digital and analog transmission

7/30/2019 Digital and Analog Transmission

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1Dr. L. Christofi Spring 2009

Lecture 3

Digital TransmissionAnalog Transmission

ACOE412

Data Communications

Spring 2009


0. Overview0. OverviewIn this lecture we will cover the following topics:

4. Digital Transmission4.1 Digital to digital conversion4.2 Analog to digital conversion4.3 Transmission modes4.4 Summary (part 4)

5. Analog Transmission5.1 Digital to analog conversion5.2 Analog and digital5.3 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 byIn this section, we see how we can represent digital data byusing digital signals. The conversion involves threeusing digital signals. The conversion involves threetechniques:techniques: line codingline coding andand scramblingscrambling . Line coding is. Line coding isalways needed; scrambling may or may not be needed.always needed; scrambling may or may not be needed.

Line CodingLine Coding SchemesScrambling

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


Line coding and decodingLine coding is the process of converting binary data i.e. asequence 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 totransmit a symbol

The bit rate define sthe maximum number of bits persecond

The relationship between pulse rate and bit rate is

BitRate = PulseRate x log 2L

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. Findthe Pulse Rate and Bit Rate.

Solution

Example

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

Bit Rate = Pulse Rate x log 2L =1000 x log 24 = 2000bps


Effect of lack of synchronization


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In a digital transmission, the receiver clock is 0.1% faster thanthe sender clock. How many extra bits per second does thereceiver receive if the data rate is 1 kbps? How many if thedata rate is 1 Mbps?

SolutionAt 1 kbps, the receiver receives 1000(1+0.001) = 1001 bpsinstead 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 timinginformation in the data being transmitted. This can beachieved if there are transitions in the signal that alertthe receiver to the beginning, middle or end of pulse.If the receivers 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 valueof the bit.

In NRZ-I the inversionor 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, notbetween bits: bit 1 = high-to-zero bit 0 = low-to-zero

Main disadvantage of RZ encoding is that it requires twosignal changes to encode 1 bit => occupies more bandwidth


Polar biphase: Manchester anddifferential 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 binaryAs we have seen, line coding can be used to convert binarydata to a digital signal. Sometimes however, our data isdata to a digital signal. Sometimes however, our data isanalog, such as audio. If we want to store the audio byanalog, such as audio. If we want to store the audio byrecording 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 ModulationPulse 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 pulsesbased on the results of sampling. Sampling means measuring the amplitude of the signal at equalintervals However PAM is not useful to data communications because eventhough it translates the original waveform to a series of pulses, thesepulses are still of any amplitude (still an analog, not a digital signal). Tomake 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 ratemust be

at least 2 times the highest frequency contained in thesignal.

Note


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Nyquist sampling rate for low-pass andbandpass signals


For an intuitive example of the Nyquist theorem, let ussample a simple sine wave at three sampling rates: f s = 4f(2 times the Nyquist rate), f s = 2f (Nyquist rate), andfs = f (one-half the Nyquist rate). Figure in the next slideshows the sampling and the subsequent recovery of thesignal.

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 sameapproximation, but it is redundant and unnecessary.Sampling below the Nyquist rate (part c) does not producea 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 byassuming a maximum frequency of 4000 Hz. The samplingrate therefore is 8000 samples per second.

Example


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A complex low-pass signal has a bandwidth of 200 kHz. Whatis the minimum sampling rate for this signal?

SolutionThe bandwidth of a low-pass signal is between 0 and f, where fis the maximum frequency in the signal. Therefore, we cansample 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. Whatis the minimum sampling rate for this signal?

SolutionWe cannot find the minimum sampling rate in this casebecause 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 sampledsignal


We want to digitize the human voice. What is the bit rate,assuming 8 bits per sample?

SolutionThe human voice normally contains frequencies from 0 to4000 Hz. So the sampling rate and bit rate are calculated asfollows:

Example


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Components of a PCM decoder


We have a low-pass analog signal of 4 kHz. If we send theanalog signal, we need a channel with a minimum bandwidthof 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 beThe transmission of binary data across a link can beaccomplished in either parallel or serial mode. In parallelaccomplished in either parallel or serial mode. In parallelmode, multiple bits are sent with each clock tick. In serialmode, multiple bits are sent with each clock tick. In serialmode, 1 bit is sent with each clock tick. While there is onlymode, 1 bit is sent with each clock tick. While there is onlyone way to send parallel data, there are two subclasses ofone way to send parallel data, there are two subclasses ofserial transmission: asynchronous and synchronous.serial transmission: asynchronous and synchronous.

Parallel TransmissionSerial 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 betweeneach 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 afteranother 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 sequenti ally. 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 theresponsibility of the receiver.


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

DigitalDigital --toto --analoganalog conversion is the process of changingconversion is the process of changingone of the characteristics of an analog signal based onone of the characteristics of an analog signal based onthe information in digital data.the information in digital data.

Aspects of Digital-to-Analog Conversion

Amplitude Shift KeyingFrequency Shift KeyingPhase Shift KeyingQuadrature 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 persecond.

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

spects of digital- o-analog conversion


An analog signal carries 4 bits per signal element. If 1000signal elements are sent per second, find the bit rate.

SolutionIn this case, r = 4, S = 1000, and N is unknown. We can findthe value of N from

Example


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Example

An analog signal has a bit rate of 8000 bps and a baud rateof 1000 baud. How many data elements are carried byeach signal element? How many signal elements do weneed?

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 baudrate. BW = (1+d) S baud , where d=modulation factor


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Implementation of ASK


Example

We have an available bandwidth of 100 kHz which spansfrom 200 to 300 kHz. What are the carrier frequency andthe bit rate if we modulated our data by using ASK withd = 1?

SolutionThe middle of the bandwidth is located at 250 kHz. This meansthat our carrier frequency can be at f c = 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 linkswith communication in both directions. We need to dividethe bandwidth into two with two carrier frequencies, asshown in figure below. The figure shows the positions of twocarrier frequencies and the bandwidths. The availablebandwidth for each direction is now 50 kHz, which leaves uswith 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 plusthe frequency difference between the two carriers: BW = (f c2 -fc1) + S baud


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Example

We have an available bandwidth of 100 kHz which spans from 200 to300 kHz. What should be the carrier frequency and the bit rate if wemodulated our data by using FSK with d = 1?

SolutionThis problem is similar to Example 5.3, but we are modulating by usingFSK. 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. Peakamplitude 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 thatfor ASK: BW = (1+d) S baud , 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 forQPSK. The value of d = 0.

SolutionFor QPSK, 2 bits is carried by one signal element. This meansthat r = 2. So the signal rate (baud rate) is S = N (1/r) = 6Mbaud. 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, andQPSK signals.

Solution

Example


Quadrature Amplitude Modulation is a combination ofASK 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 ofanalog conversion is the representation ofanalog information by an analog signal. One may ask whyanalog information by an analog signal. One may ask whywe 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 isModulation is needed if the medium is bandpassbandpass in naturein natureor if only aor if only a bandpassbandpass channel is available to us.channel is available to us.

Amplitude ModulationFrequency ModulationPhase Modulation



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Types of analog-to-analog modulation


Amplitude Modulation (AM)

In AM radio, the bandwidth of the modulated signal must betwice 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 be10 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 t he 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 oneor 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 baudrate.

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 ,6 th edition, Prentice Hall, 2000

F. Halsall, Data Communications, Computer Networks andOpen Systems , 4 th edition, Addison Wesley, 1995

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

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

digital and analog transmission

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