As can be seen from the figure below, I have plotted a signal before it is pulse shaped and after performing half sine pulse shaping using the formula

$p(t) = \sin(\pi t/2T_c),\ 0<t<2T_c$

$p(t) = 0, {\rm otherwise}$

enter image description here

Now I am confused about how to interpret the signal after pulse shaping. For example, between 0 and 0.002, the digital signal (the one before pulse shaping) has 5 peaks (including the peaks at 0 and 0.002). Shouldn't I expect the analog signal to also follow such pattern with 5 peaks or is this something special to the half sine pulse shaping?

If it matters, I am using OQPSK modulation with DSSS to produce the digital signal and then performing half sine pulse shaping.

  • $\begingroup$ Are the digital impulses spaced $2T_c$ seconds? Are you doing pulse shaping by convolving the sequence of impulses with $p(t)$? $\endgroup$
    – MBaz
    Sep 22, 2015 at 14:08
  • $\begingroup$ The digital impulses are spaced $T_c$ seconds and yes, I am convolving the sequence of impulses with $p(t)$ @MBaz $\endgroup$
    – smyslov
    Sep 22, 2015 at 14:21

1 Answer 1


I assume in the following that your digital signal is a sequence of width-1 spikes or, equivalently, a sequence of delta pulses.

Your half-sine pulse shaping function gives you a lagged peak for every spike. Consider just your first digital signal spike at $t = 0$. If that were all you have, then the shaped output signal would be $$g(t) := f(t) \star p(t) = \sin(\pi t / 2T_c),$$ which has a peak at $t = T_c$. That is, the original signal had a peak at $t = 0$ and the shaped signal has a peak at $T_c$. Is this the behavior you intended?

If you only had your first two peaks, then the shaped signal would be the sum $$ g(t) = \sin(\pi t / 2T_c) + \sin(\pi t / 2T_c - T_c).$$ We can use standard trig identities to rewrite this. Set $\theta = \pi t / 2T_c$ and $\phi = \theta - T_c$, we have $\theta + \phi = 2\theta - T_c$ and $\theta - \phi = T_c$, so we can write $$\sin(\theta) + \sin(\phi) = 2\sin(\frac{\theta + \phi}{2}) \cos(\frac{\theta - \phi}{2}) = 2\sin(\theta - T_c/2)\cos(T_c/2).$$ This function has only one peak, at $\theta = \pi/2 + T_c/2$. In other words, two peaks in the digital signal correspond to only one peak in the shaped signal. I think this is the behavior which surprised you.

Generally speaking, and without doing the math, I'd expect $N$ (edit) consecutive same-sign peaks in the digital signal to give rise to $N-1$ peaks in the shaped signal.

edit: Without doing the math, I'm not very sure what I expect out of a positive peak followed by a negative peak. I expect you'll get a positive shaped peak followed by a negative shaped peak.

  • $\begingroup$ Generally speaking, and without doing the math, I'd expect N peaks in the digital signal to give rise to N−1 peaks in the shaped signal. Going by this assumption, between 0 and 0.006 in the figure, the digital signal has 13 peaks and I should expect 12 peaks in the shaped signal, whereas I just have 9 peaks @Austin A. $\endgroup$
    – smyslov
    Sep 22, 2015 at 15:29
  • $\begingroup$ @smyslov, I believe that Austin meant 'consecutive' peaks. For instance, you have 8 consecutive negative symbols but only 7 peaks in the pulse-shaped signal. $\endgroup$
    – MBaz
    Sep 22, 2015 at 15:31
  • $\begingroup$ @MBaz indeed I did. Thank you for pointing it out! I edited my answer. $\endgroup$
    – Austin A.
    Sep 22, 2015 at 15:37
  • $\begingroup$ can I safely assume that my pulse shaping is correct? @AustinA. $\endgroup$
    – smyslov
    Sep 23, 2015 at 7:34
  • $\begingroup$ @smyslov your graph certainly looks as though you correctly convolved your function $p(t)$ with your digital signal, yes. If you have confidence that the computation is correct, does that help you with your original question of interpreting it? $\endgroup$
    – Austin A.
    Sep 23, 2015 at 14:02

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