1 Intro Digital Communication

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    Digital Communication

    Introduction

    By- D. Sarma

    Asst. Prof., Dept. Of ECE,DBCET

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    Why Digital Communication

    Easy to regenerate. Less subject to distortion- Since in two

    operated in either on or off . Digital circuits are more reliable and of lower

    cost. Provides encryption and privacy. Finally signal processing.

    Non Graceful Degradation

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    Basic Flow of Digital Communication

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    Some milestones in the history of electrical communications

    Versi on

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    Milestones in the history of electrical communicationscontd .

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    Electromagnetic bands with typical applications

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    Electromagnetic bands contd..

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    A few important frequency bands

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    Some Carrier frequency values and nominal bandwidth that may be

    available at the carrier frequency

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    Typical power losses duringtransmission through a few media

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    The PCM system

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    Introduction Two basic operations in the conversion of analog signal

    into the digital.Time discretization

    Amplitude discretization.In the context of PCM, The first is accomplished with the sampling operation The second by means of quantization. PCM involves another step,

    Conversion of quantized amplitudes into a sequence of simpler pulse patterns (usually binary) - code words.

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    Block Diagram

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    m(t ) is the information bearing messagesignal that is to be transmitted digitally.

    m(t ) is first sampled and then quantized. The quantizer converts each sample to one of

    the values that is closest to it from among apre-selected set of discrete amplitudes.

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    4.15

    Sampling Analog signal is sampled every T S secs. Ts is referred to as the sampling interval. f s = 1/Ts is called the sampling rate or sampling

    frequency.

    According to the Nyquist theorem, thesampling rate must be

    at least 2 times the highest frequencycontained in the signal.

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    Let a signal x (t ) be band limited to W Hz; that

    is, X (f ) = 0 for f > W . Let X(nTs ) = X(t ) t=nTs , - < n < represent

    the samples of x (t ) at uniform intervals of Ts

    seconds. If Ts 1/ 2W , than it is is possible to

    reconstruct x (t ) exactly from the set of

    samples, {X (n Ts )} .

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    Types of sampling There are 3 sampling methods:

    Ideal - an impulse at each sampling instant Natural - a pulse of short width with varying amplitude Flattop - sample and hold, like natural but with single

    amplitude value The process is referred to as pulse amplitude

    modulation PAM and the outcome is a signal withanalog (non integer) values

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    Ideal impulse sampling

    Consider an arbitrary low pass signal x (t )shown in Fig 3 . Let

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    where ( t ) is the unit impulse function of Xs (t) , shown in red in Fig 3. consists of a

    sequence of impulses; the weight of the impulse at t = nTs is equal to

    x (nTs ) . X s (t ) is zero between two adjacent impulses.

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    Fig 3:

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    It is very easy to show in the frequencydomain that Xs (t ) preserves the completeinformation of x (t ).

    X s (t ) is the product of x (t ) and (t n Ts)hence the corresponding Fourier relation isconvolution.

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    That is,

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    we see that Xs (f ) is a superposition of X (f )and its shifted versions (shifted by multiples of fs , the sampling frequency) scaled by 1/ Ts

    Let X (f ) be a triangular spectrum as shown inFig. 4(a).

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    Fig 4:

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    it is obvious that we can recover x (t ) from Xs(t ) by passing Xs (t ) through an ideal low

    pass filter with gain Ts and bandwidth W asshown bellow.

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    Of course, with respect to Fig. 4(b), whichrepresents the over-sampled case,

    reconstruction filter can have some transitionband which can fit into the gap between f = W and f = (fs W) .

    when fs < 2W , (under-sampled case) we seethat spectral lobes overlap resulting in signaldistortion, called aliasing distortion. In this

    case, exact signal recovery is not possible

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    Sampling with a rectangular pulsetrain

    Let yp (t ) represent the periodic rectangularpulse train as shown in Fig.5(b).

    Let Xs (t ) =[ x (t )] yp (t ) Then, Xs (f ) = X (f ) Yp (f ) Therefore

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    4.30

    Figure : Three different sampling methods for PCM

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    4.31

    Quantization Sampling results in a series of pulses of varying

    amplitude values ranging between two limits: amin and a max.

    The amplitude values are infinite between thetwo limits.

    We need to map the infinite amplitude valuesonto a finite set of known values.

    This is achieved by dividing the distance betweenmin and max into L zones , each of height= (max - min)/L

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    4.32

    Quantization Levels

    The midpoint of each zone is assigned avalue from 0 to L-1 (resulting in L values)

    Each sample falling in a zone is thenapproximated to the value of the midpoint.

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    4.33

    Quantization Error When a signal is quantized, we introduce an error

    - the coded signal is an approximation of theactual amplitude value.

    The difference between actual and coded value(midpoint) is referred to as the quantization error.

    The more zones, the smaller which results insmaller errors.

    BUT, the more zones the more bits required toencode the samples -> higher bit rate

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    4.34

    Quantization Error and SN Q R Signals with lower amplitude values will suffer

    more from quantization error as the error range:/2, is fixed for all signal levels.

    Non linear quantization is used to alleviate thisproblem. Goal is to keep SN Q R fixed for all samplevalues.

    Two approaches: The quantization levels follow a logarithmic curve.

    Smaller s at lower amplitudes and larger s at higheramplitudes.

    Companding: The sample values are compressed at thesender into logarithmic zones, and then expanded atthe receiver. The zones are fixed in height.

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    4.35

    Bit rate and bandwidth requirements of

    PCM The bit rate of a PCM signal can be calculated form the

    number of bits per sample x the sampling rate

    Bit rate = n b x f s The bandwidth required to transmit this signal depends

    on the type of line encoding used. Refer to previoussection for discussion and formulas.

    A digitized signal will always need more bandwidth thanthe original analog signal. Price we pay for robustness andother features of digital transmission.

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    4.36

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

    SolutionThe human voice normally contains frequencies from 0to 4000 Hz. So the sampling rate and bit rate arecalculated as follows:

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    4.37

    PCM Decoder

    To recover an analog signal from a digitized signalwe follow the following steps:

    We use a hold circuit that holds the amplitude value of a pulse till the next pulse arrives. We pass this signal through a low pass filter with a

    cutoff frequency that is equal to the highest frequencyin the pre-sampled signal.

    The higher the value of L, the less distorted asignal is recovered.

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    4.38

    We have a low-pass analog signal of 4 kHz. If we send theanalog signal, we need a channel with a minimumbandwidth 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.

    A brief aside about ADCs

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    39

    0000

    1111

    1110

    1101

    1100

    1011

    1010

    1001

    0001

    0010

    0011

    0100

    0101

    0110

    0111

    A brief aside about ADCs

    0000 0110 0111 0011 1100 1001 1011

    Numbers passed from ADC to computer to represent analogue voltage

    ADCs are used to convert an analogue input voltage into a number that can beinterpreted as a physical parameter by a computer.

    Resolution= 1 part in 2 n

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    THANK YOU