# EE LECTURE

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```					                      Today‟s Schedule
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• Reading: Lathi 11.5,13.1
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11100   • Mini-Lecture 1:
00100
0100       – Go over Quiz 2
0011
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1101   • Mini-Lecture 2:
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– Bandpass Random Processes -Equivalent filters
• Mini-Lecture 3:
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– Optimal Threshold Detection
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Lecture 15           Dickerson EE422           1
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1111        Bandpass Random Process
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• What happens to a signal at a receiver?
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How does the PSD of the signal after a
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BPF correspond to the signal before the
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0111     BPF?
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01010                                Sx(w)
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00011
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-wc       0             wc   w
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Lecture 15         Dickerson EE422            2
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1111                           BPF System
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• Bandpass random process can be written as:
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x t   xc t  cosw0t  xs t  sin w0t

• With the impulse response: h t   2h0 t  cosw0t
11100
00100
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0011                2cos(wct+q)                               cos(wct+q)
0111
1101                                                      xc(t)
01010                                Ideal LPF                     x
x
H0(w)
10010
10111
1101   x(t)                                                                    x(t)
1111                 2sin(wct+q)                            sin(wct+q)     +
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00011
Ideal LPF            xs(t)
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x                                         x
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01010
H0(w)
Lecture 15                       Dickerson EE422                               3
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1111                    Impulse Response
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• Impulse Response
00001         H0(w)                                         h  t   2h0 t  cosw0t
10100                            1
11100
00100                                                 • Transfer Function
0100                                                  H w   H0 w  wc   H0 w - wc 
0011                        2pB
0111
1101                                                 • So, xc(t) and xs(t) are low-
01010            H (w)               4pB
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pass random processes,
10111                        1                          what else can be deduced?
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1111                                                 • Assume theta is uniformly
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00011     -wc            0           wc                 distributed phase noise
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Lecture 15                        Dickerson EE422                               4
10111
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1111    PSD of BP Random Processes
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• PSD of xc(t) and xs(t)
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 S x w  w c   S x w - w c 
                                         w  2p B
S xc w   S xs w   
00001

w  2p B
10100
11100                              0

00100
0100
0011    Sx(w)
0111
1101                     -wc                                         wc         w
01010   Sx(w-wc)                                           LPF
10010
10111   +Sx(wwc)
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1111                 -2wc                        0                         2wc w
1010
00011   Sxc(w) or Sxs(w)
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01010
-2wc                    0                   2wc         w
Lecture 15                     Dickerson EE422                              5
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Mean and variance of narrowband
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noise
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1000
0000    • In-phase and quadrature components have
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• In-phase and quadrature components of
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narrowband noise are zero-mean
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01010       – Noise comes original signal being passed
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through a narrowband linear filter
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1111   • Variance of the processes is the same
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(area under PSD same) ________ ________ ________
xc  t   xs  t   x  t 
00011                             2        2        2
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01010
Lecture 15          Dickerson EE422                               6
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1111                Properties
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• If the narrowband noise is Gaussian, then
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the in-phase (nI) and quadrature (nQ) are
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jointly Gaussian.
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• If the narrow band noise is wide-sense
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stationary (WSS), then the in-phase and
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01010
Lecture 15       Dickerson EE422              7
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1111                  Cross Correlations
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• Definition Cross Correlation
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 
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Rxy  t1 , t2   x t y t    x1 y2 f xy  x1 , y2  dx1dy2
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00100
0100                           1   2  - -
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• Definition Jointly Stationary
Rxy t1, t2   Rxy t2 - t1   Rxy t 
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00011
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Lecture 15                    Dickerson EE422                     8
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1111      Jointly Stationary Properties
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Rxy t   Rxy  -t 
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• Properties
Rxy t   Rx  0  Ry  0 
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00001
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00100
Rxy t   1  Rx  0   Ry  0  
2                      
0100
• Uncorrelated:                 __________________
Rxy t   x  t  y  t  t   x y
0011
0111
1101

• Orthogonal: Rxy t   0
01010
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1101
1111   • Independent: if x(t1) and y(t2) are independent
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00011
(joint distribution is product of individual
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distributions)
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Lecture 15            Dickerson EE422                    9
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1111                  Activity
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• Working with a partner come up with a list
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of communications systems that will need
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bandpass analysis for performance
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0111     assessment.
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• The PSD of a BP white noise process is
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1101     N/2. What is the PSD and variance of the
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in-phase and quadrature components?
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Lecture 15       Dickerson EE422          10
10111
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1111    Example White noise process
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• The PSD of a BP white noise process is
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N/2. What is the PSD and variance of the
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in-phase and quadrature components?
n t   nc t  cos wct  ns t  sin wct
0011
0111
1101

 S n  w  w c   S n w - w c 
01010
                                         w  2p B
S nc w   S ns w   
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1101
0
                                         w  2p B
1111
1010
00011                                          N
               w  2p B
10000                  S nc w   S ns w   
10110
01010                                          0
               w  2p B
Lecture 15                      Dickerson EE422                              11
10111
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1111   Variance of White Noise Process
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0000    From the SNR calculations, it is clear that:
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fc  B   N
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n  2
2
df  2 NB  nc 2  ns 2
fc - B   2
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1111
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Lecture 15                      Dickerson EE422            12
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• Signal is a sinusoid mixed with narrow-band
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additive white Gaussian noise (AWGN)
00100
0100                  y t   A cosw ct     nt 
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• Can be written as:
yt       A  nc t coswct     ns t sinwct   
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 E t coswct  t    
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E t        A  nc t 2  ns 2 t 
ns t 
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t 
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 - tan -1
Dickerson EE422t 
01010
Lecture 15                        A  nc                       13
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1111                Example (continued)
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• In-phase and Quadrature terms of noise
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Gaussian with variance s2
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• In a similar transformation to that used for
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calculating the dart board example, the
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joint density can be found in polar
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E -E 2 -2 AE cosq  A2 / 2s 2
f E E ,q 
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00011
       e
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10110                         2ps 2
01010
Lecture 15           Dickerson EE422                 14
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1111                Marginal Density of E
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• Rician Density
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p
f E E    f E E ,q dq
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00100                  -p
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E               1 p - AE cosq / s 2 
 2p -pe
- E 2  A 2 / 2s 2
                                             dq 
0011
0111                         e
1101
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s   2
                         
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E -E 2  A2 / 2s 2  AE 
 2e
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1101                                        I0  2 
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s                       s 
00011
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• Approaches a Gaussian if A>>s
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Lecture 15                     Dickerson EE422                     15
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Digital Communications Systems
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in Noise
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• Analog Comm: Goal is to reproduce the
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waveform accurately
00100
0100         – Figure of Merit: output signal to noise ratio
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• Digital Communications: Goal is to decide
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– Figure of Merit: Probability of error in making
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this decision at the receiver
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Lecture 15           Dickerson EE422                  16
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1111     Detection: Bipolar Signaling
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1000
0000    • If “1”, send p(t) s(t)        +       Peak
Detector
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10100   • If “0”, send –p(t)
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• Received signal:             n(t)
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r(t) = +/- p(t) + n(t)
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01010   • Optimal threshold?
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• Noise is additive,
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Gaussian noise
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Lecture 15          Dickerson EE422                   17
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PE Calculations
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0000    • P(1/0) – prob of
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00100     „0‟ sent
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0011       – P(n>Ap)
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• P(0/1) – prob of
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detecting a „0‟ when
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„1‟ sent
1010        – P(n<-Ap)
 - Ap      Ap 
P   1  1 - Q 
00011
10000                                              Q    
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01010
Lecture 15            Dickerson EE422                  18
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1111                        Bipolar
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• Total Error Probability
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1                   Ap             Ap 
PE   P  mi  P   mi   Q 
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 P 1  Q      P  0
 sn             sn 
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0011
i 0
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• If P(1)=P(0)=0.5                 Q(x)
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 Ap                 0.5
PE  Q       Qr
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1111
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00011                sn 
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10110                                                          x
01010
Lecture 15             Dickerson EE422                     19
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1111                 Minimize PE
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1100
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0000    • To minimize PE, must maximize r since Q
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10100     decreases monotonically as r increases
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00100                        Ap 
0100
PE  Q       Qr
 sn 
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• Ap is the signal amplitude and sn is the rms
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noise.
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1010    • Goal: filter signal to enhance signal and
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10000     reduce noise power
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Lecture 15         Dickerson EE422          20
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1111                       Linear Filtering
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x(t)=+/- p(t)+n(t)       Linear Network          y(t)=+/- po(t)+no(t)
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11100                X(f)                                          Y(f)
00100                                   H(f)
0100
0011
0111   • Recall:
y t   h t   x t   Y  f   H  f  X  f 
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S y  f   H  f  Sx  f 
2
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00011
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Lecture 15                   Dickerson EE422                        21
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po 2  tm 
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• The goal is maximize               r2 
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2
o
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00100
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+/- p(t)+n(t)                                         t=tm
0111                     h(t)                                        Threshold
1101                     H(f)
01010                                                                  Detector
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p(t)+n(t)                                     p(t)+n(t)
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p(t)
00011    p(t)
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01010
Lecture 15                   Dickerson EE422                             22
10111
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1111     Signal and Noise Calculation
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po  t   F -1 P w  H w 
0100
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• Signal output:
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1 
po  tm          P w  H w e jw tm dw
2p -
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00100
0100    • Output noise power or variance
0011                       ____
1                     
       S n  w  H  w  dw
2
s n 
0111                 2         2
1101
2p
no      o              -
01010
10010
10111   • So, the ratio becomes:
1101                                                                     2
 P w  H w e jwtm dw 


 -
1111


1010                            ____
00011
10000                      r n 
2      2                      

2p  Sn w  H w  dw
o                     2
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01010
-
Lecture 15                         Dickerson EE422                   23
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1111                  Next Time
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• Reading: Lathi 13.1, 13.2
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10100
11100   • Mini-Lecture 1:
00100
0100       – Optimum Threshold Detection
0011
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1101   • Mini-Lecture 2:
01010
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10111
– Optimum Binary receivers
• Mini-Lecture 3:
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1111
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00011
10000
– Optimum Binary receivers
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Lecture 15         Dickerson EE422   24

```
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 views: 4 posted: 3/23/2011 language: English pages: 24