communication system eeeb453 chapter 6 digital modulation intan shafinaz mustafa dept of electrical...

COMMUNICATION SYSTEM EEEB453Chapter 6

DIGITAL MODULATION

Intan Shafinaz MustafaDept of Electrical Engineering

Universiti Tenaga Nasionalhttp://metalab.uniten.edu.my/~shafinaz

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Digital-to-Analog Conversion Modulating a digital signal

Eg. Sending computer data through public telephone line

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Digital-to-Analog Conversion

B Mbps B MHz

Process of changing one of the characteristics of an analog signal based on the information in digital signals (0’s and 1’s).

Modulation involved switching (known as keying) between short bursts of different signals to transmit the encoded message.

A general carrier wave may be written:

Modulation methods based on varying the amplitude, A, frequency, f and phase, to transmit digital data is known as Amplitude Shift Keying (ASK), Frequency Shift Keying (FSK) and Phase Shift Keying (PSK) respectively.

Example – Tx of digital data over telephone wire (modem)

ftAtC 2sin

Information Capacity, Bits, Bit Rate, Baud and M-ary Encoding.

Information Capacity – is a measure of how much information can be propagated through a communications system and is a function of BW and Tx line.

It represents the number of independent symbols that can be carried through a system in a given unit of time.

Hartley’s law is

I B x tWhere I = information capacity (bps)

B = bandwidth (hertz)t = transmission time (sec)

M-ary Encoding

M-ary is a term derived from the word binary. M simply represents a digit that corresponds to the number of

condition, levels or combinations possible for a given number of binary variables.

For example, a digital signal with four possible conditions is an M-ary system where M = 4.

The number of bits necessary to produce a given number of condition is expressed mathematically as

N = log 2 M

Where N = number of bits necessary M = Number of conditions, levels or combination

possible with N bits Rearranged the equation to express the number of conditions possible

with N bits as2N = M

Example, with one bit, only 21 = 2 conditions are possible, with two bits, 22 = 4 conditions are possible etc..

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Bit Rate, Baud Rate and Minimum BW Two basic aspects of digital-to-analog modulation; bit and baud rate. Bit rate - number of bits per second (rate at which bit changes,

bps). Computer Efficiency – how long it takes to process each piece of

information (time to send) Baud rate – number of signal units per second (rate at which signal

element changes). Also called modulation rate or symbol rate. Data Transmission Efficiency – how efficient we can move those

data from place to place. Analogy – In transportation, baud ≈ car and bit ≈ passenger

A car can carry one or more passengers. If 1000 cars go from one point to another, carrying only 1 passenger (i.e the driver), then 1000 passengers are transported.

However, if each car carries four passengers (carpooling), then 4000 passengers are transported.

Note that the number of cars not the number of passengers, determine the traffic, and therefore, the need for wider highways.

Similarly, the number of bauds determines the required bandwidth, not the number of bits.

Bit Rate, Baud Rate and Minimum BW According to H. Nyquist, binary digital signals can be propagated

through an ideal noiseless transmission medium at a rate equal to two times the bandwidth of the medium.

The minimum theoretical bandwidth necessary to propagate a signal is called the minimum Nyquist bandwith or the minimum Nyquist frequency.

Thus, fb = 2B

Where fb = bit rate (bps)

B = ideal Nyquist bandwidth (hertz) Mathematically, Baud = (baud per second)

where ts = time of one signaling element (second)N = number of bits per signal elementfb = bit rate (bps)

In analog Tx of digital data, the baud rate is less than or equal to the bit rate.

- In binary system such as binary FSK and binary PSK, baud and bits per second are equal. However, in higher-level system such as QPSK and 8-PSK, bps is always greater than baud.

N

f

t

b

s

1

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Bit Rate, Baud rate and Minimum BW

Term Units Definition

Data element Bits A single binary one or zero

Data rate Bits per second (bps) The rate at which data elements are transmitted.

Signal element Digital: a voltage pulse of constant amplitudeAnalog: a pulse of constant frequency, phase and amplitude

That part of a signal that occupies the shortest interval of a signaling code

Signaling rate or modulation rate

Signal elements per second (baud) The rate at which signal elements are transmitted.

Summary of the Terms:

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Example 1 – An analog signal carries four bits in each signal element. If 1000 signal elements are sent per second, find the baud rate and bit rate. Example 2 – The bit rate of a signal is 3000. If each signal element carries 6 bits, what is the baud rate?

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Types of digital-to-analog conversion

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Amplitude Shift Keying (ASK) or On-Off

Keying (OOK) A simple version of amplitude modulation used for digital

modulation. Both freq and phase remain constant while the amplitude

changes. Uses logic levels in the data to control the amplitude of the

carrier wave. ‘1’ for high amplitude (switch ON) ‘0’ for low amplitude (switch OFF).

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Keying (OOK)

Basic implementation of Binary ASK

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Keying (OOK) ASK Modulator

- The modulator cct has 2 inputs:1. data to be transmitted2. high freq carrier sinewave

At the Tx, let the input of a data stream is 0110001011

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Keying (OOK) At the Rx, the data stream need to extracted:

Step 1 – Rectify the input ASK waveform to contain only +ve signal but it will still contain unwanted carrier wave component.

Step 2 – Pass through a LPF to remove the carrier component.

Step 3 – Pass through a voltage comparator to get a true copy of the original data stream

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Amplitude Shift Keying (ASK)

Example 3 – Determine the baud and minimum bandwidth necessary to pass a 10 kbps binary signal using amplitude shift keying.

Solution

For ASK, N=1

Bmin = fb/N

Bmin = 10k/1 = 10kHz

Baud = fb/N = 10kbaud/sec

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Amplitude Shift Keying (ASK)

Example 4 – Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate?

Solution

In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000.

The baud rate and the bit rate are also the same for ASK, the bit rate is 5000bps.

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Frequency Shift Keying (FSK)

Use logic levels in the data to control the frequency of the carrier wave.

Data = ‘1’ for high frequency Data = ‘0’ for low frequency

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FSK Bit Rate, Baud and Bandwidth

It can be seen that the time of one bit (tb) is the same as the time the FSK output is a mark or a space frequency (ts).

Thus the bit time equals the time of an FSK signaling element and the bit rate equals the baud.

The baud for binary FSK can also be determined by substituting N = 1,

baud = b

bb ff

N

f

1


The minimum bandwidth for FSK is given as

And since Then the minimum bandwidth can be approximated as

(**)

Where B = minimum Nyquist bandwidth (hertz)f = frequency deviation (|fm - fs|) (hertz)

fb = input bit rate (bps)

)()( bmbs ffffB

bms fff 2

fff ms 2

)(2 bffB

Note that equation (**) resembles Carson’s rule for determining the approximate bandwidth for an FM wave. The only difference in the two equations is that, for FSK, the bit rate, fb is substituted for the modulation signal freq fm.

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Example 5 – A binary FSK with a mark frequency of 49kHz, a space frequency of 51kHz and an input bit rate of 2kbps, determine

a. The peak frequency deviationb. The minimum bandwidthc. Baud

Solution

a. From 2f = |fm - fs|,

and f = |49kHz – 51kHz|/2 = 1kHz

b. Min BW, B = 2(f + fb) = 2(1000+2000) = 6kHz

c. For FSK, N = 1, the baud is fb/N =2000/1 = 2000

FSK Bit Rate, Baud and Bandwidth Bessel function can also be used to determine the

approximate bandwidth for an FSK wave.The fastest rate of change i.e highest fundamental freq occurs when alternating 1s and 0s are occuring.

Therefore,

Where

fa = highest fund freq (hz)

fb = input bit rate (bps)

2b

af

f

The formula used for modulation index in FM is also valid for FSK, thus

af

fh

(unitless)

Where h = FM modulation index called h-factor in FSK

f = peak freq deviation (Hz)

FSK Bit Rate, Baud and Bandwidth The worst-case modulation index (deviation ratio) yields the

widest BW. The widest BW occurs when both the freq deviation and the

modulating signal freq are at their maximum values, thus

or

Thus, bandwidth,

B = 2(n x fa)

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Example 6 – A binary FSK with a mark frequency of 49kHz, a space frequency of 51kHz and an input bit rate of 2kbps, using Bessel table, determine

a. The modulation index, hb. The bandwidth

Solution

a. h = |49kHz – 51kHz|/2kbps = 1

b. From a Bessel table, for modulation index of 1, n = 3,

then B = 2(3 x 1000)

= 6kHz

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There are many different ways of generating an FSK waveform.

One way is by combining 2 different ASK waveform/modulator.

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Lets assume that the above data stream is applied to an ASK modulator using the higher freq as the carrier. The resulting output:

Inverting the original data stream:

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This inverted data stream will be the input to another ASK modulator using a lower carrier freq - the original data 0 periods filled with a lower freq carrier.

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Summing amplifier is used to add the two ASK waveforms:

Output from Modulator 1Output from Modulator 2

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Advantage of FSK over ASK – higher reliability in term of data accuracy.

Disadvantage – requires higher BW (the actual increase depends on the 2 freqs used). The higher the freq and the more they differ from each other, the wider the BW required.

Multiple FSK (MFSK) A signal that is more bandwidth efficient, but also more

susceptible to errors is multiple FSK (MFSK), in which more than two frequencies are used. (i.e each signaling element represents more than one bit)

The transmitted MFSK signal for one signal element time can be defined as follows:

Where fc = the carrier frequency

∆f = the difference frequencyM = number of signal element (2n)n = number of bits per signal element

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MitfAts ii 1,2cos)(

fMiff ci )12(

Example 7– With fc = 350kHz, ∆f = 20kHz, and M=8 (n=3), the following frequency assignments for each of the eight possible 3-bit data combinations:

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fi Frequency Assignment

Freq (kHz)

f1

f2

f3

f4

f5

f6

f7

f8

Example – Figure shows an example of MFSK with M = 4. An input bit streams of 20 bits is encoded 2 bits at a time, with

each of the four possible 2-bit combinations transmitted as a different frequency.

The display in the figure shows the frequency transmitted (y-axis) as a function of time (x-axis).

Each column represents a time Ts in which a single 2-bit signal element is transmitted.

The shaded rectangle in the column indicates the frequency transmitted during that time unit.

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Phase Shift Keying (PSK) In PSK, the phase of the carrier is shifted to represent data. Two-Level PSK(Binary PSK) - BPSK

Uses two phases (0 and 180°) to represent the two binary digits.

The resulting transmitted signal for one bit time is:

0)2cos(

1)2cos(

)2cos(

)2cos()(

binarytfA

binarytfA

tfA

tfAts

c

c

c

c

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Phase Shift Keying (PSK) Example: binary 1 is represent with a phase 0°, while binary

0 is represented with a phase of 180°. PSK is equivalent to multiplying the carrier by +1 when the

info is 1, and by -1 when the info is 0.

binary ‘1’ binary ‘0’

Bipolar NRZ

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Phase Shift Keying (PSK) PSK Transmitter – same modulator as in ASK system

Since a sinewave is symmetrical, it is impossible for the Rx to know whether signal is inverted form or not.

Thus need to apply some data conditioning to the incoming stream to convert it to a form which recognizes logic levels by changes that occur and not by the absolute levels.

One such code is bipolar NRZ. The amplitude of the carrier is controlled by the bipolar signal on

the modulation input. When the signal goes negative, the sinewave inverts.

PSK Bit Rate, Baud and Bandwidth

Mathematically, the output of a BPSK modulator is proportional to

Solving for the trig identity for the product of two sine function,

Thus, the minimum double-side Nyquist BW, B is

But ,

then

)]2[sin()]2[sin( tftf ca outputBPSK

tfftff acac )(2cos[21

)(2cos[21

lsfusf ffB

aacac fffff 2)()(

2b

af

f

bb f

fB

22

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PSK Bit Rate, Baud and Bandwidth

Example 8 – For a BPSK modulator with a carrier frequency of 70 MHz and an input bit rate of 10Mbps,

a. determine the maximum and minimum upper and lower side frequencies

b. Draw the output spectrumc. Determine the minimum bandwidthd. Calculate the baud.

Solution

a.

b. Output spectrum

c.

d.

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Quadrature PSK (QPSK) The term “quadrature” implies that there are four possible

phases (4-PSK) which the carrier can have at a given time. The pair of bits represented by each phase is called dibit. The rate of change (baud) in this signal determines the

signal bandwidth. BUT the throughput or bit rate for QPSK is twice the baud

rate.

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QPSK = 4-PSK

Assumption

Modulation for Data Communication: QPSK

One way to increase the binary data rate while not increasing the bandwidth required for the signal transmission is to encode more than 1 bit per phase change.

In the system known as quadrature, quarternary, or quadra phase PSK (QPSK or 4-PSK), more bits per baud are encoded, the bit rate of data transfer can be higher than the baud rate, yet the signal will not take up additional bandwidth.

In QPSK, each pair of successive digital bits in the transmitted word is assigned a particular phase.

Each pair of serial bits, called a dibit, is represented by a specific phase.

Quadrature PSK modulation. (a) Phase angle of carrier for different pairs of bits. (b)

Phasor representation of carrier sine wave. (c) Constellation diagram

of QPSK.

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QPSK Bit Rate, Baud and Bandwidth

Example 9 – The CCITT V.22 (like Bell 212A) modem uses QPSK to send data at 1200 bits per second; What is the baud rate.

The CCITT V.22 (like Bell 212A) modem uses QPSK to send data at 1200 bits per second; however, the phases change only 600 times per second, conveying two bits per change - this is a 600 baud modem.

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Example 10 – For a QPSK system, with the following input bit sequence 100010101101, an input bit rate equal to 20Mbps,

a. Draw the QPSK modulated waveform, state assumption used.

b. Determine the minimum bandwidth required.

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Solution eg. 9 – 100010101101

a.

Assumption:

b. B = fb/N

B = 20M/2 = 10MHz.

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QUADRATURE AMPLITUDE MODULATION

-Uses both amplitude and phase modulation of the carrier; not only are different phase shifts produced but also the amplitude of the carrier is varied

8 – QAM

Is an M–ary encoding technique where M=8

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8-QAM

Figure : 8-QAM Modulator a) Truth Table b)phasor diagram c) Constellation diagram

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8-QAM

Figure : Output phase and amplitude versus-time realtionship for 8-QAM

Min. bandwidth required = fb/3

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Quadrature Amplitude ModulationA combination of ASK and PSK: both phase and amplitude varied

#amplitude shifts << #phase shiftsLower susceptible to noise than ASK, higher bit rate than PSK

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8-QAM

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16-QAM

Better susceptible to noise because- Not all possibilities are used- Sometimes, Amp and Phase have a relationship

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Example 12 - Calculate the bit rate and baud rate of BPSK and 8-PSK if the bit duration, Tb and signaling time, Ts are both 0.01msec.

BPSKBit rate = baud rate

= 1/0.01 = 100kbps

Or 100kbaud/sec

8-PSKbaud rate = 1/0.01

= 100kbaud/s

Then bit rate, fb = 3x100k=300kbps

MORE EXAMPLE :

Example 13: An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many number of conditions ?

Example 14: We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies) and the baud rate

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Example 15 - For a 16-QAM modulator with input data rate fb =10 Mbps, and a carrier frequency of 70 MHz, determine the minimum bandwidth and the baud rate

Example 16 – For the following modulation schemes, construct a table showing the number of bits encoded, number of output conditions, min. bandwidth, and baud for an information data rate of 12 kbps; QPSK, 8-PSK, 8-QAM,16-PSK and 16 QAMModulation n M B(Hz) baud

QPSK

8-PSK

8-QAM

16-PSK

16 QAM

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Bandwidth Efficiency Used to compare the performance of one digital

modulation technique to another Ratio of the transmission bit rate to the minimum

bandwidth required for a particular modulation scheme.

)(.min

)(

Hzbandwidth

bpsonbitratetransmissiB

Example 17 – For an 8-PSK system, operating with an information bit rate of 24kbps, determine:

a) baud rateb) min. bandwidthc) bandwidth efficiency

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Summary

ASK demodulation: only the presence or absence of a

sinusoid in a given time interval needs to be determined advantage: simplicity disadvantage: ASK is very susceptible to noise

interference–noise usually (only) affects the amplitude, therefore ASK is the modulation technique most affected by noise

application: ASK is used to transmit digital data over optical fiber

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Summary

FSK demodulation: demodulator must be able to determine

which of two possible frequencies is present at a given time

advantage: FSK is less susceptible to errors than ASK –receiver is looking for specific frequency changes over a number of intervals, so voltage (noise) spikes can be ignored

disadvantage: FSK spectrum is 2x ASK spectrum application: over voice lines, in high-frequency radio

transmission, etc.

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Summary

PSK demodulation: demodulator must be able to determine

the phase of received sinusoid with respect to some reference phase

advantage: (i) PSK is less susceptible to errors than ASK, while it requires/occupies the same bandwidth as ASK; (ii) more efficient use of bandwidth (higher data-rate) are possible.

disadvantage: more complex signal detection / recovery process, than in ASK and FSK.

Exercise – past year Qs1. There are quite a number of signaling techniques used in digital pulse modulation. Explain

the difference between unipolar NRZ signaling and unipolar RZ signaling. [2 marks]2. Given a binary data stream of 11101010, draw the pulse train for unipolar NRZ signaling

and unipolar RZ signaling. [3 marks]3. What is the major disadvantage in using NRZ encoding? How does RZ encoding attempt

to solve the problem? [2 marks]4. For a data stream 10011101, draw the output waveform for Manchester signaling

technique. Assume the output level is initially LOW. [2 marks]5. An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second,

find the baud and the bit rate. [2 marks]6. Frequency Shift Keying (FSK) is a modulation technique where the frequency of the

carrier signal is modulated with respect to the input data stream. There are several methods in generating FSK waveforms. Given a data bit stream of 1011000110, explain fully with the help of diagrams, how we can generate the FSK waveform for the data stream given. [9 marks]

7. What is baud rate? [1 mark]8. Calculate the bit rate and baud rate of BPSK and 8-PSK if the bit duration, Tb and

signaling time, Ts are both 0.01msec. [2 marks] 9. For a data stream 1010 1110, draw the output waveform for Manchester signaling

technique. Assume the output level is initially LOW. [2 marks]

communication system eeeb453 chapter 6 digital modulation intan shafinaz mustafa dept of electrical...

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