Multipath fading and reflections by dfhdhdhdhjr

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									Multipath fading and reflections




The signal takes many paths to the destination. The
propagation delay along each path is different.
How many meters difference gives you 0.00001 seconds
of delay difference?
           Effects of Multipath
            Fading/Reflection
•   “Ghost” on TV.
•   GPS – incorrect position calculation.
•   Frequency Selective Fading.
•   Intersymbol interference.
                    Intersymbol Interference
                                                    Suppose that there are two paths.
                                                    The shorter path has length d1, the longer
                                                    path has length d2.
                                                    What is the difference in propagation delay
                                                    between the two paths?
   Symbols
                    3               2
received over
  the shorter
                            1               0
     path                                               How big does  have to be to so that
                                                        the 3 of the longer path arrives
   Symbols              3               2
received over                  1
                                                        exactly when the 0 arrives on the
                                                        shorter path?
  the longer                                    0
      path                                              How big must the difference in paths
                                                        be for this to happen?
   Received
   signal – a
 combination
   of the two
    signals
                       Intersymbol Interference
                       Suppose we use differential phase shift keying to transmit 3 2 1 0

                                                     P ( t)    0 if t  T
                                                                  3   if ( t  T) ( t  2 T)
                                                                       
s t   sin 2f ct  P t                                    2 
                                                                  3     if ( t  2 T) ( t  3 T)
                                                                           
                                                                 2         
                                                                 3             
                                                                       if t  3 T
                                                                 2             2



                               ?       3        2                 1                0
                       1
                   1




          f ( t)       0




             1        1
                           0       1        2               3                4               5
                           0                     t                                           5
                      Intersymbol Interference
                              1
                      1



       f ( t)

       f  t
                 T           0
                 
                5



                  1          1
                                  0         0.5       1         1.5           2       2.5   3
                                  0                             t                           3


                          1
             0.951




f ( t) f  t
                 T
                 
                5       0
        2



          0.951          1
                              0       0.5         1       1.5       2   2.5       3
                              0                           t                       3
                            Phasor
addition of sine waves with the same frequency
                                               magnitude          phase
           x t   a sin wt           x= a 
   x t   a sin wt     b sinwt        x = a  +b                    

              x                  a         = (a cos() , a sin())

                a                 b    = (b cos() , b sin())
     b                          x = (b cos()+acos() , b sin()+asin())
                                      = M A
                                                 b sin   a sin  
                                      A  arctan
                                                 b cos   a cos  
                                                                        
                                                                       
                                M    a sin   b sin 2  a cos   b cos 2
                                                           P ( t)        0 if t  T
                                                                            3   if ( t  T) ( t  2 T)
s t   sin 2f ct  P t                                              
                                                                           2 
                                                                                  

                                                                            3     if ( t  2 T) ( t  3 T)
                                                                                     
                                                                           2         
x t   s(t )  s(t  )
                      T
                                                                           3             
                                                                                 if t  3 T
                      5                                                    2             2


                            sin(P (t))  sin P  t  T   
                                                      
              A( t)  atan
                                               5  
                                                             
                                               P t  T   
                            cos(P (t))  cos           
                                               5  
                  2
         1.571

                  1
         A ( t)
                  0

         0.785   1
                      0       0.5         1          1.5               2           2.5          3
                      0                               t                                         3
                                                                                   R  10

                                                   1                                                    5


                              f ( t) f  t 
                                            T                                                A ( t)
                                         R
                                                 0
                                                                                             Q ( t)
                                                                                                        0
                                        2


                                                   1                                                    5
                                                       0   1               2       3                        0       1                   2       3
                                                                   t                                                        t



                                                                                   R3

                                                   1                                                    5


                              f ( t) f  t 
                                            T                                                A ( t)
                                         R
                                                 0
                                                                                             Q ( t)
                                                                                                        0
                                        2


                                                   1                                                    5
                                                       0   1               2       3                        0       1                   2       3
                     T
  x t   s(t )  s t  
                                                                   t                                                        t


                     R                                                               R2

                                                   1                                                        5


                               f ( t) f  t 
                                             T                                                 A ( t)
                                          R
                                                 0
                                                                                               Q ( t)
                                                                                                            0
                                         2


                                                   1                                                        5
                                                       0       1               2       3                        0       1                   2       3
                                                                       t                                                            t
                                                                                   R1

zero amplitude!                                    1                                                    5


                              f ( t) f  t 
                                            T                                                A ( t)
                                         R
                                                 0
                                                                                             Q ( t)
                                                                                                        0
                                        2


                                                   1                                                    5
                                                       0   1               2       3                        0       1                   2       3
                                                                   t                                                            t
  Intersymbol Interference
ISI can be avoided by making the baud rate small.

If the baud rate is 11MHz (802.11b), how much delay will cause
complete ISI interference?

How much path length difference will cause complete ISI
interference?

•In suburban areas, multiple signals arrive with timing
differences up to 25microsec.

•Indoors, timing differences up to 300ns.

•What is the max baud rate so that complete ISI occurs
Frequency Selective Fading

The received signal is made up of many different,
slightly delayed, versions of the same signal.




           What is going on at these frequencies?
           Frequency Selective Fading
             xt   s(t )  st  D 
              sin2f ct  Pt   sin2f c t  D   Pt 

f c  80                            1                                5


              f ( t) f  t
                               T
T 1
                                                            A ( t)
                               
                              R   0
                                                            Q ( t)
                                                                     0
                      2


R  160                             1
                                        0   1       2   3
                                                                     5
                                                                         0   1        2        3
                                                t                                 t




                 Phase is ok, but zero amplitude!                                2f c D  
            For what values of T/R does this happen?                                 1
                                                                                 D
                                                                                    2 fc
Frequency Selective Fading

     Indoor impulse responds
          Frequency Selective Fading
               delay = 1/(2*fc)

               Suppose fc = 2.4GHz.
               Delay = 0.2 ns
               Distance = 0.2ns * 0.3m/ns = 0.06m (6cm)!!!!

So very small differences in path length cause very big changes in signal.

Frequency selective fading is be mitigated by
• Using spread spectrum. Thus multiple frequencies are simultaneously used.
  If a few frequencies suffer attenuation the others might not. (Used in
  802.11b)
• Channel estimation and adaptation (used in GSM cell phones)
• Use many narrow band frequencies. Then the good ones should work (like
  spread spectrum). Used in 802.11a
          Frequency Selective Fading
   Now suppose that there are many paths, each with a different delay.

   Then the received signal is:
                      Q                                     The I component is modeled as a normally
                                                            distributed random variable.
                                                            The Q component is modeled as a normally
                                                            distributed random variable.
                                              I             Both have zero mean and the same variance.

                                                            Then, the amplitude is a Raleigh random
received signal                                             variable and the phase is uniform between 0
                                                            and 2.

                                                            This is called a Raleigh channel.
          Hence, the result is that the amplitude and phase are random.
          If they vary slowly, then the channel is called a slowly fading channel (indoors).
          If the channel varies quickly, it is a fast fading channel (driving with cell phone).
          If the channel changes too fast, then changes in phase and amplitude cannot be detected.
                Effect of Movement
If the receiver or transmitter are moving, then the channel will vary.
Hence, the I and Q components will vary with time.



                                  Here is a plot of the magnitude of fading
                                  as a function of time and frequency.

                                  In this case, the channel does not change
                                  much over time. It is a slowly fading
                                  channel.
                Effect of Movement
If the receiver or transmitter are moving, then the channel will vary.
Hence, the I and Q components will vary with time.



                                    Here is a plot of the magnitude of fading
                                    as a function of time and frequency.

                                    In this case, the channel does not vary with
                                    frequency, it only varies over time.
                Effect of Movement
If the receiver or transmitter are moving, then the channel will vary.
Hence, the I and Q components will vary with time.




                                   Here is a plot of the magnitude of fading
                                   as a function of time and frequency.

                                   In this case, the channel varies both in time
                                   and frequency.
                      Doppler Effect
• When the receiver or transmitter are moving, the frequency is
  shifted by f = v/ cos(), v is velocity and  is wave length

                                          


                            v
  The maximum shift is f m  f c        c is the speed of light.
                            c

 If the the signal is sent at fc and
 passed through a fading channel, the
 spectrum of the received signal is:


         Thus, not only one frequency is received, but many.
             Doppler Effect
• To mitigate the Doppler effect:
  – Use low frequencies
  – Transmit in bursts so the channel is constant
    during the burst.
  – Include training sequences on each frame so the
    channel can be re-estimated for each
    transmission.
  – Do move – indoor use only
          Rician Channel Model
• A Raleigh channel assumes that all the paths arrive with
  random amplitude.
• A Rician channel assumes that there is a line of sight
  component that has much larger amplitude.

								
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