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
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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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01010                                Sx(w)
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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
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 0011                2cos(wct+q)                               cos(wct+q)
 0111
 1101                                                      xc(t)
01010                                Ideal LPF                     x
                         x
                                       H0(w)
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 1101   x(t)                                                                    x(t)
 1111                 2sin(wct+q)                            sin(wct+q)     +
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                                      Ideal LPF            xs(t)
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                          x                                         x
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                                        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
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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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        Lecture 15                        Dickerson EE422                               4
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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   
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                                                                            w  2p B
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                                   
00100
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 0011    Sx(w)
 0111
 1101                     -wc                                         wc         w
01010   Sx(w-wc)                                           LPF
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 1111                 -2wc                        0                         2wc w
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00011   Sxc(w) or Sxs(w)
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                          -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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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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          (area under PSD same) ________ ________ ________
                                              xc  t   xs  t   x  t 
00011                             2        2        2
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        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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        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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        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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                       Rxy t   1  Rx  0   Ry  0  
                                  2                      
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        • Uncorrelated:                 __________________
                          Rxy t   x  t  y  t  t   x y
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        • Orthogonal: Rxy t   0
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 1111   • Independent: if x(t1) and y(t2) are independent
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          (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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        • 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
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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
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                                 S n  w  w c   S n w - w c 
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                                                                         w  2p B
        S nc w   S ns w   
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                                0
                                                                         w  2p B
 1111
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00011                                          N
                                                              w  2p B
10000                  S nc w   S ns w   
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01010                                          0
                                                              w  2p B
        Lecture 15                      Dickerson EE422                              11
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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
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                                          df  2 NB  nc 2  ns 2
                               fc - B   2
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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
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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 
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        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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                                   e
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        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 
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                        s   2
                                                                      
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                         E -E 2  A2 / 2s 2  AE 
                       2e
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                        s                       s 
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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
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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
01010
         Lecture 15           Dickerson EE422                  16
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 1111     Detection: Bipolar Signaling
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0000    • If “1”, send p(t) s(t)        +       Peak
                                                           Detector
00001                                         Detect and
10100   • If “0”, send –p(t)
11100                                          Sample
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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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        • 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 
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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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              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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00011                sn 
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01010
        Lecture 15             Dickerson EE422                     19
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 1111                 Minimize PE
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0000    • To minimize PE, must maximize r since Q
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                     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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        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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            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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 1111                   Signal Amplitude
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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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        +/- p(t)+n(t)                                         t=tm
 0111                     h(t)                                        Threshold
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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
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 1111     Signal and Noise Calculation
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                                        po  t   F -1 P w  H w 
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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
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 0011                       ____
                          1                     
                                                   S n  w  H  w  dw
                                                                   2
                  s n 
 0111                 2         2
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                         2p
                      no      o              -
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10111   • So, the ratio becomes:
 1101                                                                     2
                                   P w  H w e jwtm dw 
                                             

                                   -
 1111

                                                           
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00011
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                            2      2                      
                                        
                                   2p  Sn w  H w  dw
                                  o                     2
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                                                    -
         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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11100   • Mini-Lecture 1:
00100
0100       – Optimum Threshold Detection
 0011
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 1101   • Mini-Lecture 2:
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           – Optimum Binary receivers
        • Mini-Lecture 3:
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00011
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           – Optimum Binary receivers
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         Lecture 15         Dickerson EE422   24

				
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