Modulator

From the above block diagram, the expression for BFSK signal is given by

S (t )  2Ps PH (t ) cos(2 f H t )  2Ps PL (t ) cos(2 f Lt ) When b(t) =1; PH(t)=1 & PL(t)=0; S (t )  2Ps cos(2 f H t ) When b(t)=0; ; PL(t)=1 & PH(t)=0; S (t )  2Ps cos(2 f Lt )

BFSK signal Prepared by Venkata Satish N

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COHERENT DETECTION OF BINARY FSK

The expression of BFSK is given by

S (t )  2Ps PH (t ) cos(2 f H t )  2Ps PL (t ) cos(2 f Lt ) The detectors consists of two correlators that are individually tuned to two different carrier frequencies to represent symbols 1 and 0. A correlator consists of a multiplier followed by an integrator. Then the received binary FSK signal is applied to the multipliers of both the correlators. To the other input of the multipliers, carriers with frequency fH & fL are applied as shown in figure below. The multiplied output of each multiplier is subsequently passed through integrators generating outputs X 1 and X2 in the two paths.

Coherent Detection of BFSK The outputs of the two integrators are then fed to the decision making device. The decision making device is essentially a comparator which compares the output X1 and output X2.

If X1 > X2 then decision making device makes a decision in the favour of symbol 1.

If X1 < X2, then decision making device makes a decision in the favour of symbol 0.

Note: In this coherent detection process, it is required to generate synchronized (coherent) carriers of frequencies fL & fH. Note: Upper correlator recovers PH (t ) and lower correlator recovers PL (t ) . When upper correlator recovers PH (t ) , lower correlator produces null output & vice versa.

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NON-COHERENT DETECTION OF BINARY FSK

Binary FSK waves may be demodulated non-coherently using envelope detector.

The received FSK signal is applied to a bank of two band pass filters, one tuned to frequency fH & other tuned to fL. Each filter is followed by an envelope detector. The resulting outputs of the two envelope detectors are sampled and the compared with each other.

Non-Coherent Detection of Binary FSK

A decision is made in favour of symbol 1 if the envelope detector output derived from the filter tuned to frequency fH is larger than that derived from the second filter.

Otherwise, a decision is made in favour of the symbol 0.

SIGNAL SPACE DIAGRAM OF BFSK SIGNAL

The BFSK signal is given S (t )  2Ps PH (t ) cos(2 f H t )  2Ps PL (t ) cos(2 f Lt )

S (t )  PT s b PH (t )

2 2 cos(2 f H t )  PT cos(2 f Lt ) s b PL (t ) Tb Tb

S (t )  PT s b PH (t )1 (t )  PT s b PL (t )2 (t ) Prepared by Venkata Satish N

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Where 1 (t ) & 2 (t ) are orthogonal carriers (also unit energy carriers) over the period Tb. Because in one bit interval of the input signal, 1 (t ) & 2 (t ) have integral number of cycles. i.e f H  mfb & f L  nfb .

If the carriers are orthogonal then the distance dmin is maximum. As probability of error depends on dmin, maximizing the distance dmin decreases the error rate in BFSK modulation scheme.

When symbol 1 is transmitted, modulated carrier BFSK has level of symbol 0 is transmitted, modulated carrier BFSK has level of

PT and when s b

PT s b .

Note that there are two signal points in the signal space. The distance between these two points may be evaluated as under: 2 2 d 2  ( PT s b )  ( PT s b)

d  2 PT s b  1.414 PT s b

Signal Space diagram (Constellation) of BFSK signal

Note: dmin = dmin =2

PT s b in Binary ASK PT s b in Binary PSK

dmin = 1.414 PT s b in Binary FSK

 d min ( BPSK )  d min ( BFSK )  d min ( BASK )  Pe ( BPSK )  Pe ( BFSK )  Pe ( BASK ) Prepared by Venkata Satish N

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SPECTRUM OF BFSK SIGNALS

The BFSK signal is given S (t )  2Ps PH (t ) cos(2 f H t )  2Ps PL (t ) cos(2 f Lt ) . Let us compare this equation with BPSK equation which is written below: SBPSK=b(t) cos (2πfct) It may be noted that this equation is identical to BFSK equation. In BPSK equation, b(t) is a polar signal where in BFSK, the similar coefficients P H(t) or PL(t) are unipolar. Hence let us convert these coefficients in bipolar form as under:

1 1 PH(t) =  PH 1 (t ) 2 2 and PL (t ) 

1 1 1  PL (t ) 2 2

Where PH 1 (t ) & PL1 (t ) will be bipolar (i.e., +1 or -1).

Substituting these values in the above equation, we obtain

1 1 1 1 S (t )  2 Ps [  P I H (t )]cos(2 f H t )  2 Ps [  P I L (t )]cos(2 f Lt ) 2 2 2 2 S (t ) 

Ps P P P cos(2 f H t )  s P I H (t ) cos(2 f H t )  s cos(2 f Lt )  s P I L (t ) cos(2 f Lt ) 2 2 2 2

Term

Fourier Transform s( f )

Power Spectral density

s( f )

2

cos(2 f H t )

 ( f  fH )

 ( f  fH )

P I H (t ) cos(2 f H t )

Sa( ( f  f H )Tb )

Sa 2 ( ( f  f H )Tb )

cos(2 f Lt )

 ( f  fL )

 ( f  fL )

P I L (t ) cos(2 f Lt )

Sa( ( f  f L )Tb )

Sa 2 ( ( f  f L )Tb )

Power spectral density of BFSK signal consists of impulses located at fL & fH and two main lobes (sampling functions) located symmetrically about fL & fH of widths 2fb.

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Power Spectral Density of BFSK signal Here the values fL & fH were chosen such that the bandwidth required by BFSK is minimum. i.e f H ď&#x20AC;­ f L ď&#x20AC;˝ 2 fb BANDWIDTH OF BFSK SIGNAL

Width of one lobe is 2fb. the two main lobes due to fH & fL are placed such that the total width due to both main lobes is 4 fb. Therefore, we have Bandwidth of BFSK= 2 fb + 2 fb = 4 fb if we compare this bandwidth with that of BPSK, we note that BW (BFSK) = 2*BW (BPSK)

SALIENT FEATURES OF BFSK

i.

BFSK is relatively easy to implement

ii.

It has better noise immunity than ASK. Hence the probability of error free reception of data is high.

DRAWBACK OF BFSK

The major drawback is its high bandwidth requirement. Therefore, FSK is extensively used in low speed modems having bit rates below 1200 bits/sec.

Note: As there are no abrupt phase changes in BFSK, side lobe levels are minimum in the spectrum of BFSK signal. Hence bandpass filtering is not required to remove side lobes.

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COMPARISON BETWEEN BASK, BPSK & BFSK

Parameter of

Binary ASK

Binary FSK

Binary PSK

Comparison Variable Characteristic

Amplitude

Frequency

Phase

Bandwidth

2 fb

4 fb

2 fb

Noise immunity

Low

Medium

High

Probability of error

High

Medium

Low

Performance in

Poor

Better than ASK

Best of three

System Complexity

Simple

Moderately Complex

Very Complex

Bit rate

Suitable upto

Suitable upto about

Suitable for

100 bits/sec

1200 bits/sec

high bit rates

Envelope (Non

Envelope(Non

Coherent

coherent )Detection

coherent )detection

detection

1.414 PT s b

2 PT s b

presence of noise

Demodulation method

Distance b/w signal

PT s b

space points dmin

From the comparison table we can conclude that BPSK offers more advantages than other modulation schemes. One difficulty lies in BPSK scheme is requirement of coherent detector. Due to this, system complexity increases.

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DIFFERENTIAL PHASE SHIFT KEYING DPSK does not need a synchronous carrier at the demodulator. The input sequence of binary bits is modified such that the next bit depends upon the previous bit.

Therefore, in the receiver, the previous received bits are used to detect the present bit.

GENERATION OF DPSK

Thus in order to eliminate the need for phase synchronization of coherent receiver with PSK, a differential encoding system can be used with PSK. The digital information content of the binary data is encoded in terms of signal transitions.

A symbol 0 is used to represent transition in a given binary sequence and a symbol 1 is used to indicate no transition.

This new signaling scheme which combines differential encoding with phase shift keying (PSK) is known as differential phase shift keying.

Schematic Diagram

Block Diagram of DPSK Transmitter

The data stream b (t) is applied to the input of the encoder. The output of the encoder (differential data) is applied to one input of the product modulator. To other input of this product modulator, a sinusoidal carrier of fixed amplitude and frequency is applied.

The relation between the binary sequence and its differential encoded version is illustrated in the following table for a assumed data sequence 0 0 1 0 0 1 0 0 1 1.

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Encoding has been done in such a way that transition in the given binary sequence with respect to the previous encoded bit is represented by symbol 0 and no transition by symbol 1. i.e., the logic function performed by encoder is XNOR logic function.

dk  bk  dk 1 It may be noted that an extra bit has been arbitrarily added as an initial bit. We can choose 0 or 1 as an initial bit for the encoded sequence. The phase of the generated DPSK signal has been shown in the third row of tables below.

Binary data bk

0

0

1

0

0

1

0

0

1

1

Differential encoded data d k

1*

0

1

1

0

1

1

0

1

1

1

Phase of DPSK

0

π

0

0

π

0

0

π

0

0

0

Phase of shifted DPSK signal

0

π

0

0

π

0

0

π

0

0

Phase comparison output

-

-

+

-

-

+

-

-

+

+

Detected binary sequence

0

0

1

0

0

1

0

0

1

1

0

0

1

0

0

1

0

0

1

1



Arbitrary starting reference bit 1

Binary data bk Differential encoded data d k

0*

1

0

0

1

0

0

1

0

0

0

Phase of DPSK

π

0

π

π

0

π

π

0

π

π

π

Phase of shifted DPSK signal

π

0

π

π

0

π

π

0

π

π

Phase comparison output

-

-

+

-

-

+

-

-

+

+

Detected binary sequence

0

0

1

0

0

1

0

0

1

1



Arbitrary starting reference bit 0

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DETECTION OF DPSK

For the detection of the differentially encoded PSK, we can use the receiver arrangement as shown in figure below.

Block Diagram of DPSK Receiver

The received DPSK signal is applied to one input of the multiplier. To other input of the multiplier, a delayed version of the received DPSK signal by the time interval T b is applied (it has been shown in 4th row of the table).

The output of the difference is proportional to cos  , here  is the difference between the carrier phase angle of the received DPSK signal and its delayed version, measured in the same bit interval. The phase angles of the DPSK signal and its delayed version have been shown in 3rd and 4th rows respectively.

When Φ=0, the integrator output is positive whereas when Φ=π, the integrator output is negative. By comparing the integrator output with a decision level of zero volt.

The decision can reconstruct the binary sequence by assigning a symbol 0 for negative output and symbol 1 for positive output. The reconstructed data is shown in the last row of the table.

Reconstruction is invariant with the choice of the initial bit in the encoded data.

Bandwidth of DPSK signal Bandwidth calculation is made by using power spectral density of it. The power spectral density of DPSK signal is same as that of BPSK signal. The only difference b/w them is the symbol duration. The symbol duration of BPSK scheme is T b where as for DPSK scheme symbol duration is given by 2T b. As the generation of each symbol in differential data depends on the present bit and previous encoded bit, the symbol duration is equivalent to Ts = 2Tb. Prepared by Venkata Satish N

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BW=

ECE Department

2 2 = ď&#x20AC;˝ fb Ts 2Tb

Hence minimum bandwidth in DPSK is equal to fb. i.e. maximum baseband signal frequency.

ADVANTAGES

i.

DPSK does not need carrier at the receiver end. This means that the complicated circuitry for generation of local carrier is not required.

ii.

The bandwidth requirement of DPSK is reduced as compared to that of BPSK.

DRAWBACKS

Because DPSK uses two successive bits for its reception, error in the first bit creates error in the second bit. Therefore, error propagation in DPSK is more. On other hand, in BPSK single bit can go in error since detection of each bit is independent. i.

The probability of error of DPSK is higher than that of BPSK.

ii.

Noise interference in DPSK is more.

Note: In DPSK, previous bit is used to detect next bit. Hence, if error is present in previous bit, detection of next bit can also go wrong. Hence error is created in next bit also. Therefore, the tendency of appearing errors in pairs in DPSK. Note: Geometrical representation of DPSK signal is same as that of BPSK.

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QUADRATURE PHASE SHIFT KEYING

In communication systems, we have two main resources. There are the transmission power and the channel bandwidth. The channel bandwidth depends upon the bit rate or signaling rate fb. if two or more bits are combined in some symbols, then the signaling rate will be reduced. This reduces the transmission channel bandwidth. Hence because of grouping of bits in symbols, the transmission channel bandwidth can be reduced. In quadrature phase shift keying (QPSK), two successive bits in the data sequence are grouped together. This reduces the bit rate or signaling rate and thus reduces the bandwidth of the channel.

In case of BPSK, we know that when symbol changes the level, the phase of the carrier is changed by 1800. Because, there were only two symbols in BPSK, the phase shift occurs in two levels only.

However, in QPSK, two successive bits are combined. Infact, this combination of two bits forms four distinct symbols. When symbol is changed to next symbol, then the phase of the carrier is changed by 900.

GENERATION OF QPSK (OFFSET QPSK)

Figure below shows the block diagram of QPSK transmitter. 

Input binary sequence is first converted to a bipolar NRZ type of signal. This signal is denoted by b (t). It represents binary „1‟ by +1 V and binary „0‟ by -1 V.



In QPSK, we parallelize the bit stream so that every two incoming bits are split up. The demultiplexer divides b (t) into two separate bit streams of the odd numbered bits bo (t) and even numbered bits be (t) so that two successive bits of b (t) can be applied to the modulator simultaneously. The symbol duration of both of these odd and even numbered sequences is 2T b.



It may be observed that first even bit occurs after the first odd bit. Hence even numbered bit sequence be (t) starts with the delay of one bit period due to the first odd bit. This delay of Tb is known as offset. This shows that the change in levels of be (t) and bo(t) cannot occur at the same time due to offset.

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Block Diagram of QPSK Transmitter



The bit stream bo (t) modulates the carrier Ps cos(2 f ct ) and be (t) modulates the carrier Ps sin(2 f ct ) . These carriers are also known as quadrature carriers.



The two modulated signals can be written as, Se (t) = be (t ) Ps sin(2 f ct ) So (t) = bo (t ) Ps cos(2 f ct ) Hence Se (t) & So (t) are basically BPSK signals. The only difference is that T=2Tb here.



The output of the adder is QPSK signal and it is given by S (t) =So (t) +Se (t) s(t )  bo (t ) Ps cos(2 f ct )  be (t ) Ps sin(2 f ct )

Case 1: if bo (t )  1(1V ) & be (t )  0( 1V ) s(t )  1 Ps cos(2 f ct )  (1)  Ps sin(2 f ct )

s (t )  2 Ps cos(2 f ct ) 

1 1  2 Ps sin(2 f ct )  2 2

s (t )  2 Ps cos(2 f ct )  cos(450 )  2 Ps sin(2 f ct )  sin(450 ) s (t )  2 Ps cos(2 f ct  450 )

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Case 2: if bo (t )  0(1V ) & be (t )  0( 1V ) s(t )  1 Ps cos(2 f ct )  (1)  Ps sin(2 f ct )

1 1  2 Ps sin(2 f ct )  2 2 3 3 s (t )  2 Ps cos(2 f ct )  cos( )  2 Ps sin(2 f ct )  sin( ) 4 4 3 s (t )  2 Ps cos(2 f ct  ) 4 s (t )  2 Ps cos(2 f ct ) 

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Case 3: if bo (t )  0(1V ) & be (t )  1( 1V ) s(t )  1 Ps cos(2 f ct )  (1)  Ps sin(2 f ct )

1 1  2 Ps sin(2 f ct )  2 2 5 5 s (t )  2 Ps cos(2 f ct )  cos( )  2 Ps sin(2 f ct )  sin( ) 4 4 5 s (t )  2 Ps cos(2 f ct  ) 4 s (t )  2 Ps cos(2 f ct ) 

Case 4: if bo (t )  1(1V ) & be (t )  1( 1V ) s(t )  1 Ps cos(2 f ct )  1 Ps sin(2 f ct )

1 1  2 Ps sin(2 f ct )  2 2 7 7 s (t )  2 Ps cos(2 f ct )  cos( )  2 Ps sin(2 f ct )  sin( ) 4 4 7 s (t )  2 Ps cos(2 f ct  ) 4 s (t )  2 Ps cos(2 f ct ) 

Symbol

Phase of the carrier

10

450

00

1350

01

2250

11

3150

Thus the phase shift in QPSK signal for symbol to symbol change is 900.

DETECTION OF QPSK



This is synchronous reception. Hence the coherent carrier is to be recovered from the received signal S (t). The received signal S (t) is first raised to its 4 th power, i.e., S4(t). After that, it is allowed to pass through a bandpass filter (BPF) which is centered around 4fc. The output of the bandpass filter is a coherent carrier of frequency 4fc. This is divided by 4 and it provides two coherent quadrature carriers. i.e., cos(2 fct ) & sin(2 f ct) .



These coherent carriers are applied to two synchronous demodulators. These synchronous demodulators consist of multiplier and an integrator.

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

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The incoming signal is applied to both the multipliers. Here, the integrator integrates the product signal over two bit interval. The upper correlator recovers the even ordered sequence and lower correlator recovers the odd ordered sequence as follows.

Now, let us consider the product signal at the output of the upper multiplier, i.e.,

s(t )sin(2 fct )  {bo (t ) Ps cos(2 f ct )  be (t ) Ps sin(2 f ct )}sin(2 f ct ) This signal is integrated by the upper integrator over 2T b interval.

Therefore, we have (2 k  2)Tb



s(t )sin(2 f ct )dt 

2 kTb



(2 k  2)Tb



{bo (t ) Ps cos(2 f ct )  be (t ) Ps sin(2 f ct )}sin(2 f ct )dt

2 kTb

(2 k  2)Tb



bo (t ) Ps cos(2 f ct )sin(2 f ct )dt 

2 kTb



2 kTb

 0



be (t ) Ps sin 2 (2 f ct )dt

2 kTb

(2 k  2)Tb



(2 k  2)Tb

1 bo (t ) Ps sin(4 f ct )dt  2

(2 k  2)Tb



2 kTb

(2 k  2)Tb



2 kTb

1 be (t ) Ps {1  cos(4 f ct )}dt 2

1 be (t ) Ps dt  be (t ) Ps Tb 2

Similarly odd sequence is recovered from lower multiplier and integrator combination.

These odd and even sequences are combined by multiplexer to generate binary sequence.

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SIGNAL SPACE DIAGRAM ( Phasor representation) OF QPSK SIGNAL

Symbol

If

QPSK signal

QPSK signal interms of unit energy signal

10

s(t )  2Ps cos(2 f ct  450 )

00

s(t )  2 Ps cos(2 f ct 

01

11

s(t )  PT s s{

2 cos(2 f ct  450 )} Ts

3 ) 4

s(t )  PT s s{

2 cos(2 f ct  1350 )} Ts

s(t )  2 Ps cos(2 f ct 

5 ) 4

s(t )  PT s s{

2 cos(2 f ct  2250 )} Ts

s(t )  2 Ps cos(2 f ct 

7 ) 4

s(t )  PT s s{

2 cos(2 f ct  3150 )} Ts

2 cos(2 f c t ) (unit energy carrier) is represented by phasor  (t ) , then QPSK signal Ts

can be geometrically represented as shown below.

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Minimum distance b/w signal space points dmin

2 2 ( PT s s )  ( PT s s)

From the above signal space diagram d min =

2 PT s s

= For QPSK, Ts = 2Tb

dmin =

2 Ps  2Tb = 2

PT s b

Thus dmin of QPSK is same as that of dmin of BPSK. It shows that noise immunities of BPSK and QPSK are same.

SPECTRUM OF QPSK SIGNAL

In BPSK, the input sequence (Bipolar NRZ ) is of bit duration T b. Power spectral density of such a waveform can be given as s(f)= S (t )  Vb 2Tb [  PT s b[

sin( fTb ) 2 ]  fTb

sin( fTb ) 2 ]  fTb

Where Ps is the power of the signal over (0, T b) In QPSK, the input sequences (Bipolar NRZ) of be (t) & bo (t) are of bit duration 2Tb. Then spectrum of QPSK (shape) is similar to that of BPSK except in the width of baseband signal spectrum (that is due to symbol duration).

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The power spectral densities of be (t) & bo (t) are given by Se (t )  PT s s[

sin( fTs ) 2 ]  fTs

So (t )  PT s s[

sin( fTs ) 2 ]  fTs

The baseband power spectral density of QPSK signal equals the sum of the individual power spectral densities of be (t) & bo (t) i.e., S H (t )  2 PT s s[

sin( fTs ) 2 ]  fTs

Bandwidth for QPSK signal

Base band Power spectral density of QPSK signal BW= Highest frequency – Lowest frequency in the main lobe  fc  

1 1  ( fc  ) Ts Ts

2 2   fb Ts 2Tb

Hence the bandwidth of QPSK signal is equal to bit rate fb, where as in BPSK, BW is 2fb.

Advantages of QPSK i.

For the same bit error rate, the bandwidth required by QPSK is reduced to half as compared to BPSK.

ii.

Noise immunity is more

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iii.

Probability of error is less than BASK & BFSK signaling schemes

iv.

Noise interference is less

Disadvantages of QPSK i.

Providing synchronization in the reception of QPSK is very difficult.

ii. Discontinuity of PSK signal creates high frequencies as side effects. iii. An abrupt phase change in QPSK signal creates significant side lobes. Hence band pass filtering is required to avoid inter channel interference. The solution to this problem is MINIMUM SHIFT KEYING. M-ary Transmission In digital modulations, instead of transmitting one bit at a time, we transmit two or more bits simultaneously. This is known as M-ary transmission. This type of transmission results in reduced channel bandwidth.

COMPARISON of BPSK, QPSK & M-ary PSK

Parameter of

BPSK

QPSK

M-ary PSK

No of symbols

2 (symbol 0 &1 )

4 (Symbols 00, 01, 10,11)

M-Symbols

No of

1 bit

2 bits

N bits

the carrier

2  2

2   4 2

2 (M=2N) M

Symbol

Ts =Tb

Ts =2Tb

Ts=NTb

Comparison

bits/symbol Phase shift in

Duration

(2m  1) ) 4

Expression

s(t )  b(t ) Ac cos(2 f ct )

Bandwidth

2fb

fb

2fb/N

dmin

2 psTb

2 psTb

2 PT s s sin(

s(t )  2 Ps cos(2 f ct 

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s(t )  2 Ps cos(2 f ct 

 M

)

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M-ary PSK Transmitter



The serial to parallel converter forms a symbol of N successive bits. That is the output of serial to parallel converter is „N‟bit word.



The digital to analog converter output remains unchanged till last N th bit received.

Then depending upon the input „N‟ bits, the output of D/A converter is defined.

This output of D/A converter remains unchanged till last bit is received.



The voltage m (t) is applied to modulator. This modulator modulates the phase of sinusoidal carrier depending upon the amplitude of symbol m (t).

M-ary PSK Receiver



M-ary PSK signal given to coherent detectors. Each Coherent detector consists of multiplier followed by a integrator. According to the phase of M-ary PSK signal in each symbol duration, phase detector recovers the analog signal.



This analog signal is converted to digital signal by applying through A/D converter. The analog to digital converter output is the N-bit symbol in parallel.



These bits are then converted to serial bit stream by parallel to serial converter.

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Power Spectral density of M-ary PSK The power spectral density of M-ary PSK is given by S ( f )  2 PT s s[

sin( fTs ) 2 ]  fTs

Where Ts = NTb.

BW= Highest frequency – Lowest frequency in the main lobe =

2f 2 2   b Ts NTb N

Signal Space diagram or geometrical representation of M-ary PSK signals Figure shows the signal space diagram of M-ary PSK. All signal space points lies on a circle of radius

P sTb and any two successive signal space points phase differed by

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2 .

M

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For M=8, the signal space diagram for M-ary PSK is shown below. In the figure, the distance between signal point S1 and signal point S2 can be obtained by considering the triangle followed by S1OA. The distance between S1 and S2 is denoted by d12.

Signal Space representation of 8-PSK scheme Distance S1A = Distance S2A =

d min 2

By standard relation of right angle triangle sin(

 M

)

Dis tan ceS1 A Dis tan ceOS1

d min sin( )  2 M PT s s



i.e. d min  2 PT s s sin(

 M

) where Ts = NTb & M =2N

Let us verify the result for M =4, i.e 4-PSK scheme. d min  2 PT s s sin(

 M

)



d min  2 Ps 2Tb sin( ) 4 = 2 PT s b Let us find the result for M =8, i.e 8-PSK scheme d min  2 PT s s sin(

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



)  2 Ps 3Tb sin( ) M 8

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Comparison b/w BFSK and M-ary FSK Parameter of Comparison

BFSK

M-ary FSK

No of symbols

2 (symbol 0 &1 )

M-Symbols

No of bits/symbol

1 bit

N bits

Symbol Duration

Ts =Tb

Ts=NTb

Output frequencies

fL & fH

f0, f1,f2,………fM-1

Expression

S (t )  2Ps cos(2 f mt )

S (t )  2Ps cos(2 f mt )

For m=0,1

m=0, 1,……..M-1

4fb if fH-fL=2fb

2 fs M  2

Bandwidth

1 N 1 N 2 2 2 Ts NTb

M-ary FSK Transmitter

M-ary FSK transmitter



The „N‟ successive bits are presented in parallel to digital to analog converter.

These „N‟ bits form a symbol at the output of digital to analog converter. There will be a total 2N =M possible symbols. The symbol is presented every T s =NTb period.

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The output of digital to analog converter is given to a frequency modulator. Thus depending upon the value of symbol, the frequency modulator generates the output frequency.



For every symbol, the frequency modulator produces different frequency output.

This particular frequency signal remains at the output for one symbol duration.



Thus for M symbols, there are M frequency signals at the output of modulator.

Thus the transmitted frequencies are f0, f1, f2,…….& fM-1.

M-ary FSK Receiver 

The M-ary FSK signal is given to the set of „M‟ bandpass filters. The center frequencies of those filters are f0, f1,f2, …….fM-1.

These filters pass their

particular frequency and alternate others. 

The envelope detector outputs are applied to a decision device. The decision device produces its outputs are applied to a decision device. The decision device produces its output depending upon the particular symbol; only one envelope detector will have higher output. The outputs of other detectors will be very low.



The output of the decision device is given to „N‟ bit analog to digital converter.

The analog to digital converter output is the „N‟ bit symbol in parallel. These bits are then converted to serial bit stream by parallel to serial converter.

M-ary FSK Receiver

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POWER SPECTRAL DENSITY OF M-ARY FSK

We know that for M symbol f0, f1,f2, …….fM-1 frequencies are used for transmission. The probability of error is minimized by selecting those frequencies such that transmitted signals are mutually orthogonal. If those frequencies are selected as successive even harmonics of symbol frequency fs, then transmitted signals will be orthogonal.

Let‟s say that the lowest carrier frequency f0 is the kth harmonic of symbol frequency i.e., f0 =kfs, then the other frequencies will be, f1=(k+2)fs, f2=(k+4)fs ….etc. Thus every frequency is separated by 2fs from its nearest carriers. Figure shows the power spectral density of M-ary FSK. Band width required for the transmission of M-ary FSK is 2 f s M  2

1 N 1 N 2 2 2 Ts NTb

 2 N 1

fb N

Signal Space Diagram of M-ary FSK Here S0, S1,…………SM-1 are mutually orthogonal signals for M-symbols.

S0  PT s b 0 (t ) S1  PT s b 1 (t ) S 2  PT s b 2 (t )

….

SM 1  PT s b M 1 (t ) The orthogonal carriers

0 (t ), 1 (t ), 2 (t )..............., M 1 (t )

can be represented as

follows.

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0 (t ) 

2 cos(2 f 0t ) Ts

1 (t ) 

2 cos(2 f1t ) Ts :

M 1 (t ) 

2 cos(2 f M 1t ) Ts

The figure below shows signal space diagram for M-ary FSK for M=3

In

the

signal

space

diagram, 0 (t ), 1 (t ), 2 (t )..............., M 1 (t ) form

mutually

perpendicular axes. For simplicity of understanding we will consider M=3.

Then the three carriers

0 (t ), 1 (t ), 2 (t )

will form three axes. Then the signals S0(t), S1 (t)

and S2(t) will be represented by vectors of length

PT s s

along those axes.

Distance between signal points: Distance between the signal space points is dmin = =

2 2 ( PT s s )  ( PT s s)

2 PT s s

This equation gives the minimum distance between any two signal points. This relation holds for „M‟ signal points since all axes are perpendicular to each other.

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MINIMUM SHIFT KEYING In QPSK, the phase changes by 900 (OQPSK). This creates abrupt amplitude variations in the waveform. Because of the abrupt amplitude variations, side lobes are having significant levels. These side lobes may interfere with the main lobe of other adjacent channels causing inter channel interference. Hence band pass filtering is required to avoid side lobes but these filter have other side effects for example say, it alters the amplitude of the waveform.

MSK overcomes these problems. In MSK, the output waveform is continuous in phase hence there are no abrupt changes in amplitude. These side lobes of MSK are very small hence band pass filtering is not required to avoid inter channel interference.

Figure shows the waveform of MSK. The binary bit sequence at the top.

Fig.a shows the corresponding NRZ waveform b(t). From b(t), two waveforms are generated for odd and even bits. be (t) represent even bits (Fig.c) and bo (t) represent odd bits (Fig.b).

The duration of each bit in bo (t) or be (t) is 2Tb, where as it is Tb in b(t).The waveforms b0 (t) and be (t) have an offset of Tb. this offset is essential in MSK. Two waveforms The waveform of

cos(2

sin(2

fb f t) cos(2 b t ) 4 and 4 are generated as shown in fig.(d).

sin(2

fb t) 4 passes through zero at the end of symbol time in be (t) and

fb t) 4 passes through zero at the end of symbol in b0 (t).

be (t) is multiplied by

sin(2

fb f t) cos(2 b t ) 4 and bo (t) is multiplied by 4 .

Those product waveforms are shown in fig.(e) and (f).

The transmitted MSK signal is represented as,

s(t )  2 Ps be (t )sin(2

fb f t ) cos(2 f ct )  2 Ps bo (t ) cos(2 b t )sin(2 f ct ) 4 4

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Sri Sai Aditya Institute of Science & Technology That is the product signals be (t )sin(2

ECE Department

fb f t ) and bo (t ) cos(2 b t ) modulate the 4 4

quadrature carriers of frequency fc.

After the rearrangements

S (t )  2 Ps {

bo (t )  be (t ) f b (t )  be (t ) f }sin 2 ( f c  b )t  2 Ps { o }sin 2 ( f c  b )t 2 4 2 4

S (t )  2Ps CH (t )sin 2 f H t  2Ps CL (t )sin 2 f Lt Where CH (t )  and f H  f c 

bo (t )  be (t ) b (t )  be (t ) & CL (t )  o 2 2

fb f & f L  fc  b 4 4

Case 1: when bo (t )  1(1v) & be (t )  1(1v) ; then CH(t) =1 &CL(t)=0 MSK signal is S (t )  2Ps sin 2 f H t Case 2: when bo (t )  0(1v) & be (t )  0(1v) ; then CH(t) =-1 &CL(t)=0 MSK signal is S (t )   2Ps sin 2 f H t  2Ps sin(2 f H t  1800 ) Case 3: when bo (t )  1(1v) & be (t )  0(1v) ; then CL(t) =1 &CH(t)=0 MSK signal is S (t )  2Ps sin 2 f Lt Case 4: when bo (t )  0(1v) & be (t )  1(1v) ; then CL(t) =-1 &CH(t)=0 MSK signal is S (t )   2Ps sin 2 f Lt  2 Ps sin(2 f Lt  1800 ) The frequencies fH & fL are chosen such that they are orthogonal over the interval T b.

For orthogonality following relation must be satisfied i.e.,

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 sin(2 f

H

ECE Department

t )sin(2 f Lt )dt  0

0

This relation will be satisfied if we have integers „m‟ and „n‟ such that 2 ( f H  f L )Tb  n

and

2 ( f H  f L )Tb  m

Let us substitute values of fH and fL in the above relations.

2 ( f c 

fb f  f c  b )Tb  n 4 4 fbTb  n  n  1

Similarly from equation above

2 ( f c 

fb f  f c  b )Tb  m 4 4 4fcTb =m

fc 

m fb 4

fb Here fc must be integer multiple of 4 .

With n=1, this equation

2 ( f H  f L )Tb  n

becomes

fH  fL 

fb 2 .

Here n=1, means the difference between fH and fL is minimum and at the same time, (MSK) they are orthogonal. Therefore this technique is called minimum shift keying (MSK).

Substituting, this value of fc in equations f H  f c  fH = fc +fb/4

=> mfb/4 + fb/4 = (m+1)fb/4

fL =fc –fb/4

=> mfb/4- fb/4 =(m-1) fb/4

fb f & f L  fc  b 4 4

for these waveforms fH & fL are calculated with m=5.

We know that if bo(t)=be(t), then the transmitted waveform is of frequency fH. And if bo(t)=-be(t), then the transmitted waveform is of frequency fL. This shows that MSK is basically FSK with reduced bandwidth and continuous phase.

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MSK generation

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MSK TRANSMITTER

Block diagram of MSK transmitter



The two sinusoidal signals sin(2πfct) and cos(2πfb/4t) are mixed.



The bandpass filters then pass only sum and difference components fc+fb/4 and fc-fb/4.



The outputs of bandpass filters are then added and subtracted such that two signals x(t) and y(t) are generated.



Signal x(t) is multiplied by √2Ps b0(t) and y(t) is multiplied by √2Ps be(t).



The outputs of the multipliers are then added to give final MSK signal. Thus the block diagram of above figure is the step to step implementation.

RECEPTION OF MSK SIGNAL

Figure shows the block diagram of MSK receiver. MSK uses synchronous detection. 

The signals x(t) and y(t) are multiplied with the received MSK signal. Here x(t) and y(t) have same values as shown in transmitter block diagram.



The outputs of the multipliers are b0(t) and be(t). The integrators integrate over the period of 2Tb.



For the upper correlator, the sampling switch samples output of integrator at t = (2k+1)Tb. Then the decision device recovers b0(t).

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MSK receiver block diagram



Similarly, lower correlator output is be (t). The outputs of two decision devices are staggered by Tb



The switch S3 operates at t=kTb and simply multiplexes the two correlator outputs.

Signal Space diagram of MSK signal

S (t )  2Ps CH (t )sin 2 f H t  2Ps CL (t )sin 2 f Lt S (t )  PT s s CH (t )

2 2 sin 2 f H t  PT sin 2 f Lt s s CL (t ) Ts Ts

S (t )  PT s s CH (t )H (t )  PT s s CL (t )L (t ) The carriers ΦH(t) and ΦL(t) are in quadrature. Depending on the values of CH (t) and CL (t), there will be four signal points in ΦHΦL plane.

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Power Spectral Density and Bandwidth of MSK The power spectral density of MSK signal is given by

s( f ) 

8Eb cos(2 ( f  f c )Tb ) 2 8Eb cos(2 ( f  f c )Tb ) 2 { }  2 { }  2 1  [4( f  f c )Tb ]  1  [4( f  f c )Tb ]

Power Spectral densities of MSK and QPSK

Bandwidth calculation of MSK From the PSD spectrum of MSK, BW=0.75fb-(0.75fb) =1.5fb BW of MSK is higher than that of QPSK.

Advantages & Drawbacks of MSK as compared to QPSK Advantages 1. The MSK baseband waveforms are smoother compared to QPSK. 2. MSK signal have continuous phase in all the cases, whereas QPSK has abrupt phase shift of π/2 3. MSK waveform does not have amplitude variations, whereas QPSK signals have abrupt amplitude variations.

4. The main lobe of MSK is wider than that of QPSK. Main lobe of MSK contains around 99% of signal energy whereas QPSK main lobe contains around 90% of signal energy.

5. Side lobes of MSK are smaller compared to that of QPSK. Hence inter channel interference is absent in MSK Prepared by Venkata Satish N

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6. To avoid inter channel interference due to side lobes, QPSK needs the band pass filtering, whereas it is not required in MSK.

Drawbacks 1. The bandwidth requirement of MSK is 1.5fb, whereas it is fb in QPSK. Actually, this cannot be said serious drawback of MSK. Because power to bandwidth ratio of MSK is more. In fact 99% of signal power can be transmitted within the bandwidth of 1.2fb in MSK. While QPSK needs around 8fb to transmit the same power.

2. The generation and detection of MSK is slightly complex. Because of incorrect synchronization, phase jitter can be present in MSK. This degrades the performance of MSK.

Important Questions 1. Comparison of BASK, BPSK & BFSK modulation schemes 2. Explain the working of BPSK modulator & Demodulator with block diagrams?

3. Explain the coherent & NON coherent detection process of BFSK signals?

4. Explain with neat block diagram the generation and recovery of DPSK signals.

5. Explain the working offset QPSK transmitter and receiver with neat block diagrams?

6. Draw the constellation diagrams for the modulation schemes BASK, BPSK, BFSK, & QPSK. 7. Explain M-ary PSK signaling scheme. Draw the signal space representation of M-ary PSK for M=8.

8. Explain M-ary FSK signaling scheme.

9. (a) Draw the signal space representation of MSK. (b) Show that in a MSK signaling scheme, the carrier frequency in integral multiple of „fb/4‟ where „fb‟ is the bit rate. (c) Bring out the comparisons between MSK and QPSK. Prepared by Venkata Satish N

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