data comm presentation

8/2/2019 Data Comm Presentation

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Digital To Analog Conversion


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Digital-to-analog conversion is the process of changing

one of the characteristics of an analog signal based on

the information in digital data.


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Types of Digital to Analog Conversion


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Digital to Analog Conversion

Digital data needs to be carried on an analogsignal.

A carrier signal (frequency fc) performs thefunction of transporting the digital data in an

analog waveform. The analog carrier signal is manipulated to

uniquely identify the digital data being carried.


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Aspects of Digital to Analog Conversion

Data element and signal elementData element is the smallest piece of information to be

exchanged i.e. bits.

Signal elements is the smallest unit of a signal

Data rate and signal rateS=N*1/r => S


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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.


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Bandwidth:

Bandwidth is proportional to the signal rate except

in FSK where a difference between the carriersignals needs to be added.

Carrier Signal:

In analog transmission, the sending deviceproduces a high frequency signal which acts asthe base signal for information signal. This isknown as carrier signal

The receiving device must be tuned to this carrierfrequency.


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Amplitude Shift Keying

Here the amplitude of the signal is varied tocreate the signal elements while frequencyand phase are kept constant.

Types: Binary ASK (BASK) and MultilevelASK


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Binary ASK Here ASK is implemented with only two levels Also known as on-off keying(OOK).

Peak amplitude of one signal level is 0, while for theother its same as amplitude of carrier frequency.


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Bandwidth for ASK:

Bandwidth is proportional to signal rate.

But there is another factor involved, called d, whichdepends on modulation and filtering process.

Value for d ranges between 0 and 1

B = (1+d) * S

It is clear from the formula that minimum required

bandwidth is S and maximum allowed value is 2S.


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Implementation of Binary ASK:

Here we have a Unipolar NRZ signal on the carrier signalfrom the oscillator.

When amplitude of signal is 1 amplitude of carrier

frequency is retained. When amplitude of signal becomes 0 then amplitude of

carrier frequency is also made zero.


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Example

We have an available bandwidth of 100 kHz whichspans 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.

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.


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Multilevel ASK:

In multilevel ASKwe can also have more than twolevels.

Like 4,8,16 and so on using 2,3,4 or more bits at a

time.

Not implemented with pure ASK rather it is

implemented with QAM.


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Frequency Shift Keying Here the frequency of the carrier signal is varied to

represent the data.

Frequency is constant during the duration of onesignal element but changes the next signal element if

the data element changes.

Peak amplitude and phase remain constant.

Two types: Binary FSK and Multilevel FSK


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Binary FSK:

Here two carrier frequencies say f1 and f2 are used.

We use the f1 when data element is 0 and f2 whendata element is 1.


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Bandwidth of BFSK:

From previous figure we can see that middle of onebandwidth is f1 and middle of other bandwidth is f2.

The difference between the two frequencies is 2f.

B = (1+d) * S + 2f

The minimum value for 2f must be equal to S for

proper operation of modulation and demodulation.

E l


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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?

SolutionThe midpoint of the band is at 250 kHz. We choose 2fto

be 50 kHz; this means


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Implementation of BFSK:

Two ways: noncoherent and coherent.

In noncoherent BFSK there is a discontinuity in phasewhen one signal element ends and the next begins.

In coherent the phase is continuous.


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VCO changes its frequency according to the input

voltage.

The input is unipolar NRZ.

VCO keeps its regular frequency when amplitude ofinput is zero and when input is 1 frequency isincreased.


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Multilevel FSK:

We use more than two frequencies.

Like to send 2 bits we can have 4 differentfrequencies say f1,f2,f3 and f4.

To send 3 bits we can use 8 bits and so on.

Frequencies must be 2f apart and minimum value of2f must be S.

Bandwidth is

B = (1+d) * S + (L-1)/2f

= L * S (when d=0)

E l


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Example

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), the baud rate, and the bandwidth.

Solution

We can have L = 23 = 8. The baud rate is S = 3 Mbps/3 = 1 Mbaud.

This means that the carrier frequencies must be 1 MHz apart (2f=1 MHz). The bandwidth is B = 8 1M = 8M. Figure shows the

allocation of frequencies and bandwidth.


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Phase Shift Keying(PSK)

In the phase shift keying, We vary the phase shift ofthe carrier signal to represent digital data. Both peak

amplitude and frequency remain constant as thephase changes. Thus it is more common than ASKand FSK.

Two types are: Binary PSK(BPSK) and QuadraturePSK(QPSK).

1.Binary PSK(BPSK) Here we have two signal elements one with a

phase of 0o and other with phase 180o .

It is more simple than binary ASK with oneadvantage-it is less susceptible to noise.


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Figure 5.9 Binary phase shift keying


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Implementation of Binary PSK

It is simple as that of ASK because the signalelement with phase 180o can be seen ascomplement of signal with phase 0o.

The polar NRZ signal is multiplied by the carrierfrequency;

The 1 bit is represented by a phase starting at

0o.The 0 bit is represented by a phase starting at

180o.


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Quadrature PSK(QPSK)

QPSK uses two separate BPSK modulations; one isin-phase, the other is quadrature(out-of-phase).

The incoming bits are passed through a serial toparallel converter that sends one bit to one

modulator and next bit to other modulator.

If the duration of each bit in the incoming signal is T,the duration of each bit sent to corresponding BPSKis 2T.


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Figure 5.11 QPSK and its implementation


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Example

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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 = 6MHz.


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Constellation Diagram A constellation diagram helps us to define the

amplitude and phase of a signal when we are usingtwo carriers, one in quadrature of the other.

The X-axis represents the in-phase carrier and the Y-axis represents quadrature carrier.

For each point in diagram four pieces of informationcan be deduced:-

1.The projection of point on X-axis defines the peakamplitude of in-phase component.

2.The projection of point on Y-axis defines the peakamplitude quadrature component.

3.The length of the line that connects the point to theorigin is

the peak amplitude of the signal element.

4.The angle of the line with X-axis the phase of the


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Concept of a constellation diagram


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Three constellation diagrams


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Quadrature Amplitude Modulation

It is a combination of ASK and PSK. Thus it uses two carriers one in-phase and other

quadrature with different amplitude levels for eachcarrier.

The minimum bandwidth for QAM transmission issame as that for ASK and PSK transmission.

Figure represent:

a. The simplest 4-QAM scheme using unipolar NRZsignal to modulate each carrier.

b. 4-QAM using polar NRZ, exactly same as QPSK.

c. Signals with two positive levels to modulate each ofthe two carriers.

d.16-QAM constellation of a signal with eight levels, four

positive and four negative.


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Constellation diagrams for some QAMs

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