Rappaport CH 4
Rappaport CH 4
              In built-up urban areas, fading occurs because the height of the mobile
       antennas are well below the height of surrounding structures, so there is no sin-
       gle line-of-sight path to the base station. Even when a line-of-sight exists, multi-
       path still occurs due to reflections from the ground and surrounding structures.
       The incoming radio waves arrive from different directions with different propa-
       gation delays. The signal received by the mobile at any point in space may con-
       sist of a large number of plane waves having randomly distributed amplitudes,
139
                                       a    = 21L\l
                                                      =                                 (4.1)
       and hence the apparent change in frequency, or Doppler shift, is given by fd'
       where
                                                                                        (42)
              Equation (4.2) relates the Doppler shift to the mobile velocity and the spa-
       tial angle between the direction of motion of the mobile and the direction of
       arrival of the wave. It can be seen from equation (4.2) that if the mobile is mov-
       ing toward the direction of arrival of the wave, the Doppler shift is positive (i.e.,
       the apparent received frequency is increased), and if the mobile is moving away
       from the direction of arrival of the wave, the Doppler shift is negative (i.e. the
                                                                        / /           I
                                                                      /
                                                               /
                                                                 /            I
                                                                                  /
        Figure 4.1
        Illustration of Doppler effect.
              Example 4.1
                     Consider a transmitter which radiates a sinusoidal carrier frequency of 1850
                     MHz. For a vehicle moving 60 mph, compute the received carrier frequency if
                     the mobile is moving (a) directly towards the transmitter, (b) directly away
                     from the transmitter, (c) in a direction which is perpendicular to the direction
                     of arrival of the transmitted signal.
                                                                      x
                        Therefore, wavelength k = c/fe =                              =    0.i62m
                                                                 1850x 10
                        Vehicle speed v =     60   mph = 26.82 rn/s
tIJfd = = 1850.00016MHz
= = 1849.999834MHz
                    (c) The vehicle is moving perpendicular to the angle of arrival of the transmit-
                       ted signal.
                       In this case, S = 90, cosS = 0, and there is no Doppler shift.
                       The received signal frequency is the same as the transmitted frequency of
                        1850 MHz.
                        if
                                                                              spatial position
       Figure 4.2
       The mobile radio channel as a function of time and space.
              In Figure 4.2, the receiver moves along the ground at some constant veloc-
       ity v. For a fixed position d, the channel between the transmitter and the receiver
       can be modeled as a linear time invariant system. However, due to the different
       multipath waves which have propagation delays which vary over different spa-
      tial locations of the receiver, the impulse response of the linear time invariant
      channel should be a function of the position of the receiver. That is, the channel
      impulse response can be expressed as h(d,t). Let x(t) represent the transmitted
      signal, then the received signal y(d,t) at position d can be expressed as a convo-
      lution of x (t) with h(d,t).
For a causal system, h(d, t) = 0 for t.cO, thus equation (4.3) reduces to
       From equation (4.7) it is clear that the mobile radio channel can be modeled as a
       linear time varying channel, where the channel changes with time and distance.
            Since u may be assumed constant over a short time (or distance) interval,
       we may let x(t) represent the transmitted bandpass waveform, y(t) the
       received waveform, and h(t, v) the impulse response of the time varying multi-
       path radio channel. The impulse response h (t, t) completely characterizes the
       channel and is a function of both t and t.     The variable t represents the time
       variations due to motion, whereas t represents the channel multipath delay for
       a fixed value of t - One may think of t as being a vernier adjustment of time. The
       received signal y(t) can be expressed as a convolution of the transmitted signal
       x(t) with the channel impulse response (see Figure 4.3a).
                                                                    y(t)   =
                                     (a)
                                                                      y(t) = x(t)h(t)
              ca)                               t)                , r(t)
                                                                               =
(b)
        Figure 4.3
        (a)Bandpass channel impulse response model.
        (b) Baseband equivalent channel impulse response model.
        where c(t) and .r(t) are the complex envelopes of x(t) and y(t), defined as
                                   x(t) =                                                (4.10)
                                   y(t) =                                                (4.11)
             The factor of 1/2 in equation (4.9) is due to the properties of the complex
        envelope, in order to represent the passband radio system at baseband. The low-
        pass characterization removes the high frequency variations caused by the car-
        rier, making the signal analytically easier to handle. It is shown by Couch
        [Cou93] that the average power of a bandpass signal x2(t) is equal to
        where the overbar denotes ensemble average for a stochastic signal, or time
       average for a deterministic or ergodic stochastic signal.
             It is useful to discretize the multipath delay axis 'r of the impulse response
       into equal time delay segments called excess delay bins, where each bin has a
       time delay width equat tot) + I        where t0 is equal to 0, and represents the
       first arriving signal at the receiver. Letting i = 0, it is seen that r to is equal to
                                                                                   
       the time delay bin width given by At. For convention, t0 = o, t1 = At, and
           = iAt, for i = 0 to N  , where N represents the total number of possible
                                           I
       of the first arriving multipath component, and neglects the propagation delay
       between the transmitter and receiver Excess delay is the relative delay of the
       i th multipath component as compared to the first arriving component and is
       given by t1. The maximum excess delay of the channel is given by NAt.
             Since the received signal in a multipath channel consists of a series of
        attenuated, time-delayed, phase shifted replicas of the transmitted signal, the
       baseband impulse response of a multipath channel can be expressed as
                          N-I
              h6(t, 'c) =     a1(t,               + $1(t, r))j8(t                                     (4.12)
       where a1(t, t) and         are the real amplitudes and excess delays, respectively,
       of i th multipath component at time t [Tur721. The phase term
               +       t)in (4.12) represents the phase shift due to free space propaga-
       tion of the i th multipath component, plus any additional phase shifts which are
       encountered in the channel. In general, the phase term is simply represented by
       a single variable       t) which lumps together all the mechanisms for phase
       shifts of a single multipath component within the ith excess delay bin. Note that
       some excess delay bins may have no multipath at some time t and delay t1,
       since    t) may be zero. In equation (4.12), N is the total possible number of
       multipath components (bins), and S(.) is the unit impulse function which deter-
       mines the specific multipath bins that have components at time t and excess
       delays t1. Figure 4.4 illustrates an example of different snapshots of
       h5(t, t), where t varies into the page, and the time delay bins are quantized to
       widths of At.
                     4
1(13)
1(h)
                to
                     to   t1    t2     t3      t4                   TNI
       Figure 4.4
       An example of the time varying discrete-time impulse response model for a multipath radio channel.
hb(r) = (4.13)
                                         F (t;t) (t;t) ;2
                                                       kjhb                      (4.15)
       and many snapshots of hb (t;t) are typically averaged over a local (small-
                                                   2
ing pulse p (t) to the total power received in a multipath delay profile.
       and let p (t) be zero elsewhere for all excess delays of interest. The low pass
       channel output r (t) closely approximates the impulse response hb (t) and is
       given by
                            r(t)                              (exp               .
                                                                                     p (t 
                                        =
                                                                                                                            (4.16)
                                        =
                                                                            -
             Th determine the received power at some time to, the powerjr (t0) 2 is mea-
       sured. The quantity V (t0) 12 is called the instantaneous rnultipath power delay
       profile of the channel, and is equal to the energy received over the time duration
       of the multipath delay divided by tmax That is, using equation (4.16)
tmax
                            j2
                  jr (t0)        =                                        (t0)p2 (t          dt                           (4.18)
                                        max      j'
                                                 o            k=O
                                                N    I
                                                                                                           2
                                                          2                                        T
                                                                      j
                                            1
                                 =                                         {J.                                 dt
                                        max k=0                                      66
                                                                      o
                   a wideband probing signal p(t), T66 is smaller than the delays between
       multipath components in the channel, and equation (4.18) shows that the total
      received power is simply related to the sum of the powers in the individual mul-
      tipath components, mad is scaled by the ratio of the probing pulse's width and
      amplitude, and the maximum observed excess delay of the channel. Assuming
      that the received power from the multipath components forms a random process
      where each component has a random amplitude and phase at any time t, the
      average small-scale received power for the wideband probe is found from equa-
      tion (4.17) as
                                            = EQ
                                                                                      t:    (4.19)
       In equation (4.19), E09 [.1 denotes the ensemble average over all possible val-
       ues of a1 and    in a local area, and the overbar denotes sample average over a
       local measurement area which is generally measured using multipath measure-
       ment equipment. The striking result of equations (4.18) and (4.19) is that if a
       transmitted signal is able to resolve the multipaths, then the small-scale
       received power is simply the sum of the powers received in each multipath compo-
       nent. In practice, the amplitudes of individual multipath components do not fluc-
       tuate widely in a local area. Thus, the received power of a wideband signal such
       as p (t) does not fluctuate significantly when a receiver is moved about a local
       area [Rap89].
             Now, instead of a pulse, consider a CW signal which is transmitted into the
       exact same channel, and let the complex envelope be given by c (t) = 2. Then,
       the instantaneous complex envelope of the received signal is given by the phasor
       sum
r(t) = (4.20)
                                           2
                                  jr(t)                          U01 (t, t)                (4.21)
                                                 =      :aiexP
            As the receiver is moved over a local area, the channel changes, and the
       received signal strength will vary at a rate governed by the fluctuations of a
       and     As mentioned earlier, a1 varies little over local areas, but e. will vary
               -
      greatly due to changes in propagation distance over space, resulting in large fluc-
      tuations of r (t) as the receiver is moved over small distances (on the order of a
      wavelength). That is, since r (t) is the phasor sum of the individual multipath
      components, the instantaneous phases of the multipath components cause the
      large fluctuations which typifies small-scale fading for CW signals. The average
      received power over a local area is then given by
                   a                       +a           +       + aN   1eJON
                                                                               -
                                                                                           (4 23)
                                  (       JO0
                                                                                   -J9N
                              x                  +a1g       +      +aNIe
                                                      NI N
                            E08[Pcw]                                                               (4.24)
                                           i=O        i=O &j*i
       where            is the path amplitude correlation coefficient defined to be
                                                 =                                                 (4.25)
       and the overbar denotes time average for CW measurements made by a mobile
       receiver over the local measurement area [RapS9]. Note that when
        cos              and/or
                          = 0       = 0, then the average power for a CW signal is
       equivalent to the average received power for a wideband signal in a small-scale
       region. This is seen by comparing equation (4.19) and equation (4.24). This can
       occur when either the multipath phases are identically and independently clis-
       tributed (i.i.d uniform) over [0, 2nJ or when the path amplitudes are uncorre-
       lated. The !.i.d uniform distribution of 9 is a valid assumption since niultipath
       components traverse differential path lengths that measure hundreds of wave-
       lengths and are likely to arrive with random phases. If for some reason it is
       believed that the phases are not independent, the average wideband power and
       average CW power will still be equal if the paths have uncorrelated amplitudes.
       However, if the phases of the paths are dependent upon each other, then the
       amplitudes are likely to be correlated, since the same mechanism which affects
       the path phases is likely to also affect the amplitudes. This situation is highly
       unlikely at transmission frequencies used in wireless mobile systems.
            Thus it is seen that the received local ensemble average power of wideband
       and narrowband signals are equivalent. When the transmitted signal has a
       bandwidth much greater than the bandwidth of the channel, then the multipath
       structure is completely resolved by the received signal at any time, and the
       received power varies very Little since the individual multipath amplitudes do
       not change rapidly over a local area. However, if the transmitted signal has a
       very narrow bandwidth (e.g., the baseband signal has a duration greater than
       the excess delay of the channel), then multipath is not resolved by the received
       signal, and large signal fluctuations (fading) occur at the receiver due to the
       phase shifts of the many unresolved multipath components.
              Figure 4.5 illustrates actual indoor radio channel measurements made
       simultaneously with a wideband probing pulse having Tbb = IOns, and a CW
       transmitter. The carrier frequency was 4 GHz. It can be seen that the CW signal
       undergoes rapid fades, whereas the wideband measurements change little over
       the 5?. measurement track. However, the local average received powers of both
       signals were measured to be virtually identical EHaw9l].
              Example 4.2
                     Assume a discrete channel impulse response is used to model urban radio
                     channels with excess delays as large as 100 gs and microcellular channels
                     with excess delays no larger than 4 j.&s. If the number of multipath bins is fixed
        Figure 4.5
        Measured wideband and narrowband received signals over aSk (0.375 in) measurement track inside
        a building. Carrier frequency is 4 GHz. Wideband power is computed using equation (4.19), which
        can be thought of as the area under the power delay profile.
                     at 64, find (a) at, and (b) the maximum bandwidth which the two models can
                     accurately represent. Repeat the exercise for an indoor channel model with
                     excess delays as large as 500 ns. As described in section 4.7.6, SIRCIM and
                     SMRCIM are statistical channel models based on equation (4.12) that use
                     parameters in this example.
                                                                                      9
                                                                         =   500x10
                    Similarly, for indoor channels, Ar = tN/N                              = 7.8 125 ns.
              Example 4.3
                   Assume a mobile traveling at a velocity of 10 mIs receives two multipath com-
                   ponents at a carrier frequency of 1000 MHz. The first component is assumed to
                   arrive at t = 0 with an initial phase of 00 and a power of -70 dBm, and the
                   second component which is 3 dE weaker than the first component is assumed
                   to arrive at r = pss, also with an initial phase of 00. If the mobile moves
                                           1
                   directly towards the direction of arrivgl of the first component and directly
                   away from the direction of arrival of the second component, compute the nar-
                   rowband instantaneous power at time intervals of 0.1 s from 0 5 to 0.5 S. Com-
                   pute the average narrowband power received over this observation interval.
                   Compare average narrowband and wideband received powers over the inter-
                   val.
jr(t)12
                  Now, as the mobile moves, the phase of the two multipath components changes
                  in opposite directions.
                  At t = 0.1 the phase of the first component is
                              2nd2irut2nxlQ(mJs)xO.Is
                                  A                        0.3m
                           = 20.94 rad = 2.09 rad = 120
                 Since the mobile moves towards the direction of arrival of the first component,
                 and away from the direction of arrival of the second component,       is positive,
                 and 02 is negative.
                 Therefore, at t = 0.ls,       = 120, and 2 = 120, and the instantaneous
                 power is equal to
r(t)12 =
= = 78.2pW
r(t)12 =
= = 29) pW
                   It follows that at t =                 0.4 s,       Ir(t)I2 = 78.2 2W, and at t =      0.5   s. rU)1    = 81.5
                   pW.
E UOt)12j
                                                                      tbb
                                                          '    Pulse Width =
                        Rr
                             a--                     Resolution = Pulse Width
       Figure 4.6
       Direct RF channel impulse response measurement system.
            to interference and noise, due to the wide passband filter required for multi-
       path time resolution. Also, the pulse system relies on the ability to trigger the
       oscilloscope on the first arriving signal. If the first arriving signal is blocked or
       fades, severe fading occurs, and it is possible the system may not trigger prop-
       erly. Another disadvantage is that the phases of the individual multipath
       nents are not received, due to the use of an envelope detector. However, use of a
       coherent detector permits measurement of the multipath phase using this tech-
       nique.
Tx
BW=
        Figure 4.7
        Spread spectrum channel impulse response measurement system.
             The spread spectrum signal is then received, filtered, and despread using a
       PN sequence generator identical to that used at the transmitter. Although the
       two PN sequences are identical, the transmitter chip clock is run at a slightly
       faster rate than the receiver chip clock. Mixing the chip sequences in this fashion
       implements a sliding correlator [Dix84]. When the PN code of the faster chip
        clock   catches up with the PN code of the slower chip clock, the twa chip
        sequences will be virtually identicallyaligned, giving maximal correlation. When
        the two sequences are Qot maximally correlated, mixing the incoming spread
        spectrum signal with the unsynchronized receiver chip sequence will spread this
        signal into a bandwidth at least as large as the receiver's reference PN sequence.
        In this way, the narrowband filter that follows the correlator can reject almost all
        of the incoming signal power. This is how processing gain is realized in a spread
        spectrum receiver and how it can reject passband interference, unlike the direct
        RF pulse sounding system.
              Processing gain (PG) is given as
                                           =          =
                              PG =                                                            (4.28)
                                     Rbb
        where tbb = 1/Rbb, is the period of the baseband information. For the ease of a
       sliding correlator channel sounder, the baseband information rate is equal to the
       frequency offset of the PN sequence clocks at the transmitter and receiver.
             When the incoming signal is correlated with the receiver sequence, the sig-
       nal is collapsed back to the original bandwidth (i.e., "despread"), envelope
        detected, and displayed on an oscilloscope. Since different incoming multipaths
        will have different time delays, they will maximally correlate with the receiver
        PN sequence at different times. The energy of these individual paths will pass
        through the correlator depending on the time delay. Therefore, after envelope
        detection, the channel impulse response convolved with the pulse shape of a sin-
        gle chip is displayed on the oscilloscope. Cox [Cox72] first used this method to
        measure channel impulse responses in outdoor suburban environments at 910
        MHz. Devasirvatham [Dev86], [Dev9Oa] successfully used a direct sequence
        spread spectrum channel sounder to measure time delay spread of multipath
       components and signal level measurements in office and residential buildings at
       850 MHz. Bultitude [Bul891 used this technique for indoor and microcellular
       channel sounding work, as did Landron [Lan92].
             The time resolution (Ar) of multipath components using a spread spec-
       trum system with sliding correlation is
At = = (4.29)
            In other words, the system can resolve two multipath components as long
       as they are equal to dr greater than      seconds apart. In actuality, multipath
       components with interarrival times smaller than 27', can be resolved since the
       nns pulse width is smaller than the absolute width of the triangular correlation
       pulse, and is on the order of
            The sliding correlation process gives equivalent time measurements that
       are updated every time the two sequences are maximally correlated. The time
       between maximal correlations (7') can be calculated from equation (4.30)
= Tjl = (4.30)
                                                 -r                                   (4.31)
                                                          =
        where a = transmitter chip clock rate (Hz)
                        = receiver
                            chip clock rate (Hz)
        For a maximal length PN sequence, the sequence length is
                                                      =                               (4.32)
        where n is the number of shift registers in the sequence generator [Dix841.
             Since the incoming spread spectrum signal is mixed with a receiver PN
        sequence that is slower than the transmitter sequence, the signal is essentially
        down-converted ("collapsed") to a low-frequency narrowband signal. In other
        words, the relative rate of the two codes slipping past each other is the rate of
       information transferred to the oscilloscope. This narrowband signal allows nar-
       rowband processing, eliminating much of the passband noise and interference.
       The processing gain of equation (4.28) is then realized using a narrowband filter
       (BW = 2(af3) ).
            The equivalent time measurements refer to the relative times of mu.ltipath
       components as they are displayed on the oscilloscope. The observed time scale on
       the oscilloscope using a sliding correlator is related to the actual propagation
       time scale by
S21(w)zH(o))
                                                       Inverse
                                                  DFT Processor
                                                          I
                                                                       h(t)     FT'[W(u)}
       Figure 4.8
       Frequency domain channel impulse response measurement system.
                                                 =
                                                                                                (4.35)
            The rms delay spread is the square root of the second central moment of the
        power delay profile and is defined to be
                                            =         (t)2                                      (4.36)
       where
                                        k            k
                                                                                               (4.37)
       These delays are measured relative to the first detectable signal arriving at the
       receiver at to = 0. Equations (4.35) - (4.37) do not rely on the absolute power
       level of P (t), but only the relative amplitudes of the multipath components
       within P (t). Typical values of rms delay spread are on the order of microsec-
       onds in outdoor mobile radio channels and on the order of nanoseconds in indoor
       radio channels. Table 4.1 shows the typical measured values of rms delay spread.
            It is important to note that the rms delay spread and mean excess delay are
       defined from a single power delay profile which is the temporal or spatial aver-
       age of consecutive impulse response measurements collected and averaged over
       a local area. I5rpically, many measurements are made at many local areas in
       order to determine a statistical range of multipath channel parameters for a
       mobile communication system over a large-scale area [Rap9O].
           The maximum excess delay (X dB) of the power delay profile is defined to be
       the time delay during which multipath energy falls to X dB below the mat-
85
          m
                -go
          C
          0
          C-
          w
          a     -95
          V
          S.   -ioo
          S.
         -J
         -4
          C
         U,
               -jos
         V
          5,
         .4
          U,
          U
               HO
         aSI
"5
                                                     (a)
                                                                               H
                                                                   3&Qmpath
                                                                   18   dE Atta2
                                                                       2 mV/div
                                                                   100 na/dIv
                                                                    51.7   RicE
                                                                          ns
                                                   (b)
       Fgure 4.9
       Measured mukipath power delay profUes
       a) From a 900 MHz cellular system in San Francisco [From [Rap9OJ IEEE].
       b) Inside a grocery store at 4 GHz [From [Haw9lJ  IEEE].
                    I
                                                      RMSDeIaySprsad-46.40rs
                                                      MSmumExcsuD.IayclOdBsB4ns
                        -10
-20
                        30
                                                                  I
        Figure 4.10
        Example of an indoor power delay profile; nns delay spread, mean excess delay, maximum excess
        delay (10 dB), and threshold level are shown.
                                                                                                (4.38)
                                                          t
       If the definition is relaxed so that the frequency correlation function is above 0.5.
       then the coherence bandwidth is approximately
                                                                                                   (4.39)
                                                      5
              Example 4.4
                    Calculate the mean excess delay, rms delay spread, and the maximum excess
                    delay (10 dE) for the multipath profile given in the figure below. Estimate the
                    50% coherence bandwidth of the channel. Would this channel be suitable for
                    AMPS or GSM service without the use of an equalizer?
                                            P/r)
                                     0 dB-
                                   -10 dB
-20dB
-30 dR
                                            0      12             J
                                                                        (las)
                                                   Figure E4.4
                    The second moment for the given power delay profile can be calculated
                    as
                                    (5)2k
                              = (1)        (0.1) (1)2+ (0.1) (2)2+ (0.01) (0)           2
                                                                              - 2107          .
1.21
                                                   = 146kHz
                                    =
                    Since    is greater than 30 kHz. AMPS will work without an equalizer. How-
                    ever, GSM requires 200 kHz bandwidth which exceeds       thus an equalizer
                    would be needed for this channel.
                                                                                     (4.40.a)
                                                 =   /.
             Coherence tine is actually a statistical measure of the time duration over
       which the channel impulse response is essentially invariant, and quantifies the
       similarity of the channel response at different times. In other words, coherence
       time is the time duration over which two received signals have a strong potential
       for amplitude correlation. If the reciprocal bandwidth of the baseband signal is
       greater than the coherence time of the channel, then the channel will change
       during the transmission of the baseband message, thus causing distortion at the
       receiver. If the coherence time is defined as the time over which the time correla-
       tion function is above 0.5, then the coherence time is approximately iSte94l
        wildly, and (4.40.b) is often too restrictive. A popular rule of thumb for modem
        digital communications is to define the coherence time as the geometric mean of
        equations (4.40.a) and (4.40.b). That is,
                                                             = 0.423
                                              =   f                                                       (4.40.c)
                                                                 I'm
                The definition of coherence time implies that two signals arriving with a
        time separation greater than        are affected differently by the channel. For
        example, for a vehicle traveling 60 mph using a 900 MHz carrier, a conservative
        value of    can be shown to be 2.22 ms from (4.40.b). If a digital transmission
        system is used, then as long as the symbol rate is greater than I     = 454 bps,
        the channel will not cause distortion due to motion (however, distortion could
        result from multipath time delay spread, depending on the channel impulse
        response). Using the practical formula of (4.40.c),     = 6.77 ms and the symbol
        rate must exceed 150 bits/s in order to avoid distortion due to frequency disper-
        sion.
                Ernniple 4.5
                    Determine the proper spatial sampling interval required to make small-scale
                    propagation measurements which assume that consecutive samples are highly
                    correlated in time. How many samples will be required over 10 m travel dis-
                    tance if 4 = 1900 MHz and v = 50 in/s. How long would it take to make these
                    measurements, assuming they could be made in real time from a moving vehi-
                    cle? What is the Doppler spread B0 for the channel?
                          Tc =
                    Taking time samples at less than half
                    corresponds     to a spatial sampling interval of
                                =        = SOXS6SMS
                                                        =   0.014l25m     = 1.41cm
                                                       i9oo x io6
                             BD=tm___            
                                                       3xl08
                                                                     =316.66Hz
                                              Small-Scale Fading
                               (Based on multipath time delay spread)
                                          Small-Scale Fading
                                       (Based on Doppler spread)
        Figure 4.11
        Types of small-scale fading.
                               sO)
                                                 h('z.t,)
0 i ot 0
       Figure 4.12
       Flat fading channel characteristics.
             It can be seen from Figure 4.12 that if the channel gain changes over time,
       a change of amplitude occurs in the received signal. Over time, the received sig-
       nal r (t) varies in gain, but the spectrum of the transmission is preserved. In a
       flat fading channel, the reciprocal bandwidth of the transmitted signal is much
       larger than the multipath time delay spread of the channel, and h,, (t, r) can be
       approximated as having no excess delay (i.e., a single delta function with r = 0).
       Flat fading channels are also known as amplitude uarying channels and are
       sometimes referred to as narrowband channels, since the bandwidth of the
      applied signal is narrow as compared to the channel flat fading bandwidth. Typi-
      cal flat fading channels cause deep fades, and thus may require 20 or 30 dB more
      transmitter power to achieve low bit error rates during times of deep fades as
r(r)
s(t)
                            01;
                                                 fl,0
                                                         ho, t)
                                                                  t
                                                                                 r(t)
or,
        Figure 4.13
        Frequency selective fading channel characteristics.
                                  'E        T
                                                        FlatSiow               '   Flat Fast
                              0UD
                                                         Fading                I
                                                                                    Fading
                                         tl                           .       I
                                                Frequency Selective1 Frequency Selective
                                                        Slow Fading                   Fast Fading
                                            B8
                              C                                            I
                              I                Frequency Selective I
                                                   Fast Fading
                                                                                   Frequency Selective
                                                                                      Slow Fading   --
Bd
                             0
                            -5
                      a
                   -a
                      0
                   'U
                           15
                    0.)
                      >    20
                    0)
                   -J
                           -25
                    C
                    a,
35
                           -40
                                 0                   50                     I DO        1 50                 200   250
                                                                           Elapsed Time (rns)
        Figure 4.15
       A typical Rayleigh fading envelope at 900 MHz (From 1jun93)  IEEE].
                                                                   (
                                                    fT                     r
                                                    jexp              
                                 p(r)          =   12                                                                    (4.49)
                                                           0                           (rcO)
       where a is the rms value of the received voltage signal before envelope detection,
       and      is the time-average power of the received signal before envelope detec-
       tion. The probability that the envelope of the received signal does not exceed a
       specified value R is given by the corresponding cumulative distribution function
       (CDF)
                                                                       R
                                                                                                    1        2"
                          P(R)       =   Pr(rR)                =       fp(r)dr     =                           I         (4.50)
                                                                                                        2a)
            The mean value                   rmean of the          Rayleigh distribution is given by
= = 0.4292a2
        The rms value of the envelope is the square root oTthe mean square, or lie.
              The median value of r is found by solving
                                         = I
                                         2  .v0
                                                         p(r)dr                                  (4.53)
        and is
                                             = l177a
                                             rmedian                                (4.54)
            Thus the mean and the median differ by only 0.55 dB in a Rayleigh fading
       signal. Note that the median is often used in practice, since fading data are usu-
       ally measured in the field and a particular distribution cannot be assumed. By
       using median values instead of mean values it is easy to compare different fad
       ing distributions which may have widely varying means. Figure 4.16 illustrates
       the Rayleigh pdf. The corresponding Rayleigh cumulative distribution function
       (CDF) is shown in Figure 4.17.
                                                         0.6065
                                                              a
                                              /
                                  p(r)
/ I
                          10
                                     U OBS light clutter, Site C
                          50
                                     S LOS heavy clutter, Site E
                                     * LOS light cluter, Sue D
                          20
                                                                                   U
10
      % Probability
                           5                                            .
       Signal Level                                                 .              *1
       <Abscissa                                                .                 *:
                           2                                .
                                                        .                    t.
                           1                        U
0.5
                         0.2                                                            Log-normal      dB
                                                                                        Rayleigh
                                                                                        Rician K=6 dB
                         0.1
                               -30            -20                           -10               0              10
        Figure 4.17
        Cumulative distribution for three small-scale fading measurements and their fit to Rayleigh, Ricean,
        and log-normal distributions [From [RapS9]  IEEEI.
                                              2a
                        p(r)   = jL..e
                                                                                                     (455)
                                              0  for(rcO)
              The parameter A denotes the peak amplitude of the dominant signal and
          (.) is the modified Bessel function of the first kind and zero-order. The Ricean
       distribution is often described in terms of a parameter K which is defined as the
       ratio between the deterministic signal power and the variance of the multipath.
       ItisgivenbyK =A2 /(2a) or,intermsofdB
                                        2
                                                            A2
                                    K(dB)          =              dB                                 (456)
                                                            2a
            The parameter K is known as the Ricean factor and completely specifies
       the Ricean distribution. As A *0, K *        dB, and as the dominant path
       decreases in amplitude, the Ricean distribution degenerates to a Rayleigh distri-
       bution. Figure 4.18 shows the Ricean pdf The Ricean CDF is compared with the
       Rayleigh CDF in Figure 4.17.
K=-xdB
                                                       dB
                                p(r):
                                        /
        areas, it assumes the existence of a direct path between the transmitter and
        receiver, and is limited to a restricted range of reflection angles. Ossana's model
        is therefore rather inflexible and inappropriate for urban areas where the direct
        path is almost always blocked by buildings or other obstacles. Clarke's model
        [C1a681 is based on scattering and is widely used.
f,, = (4.57)
                                                              y
                                                                  in x-y plane
       Figure 4.19
       Illustrating plane waves arrivingat random angles.
               The vertically polarized plane waves arriving at the mobile have E and H
        field components given by
= (4.58)
= - + (4.59)
=- (4.60)
= I (4.62)
               Since the Doppler shift is very small when compared to the carrier fre-
        quency, the three field components may be modeled as narrow band random pro-
        cesses. The three components         and H,, can be approximated as Gaussian
        random variables if N is sufficiently large. The phase angles are assumed to
        have a uniform probability density funccion (pdf) on the interval (0,2it]. Based on
        the analysis by Rice [Ric481 the E-field can be expressed in an in-phase and
        quadrature form
                                        =                                                         (4.63)
        where
= E0Z C1 + (4.M)
and
= (4.65)
            Both     (t) and (t) are Gaussian random processes which are denoted
        as    and T9, respectively, at any time t. and     are uncorrelated zero-mean
        Gaussian random variables with an equal variance given by
                                                                                        (4.66)
        where the overbar denotes the ensemble average.
            The envelope of the received E-field, E/t), is given by
                                      =
                              p(r)                                                    (4.68)
where a2 = E0/2 2
= JAG(a)p(a)da (4.69)
       where AG (a)p (a)da is the differential variation of received power with angle.
       If the scattered signal is a CW signal of frequency  then the instantaneous fre-
       quency of the received signal component arriving at an angle a is obtained using
       equation (4.57)
                             [(a) =    [                   +f   =                    (4.70)
       where       is the maximum Doppler shift. It should be noted that [(a) is an even
       function of a, (i.e., /(a) = fT-a)).
             If S (I) is the power spectrum of the received signal, the differential varia-
       tion of received power with frequency is given by
                                                S(f)Id)9                            (4.71)
                                         a=         cos-1
                                                               [t'fc]                                   (4.74)
sina = (4.75)
        Substituting equation (4.73) and (4.75) into both sides of (4.72), the power spec-
        tral density S (/) can be expressed as
        where
                                   S(/) =      0,
                                                               Vft!        >fm                          (4.77)
             The spectrum is centered on the carrier frequency and is zero outside the
       limits of  fm Each of the arriving waves has its own carrier frequency (due to
       its direction of arrival) which is slightly offset from the center frequency. For the
       case of a vertical A/4 antenna (G(a) =                              1.5),   and a uniform distribution
       p(a) =      1/2n over 0 to 2E ,the output spectrum is given by (4.76) as
                                           =                   1.5
                                  5E (f)                                                               (4.78)
                                    
                                                           f     U
                                                                                 I
                                I&fm                                     fcilm
         Figure 4.20
         Doppler power spectrum for an unmodulated CW carrier [From [Gan721]  IEEE].
                                                              (1)2]
                                                =   8fmK[1I
        where K [.} is the complete elliptical integral of the first kind. Equation (4.79) is
        not intuitive and is a result of the temporal correlation of the received signal
        when passed through a nonlinear envelope detector. Figure 4,21 illustrates the
       baseband spectrum of the received signal after envelope detection.
             The spectral shape of the Doppler spread determines the time domain fad-
       ing waveform and dictates the temporal correlation and fade slope behaviors.
       Rayleigh fading simulators must use a fading spectrum such as equation (4.78)
       in order to produce realistic fading waveforms that have proper time correlation.
0dB
-1dB
-2dB
-3dB
-5dB
2 -6dB
-7dB
                        -8dB
                           0.001           0.01               0.1                1.0 2.0           10
                                                                     1/fm
        Figure 421
        Baseband power spectral density of a CW Doppler signal after envelope detection.
                            C
                        a
                        C
sin 2irQ
(a)
sin 2nf0t
                                                           (b)
       Figure 4.22
       Simulator using quadrature amplitude modulation with (a) RY Doppler filter and
                                                            (b) baseband Doppler filter.
       sian noise components are actually a series of frequency components (line spec-
       trum from -4,            to   In)' which are equally spaced and each have a complex
       Gaussian weight. Smith's simulation methodology is shown in Figure 4.23.
             'lb implement the simulator shown in Figure 4.23, the following steps are
       used:
       (1) Speci5r the number of frequency domain points (N) used to represent
           JSE (I') and the maximum Doppler frequency shift                     The value used for
           N is usually a power of 2.
       (2) Compute the frequency spacing between adjacent spectral lines as
           11= 24,/(N_ I). This defines the time duration of a fading waveform,
           T = l/Af.
      (3) Generate complex Gaussian random variables for each of the N/2 positive
          frequency components of the noise source.
      (4) Construct the negative frequency components of the noise source by conju-
        Figure 4.23
       Frequency domain implementation of a Rayleigh fading simulator at baseband
                    Signal                    s(t)
                    under test
                                                                                          nt)
        Figure 4.24
        A signal may be applied to a Rayleigh fading simulator to determine performance in a wide range of
        channel ccnditions. Both flat and frequency selective fading conditions may be simulated, depending
        on gain and time delay settings.
NH = = (4.80)
        where r is the time derivative of r (t) (i.e., the slope), p (R, r') is the joint den-
        sity function of r and f at r = R,          the maximum Doppler frequency and
                                                   f,,, is
         p = R/Rrms is the value of the specified level R, normalized to the local rms
        amplitude of the lading envelope EJakl4]. Equation (4.80) gives the value of NR,
        the average number of level crossings per second at specified R. The level cross-
        ing rate is a function of the mobile speed as is apparent from the presence of fm
        in equation (4.80). There are few crossings at both high and low levels, with the
        maximum rate occurring at p = I / ,fi, (i.e., at a level 3 dB below the rms level).
        The signal envelope experiences very deep fades only occasionally, but shallow
        fades are frequent.
              Example 4.6
                    For a Rayleigh fading signal, compute the positive-going level crossing rate for
                    p = 1, when the maximum Doppler frequency          ) is 20 Hz. What is the max-
                    imum velocity of the mobile for this Doppler frequency if the carrier frequency
                    is 900 MHz?
             The average fade duration is defined as the average period of time for which
       the received signal is below? a specified level R. For a Rayleigh fading signal, this
       is given by
t = jrPr[rR] (4.81)
       where Pr [r R] is the probability that the received signal r is less than R and
       is given by
Fr[rR] = (4.82)
       where    is the duration of the fade and T is the observation interval )f the fad-
       ing signal. The probability that the received signal r is less than the threshold
       R is found from the Rayleigh distribution as
         where p (r) is the pdf of a Rayleigh distribution. Thus, using equations (4.80),
         (4.81), and (4.83), the average Vade duration as a function of p and fm can be
         expressed as
                                                             =   e
                                                                                                 (4.84)
                                                                 PIm
               The average duration of a signal fade helps detennine the most likely num-
        ber of signaling bits that may be lost during a fade. Average fade duration pnma-
        rily depends upon the speed of the mobile, and decreases as the maximum
        Doppler frequency fm becomes large. If there is a particular fade margin built
        into the mobile communication system, it is appropriate to evaluate the receiver
        performance by determining the rate at which the input signal falls below a
        given level B, and how long it remains below the level, on average. This is use-
        ful for relating SNR during a fade to the instantaneous BElt which results.
               Example 4.7
                    Find the average fade duration for threshold levels p =     0.01, p = 0.1,    and
                        p = I, when the Doppler frequency is 200 Hz.
                               =                   l
                                                           =200ps
                                      0. 1) 200
                    For p =     I
3.43ms
             Example 4.8
                   Find the average fade duration for a threshold level of p = 0.707 when the
                   Doppler frequency is 20 Hz. For a binary digital modulation with bit duration
                   of 50 bps, is the Rayleigh fading slow or fast? What is the average number of
                   bit errors per second for the given data rate. Assume that a bit error occurs
                   whenever any portion of a bit encounters a fade for which p c 0.1
                                               0 707
                                           e           I
                                                                183
                       For a data rate of 50 bps, the bit period is 20 ms. Since the bit period is greater
                       than the average fade duration, for the given data rate the signal undergoes
                       fast Rayleigh fading. Using equation (4.84), the average fade duration for
                       p    0.1 is equal to 0.002 s. This is less than the duration of one bit. Therefore,
                      only one bit on average will be lost during a fade. Using equation (4.80), the
                      number of level crossings for p = 0.1 is N,. = 4.96 crossings per seconds.
                      Since a bit error is assumed to occur whenever a portion of a bit encounters a
                      fade, and since average fade duration spans only a fraction of a bit duration,
                      the total number of bits in error is 5 per second, resulting in a BER = (5/50) =
                      0.1.
a2exp(j42)
        Figure 4.25
        Two-ray Rayleigh fading model.
hb (t, X1,
S Sm,
       tions from walls and ceilings and random scattering from inventory and equip-
       ment.
            By analyzing the measurements from fifty local areas in many buildings, it
       was found that the number of multipath components,           arriving at a certain
       location is a function ofX,    and      and almost always has a Gaussian distri-
       bution. The average number of multipath components ranges between 9 and 36,
       and is generated based on an empirical fit to measurements. The probability that
       a multipath component will arrive at a receiver at a particular excess delay T4 in
       a particular environment Sm is denoted as         (Ti, 5m) This was found from
       measurements by counting the total number of detected multipath components
       at a particular discrete excess delay time, and dividing by the total number of
       possible multipath components for each excess delay interval. The probabilities
       for multipath arriving at a particular excess delay values may be modeled as
       piecewise functions of excess delay, and are given by
                                                T
                                          1fl67
                                                    (T.II0)
                                 = <0.65
                                                        4
                                                            360
                                                                  (110   nscT.<200 ns)            (4.88)
                  for LOS
                                                   (T.200)
                                     [0.22                       (200 ns   cT <500 ns)
                                         008+O67exp                         (IOOnscT,<SOOns)
       where        corresponds to the LOS topography, and     corresponds to obstructed
       topography. SIRCIM uses the probability of arrival distributions described by
       equation (4.88) or (4.89) along with the probability distribution of the number of
       multipath components,         (X, 5m' Pa), to simulate power delay profiles over
       small-scale distances. A recursive algorithm repeatedly compares equation (4.88)
       or (4.89) with a uniformly distributed random variable until the proper         is
      generated for each profile [Hua9l], [Rap9la].
            Figure 4.26 shows an example of measured power delay profiles at 19 dis-
      crete receiver locations along a 1 m track, and illustrates accompanying narrow-
      band information which SIRCIM computes based on synthesized phases for each
      multipath component [Rap9laJ. Measurements reported in the literature pro-
      vide excellent agreement with impulse responses predicted by SIRCIM.
            Using similar statistical modeling techniques, urban cellular and microcel-
      lular multipath measurement data from [Rap9O}, [Sei9l], [Sei92a] were used to
      develop SMRCIM. Both large cell and microcell models were developed. Figure
        Figure 4.26
        Indoor wide band impulse responses simulated by SIRCIM at 1.3 GHz. Also shown are the distribu-
        tions of the ntis delay spread and the narrowband signal power distribution. The channel is simu-
        lated as being obstructed in an open-plan building, T-R separation is 25 m. The rms delay spread is
        137.7 ns. All multipath components and parameters are stored on disk [From [Rap93a]  IEEE).
       4.8 Problems
           4.1 Draw a block diagram of a binary spread spectrum sliding correlator multipath
               measurement system. Explain in words how it is used to measure power delay
                   profiles.
                    (a) If the transmitter chip period is 10 ns, the PN sequence has length 1023,
                        and a 6 0Hz carrier is used at the transmitter, find the time between
                        maximal correlation and the slide factor if the receiver uses a PN
                        sequence clock that is 30 kI-Iz slower than the transmitter.
                    (b) If an oscilloscope is used to display one complete cycle of the PN sequence
                        (that is, if two successive maximal correlation peaks are to be displayed
                        on the oscilloscope), and if 10 divisions are provided on the oscilloscope
                        time axis, what is the most appropriate sweep setting (in secorids/divi-
       Figure 4.27
       Urban wideband impulse responses simulated by SMRCIM at 1.3 GHz. Also shown are the distribu-
       tions of the rms delay spread and the narrowband fading. T-R separation is 2.68 km. The ruts delay
       spread is 3.8     All multipath components and parameters are saved on disk. [From [Rap93a] 
       IEEE
                           sion) to be used?
                     (c)   What is the required IF passband bandwidth for this system? How is this
                           much better than a direct pulse system with similar time resolution?
            4.2    If a particular modulation provides suitable BELt performance whenever
                          0.!, determine the smallest symbol period   (and thus the greatest
                   symbol rate) that may be sent through RF channels shown in Figure P4.2,
                   without using an equalizer.
            4.3    For the power delay profiles in Figure P4.2, estimate the 90% correlation and
                   50% correlation coherence bandwidths.
            4.4 Approximately how large can the rms delay spread be in order for a binary
                modulated signal with a bit rate of 25 kbps to operate without an equalizer?
                What about an 8-PSK system with a bit rate of 75 kbps?
            4.5 Given that the probability density function of a Rayleigh distributed envelope
                            P/t)                                          PJr)
                   0dB:                             indoor
                                                                    0dB
                 -10dB                                                     -
2OdB -20dB-
-30dB- -30dB-
                           0              50      75    100 excess        0        5       10
                                    (a)                     delay (ns)                 (b)
                               Figure P4.2: Two channel responses for Problem 4.2
                                        ta)
                                     (in mis)
30/.
                                                0   10                      90    100
                                                                                 time (s)
                                Figure P4.8 Graph of velocity of mobile.
              4.9 For a mobile receiver operating at frequency of 860 MHz and moving at 100
                      kmIhr
                      (a) sketch the Doppler spectrum if a CW signal is transmitted and indicate
                           the maximum and minium frequencies
                      (b) calculate the level crossing rate and average fade duration if p = 20 dB-
              4.10   For the following digital wireless systems, estimate the maximum rms delay
                     spread for which no equalizer is required at the receiver (neglect channel cod-
                     ing, antenna diversity, or use of extremely low power levels).
                     System           RF Data Rate Modulation
                     USDC             48.6 kbps           ,ti4DQPSK
                     GSM              270.833 kbps        GMSK
                     DECT             1152 kbps           GMSK
             4.11 Derive the RF Doppler spectrum for a 5i8k vertical monopole receiving a CW
                  signal using the models by Clarke and Gang. Plot the RF Doppler spectrum
                  and the corresponding baseband spectrum out of an envelope detector. Assume
                  isotropic scattering and unit average received power.
             4.12 Show that the magnitude (envelope) of       sum of two independent identically
                  distributed complex (quadrature) Gaussian sources is Rayleigh distributed.
                  Assume that the Gaussian sources are zero mean and have unit variance.
             4.13 Using the method described in Chapter 4, generate a time sequence of 8192
                  sample values of a Rayleigh fading signal for
                     (a) Td =   20   Hzand(b) Id    = 200 Hz.
            4.14 Generate 100 sample functions of fading data described in Problem 4.13, and
                  compare the simulated and theoretical values of RRMS, NR, and t for p = 1,
                  0.1, and 0.01. Do your simulations agree with theory?
            4.15 Recreate the plots of the DFs shown in Figure 4.17, starting with the pdfs fo;
                 Rayleigh, Ricean, and log-normal distributions.
            4i6 Plot the probability density function and the CDF for a Ricean distribution
                 having (a) K = 10 dB and (b) K = 3 dB. The abscissa of the CDF plot should be
                 labeled in dE relative to the median signal level for both plots. Note that the
                 median value for a Ricean distribution changes as K changes.
               4.17 Based on your answer in Problem 4.16, if the median SNR is -70 dBm, what is
                    the likelihood that a signal greater than -80 dBm will be received in a Ricean
                    fading channel having (a) K = 10 dB, and (b) K = 3 dE?
               4.18 A local spatial average of a power delay profile is shown in Figure P4.18.
                                               Prtr)
                                        0 dB
                                      -10dB
                                      -20dB
-30 dB
                                               0
                                                                        (ps)
                                 Figure P4.18. Power delay profile.
                    (a) Determine the rms delay spread and mean excess delay for the channel.
                    (b) Determine the maximum excess delay (20 dB).
                    (c) If the channel is to be used with a modulation that requires an equalizer
                       whenever the symbol duration T is less than       determine the maxim urn
                       RF symbol rate that can be supported without requiring an equalizer.
                    (d) If a mobile traveling at 30 km/lu receives a signal through the channel,
                       determine the time over which the channel appears stationary (or at least
                      highly correlated).
              4.19 A flat Rayleigh fading signal at 6 GHz is received by a mobile traveling at 80
                   kmthr.
                   (a) Determine the number of positive-going zero crossings about the ms value
                       that occur over a 5s interval.
                   (b) Determine the average duration of a fade below the rms level.
                   (c) Determine the average duration of a fade at a level of 20 dB below the ms
                        value.
              4.20 Using computer simulation, create a Rayleigh fading simulator that has 3
                   independent Rayleigh fading multipath components, each having variable
                   multipath time delay and average power. Then convolve a random binary bit
                  stream through your simulator and observe the time waveforms of the output
                   stream. You may wish to use several samples for each bit (7 is a good number).
                   Observe the effects of multipath spread as you vary the bit period and time
                   delay of the ehannel.
              4.21 Based on concepts taught in this chapter, propose methods that could be used
                   by a base station to determine the vehicular speed of a mobile user. Such meth-
                   ods are useful for handoff algorithms.