Orthogonal Codes for Band Limited Channel
Let $p_k(t) = p(t-kT)$ an orthonormal signal set for integer $k$ and $T&gt;0$. In other words, we require that:

$$
\int_{-\infty}^\infty p(t-\alpha T) p(t - \beta T) dt = 
\begin{cases}0, \text{  if $\alpha \neq \beta$}\\1, \text{  if $\alpha = \beta$}\end{cases} $$
The most common example of an orthonormal set is the square-root raised cosine pulse (SRRC). After matched filtering, SRRC pulses become raised-cosine pulses, which have zero ISI. 

Let $s_1(t) = \sum_m a_m p_m(t)$ and $s_2(t) = \sum_m b_m p_m(t)$, where $a_m, b_m \in \mathbb{R}$ (the result below can be easily extended to the complex case).

The dot product $s_1(t) \cdot s_2(t)$ is

$$
\begin{eqnarray*}
\int_{-\infty}^\infty s_1(t) s_2(t) dt &amp;=&amp; \int_{-\infty}^\infty \left( \sum_m a_m p_m(t) \right) \left( \sum_m b_m p_m(t) \right) dt \\
&amp;=&amp; \int_{-\infty}^\infty \sum_m a_m b_m p^2_m(t) dt \\
&amp;=&amp; \sum_m a_m b_m.
\end{eqnarray*}
$$
In the second step, I used the fact that $\int p_m(t) p_n(t) dt = 0$ if $m \neq n$. In the third step, I used the fact that $p(t)$ has energy equal to one.

Then, the dot product is zero only when $\sum_m a_m b_m = 0$. Note that the assumptions stated above about $p(t)$ are crucial. If you use a pulse that does not meet these conditions (for example, a raised-cosine pulse instead of square-root RC), then the dot product $s_1(t) \cdot s_2(t)$ will not be zero even if the sequenes $a_m,b_m$ are orthogonal.