I'm a little confused about pulse shaping I know that pulse shaping is a filter used to convert a signal that needs to be transmitted from a rectangular pulse signal into a signal that similar to Sinc. But how can this filter be split between the transmitter, channel, and receiver ?! Why isn't this filter design only on the transmitter side? Also, why not just use modulation (such as ASK) instead of filtering (pulse shaping), so the basis sine wave(of ASK) has limited bandwidth that can be passed through the channel without spreading and affecting the adjacent wave/bit(no ISI)?


2 Answers 2


To add a simplified explanation to the excellent mathematical description jithin has provided:

Pulse shaping is done specifically to limit the occupied bandwidth of the transmitted signal. If you sent a rectangular pulse (such as the OP's ASK example), even though the carrier itself is narrow band, the occupied spectrum is given by the Fourier Transform of the pulse (which is a Sinc function with the null to null bandwidth of the main lobe $B =1/T$ where B is the bandwidth in Hz and T is the duration of the pulse (symbol). Further the sidelobes have peak values that go down relatively slowly (at a rate of $1/f$)-so this takes up a LOT of bandwidth, unnecessarily. This is the ONLY reason for pulse shaping.

Adding pulse shaping can cause inter-symbol inteference since restricting bandwidth in frequency increases the time response from each symbol. However, we can choose pulse shapes that have minimum or zero intersymbol inteference at the correct sampling locations of the adjacent symbols.

We could implement all the pulse shaping in the trasnmitter, but optimum reception in AWGN (additive white Gausssian noise) conditions requires the use of a matched filter in the receiver. Therefore if we factor the pulse shaping into the cascade of two identical filters (which is essentially a polynomial square-root of the original filter) we can still achieve most of the spectral containment we desire AND get a matched filter in the receiver.

Please also see this answer for further details:

Why root raised cosine filter can eliminate intersymbol interference (ISI) ?

  • $\begingroup$ what I understood, due to bandlimited channel(also multiple paths) intersymbol interference occurs, pulse shaping also bandlimited(shaping) so cause ISI but with zero ISI at the sampling point, the pulse shaping split to use the part in the receiving side as match filter too is that correct? $\endgroup$ Apr 9, 2020 at 20:56
  • $\begingroup$ Yes and see the other post if you haven't as it shows specifically how that zero ISI results. In order to constrain the bandwidth the pulse shape in time has to stretch out (time frequency duality-- narrow in time, wide in frequency and vice versa), so the other post demonstrates how to have a very long pulse shape in time while still having zero ISI. If you factor the two filters into two, you don't get that zero ISI until you cascade them, so you can effectively cascade by having one in the transmitter and one in the receiver. $\endgroup$ Apr 9, 2020 at 21:02

Pulse Shaping Filter, helps to control the Bandwidth of your transmission but they are designed such that when you do Matched Filtering at the receiver, they eliminate Inter Symbol Interference caused by pulse shaping at the Transmitter. If you consider the transmission model $x(t) = \sum x[nT]p(t-nT)$. For now consider no effect of the channel. That is, $h(t) = 1$. At the receiver you do matched filtering and sample at $T$ intervals. $y[k] = y(t)|_{t=kT} = \sum x[nT]p(t-nT)*p(T-t)|_{t=kT}$.

This is equivalent to multiplying $(\sum x[nT]p(t-nT)*p(T-t))$ with $\sum \delta(t-kT)$ - sampling at intervals of $T$.

For NO ISI Condition, we want $$ p(t-nT)*p(T-t)\sum\delta(t-kT) = \delta(t-nT) $$ In that case $y(t)|_{t=kT} = \sum x[nT]\delta(t-nT) = x[nT]$. The fourier transform equivalent of this is $$ (P(f) \times P^*(f)) * \sum \delta(f-k/T) = 1\\ \sum|P(f-k/T)|^2 = 1 $$ There are many pulse shapes whose summation of shifted square of magnitude equals constant. For example - square, triangle, raised cosine etc. But of course, the most practical one to implement is raised cosine. We take the square root of it at the transmitter as the pulse shaping filter, and the same matched at the receiver. Hence it has roles both at the transmitter and receiver - to control bandwidth and matched filtering.

If the channel were a multi-tap instead of earlier simple channel, matched filtering with only $p(T-t)$ will not be perfect. In that case, what you will receive is $\sum x[nT]p(t-nT)*h(t)$. But since you do not have estimate of channel at the initial stage, you can only do sub optimal matched filtering with $p(T-t)$. So your received samples will be $y(kT) = \sum x[nT]p(t-nT)p(T-t)*h(t-kT)|_{t=kT} = \sum x[nT]h(kT-nT)$. You will need to eliminate the effect of channel later using equalization techniques.

  • $\begingroup$ is pulse shaping causes ISI by ? or it used essentially to overcome ISI? so filter at receiver used as part of pulse shaping and at the same time used as match filtering? $\endgroup$ Apr 9, 2020 at 18:26
  • $\begingroup$ @Computer_guy11 yes you are right. Pulse shaping causes ISI because we are limiting it's bandwidth. So in time domain it will ideally extend forever. But still we can truncate to 4 or 5 times symbol time without any visible degradation. ISI is caused by channel also which is compensated after matched filtering. $\endgroup$
    – jithin
    Apr 10, 2020 at 1:46

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