1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '15 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 '15 at 14:21
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '15 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 '15 at 15:31
  • $\begingroup$ @MBaz indeed I did. Thank you for pointing it out! I edited my answer. $\endgroup$ – Austin A. Sep 22 '15 at 15:37
  • $\begingroup$ can I safely assume that my pulse shaping is correct? @AustinA. $\endgroup$ – smyslov Sep 23 '15 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 '15 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.