UNIT - 3

McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004

UNIT - 3

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


Analog Transmission & Multiplexing

Digital – to – Analog conversion, Analog – to – analog conversion, Multiplexing, Spread Spectrum.


AnalogTransmission


Modulation of Digital Data


Digital-to-analog modulation

Modulation

Converting digital signals to analog signals is called modulation.

Example

Transmition of data from one computer to another computer using

telephone line. The digital data is modulated on an analog signal.


Types of digital-to-analog modulation

Modulating techniques are

1) Amplitude Shift Keying (ASK)

2) Frequency Shift Keying (FSK)

3) Phase Shift Keying (PSK)

4) Quadrature Amplitude Modulation (QAM)


Bit rate is the number of bits transmitted per second.

Baud rate is the number of signal elements transmitted per second. A signal element carries one or more bits.

Baud rate = The bit rate / Number of bits

represented by each signal unit.

Baud rate is less than or equal to the bit rate.

Bit Rate & Baud Rate (Signal elements)


Example Example

An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate

SolutionSolution

Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s)Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps


Example Example

The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate?

SolutionSolution

Baud rate = Bit rate / Number of bits represented by each

signal unit. Baud rate = 3000 / 6 = 500 baud/s Baud rate = 3000 / 6 = 500 baud/s



Example

An analog signal has a bit rate of 8000 bps & a baud rate of 1000 baud. How many data elements are carried by each signal element?. How many signal elements do we need?.

Solution

Baud rate, S= 1000, bit rate, N = 8000, Data Elements, r=?,

Signal Elements, L=?

S = N/r

r = N/S

r= 8000/1000=8 bits /baud.

r = log 2L

L = 2r

L = 28

L= 256www.bookspar.com | Website for students |

VTU NOTES


Carrier SignalIn analog transmission, the sending device produces a high frequency signal that acts as a basis for the information signal . This base signal is called as carrier signal or carrier frequency.

The receiving device is tuned to the frequency of the carrier signal of the sender.

The digital information then modulates the carrier signal by modifying one or more of its characteristics( amplitude, frequency or phase). This modification is called modulation ( or Shift key).


Amplitude Shift Keying (ASK / BASK)

In ASK, the frequency & phase of the carrier signal remain constant and amplitude changes to represent the bit 1 or 0. A Zero amplitude represents bit 0 and max amplitude represent bit 1. A bit duration is the period of time that defines 1 bit. The amplitude of the bit during each duration is constant. In ASK baud rate and bit rate are same. ASK is susceptible for noise.


Relationship between baud rate and bandwidth in ASK

Bandwidth of a signal is the total range of frequencies occupied by that signal. The most significant frequencies are between

fc – N baud /2 + fc +Nbaud /2 , fc is the carrier frequency. In ASK,

BW = (1+d) x Nbaud, where Nbaud is the baud rate, d is the factor related to the modulation process.


Example Example

Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex.

SolutionSolution

In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.


Example Example

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

SolutionSolution

In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps.


Example Example

Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions.

SolutionSolution

For full-duplex ASK, the bandwidth allocated for each direction is BW = 10000 / 2 = 5000 HzThe carrier frequencies can be chosen at the middle of each band,

fc (forward) = 1000 + 5000/2 = 3500 Hzfc (backward) = 11000 – 5000/2 = 8500 Hz


Solution to Example

In data communication we use full duplex links with communication in both directions. Then we divide the bandwidth into two with two carrier frequencies.


Frequency Shift Keying ( FSK /BFSK)

In FSK, the Amplitude & phase of the carrier signal remain constant and frequency changes to represent the bit 1 or 0.The frequency of the signal during each bit duration is constant.


Relationship between baud rate and bandwidth in FSK

FSK spectra is a combination of two ASK spectra centered on fc0 and fc1. Frequencies are between fc0 – N baud /2 & fc1 +Nbaud /2.

B.W = (fc1 +Nbaud /2) – ( fc0 - Nbaud /2 )= fc1- fc0 +Nbaud

The bandwidth required for FSK transmission = the frequency shift+ baud rate of the signal


Example Example

Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz.

SolutionSolution

For FSK BW = baud rate + fBW = baud rate + fc1c1 f fc0c0

BW = bit rate + fc1 BW = bit rate + fc1 fc0 = 2000 + 3000 = 5000 Hz fc0 = 2000 + 3000 = 5000 Hz


Example Example

Find the bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode.

SolutionSolution

Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = Baud rate + fcBW = Baud rate + fc11 fc fc00

Baud rate = BW Baud rate = BW (fc (fc11 fc fc00 ) = 6000 ) = 6000 2000 = 4000 2000 = 4000But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.



Example Jan / Feb 2005

Derive the expression for the bandwidth of a Frequency Shift Keying ( FSK ) signal And hence determine the maximum bit rate of transmission, if the bandwidth of the medium is 12,000 Hz and the difference between the two carrier signal is 2000 Hz. Assume that the transmission is full-duplex mode.


Phase Shift Keying ( PSK/BPSK ) In PSK, the Amplitude & frequency of the carrier signal remain constant and phase changes to represent the bit 1 or 0.The phase of the signal during each bit duration is same. In the figure a phase of 00 represents bit 0 and a phase of 180 0

represents bit 1. Bandwidth & Baud rate are same.


PSK constellation OR Phase – state diagram

The PSK is also called as 2-PSK, or BPSK, because two different phases ( 0 & 180 degrees) are used.

PSK Constellation or Phase-State diagram

Relation between phases & bits.


The 4-PSK ( Quadrature PSK) method OR QPSK

QPSK use four variations ( 0,90,180 & 270). Each phase shift represent 2 bits.


The 4-PSK characteristics

Constellation Diagram


The 8-PSK characteristics

8-PSK has eight variations with a constant shift of 45 0 . Each shift represent 3 bits of data.


Example Example

Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate?

SolutionSolution

For PSK the baud rate is the same as the bandwidth. There fore, baud rate is 5000. In 8-PSK the bit rate is 3 times the baud rate.So, the Bit rate = 5000 x 3 = 15,000 bps.


Quadrature amplitude modulation is a combination of ASK and PSK.

Note:Note:

Quadrature Amplitude Modulation


The 4-QAM and 8-QAM constellations

Phase change along x-axis, Amplitude change along y-axis.

In QAM the number of phase shift is greater than the number of amplitude shift.

Time - Domain Plots



16-QAM constellations

4 3


Example Example

A constellation diagram consists of 8 equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate?

SolutionSolution

The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore,the baud rate is = 4800 / 3 = 1600 baud


ExampleExample

Compute the bit rate for a 1000-baud 16-QAM signal.

SolutionSolution

A 16-QAM signal has 4 bits per signal unit. Because, log216 = 4.

Thus, (1000)(4) = 4000 bps

Example Example

Compute the baud rate for a 72,000-bps 64-QAM signal.SolutionSolution A 64-QAM signal has 6 bits per signal unit. Because, log 2 64 = 6. Thus, 72000 / 6 = 12,000 baudwww.bookspar.com | Website for students |

VTU NOTES


Example January/February 2005

Calculate the number of Levels (Signal ) required to transmit the maximum bit rate. What is the BAUD rate.

Example July/August 2005

Explain Quadrature Amplitude Modulation. What is the advantage?


Telephone Modems

Modem Standards

Telephone lines can carry frequencies between 300 and 3300 Hz, giving a bandwidth of 3000 Hz. The effective bandwidth of a telephone line being used for data transmission is 2400 Hz, covering the range 600 – 3000 Hz.

This bandwidth is called as base bandwidth. During data transmission, any bandwidth is to be modulated to base bandwidth. The device used for the purpose is called MODEM. www.bookspar.com | Website for students |

VTU NOTES

Telephone line bandwidthThe signal Bandwidth must be smaller than the cable bandwidth.


We need to modulate the voice signal to use data bandwidth. Devices used to do so are

called MODEMS

Modem stands for modulator/demodulator.

Modulator creates a band-pass analog signal from binary data.

Demodulator recovers the binary data from the modulated signal.

Note:Note:


Modulation/demodulation

A modulator converts a digital signal into an analog signal using ASK, FSK, PSK or QAM. A demodulator converts an analog signal into a digital signal.



Modem Standards

Most popular modems available are based on V – series standards defined by International Telecommunication Union- Telecommunication Standardization sector( ITU – T)

1) V.32 - This uses both modulation and encoding technique. The data stream is divided into 4-bit sections with a baud rate of 2400. The resulting speed is 4 x 2400 = 9600 bps.

2) V.32bis - It supports 14,400 bps transmission.

3) V.34bis – It supports a bit rate of 28,800

4) V.90 ( 56 k ) – It has a transmission capacity of 56,000 bps. Used for internet communication.


Modulation of Analog Modulation of Analog SignalsSignals


Analog-to-analog modulation

Why modulation is needed?

In Radio transmission, Govt assigns a narrow bandwidth to each radio station. The analog signal produced by each station is a low-pass signal. To be able to listen, the low-pass signal need to be modulated before transmission.


Types of analog-to-analog modulation

AM – Amplitude Modulation

FM – Frequency Modulation

PM – Phase Modulationwww.bookspar.com | Website for students |

VTU NOTES

Amplitude Modulation

In amplitude modulation, the carrier signal is modulated so that its amplitude varies with the changing amplitude of modulating signal. The frequency and phase of the carrier signal remain the same.


Amplitude modulation


AM Bandwidth


AM Band allocationAM stations are allowed the carrier frequencies between 530 kHz to 1700 kHz. Each stations fc must be separated from other at least by 10 kHz to avoid B/W over lapping.


Example Example

We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM?

SolutionSolution

An AM signal requires twice the bandwidth of the original signal:

BW = 2 x 4 KHz = 8 KHz


Frequency Modulation (FM)In Frequency Modulation, the Frequency of the Carrier Signal is modulated to that of the modulating signal. The Amplitude and Phase of the Carrier signal remains constant.

FM Bandwidth

The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BWt = 10 x BWm.


Frequency modulation


FM bandwidth


FM stations are allowed Carrier frequencies between 88 MHz & 108 MHz. Stations must be separated by at least 200 kHz to keep B/W overlapping.


FM band allocation


Example Example

An audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM?

SolutionSolution

An FM signal requires 10 times the bandwidth of the original signal: BW = 10 x 4 MHz = 40 MHz


Multiplexing



Three users at a location communicate with three other users using three separate lines. This arrangement completely dedicates network resources. This becomes inefficient and expensive as the number of users increases.



Multiplexing is introduced when a transmission link has sufficient bandwidth to carry several connection lines.

In the fig, a multiplexer combines the signals from three users in a single transmission link.



Multiplexing

It is a technique that allows simultaneous transmission of multiple signals across a single data link.

In multiplexed system n - lines share the bandwidth of one link.



Link, Channel & Line

Link – means physical path.

Channel – means the portion of the link that carries a transmission.

Line-means portion of the channel.


Dividing a link into channels

In the Fig, 04 lines direct their transmission stream to a Multiplexer. MUX combines them into a single stream and transmitted.

At the receiving, the stream is fed into a Demultiplexer. The DEMUX separates the stream back into its components and directs them into corresponding lines.



There are different approaches of multiplexing the information from multiple connections to a single transmission line.


Categories of multiplexingThere are 3 types of Multiplexing.

1) Frequency Division Multiplexing ( FDM )

2) Wave Division Multiplexing (WDM)

3) Time Division Multiplexing ( TDM )


1) Frequency Division Multiplexing ( FDM)

In FDM, Signals generated by each sending device modulate different carrier frequencies. These modulated signals are then combined into a single composite signal that can be transported by the link.

Guard Band – is the strip of unused Bandwidth to prevent signals from overlapping.

In the figure, the transmission path is divided into three channels.


FDM processEach telephone generates a signal of similar frequency . Inside the multiplexer, these similar signals are modulated onto different carrier frequencies f1, f2, f3. The resulting modulated signals are then combined in to a single composite signal that is sent out over a media.


FDM demultiplexing example

The demultiplexer uses a series of filters to decompose the multiplexed signal into its constituent component signals. The individual signals are then passed to a demodulator that separates them from their carriers and passes them to the receivers.



Example VTU 2006

Define Multiplexing. Explain FDM.


Example-1

Assume that a voice channel occupies a bandwidth of 4 KHz. We need to combine three voice channels into a link with a bandwidth of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency domain with out the use of guard bands.

SolutionModulate each of the three voice channels to a different bandwidth.The diagram is as shown in Figure


Figure


Example-2Example-2

Five channels, each with a 100-KHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 KHz between the channels to prevent interference?

SolutionSolution

For five channels, need at least four guard bands. There fore, minimum required bandwidth = 5 x 100 + 4 x 10 = 540 KHz. The diagram is as shown in Figure


Figure


Example-3 Example-3

Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM

SolutionSolution

The satellite channel is analog. Divide 1MHz into four channels, each channel having a 250-KHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits are modulated to 1 Hz. One solution is 16-QAM modulation. Figure shows one possible configuration.


Example


The FDM hierarchy

To maximize the efficiency, Telephone connections use hierarchy of multiplexed signals from low-band width lines to high bandwidth lines. The hierarchy is made up of Groups, Super groups, Master groups, and Jumbo group.



FDM was introduced in the year 1930s. The basic analog multiplexer combines 12 voice channels in one line. Each voice signal occupies about

4 kHz. The channels are assigned with 4 kHz of bandwidth to provide guard bands between channels. The multiplexer modulates each voice signal so that it occupies a 4 kHz slot in the band between 60 & 108 kHz. The combined signal is called a group. A collection of 5 groups is called Super group. A In a super group 5 groups, each of bandwidth 48 kHz, into the frequency band from 312 to 552 kHz.

10 super groups be multiplexed to master group of 600 voice signals. This occupies a band 564 to 3084 kHz.

Examples of FDM

Broadcast radio- FM, AM, & television with 10 kHz, 200 kHz, & 6 MHz

Cellular Phones- A pool of frequency slots of 30 kHz each are shared by the users with in a geographic cell.


WDM

Wave Division Multiplexing


Wave Division Multiplexing

WDM is used in the transmission of OFC networks. OFC has high data rate capacity.

Each telephone generate signals of similar frequency . Inside the Wave division multiplexer, these signals are modulated onto different narrow band carrier frequencies f1, f2 , f3. The resulting modulated signals are then combined into a single Optical signal of wider band of light and transported over OFC channel.


Prisms in WDM multiplexing and demultiplexing

A prism has the property of Combining and splitting of light sources. A prism bends a beam of light based on the angle of incidence and frequency. Using this technique, a multiplexer can be made to combine several input beams into one output beam of a wider band of frequencies. A demultiplexer can also be made to reverse the process.


TDMTDM

Time – Division Multiplex



TDM is a digital multiplexing technique.

Each connection occupies a portion of time in the link.

If the source signal is an analog, it can be changed to digital data and then multiplexed using TDM.

In synchronous TDM, the data flow of each input connection is divided into units. Each unit occupies one input time slot. Combination of units is called a frame.


Time – Division Multiplex (TDM)

In the figure, connections 1,2,3, & 4 occupy a portion of the link.


TDM frames - For n- input connections, & T- input time, each connection gets a time slot of T/n. During that time a connection gets a chance of transmitting data for T/n of time, called one unit. The link combines these units into a frame.


ExampleExample

Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find (i) the duration of 1 bit before multiplexing, (ii) the transmission rate of the link, (iii) the duration of a time slot, and (iv) the duration of a frame?

Solutioni) The duration of 1 bit = 1/ (1 Kbps) = 0.001 s = 1 msii) The rate of the link= 4 times the rate of connection= 4 Kbps.iii) The duration of each time slot= 1/4 the bit duration = 1/4 ms = 250 s. iv) The duration of a frame= ¼ +1/4 + 1/4 +1/4 = 1 ms.



Example VTU Jan 2006

In a TDM system, four channels of 1 kbps each are multiplexed together. Each

time slot , called as a unit, carries 1 bit of information. Data from each time

slot is transmitted as a frame. Find: i) Transmission rate of the output link of

the multiplexer. Ii) The duration of a frame.


Interleaving in TDM

TDM can be visualized as two fast rotating switches, one on the multiplexing side and the other on the demultiplexing side. The switches are synchronized and rotate at the same speed but in the opposite directions. On the multiplexing side, as the switch opens in front of a connection, that connection has the opportunity to send a unit onto the path. This process is called interleaving. On the demultiplexing side, as the switch opens in front of a connection, that connection has the opportunity to receive a unit from the path.


Example

Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate of the link.

Solution (Next slide)



1) Each Frame carries 1 byte from each channel.

So, the size of each frame = 4 bytes = 32 bits.

2) Duration of a Frame = 1/ 100 bytes per sec = 0.01 S

3) Frame rate = 4 x100/4 = 100 frames.

4) Bit rate = Number of frames transmitted per sec = 100 frames = 100 x 32 = 3200 bps.


Example: A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate, the frame duration, bit rate &the bit duration?Solution (Next slide)Solution (Next slide)



Each Channel send 100 kbps = 100,000 bits / s

The link carries = 100,000/ 2= 50,000 frames per Sec.

• Frame rate = 50,000

• Frame duration = 1/ 50,000 = 20 micro Sec

• Bit rate = 50000 x 8 (frame size ) = 400,000 bits = 400 Kbps

• Bit duration = 1/ 400,000 = 2.5 micro Sec



Example VTU July 2007

Define Multiplexing. How does TDM multiplex & Demultiplex the signals.

Exam: VTU July 2007

Four sources each create 250 characters per second. If the interleaved unit is a character and one synchronization bit is added to each frame, find

i) the data rate of each source ii) Duration of each character in each source iii) The frame rate iv) Duration of each frame v) Number of bits in each frame vi) The data rate of the link.


Framing bits & Synchronization If the multiplexer and demultiplexer are not synchronized, a bit belongs to one channel may be received by the wrong channel. For this reason, one or more synchronization bits are added to the beginning of each frame. These bits are called framing bits. These framing bits follow a pattern (Alternating 0 & 1) from frame to frame. This helps the demultiplexer to synchronize with the incoming stream , & it can separate time slots accurately.


ExampleExample

We have four sources, each creating 250 characters per second. If the interleaved unit is a character and one synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link.

SolutionSolution

See next slide.


Solution (continued)Solution (continued)

We can answer the questions as follows:

1. The data rate of each source is 2000 bps = 2 Kbps.

2. The duration of a character = 1/250 s = 4 ms.3. The frame rate = 250 frames per second.4. The duration of each frame = 1/250 s, or 4 ms. 5. No. of bits in each frame = 4 x 8 + 1 = 33 bits.6. The data rate of the link = 250 x 33= 8250 bps.



BIT PADDING

Suppose device A could use one time slot, while the faster device B could use two. Since the time slot is fixed, the different data rate must be integer multiples of each other.

When the speeds are not integer multiples of each other, they can be made to behave as if they were, by a technique called Padding.

For example, One device with a bit rate of 2.55 times that of the other devices. By adding enough bits , make three times the other devices. The extra bits are then discarded by the demultiplexer.


Digital Signal hierarchy

Lower order multiplexed signals are called Tributaries.www.bookspar.com | Website for students | VTU NOTES

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UNIT - 3