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  • EE456 Digital CommunicationsProfessor Ha Nguyen

    September 2015

    EE456 Digital Communications 1

  • Introduction to OrthogonalFrequency-Division Multiplexing

    (OFDM)

    EE456 Digital Communications 2

  • Spectrum of M -FSK

    f1f Mf2f 3f 1Mf

    f

    For FSK with N = 2 frequencies, only one of N frequencies is activated overone symbol duration of Ts = Tb, where Tb is the bit duration. What frequencythat is activated over any symbol duration is determined by the mapping from information bits to the frequency value.

    FSK is not a spectral-efficient modulation scheme!

    Why not using all the carriers to carry information at the same time since theyare orthogonal? This leads to OFDM (orthogonal frequency-division multiplexing)technique.

    EE456 Digital Communications 3

  • OFDM (Orthogonal Frequency-Division Multiplexing)

    Bandwidth N

    NW N f

    T =

    0f 1Nf 1f 2f

    1

    NTf =

    2Nf

    )( fH

    2 /N

    T

    In OFDM, the spectrum is divided into overlapping but orthogonal subcarriers.Each sub-carrier is independently modulated by QAM. The minimum subcarrierseparation is 1/TN , where TN is the OFDM symbol length.

    OFDM can be simply looked upon as a combination of amplitude, phase andfrequency modulation techniques.

    EE456 Digital Communications 4

  • Key Features of OFDM

    The data rate on each of the subchannels is much less than the total data rate,and the corresponding subchannel bandwidth is much less than the total systembandwidth.

    The number of subchannels can be chosen so that each subchannel has abandwidth small enough so that the frequency response over each subchannelsfrequency range is approximately constant. This ensures that inter-symbolinterference (ISI) on each subchannel is small.

    The subchannels in OFDM need not be contiguous, so a large continuous blockof spectrum is not needed for high rate transmission.

    The modulation formats on different subchannels need not be the same. In factone may adaptively choose different modulation schemes according to theinstantaneous quality of the subchannels.

    The most attractive feature of OFDM is that its modulator and demodulator canbe efficiently implemented with DSP.

    The main technical issues that impair performance of OFDM are frequency offsetand timing jitter, which degrade the orthogonality of the subchannels.

    Having QAM signals transmitted simultaneously over all carriers makes thepeak-to-average power ratio (PAPR) of the OFDM signal significantly higherthan that of single-carrier QAM signal. This is a serious problem when nonlinearamplifiers are used.

    EE456 Digital Communications 5

  • OFDM Viewed and Implemented as a Multiple QAM Systems

    (2 )e n n

    j f t +

    bitsInfor.

    bpsb

    r

    ( ) ( ) ( )s t h t t + w

    Nt kT=

    (2 )e n n

    j f t +

    Detected

    infor. bits

    EE456 Digital Communications 6

  • Review of DFT/IDFT

    Let x[n], 0 n N 1, be a DT sequence. The N-point DFT of x[n] is

    DFT{x[n]} = X[i] ,

    N1

    n=0

    x[n]ej2niN , 0 i N 1.

    Given X[i], 0 i N 1, the sequence x[n] can be recovered using the IDFT:

    IDFT{X[i]} = x[n] ,1

    N

    N1

    i=0

    X[i]ej2niN , 0 n N 1.

    When N is a power of two, the DFT and IDFT can be efficiently performed usingthe fast Fourier transform (FFT) and inverse FFT (IFFT) algorithms.When the discrete-time sequence x[n] is passed through a discrete-time lineartime-invariant system whose impulse response is h[n], the output y[n] is thediscrete-time convolution of the input and the channel impulse response:

    y[n] = h[n] x[n] = x[n] h[n] =

    k=

    h[k]x[n k]

    The N-point circular convolution of x[n] and h[n], both with length N , is

    y[n] = h[n] x[n] = x[n] h[n] =

    N1

    k=0

    h[k]x[(n k) mod N ].

    Circular convolution in time leads to multiplication in frequency:

    Y [i] = DFT{y[n] = x[n] h[n]} = X[i]H[i] , 0 i N 1

    EE456 Digital Communications 7

  • Example of Circular Convolution and FFT/IFFT in Matlab

    EE456 Digital Communications 8

  • Implementation of OFDM with DFT/IDFT

    st nT=

    st nT=

    IDFT

    or

    IFFT

    Add

    cyclic

    prefix,

    and

    parallel-

    to-serial

    converter

    p(t)

    bits/secbr

    [0]X

    [1]X

    [ 1]X N

    [0]x

    [1]x

    [ 1]x N

    ( )Qx t

    ( )0cos 2 f t

    ( )0cos 2 f t

    Remove

    prefix,

    and

    serial-to-

    parallel

    converter

    [0]y

    [1]y

    [ 1]y N

    [0]Y

    [1]Y

    [ 1]Y N

    FFT

    or

    DFT

    P/S

    Con

    verte

    r

    p(t)( )Ix t

    ( )0sin 2 f t

    ( )s t

    ( )tr

    ( )0sin 2 f t

    [ ], [ 1], , [ 1]x x x N +

    ( )p t

    ( )p t

    [ ], [ 1], , [ 1]y y y N +

    (a) Transmitter

    (b) Receiver

    bits

    M0-QAM

    modulator

    S/P

    Converter

    M1-QAM

    modulator

    MN-1-QAM

    modulator

    M0-QAM

    demodulator

    M1-QAM

    demodulator

    MN-1-QAM

    demodulator

    EE456 Digital Communications 9

  • How Does the Cyclic Prefix Work in OFDM?

    Equivalent

    discrete-time channel

    { }0

    [ ]n

    h n

    =

    , [ ], [ 1], , [ 1],x x x N + , [ ], [ 1], , [ 1],y y y N +

    [ ], [ 1], , [ 1]x N x N x N + [0], [1], [2], , [ 1]x x x x N [ ], [ 1], , [ 1]x N x N x N +

    Append last symbols to the front

    Cyclic prefix (CP) of length Original signal sequence of length N

    One has x[n] = x[n mod N ] for n N 1, which also means thatx[n k] = x[(n k) mod N ] for n k N 1.

    Given x[n] is the input of the channel (ignoring noise), the channel outputbetween 0 n N 1 can be computed as:

    y[n] = x[n] h[n] =

    k=0

    h[k]x[n k] =

    k=0

    h[k]x[(n k) mod N ] = x[n] h[n]

    where the third equality follows from the fact that, for 0 k ,x[n k] = x[(n k) mod N ] for 0 n N 1.

    Taking into account AWGN, the DFT of the channel output yields

    Y [i] = DFT{y[n] = (x[n] + w[n]) h[n]} = X[i]H[i] +W [i], 0 i N 1,

    EE456 Digital Communications 10

  • [ ], [ 1], , [ 1]x N x N x N + [0], [1], [2], , [ 1]x x x x N [ ], [ 1], , [ 1]x N x N x N +

    Append last symbols to the front

    Cyclic prefix (CP) of length Original signal sequence of length N

    [0], , [ 1]y y N

    N

    [0], , [ 1]y y N

    N

    [0], , [ 1]y y N

    N

    CP CP CPData block Data blockData block

    [0], , [ 1]x x N [0], , [ 1]x x N [0], , [ 1]x x N

    An OFDM symbol is basically a super-symbol obtained by multiplexing manyM -QAM symbols in a complicated manner. The length of a super-symbol (TN )becomes longer and hence more resistent to multipath effect.

    One can also use zero padding to create a guard interval between consecutiveOFDM symbols, hence avoiding ISI.

    EE456 Digital Communications 11

  • OFDM System with the Equivalent Discrete-Time Multipath Channel

    [0]X

    [1]X

    [ 1]X N

    [0]x

    [1]x

    [ 1]x N

    [0]y

    [1]y

    [ 1]y N

    [0]Y

    [1]Y

    [ 1]Y N

    [ ], [ 1], , [ 1]x x x N + [ ], [ 1], , [ 1]y y y N +

    { }0

    [ ]n

    h n

    =

    It was shown that the input of the IFFT block in the transmitter and the outputof the FFT block in the receiver is related by Y [i] = X[i]H[i] +W [i],0 i N 1, where W [i] is Gaussian noise component.

    The use of CP and IFFT/FFT decomposes the wideband channel H(f) into a setof narrowband orthogonal subchannels, each with a different QAM symbol.

    The demodulator needs to know the channel gains H[i] to recover the originalQAM symbols by dividing out these gains: X[i] = Y [i]/H[i]. This process iscalled frequency equalization.

    The benefits of adding a cyclic prefix come at a cost. Since symbols are addedto the input data blocks, there is an overhead of /N and a resulting data-ratereduction of N/( +N). The transmit power associated with sending the cyclicprefix is also wasted since this prefix consists of redundant data.

    EE456 Digital Communications 12

  • 16-QAM vs. Constellation of Time Samples (plotted with 4 randomOFDM symbols, i.e., 4N = 64 samples)

    1 0 1

    2

    1

    0

    1

    2

    16-QAM of X [i], PAPR 2.55 dB

    0.5 0 0.5

    0.5

    0

    0.5

    1

    x[n], N = 16, PAPR 7.96 dB

    EE456 Digital Communications 13

  • Constellation of Time Samples (plotted with 4 random OFDM symbols)for N = 64 and N = 128

    0.5 0 0.5

    0.5

    0

    0.5

    1

    x[n], N = 64, PAPR 11.91 dB

    0.5 0 0.5

    0.5

    0

    0.5

    1

    x[n], N = 128, PAPR 14.33 dB

    The peak powers appear to be the same, but the average power of the rightconstellation is much smaller than that of the left one, giving rise to a higher PAPR.

    EE456 Digital Communications 14

  • Communication Services using OFDM

    Wireless Wireline

    IEEE 802.11a, g, n (WiFi) Wireless LANsADSL and VDSL broadband accessvia POTS copper wiring

    IEEE 802.15.3a Ultra Wideband (UWB) Wireless PANMoCA (Multi-media over CoaxAlliance) home networking

    IEEE 802.16d, e (WiMAX), WiBro,PLC (Power Line Communication)

    and HiperMAN Wireless MANsIEEE 802.20 Mobile BroadbandWireless Access (MBWA)DVB (Digital Video Broadcast) terrestrial TVsystems: DVB -T, DVB -H, T-DMB, and ISDB-TDAB (Digital Audio Broadcast) systems:EUREKA 147, Digital Radio Mondiale,HD Radio, T-DMB, and ISDB-TSBFlash-OFDM cellular systems3GPP UMTS & [email protected] LTE (Long-Term Evolution),and 4G

    EE456 Digital Communications 15

  • Multipath Problem in High-Speed Wireless Transmission

    Example: Consider the symbol rate of 106 symbols/sec The receiver expects aspecific symbol within a window of 1 s. If multi-path delays the signal by more than1 s (easily happen in real propagation environment), then the receiver will alsoreceive the symbol in the next symbol period, causing inter-symbol-interference (ISI),hence severe performance degradation.

    EE456 Digital Communications 16

  • Application Example of OFDM: Wireless LAN (Wi-Fi) Standards

    The IEEE 802.11a Wireless LAN standard, which occupies 20 MHz of bandwidthin the 5 GHz unlicensed band, is the first version of 802.11 family that is basedon OFDM. Modulation choices are BPSK, QPSK, 16-QAM and 64-QAM. Themaximum bit rate is 54 Mbps.

    The 802.11g standard is virtually identical, but operates in the smaller and morecrowded 2.4 GHz unlicensed ISM band.

    IEEE 802.11ac, released in December 2012, also operates in the 5 GHz band. Ituses a wider RF bandwidth (80 or 160 MHz), multiple-input multiple-output(MIMO) technology (up to 8 MIMO streams), high-density modulation (up to256 QAM) and could deliver a maximum data rate close to 7Gbps (on eight256-QAM channels, each delivering 866.7Mbps).

    The latest development of 802.11 standard is 802.11ad, which operates in thetri-band 2.5/5.0/60 GHz.

    EE456 Digital Communications 17

  • The 802.11a StandardN = 64 subcarriers: 48 are for data, the outer 12 are zeroed to reduce adjacent channelinterference, and 4 carriers used as pilot symbols for channel estimation and synchronization.The CP has = 16 samples. The length of each OFDM symbol is 80 samples.The transmitter gets periodic feedback from the receiver about the packet error rate, anduses this information to pick an appropriate error correction code and modulation scheme.The error correction code is a convolutional code with one of three possible code rates:rc = 1/2, 2/3, or 3/4. The modulation types are BPSK, QPSK, 16-QAM, or 64-QAM.The bandwidth B (and sampling rate 1/Ts) is 20 MHz. There are 64 subcarriers evenlyspaced, therefore the subcarrier bandwidth is:

    BN =20 MHz

    64= 312.5 KHz.

    Since = 16 and 1/Ts = 20MHz, the maximum delay spread for which ISI is removed is

    Tm < Ts =16

    20 MHz= 0.8 sec,

    The symbol duration per subchannel is

    T(+CP)N

    = TN + Ts = (N + )Ts = 80Ts =80

    20 106= 4 sec

    The data rate per subchannel is log2 M/T(+CP)N

    . Thus, the minimum data rate for thissystem, corresponding to BPSK (1 bit/symbol), an r = 1/2 code, and taking into accountthat only 48 subcarriers actually carry information data, is given by

    (rb)min = 48 sub. 1/2 bit

    coded bit

    1 code bit

    sub. symbol

    1 sub. symbol

    4 106= 6Mbps

    The maximum data rate corresponds to 64-QAM and r = 3/4 code:

    (rb)max = 48 sub. 3/4 bit

    coded bit

    6 code bit

    sub. symbol

    1 sub. symbol

    4 106= 54Mbps

    EE456 Digital Communications 18

  • Multipath Channels Simulated in the OFDM Lab (Fs = 32 MHz)

    0 5 10 150

    0.5

    1

    1.5

    Frequency (MHz)

    Magnituderesponse

    Echo delay = 0, Echo gain = 0

    0 5 10 150

    0.5

    1

    1.5

    Frequency (MHz)

    Magnituderesponse

    Echo delay = 8, Echo gain = 0.5

    0 5 10 150

    0.5

    1

    1.5

    Frequency (MHz)

    Magnituderesponse

    Echo delay = 16, Echo gain = -0.33333

    0 5 10 150

    0.5

    1

    1.5

    Frequency (MHz)

    Magnituderesponse

    Echo delay = 32, Echo gain = -0.25

    For a single-echo channel where the echo delay is d samples and the echo gain is ,the squared magnitude frequency response is |H(ej)|2 = (1 + 2) + 2 cos(d).

    EE456 Digital Communications 19