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Baseband Data Transmission II
Reference
  – Chapter 4.4-4.5, S. Haykin, Communication Systems,
    Wiley.




                                                         E.1
Introduction
Introduction
   – Intersymbol interference (ISI) is different from noise in
     that it is a signal-dependent form of interference that
     arises because of deviations in the frequency response of
     a channel from the ideal channel.
      • This non-ideal communication channel is also called
        dispersive

   – The result of these deviation is that the received pulse
     corresponding to a particular data symbol is affected by
     the tail ends of the pulses representing the previous
     symbols and the front ends of the pulses representing the
     subsequent symbols.                                      E.2
Introduction

– Example, for a binary PAM system without matched filter
            1     0        1



                                                    t




                      Sample points                  E.3
Introduction
– Two scenarios
   • The effect of ISI is negligible in comparison to that of
     channel noise.
         use a matched filter, which is the optimum linear
        time-invariant filter for maximizing the peak pulse
        signal-to-noise ratio.

   • The received signal-to-noise ratio is high enough to
     ignore the effect of channel noise (For example, a
     telephone system)
         control the shape of the received pulse.

                                                          E.4
Intersymbol Interference
 Consider a binary system, the incoming binary sequence
 {bk } consists of symbols 1 and 0, each of duration Tb .
 The pulse amplitude modulator modifies this binary
 sequence into a new sequence of short pulses
 (approximating a unit impulse), whose amplitude ak is
                                    + 1 if bk = 1
 represented in the polar form ak = 
                                    − 1 if bk = 0

{bk }    Pulse-     {ak }                   s (t )             xo (t )     x(t )
                            Transmit
        amplitude                                    Channel
                            filter g (t )              h(t )
        modulator
                                                                     w(t ) White noise
                                                                                   E.5
Intersymbol Interference
The short pulses are applied to a transmit filter of
impulse response g(t), producing the transmitted signal
              s (t ) = ∑ ak g (t − kTb )
                        k

 The signal s (t ) is modified as a result of transmission
 through the channel of impulse response h(t ) . In
 addition, the channel adds random noise to the signal.


{bk }    Pulse-     {ak }                   s (t )             xo (t )     x(t )
                            Transmit
        amplitude                                    Channel
                            filter g (t )              h(t )
        modulator
                                                                     w(t ) White noise
                                                                                   E.6
Intersymbol Interference
The noisy signal x(t ) is then passed through a receive
filter of impulse response c(t ) .The resulting output y (t )
is sampled and reconstruced by means of a decision
device.
     x(t )                  y(t)                       1 if y > λ
             Receive                        Decision
             filter c(t )                    device    0 if y < λ
                       Sample at ti = iTb
                                               λ
The receiver output is
                 ∑
     y (t ) = µ a k p (t − kTb ) + n(t )
                  k
where µp (t ) = g (t ) ⊗ h(t ) ⊗ c(t ) and µ is a constant.
                                                                    E.7
Intersymbol Interference
The sampled output is
                 ∑
     y (t i ) = µ ak p[(i − k )Tb ] + n(t i )
                   k
                                                                 ----- (1)
              = µa i + µ   ∑a
                           k
                                  k   p[(i − k )Tb ] + n(t i )
                           k ≠i


    µai : contribution of the ith transmitted bit.

    µ ∑ ak p[(i − k )Tb ] :
       k
       k ≠i
      the residual effect of all other transmitted bits.
      (This effect is called intersymbol interference)
                                                                             E.8
Distortionless Transmission
In a digital transmission system, the frequency
response of the channel h(t ) is specified.

We need to determine the frequency responses of the
transmit g (t ) and receive filter c(t ) so as to reconstruct
the original binary data sequence
     {bk }    Pulse-       {ak }                    s (t )             xo (t )       x(t )
                                    Transmit
             amplitude                                       Channel
                                    filter g (t )              h(t )
             modulator
                                                                             w(t ) White noise
             x(t )                  y(t)                                1 if y > λ
                     Receive                            Decision
                     filter c(t )                        device         0 if y < λ
                               Sample at ti = iTb
                                                             λ                               E.9
Distortionless Transmission

The decoding requires that
                      1 i = k
     p (iTb − kTb ) =                …..(2)
                      0 i ≠ k
(If this equation is satisfied and S/N is large,
 equation (1) becomes y (ti ) = µai )

It can be shown that equation (2) is equivalent to
       ∞

     ∑ P( f − n / T ) = T
     n = −∞
                   b     b       ….. (3)


                                                     E.10
Distortionless Transmission

– Example
            1    0        1



                                          t

        p (t )



                     Sample points        E.11
Distortionless Transmission

– Example                   p( f )
                               Tb
                                       1 / Tb
                                                      f


                                                  ∞
        p ( f − 1 / Tb )   p ( f − 2 / Tb )     ∑ p( f − n / T ) = T
                                                              b      b
                                                n = −∞




                                 f
                                                              E.12

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  • 1. Baseband Data Transmission II Reference – Chapter 4.4-4.5, S. Haykin, Communication Systems, Wiley. E.1
  • 2. Introduction Introduction – Intersymbol interference (ISI) is different from noise in that it is a signal-dependent form of interference that arises because of deviations in the frequency response of a channel from the ideal channel. • This non-ideal communication channel is also called dispersive – The result of these deviation is that the received pulse corresponding to a particular data symbol is affected by the tail ends of the pulses representing the previous symbols and the front ends of the pulses representing the subsequent symbols. E.2
  • 3. Introduction – Example, for a binary PAM system without matched filter 1 0 1 t Sample points E.3
  • 4. Introduction – Two scenarios • The effect of ISI is negligible in comparison to that of channel noise. use a matched filter, which is the optimum linear time-invariant filter for maximizing the peak pulse signal-to-noise ratio. • The received signal-to-noise ratio is high enough to ignore the effect of channel noise (For example, a telephone system) control the shape of the received pulse. E.4
  • 5. Intersymbol Interference Consider a binary system, the incoming binary sequence {bk } consists of symbols 1 and 0, each of duration Tb . The pulse amplitude modulator modifies this binary sequence into a new sequence of short pulses (approximating a unit impulse), whose amplitude ak is + 1 if bk = 1 represented in the polar form ak =  − 1 if bk = 0 {bk } Pulse- {ak } s (t ) xo (t ) x(t ) Transmit amplitude Channel filter g (t ) h(t ) modulator w(t ) White noise E.5
  • 6. Intersymbol Interference The short pulses are applied to a transmit filter of impulse response g(t), producing the transmitted signal s (t ) = ∑ ak g (t − kTb ) k The signal s (t ) is modified as a result of transmission through the channel of impulse response h(t ) . In addition, the channel adds random noise to the signal. {bk } Pulse- {ak } s (t ) xo (t ) x(t ) Transmit amplitude Channel filter g (t ) h(t ) modulator w(t ) White noise E.6
  • 7. Intersymbol Interference The noisy signal x(t ) is then passed through a receive filter of impulse response c(t ) .The resulting output y (t ) is sampled and reconstruced by means of a decision device. x(t ) y(t) 1 if y > λ Receive Decision filter c(t ) device 0 if y < λ Sample at ti = iTb λ The receiver output is ∑ y (t ) = µ a k p (t − kTb ) + n(t ) k where µp (t ) = g (t ) ⊗ h(t ) ⊗ c(t ) and µ is a constant. E.7
  • 8. Intersymbol Interference The sampled output is ∑ y (t i ) = µ ak p[(i − k )Tb ] + n(t i ) k ----- (1) = µa i + µ ∑a k k p[(i − k )Tb ] + n(t i ) k ≠i µai : contribution of the ith transmitted bit. µ ∑ ak p[(i − k )Tb ] : k k ≠i the residual effect of all other transmitted bits. (This effect is called intersymbol interference) E.8
  • 9. Distortionless Transmission In a digital transmission system, the frequency response of the channel h(t ) is specified. We need to determine the frequency responses of the transmit g (t ) and receive filter c(t ) so as to reconstruct the original binary data sequence {bk } Pulse- {ak } s (t ) xo (t ) x(t ) Transmit amplitude Channel filter g (t ) h(t ) modulator w(t ) White noise x(t ) y(t) 1 if y > λ Receive Decision filter c(t ) device 0 if y < λ Sample at ti = iTb λ E.9
  • 10. Distortionless Transmission The decoding requires that 1 i = k p (iTb − kTb ) =  …..(2) 0 i ≠ k (If this equation is satisfied and S/N is large, equation (1) becomes y (ti ) = µai ) It can be shown that equation (2) is equivalent to ∞ ∑ P( f − n / T ) = T n = −∞ b b ….. (3) E.10
  • 11. Distortionless Transmission – Example 1 0 1 t p (t ) Sample points E.11
  • 12. Distortionless Transmission – Example p( f ) Tb 1 / Tb f ∞ p ( f − 1 / Tb ) p ( f − 2 / Tb ) ∑ p( f − n / T ) = T b b n = −∞ f E.12