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Sec. 3–6 Intersymbol Interference 187
When H e (f) is chosen to minimize the ISI, H R (f), obtained from Eq. (3–64), is
called an equalizing filter. The equalizing filter characteristic depends on H C ( f ), the
channel frequency response, as well as on the required H e (f). When the channel consists
of dial-up telephone lines, the channel transfer function changes from call to call and the
equalizing filter may need to be an adaptive filter. In this case, the equalizing filter
adjusts itself to minimize the ISI. In some adaptive schemes, each communication session
is preceded by a test bit pattern that is used to adapt the filter electronically for the maximum
eye opening (i.e., minimum ISI). Such sequences are called learning or training
sequences and preambles.
We can rewrite Eq. (3–60) so that the rounded pulse train at the output of the receiving
filter is
w out (t) = an
a n h e (t - nT s )
(3–65a)
The output pulse shape is affected by the input pulse shape (flat-topped in this case), the transmitter
filter, the channel filter, and the receiving filter. Because, in practice, the channel filter
is already specified, the problem is to determine the transmitting filter and the receiving filter
that will minimize the ISI on the rounded pulse at the output of the receiving filter.
Example 3–13 INTERSYMBOL INTERFERENCE CAUSED BY RC FILTERING
Plot the output waveform when a channel filters a unipolar NRZ signal. Assume that the overall
filtering effect of the transmitter, channel, and the receiver is that of an RC low-pass filter (see
Fig. 2–15) where the 3 dB bandwidth is 1 Hz. Assume that the unipolar NRZ input signal has a
bit rate of R b = 1 Hz and that the data on the unipolar NRZ signal is [1 0 0 1 0 1 1 0 1 0]. Plot the
waveform at the receiver output and observe the intersymbol interference.
Solution
From Eq. (3–61), h e (t) = h(t) * h RC (t), where h RC (t) is the impulse response of the RC lowpass
filter as given by Eq. (2–146). Let the unipolar NRZ signal have a rectangular pulse shape that is
unity amplitude over the interval 0 to T b (where R b = 1>T b ) and zero for t elsewhere. Then,
h(t) =ß((t - T b >2)>T b ). T b is the duration of one bit. Convolving these two responses together,
we get
0, t … 0
h e (t) = c 1 - e -t>t 0
, 0 6 t … T b s
e -t>t 0
(e -Tb>t 0
- 1), t 7 T b
(3–65b)
Using Eq. (3–65b) in Eq. (3–65a), the solution is calculated and plotted by Example3_13.m.
From an applications point of view, when the required transmitter filter and receiving
filter transfer functions are found, they can each be multiplied by Ke -jvT d
, where K is a convenient
gain factor and T d is a convenient time delay. These parameters are chosen to make