Clarification regarding signal after pulse shaping
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.