Lecture 3 - FIT

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1Dr. L. Christofi Spring 2009

Lecture 3

Digital Transmission

Analog Transmission

ACOE412

Data Communications

Spring 2009


0. Overview0. Overview

In this lecture we will cover the following topics:

4. Digital Transmission

4.1 Digital to digital conversion

4.2 Analog to digital conversion

4.3 Transmission modes

4.4 Summary (part 4)

5. Analog Transmission

5.1 Digital to analog conversion

5.2 Analog and digital

5.3 Summary (part 5)

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4.1 DIGITAL4.1 DIGITAL--TOTO--DIGITAL CONVERSIONDIGITAL CONVERSION

In this section, we see how we can represent digital data by In this section, we see how we can represent digital data by

using digital signals. The conversion involves three using digital signals. The conversion involves three

techniques: techniques: line codingline coding and and scramblingscrambling. Line coding is . Line coding is

always needed; scrambling may or may not be needed.always needed; scrambling may or may not be needed.

Line Coding

Line Coding SchemesScrambling

Topics discussed in this section:Topics discussed in this section:


Line coding and decoding

Line coding is the process of converting binary data i.e. a

sequence of bits, into a digital signal

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Signal element versus data element


Pulse Rate vs Bit Rate

• The pulse rate defines the number of pulses per second

—A pulse is the minumum amount of time required to transmit a symbol

• The bit rate define sthe maximum number of bits per

second

• The relationship between pulse rate and bit rate is

BitRate = PulseRate x log2L

where, L is the number of data levels of the signal

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A signal has four data levels with a pulse duration of 1ms. Find

the Pulse Rate and Bit Rate.

Solution

Example

Pulse Rate = 1/1ms=1000 pulses per second

Bit Rate = Pulse Rate x log2L =1000 x log24 = 2000bps


Effect of lack of synchronization

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In a digital transmission, the receiver clock is 0.1% faster than the sender clock. How many extra bits per second does the

receiver receive if the data rate is 1 kbps? How many if the

data rate is 1 Mbps?

Solution

At 1 kbps, the receiver receives 1000(1+0.001) = 1001 bps

instead of 1000 bps.

Example

At 1 Mbps, the receiver receives 1,000,000(1+0.001)=

1,001,000 bps instead of 1,000,000 bps.


A self-synchronising digital signal includes timing information in the data being transmitted. This can be achieved if there are transitions in the signal that alert the receiver to the beginning, middle or end of pulse. If the receiver’s clock is out of synchronisation, these

alerting points can reset the clock.

Note

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Line coding schemes

Line coding


Unipolar NRZ scheme

Unipolar encoding uses only one voltage level

bit 0 = zero bit 1=high

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Polar NRZ-L and NRZ-I schemes

• Polar encoding uses two voltage levels (+ve and –ve)

• In NRZ-L encoding: bit 0 = high bit 1=low

• In NRZ-I encoding the signal is inverted if 1 is encountered


In NRZ-L the level of the voltage determines the value of the bit.

In NRZ-I the inversion or the lack of inversion

determines the value of the bit.

Note

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Polar RZ scheme

• Polar encoding uses three voltage levels (+ve, zero & –ve)

• In RZ encoding the signal changes during each bit, not between bits: bit 1 = high-to-zero bit 0 = low-to-zero

• Main disadvantage of RZ encoding is that it requires two

signal changes to encode 1 bit => occupies more bandwidth


Polar biphase: Manchester and

differential Manchester schemes

• In Manchester encoding: bit 1 = low-to-high bit 0 = high-to-low

• In Differential Manchester encoding: bit 1 = no transition bit 0 = transition

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In Manchester and differential Manchester encoding,

the transition

at the middle of the bit is used for synchronization.

Note

The minimum bandwidth of Manchester and differential Manchester is 2 times that of NRZ.

Note


Bipolar schemes: AMI and pseudoternary

• Bipolar encoding uses three voltage levels (+ve, zero & –ve)

• Alternate Mask Inversion (AMI):

bit 0 = zero bit 1 = alternating +ve and –ve pulses

• Pseudoternary:

bit 0 = alternating +ve and –ve pulses bit 1 = zero

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Summary of line coding schemes


4.2 ANALOG4.2 ANALOG--TOTO--DIGITAL CONVERSIONDIGITAL CONVERSION

As we have seen, line coding can be used to convert binary As we have seen, line coding can be used to convert binary

data to a digital signal. Sometimes however, our data is data to a digital signal. Sometimes however, our data is

analog, such as audio. If we want to store the audio by analog, such as audio. If we want to store the audio by

recording it in a computer so that we can send it digitally, recording it in a computer so that we can send it digitally,

we need to change it through a process called sampling.we need to change it through a process called sampling.

Pulse Amplitude Modulation

Pulse Code Modulation


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Pulse Amplitude Modulation (PAM)

• PAM is an analog-to-ditigal conversion method.

• It takes an analog signal, samples it and generates a series of pulses

based on the results of sampling.

• Sampling means measuring the amplitude of the signal at equal intervals

• However PAM is not useful to data communications because even

though it translates the original waveform to a series of pulses, these

pulses are still of any amplitude (still an analog, not a digital signal). To

make them digital we modify them using Pulse Code Modulation (PCM).


Components of PCM encoder

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Three different sampling methods for PCM


According to the Nyquist theorem, the sampling rate must be

at least 2 times the highest frequency contained in the signal.

Note

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Nyquist sampling rate for low-pass and

bandpass signals


For an intuitive example of the Nyquist theorem, let us

sample a simple sine wave at three sampling rates: fs = 4f

(2 times the Nyquist rate), fs = 2f (Nyquist rate), and

fs = f (one-half the Nyquist rate). Figure in the next slide shows the sampling and the subsequent recovery of the

signal.

It can be seen that sampling at the Nyquist rate can create

a good approximation of the original sine wave (part a). Oversampling in part b can also create the same

approximation, but it is redundant and unnecessary.

Sampling below the Nyquist rate (part c) does not produce

a signal that looks like the original sine wave.

Example

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Recovery of a sampled sine wave for

different sampling rates


Telephone companies use PCM to digitize voice by

assuming a maximum frequency of 4000 Hz. The sampling

rate therefore is 8000 samples per second.

Example

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A complex low-pass signal has a bandwidth of 200 kHz. What

is the minimum sampling rate for this signal?

SolutionThe bandwidth of a low-pass signal is between 0 and f, where f

is the maximum frequency in the signal. Therefore, we can

sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is therefore 400,000 samples per second.

Example


A complex bandpass signal has a bandwidth of 200 kHz. What

is the minimum sampling rate for this signal?

SolutionWe cannot find the minimum sampling rate in this case

because we do not know where the bandwidth starts or ends.

We do not know the maximum frequency in the signal.

Example

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Quantization and encoding of a sampled

signal


We want to digitize the human voice. What is the bit rate,

assuming 8 bits per sample?

Solution

The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as

follows:

Example

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Components of a PCM decoder


We have a low-pass analog signal of 4 kHz. If we send the

analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample,

we need a channel with a minimum bandwidth of 8 × 4 kHz =

32 kHz.

Example

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4.3 TRANSMISSION MODES4.3 TRANSMISSION MODES

The transmission of binary data across a link can be The transmission of binary data across a link can be

accomplished in either parallel or serial mode. In parallel accomplished in either parallel or serial mode. In parallel

mode, multiple bits are sent with each clock tick. In serial mode, multiple bits are sent with each clock tick. In serial

mode, 1 bit is sent with each clock tick. While there is only mode, 1 bit is sent with each clock tick. While there is only

one way to send parallel data, there are two subclasses of one way to send parallel data, there are two subclasses of

serial transmission: asynchronous and synchronous.serial transmission: asynchronous and synchronous.

Parallel Transmission

Serial Transmission



Data transmission modes

Asynchronous Synchronous

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Parallel transmission


Serial transmission

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In asynchronous transmission, we send 1 start bit (0)

at the beginning and 1 or more stop bits (1s) at the

end of each byte. There may be a gap between each byte.

Note

Asynchronous here means “asynchronous at the byte level,”

but the bits are still synchronized; their durations are the same.

Note

Asynchronous transmission


Asynchronous transmission

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In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the

responsibility of the receiver to group the bits.

Note

Synchronous transmission


Synchronous transmission

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4.4 SUMMARY (part 4)

• Line coding is the process of converting binary data to a digital signal.

• The number of different values allowed in a signal is the signal level. The number of symbols that represent data is the data level.

• Bit rate is a function of the pulse rate and data level.

• Line coding methods must eliminate the dc component and provide a means of synchronization between the sender and the receiver.

• Line coding methods can be classified as unipolar, polar, or bipolar.

• NRZ, RZ, Manchester, and differential Manchester encoding are the most popular polar encoding methods.

• AMI is a popular bipolar encoding method.

• Analog-to-digital conversion relies on PCM (pulse code modulation).

• PCM involves sampling, quantizing, and line coding.

• The Nyquist theorem says that the sampling rate must be at least twice the highest-frequency component in the original signal.

• Digital transmission can be either parallel or serial in mode.

• In parallel transmission, a group of bits is sent simultaneously, with each bit on a separate line.

• In serial transmission, there is only one line and the bits are sent sequentially.

• Serial transmission can be either synchronous or asynchronous.

• In asynchronous serial transmission, each byte (group of 8 bits) is framed with a start bit and a stop bit. There may be a variable-length gap between each byte.

• In synchronous serial transmission, bits are sent in a continuous stream without start and stop bits and without gaps between bytes. Regrouping the bits into meaningful bytes is the responsibility of the receiver.


5.1 DIGITAL5.1 DIGITAL--TOTO--ANALOG CONVERSIONANALOG CONVERSION

DigitalDigital--toto--analoganalog conversion is the process of changing conversion is the process of changing

one of the characteristics of an analog signal based on one of the characteristics of an analog signal based on

the information in digital data. the information in digital data.

Aspects of Digital-to-Analog Conversion

Amplitude Shift Keying

Frequency Shift KeyingPhase Shift Keying

Quadrature Amplitude Modulation


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Digital-to-analog conversion


Types of digital-to-analog conversion

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Bit rate is the number of bits per second.

Baud rate is the number of signal elements per second.

In the analog transmission of digital data, the

Baud Rate = Bit Rate / Number of bits per signal unitBaud Rate = Bit Rate / Number of bits per signal unit

Note

Aspects of digital-to-analog conversion


An analog signal carries 4 bits per signal element. If 1000

signal elements are sent per second, find the bit rate.

Solution

In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from

Example

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Example

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 signal elements do we

need?

SolutionIn this example, S = 1000, N = 8000, and r and L are unknown.

We find first the value of r and then the value of L.


Amplitude Shift Keying (ASK)

• In ASK the amplitude of the carrier is varied to represent binary 1 or 0.

Frequency and phase remain constant.

• ASK is highly susceptible to noise interference

• Minimum bandwidth required for transmission is equal to the baud

rate. BW = (1+d) Sbaud, where d=modulation factor

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Implementation of ASK


Example

We have an available bandwidth of 100 kHz which spans

from 200 to 300 kHz. What are the carrier frequency and

the bit rate if we modulated our data by using ASK with

d = 1?

SolutionThe middle of the bandwidth is located at 250 kHz. This means

that our carrier frequency can be at fc = 250 kHz. We can use

the formula for bandwidth to find the bit rate (with d = 1 and r = 1).

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Example

In data communications, we normally use full-duplex links

with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as

shown in figure below. The figure shows the positions of two

carrier frequencies and the bandwidths. The available

bandwidth for each direction is now 50 kHz, which leaves us

with a data rate of 25 kbps in each direction.


Frequency Shift Keying (FSK)

• In FSK the frequency of the carrier is varied to represent binary 1 or 0.

Peak amplitude and phase remain constant.

• FSK avoids most of problems with noise.

• The bandwidth required for FSK transmission is equal to the baud rate plus

the frequency difference between the two carriers: BW = (fc2-fc1) + Sbaud

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Example

We have an available bandwidth of 100 kHz which spans from 200 to

300 kHz. What should be the carrier frequency and the bit rate if we

modulated our data by using FSK with d = 1?

Solution

This problem is similar to Example 5.3, but we are modulating by using

FSK. The midpoint of the band is at 250 kHz. We choose 2∆f to be 50 kHz;

this means


Phase Shift Keying (PSK)

• In PSK the phase of the carrier is varied to represent binary 1 or 0. Peak

amplitude and frequency remain constant.

• This method is often called 2-PSK or binary PSK

• The minimum bandwidth required for PSK transmission is the same as that

for ASK: BW = (1+d) Sbaud, where d=modulation factor

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Implementation of binary PSK


QPSK (4-PSK) and its implementation

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Example

Find the bandwidth for a signal transmitting at 12 Mbps for

QPSK. The value of d = 0.

Solution

For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6

Mbaud. With a value of d = 0, we have B = S = 6 MHz.


Concept of a constellation diagram

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Show the constellation diagrams for an ASK, BPSK, and

QPSK signals.

Solution

Example


Quadrature Amplitude Modulation is a combination of ASK and PSK.

Note

QAM

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Constellation diagrams for some QAMs


5.2 ANALOG AND DIGITAL5.2 ANALOG AND DIGITAL

AnalogAnalog--toto--analog conversion is the representation of analog conversion is the representation of

analog information by an analog signal. One may ask why analog information by an analog signal. One may ask why

we need to modulate an analog signal; it is already analog. we need to modulate an analog signal; it is already analog.

Modulation is needed if the medium is Modulation is needed if the medium is bandpassbandpass in nature in nature

or if only a or if only a bandpassbandpass channel is available to us. channel is available to us.

Amplitude Modulation

Frequency Modulation

Phase Modulation


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Types of analog-to-analog modulation


Amplitude Modulation (AM)

In AM radio, the bandwidth of the modulated signal must be

twice the bandwidth of the modulating signal.

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AM band allocation


Frequency Modulation (FM)

In FM radio, the bandwidth of the modulated signal must be 10 times the bandwidth of the modulating signal.

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FM band allocation


Phase Modulation (PM)

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5.3 SUMMARY (part 5)

• Digital-to-analog modulation can be accomplished using the following: *Amplitude shift keying (ASK)—the amplitude of the carrier signal varies.*Frequency shift keying (FSK)—the frequency of the carrier signal varies.*Phase shift keying (PSK)—the phase of the carrier signal varies.*Quadrature amplitude modulation (QAM)—both the phase and amplitude of the carrier signal

vary.

• QAM enables a higher data transmission rate than other digital-to-analog methods.

• Baud rate and bit rate are not synonymous. Bit rate is the number of bits transmit-ted per second. Baud rate is the number of signal units transmitted per second. One signal unit can represent one or more bits.

• The minimum required bandwidth for ASK and PSK is the baud rate.

• The minimum required bandwidth (BW) for FSK modulation is BW =f c1-f c0 + N baud , where f c1 is the frequency representing a 1 bit, f c0 is the frequency representing a 0 bit, and N baud is the baud rate.

• ASK modulation is especially susceptible to noise.

• Because it uses two carrier frequencies, FSK modulation requires more bandwidth than ASK and PSK.

• PSK and QAM modulation have two advantages over ASK:*They are not as susceptible to noise.*Each signal change can represent more than one bit.

• Analog-to-analog modulation can be implemented by using the following:* Amplitude modulation (AM) * Frequency modulation (FM) * Phase modulation (PM)

• In AM radio, the bandwidth of the modulated signal must be twice the bandwidth of the modulating signal.

• In FM radio, the bandwidth of the modulated signal must be 10 times the bandwidth of the modulating signal.


References

• W. Stalling, Local and Metropolitan Area Networks, 6th edition, Prentice Hall, 2000

• F. Halsall, Data Communications, Computer Networks and Open Systems, 4th edition, Addison Wesley, 1995

• B.A. Forouzan, Data Communications and Networking, 4th edition, McGraw-Hill, 2007

• W. Stallings, Data and Computer Communications, 7th edition, Prentice Hall, 2004

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Lecture 3 - FIT