The document discusses intersymbol interference (ISI) in baseband data transmission. ISI arises from deviations in a communication channel's frequency response from ideal, causing received pulses to be affected by neighboring pulses. This can be mitigated by matched filtering to maximize signal-to-noise ratio or by controlling the received pulse shape when noise is negligible. ISI causes the sampled output to depend on neighboring transmitted bits. Distortionless transmission requires designing transmit and receive filters such that only the desired bit contributes to the sampled output.

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