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
•   “Ghost” on TV.
•   GPS – incorrect position calculation.
•   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
the shorter
1               0
path                                               How big does  have to be to so that
the 3 of the longer path arrives
Symbols              3               2
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?
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

slightly delayed, versions of the same signal.

What is going on at these frequencies?
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

Indoor impulse responds
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
Now suppose that there are many paths, each with a different delay.

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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