Baseband Digital Modulation

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    School of Electrical, Electronics and

    Computer Engineering

    University of Newcastle-upon-Tyne

    Baseband Digital Modulation

    Prof. Rolando Carrasco

    Lecture Notes

    University of Newcastle-upon-Tyne

    2007

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    Baseband digital information

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    Bit-rate, Baud-rate and

    Bandwidth

    B

    B

    1denotes the duration of the 1 bit

    Hence Bit rate =bits per second

    All the forms of the base band signalling shown transfer data at the same bit rate.

    E denotes the duration of the shortest signalling element.Baud rate is defined as the reciprocal of the duration of the shortest signalling element.

    Baud Rate =

    E

    1baud

    In general Baud Rate Bit Rate

    For NRZ : Baud Rate = Bit Rate

    RZ : Baud Rate = 2 x Bit Rate

    Bi-Phase: Baud Rate = 2 x Bit Rate

    AMI: Baud Rate = Bit Rate

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    Non Return to Zero (NRZ)

    The highest frequency occurs when the data is 1010101010.

    i.e.

    This sequence produces a square wave with periodic time E 2

    Fourier series for a square wave,

    If we pass this signal through a LPF then the maximum bandwidth would be 1/THz, i.e. to just allow the fundamental (1st harmonic) to pass.

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    Non Return to Zero (NRZ)

    (Contd)

    The data sequence 1010

    could then be completely

    recovered

    Hence the minimum channel bandwidth

    RateBaudSince

    RateBaud

    TB EE

    1

    22

    11min

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    Return to Zero (RZ)

    Considering RZ signals, the max frequency occurs when continuous 1s are transmitted.

    This produces a square wave with periodic timeE

    2

    2min

    RateBaudfB U

    If the sequence was continuous 0s, the signal would be V continuously, hence

    ''DCfL

    .

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    Bi-Phase

    Maximum frequency occurs when continuous

    1s or 0s transmitted.

    E

    1

    2min

    RateBaudfB U

    This is similar to RZ withBaud Rate = = 2 x Bit rate

    The minimum frequency occurs when the sequence is 10101010.

    e.g.

    B E

    2min

    RateBaudfB L

    In this case =

    Baud Rate = Bit rate

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    Digital Modulation and

    Noise

    The performance of Digital Data Systems is dependent on the bit error rate, BER, i.e.

    probability of a bit being in error.

    NasNbitsTotal

    EErrorsofNoP

    Digital Modulation

    There are four basic ways of sending

    digital data

    The BER (P) depends on several factors

    the modulation type, ASK FSK or

    PSK

    the demodulation method

    the noise in the system

    the signal to noise ratio

    Prob. of Error or BER,

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    Digital Modulation and

    Noise

    Amplitude Shift Keying ASK

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    Digital Modulation and

    Noise

    Frequency Shift Keying FSK

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    Digital Modulation and

    Noise

    Phase Shift Keying PSK

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    System Block diagram for

    Analysis

    DEMODULATORDETECTORDECISION

    For ASK and PSK

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    Demodulator-Detector-Decision

    FOR FSK

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    Demodulator

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    Demodulator Contd)

    TRCdesignHence

    dtVRC

    V INout

    1

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    Detector-Decision

    1V 0V- is the voltage differencebetween a 1 and 0.

    )22

    ( 21 VVVREF

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    Detector-Decision (Contd)

    ND is the noise at the Detector input.

    Probability of Error,

    DN

    erf 2212

    1

    Hence

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    0 v1v0 v

    0-

    P(v0)

    vn

    Probability density of binary signal

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    v0 v1

    2

    210

    2

    )(

    02

    1

    2

    1)(

    vv

    n evP

    )(1 nvP

    vn

    n

    vv

    vv

    e dvePn

    2

    20

    10

    2

    )(

    2

    12

    1

    Using the change of variable

    2

    0vvx n

    Probability density function of noise

    (*)

    DN2

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    22

    1

    01

    21

    vv

    dxx

    e eP

    dxezerfcz

    x

    22

    )(

    222

    1 011

    vverfcPe

    This becomes

    The incomplete integral cannot be evaluated analytically but can be recast as a

    complimentary error function, erfc(x), defined by

    Equations (*) and (**) become

    n

    vv

    vv

    e

    e

    dveP

    vverfP

    zerfzerfc

    n

    2

    21

    10

    2

    )(2

    0

    011

    2

    1

    221

    2

    1

    )(1)(

    (**)

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    It is clear from the symmetry of this problem that Pe0 is identical to Pe1 and the

    probability of errorPe, irrespective of whether a one or zero was transmitted, can

    be rewritten in terms ofv= v1v0

    221

    2

    1

    verfPe

    for unipolar signalling (0 and v)

    for polar signalling (symbol represented by voltage

    2

    v

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    Detector-Decision (Contd)

    PSKFSKASKOptimumFor

    PRK

    N

    SerfPSK

    NSerfFSK

    OOK

    N

    SerfASK

    IN

    INe

    IN

    INe

    IN

    INe

    ,,

    12

    1

    21

    21

    41

    2

    1

    dB/10inSNR10

    eP

    eP

    eP

    SNR in wattASK FSK PSK

    000.002415.848912

    00.00080.012710.0010

    0.00020.0060.03796.30968

    0.00240.0230.07913.98116

    0.01250.05650.13122.51194

    0.03750.1040.18671.58492

    0.07860.15870.23981.000

    Linear gainSNR in dB

    000.002415.848912

    00.00080.012710.0010

    0.00020.0060.03796.30968

    0.00240.0230.07913.98116

    0.01250.05650.13122.51194

    0.03750.1040.18671.58492

    0.07860.15870.23981.000

    Linear gainSNR in dB

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    Probability of Symbol Error

    1.00E-04

    1.00E-03

    1.00E-02

    1.00E-01

    1.00E+00

    0 2 4 6 8 10 12 14

    SNR in dB

    ProbabilityofSymbolError

    ASK

    FSK

    PSK

    Detector-Decision (Contd)

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    FM/ FSK Demodulation

    One form of FM/FSK demodulator is shown below

    In general VIN (t) will be

    tCosVtV INcIN )(

    IN ININ f 2Where is the input frequency (rad/sec)

    ttCosttCosV

    V

    BACosBACosCosBCosASince

    tCosVtCosVV

    tVtVV

    ININININ

    c

    x

    INcINcx

    ININx

    2

    2

    1

    )(.

    2

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    FM/ FSK Demodulation (Contd)

    INININc

    x

    ININININININc

    x

    CostCosV

    V

    ttCosttCosV

    V

    22

    22

    2

    )2(2

    )1(2

    22

    2

    2

    tCosV

    and

    tCosV

    IN

    c

    IN

    c

    i.e

    Thus there are two components

    Component (1) is at frequency 2 fINHz and component (2) is effectively a DC voltage if

    IN is constant.

    The cut-off frequency for the LPF is designed so that component (1) is removed and

    component (2) is passed to the output.

    tCos

    V

    V INc

    OUT 2

    2

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    FM/ FSK Demodulation (Contd)

    The V/F characteristics and inputs are shown below

    Analogue FM

    c

    cDCc

    mmDCout

    mmDCIN

    DCIN

    INout

    mc

    fTVf

    ftCosVVfei

    tCosVVV

    tmVV

    fVfcxmy

    Vf

    1,

    ..

    )(

    0

    0

    Modulation Index

    m

    m

    m

    c

    f

    V

    f

    f

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    FM/ FSK Demodulation (Contd)

    tnCosJVtVFM mcn

    ncs

    1

    )()(

    The spectrum of the analogue FM signal depends on and is given by

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    Digital FSK (Contd)

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    FM/ FSK Demodulation (Contd)

    Consider again the output from the demodulator IN

    c

    OUTCos

    VV

    2

    2

    4

    cT

    c

    cf

    T1

    cfThe delay is set to where and is the nominal carrier frequency

    c

    INcOUT

    ffCosVV

    42

    2

    2

    Hence

    c

    INcOUT

    ffCosVV

    22

    2

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    FM/ FSK Demodulation (Contd)

    The curve shows the demodulator F/V characteristics which in this case is non linear.

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    Practical realization of F/V process

    The comparator is LIMITERwhich is a zero crossing detector to give a digital input tothe first gate.

    This is form of delay and multiply circuit where the delay is set by C and R with

    = CR

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    Practical realization of F/V process (Contd)

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    Practical realization of F/V process (Contd)

    INf cf

    Consider now

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    Practical realization of F/V process (Contd)

    c

    INOUT

    f

    fAEV

    4 Plotting Vout versus

    INf (Assuming A=1)