Baseband Data Transmission IIReference  – Chapter 4.4-4.5, S. Haykin, Communication Systems,    Wiley.                    ...
IntroductionIntroduction   – Intersymbol interference (ISI) is different from noise in     that it is a signal-dependent f...
Introduction– Example, for a binary PAM system without matched filter            1     0        1                         ...
Introduction– Two scenarios   • The effect of ISI is negligible in comparison to that of     channel noise.         use a ...
Intersymbol Interference Consider a binary system, the incoming binary sequence {bk } consists of symbols 1 and 0, each of...
Intersymbol InterferenceThe short pulses are applied to a transmit filter ofimpulse response g(t), producing the transmitt...
Intersymbol InterferenceThe noisy signal x(t ) is then passed through a receivefilter of impulse response c(t ) .The resul...
Intersymbol InterferenceThe sampled output is                 ∑     y (t i ) = µ ak p[(i − k )Tb ] + n(t i )              ...
Distortionless TransmissionIn a digital transmission system, the frequencyresponse of the channel h(t ) is specified.We ne...
Distortionless TransmissionThe decoding requires that                      1 i = k     p (iTb − kTb ) =                …...
Distortionless Transmission– Example            1    0        1                                          t        p (t )  ...
Distortionless Transmission– Example                   p( f )                               Tb                            ...
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00e isi

  1. 1. Baseband Data Transmission IIReference – Chapter 4.4-4.5, S. Haykin, Communication Systems, Wiley. E.1
  2. 2. IntroductionIntroduction – 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. 3. Introduction– Example, for a binary PAM system without matched filter 1 0 1 t Sample points E.3
  4. 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. 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. 6. Intersymbol InterferenceThe short pulses are applied to a transmit filter ofimpulse 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. 7. Intersymbol InterferenceThe noisy signal x(t ) is then passed through a receivefilter of impulse response c(t ) .The resulting output y (t )is sampled and reconstruced by means of a decisiondevice. x(t ) y(t) 1 if y > λ Receive Decision filter c(t ) device 0 if y < λ Sample at ti = iTb λThe receiver output is ∑ y (t ) = µ a k p (t − kTb ) + n(t ) kwhere µp (t ) = g (t ) ⊗ h(t ) ⊗ c(t ) and µ is a constant. E.7
  8. 8. Intersymbol InterferenceThe 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. 9. Distortionless TransmissionIn a digital transmission system, the frequencyresponse of the channel h(t ) is specified.We need to determine the frequency responses of thetransmit g (t ) and receive filter c(t ) so as to reconstructthe 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. 10. Distortionless TransmissionThe 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. 11. Distortionless Transmission– Example 1 0 1 t p (t ) Sample points E.11
  12. 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
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