Ideal Bandlimited Impulse Response

Since we are working with sampled data, the ideal single-side-band filter impulse response is

$$h(n) = \hbox{\sc IDTFT}_n(u) \isdef \frac{1}{2\pi} \int_{-\pi}^{\pi}u(\omega)e^{j\omega n} d\omega$$

where $u(\omega)$ is a unit step function in the frequency domain:

$$u(\omega) \isdef \left\{\begin{array}{ll} 1, & \omega\ge 0 \\ [5pt] 0, & \omega<0 \\ \end{array} \right.$$

We can evaluate the IDTFT directly as follows:

$$\begin{eqnarray*} h(n) &=& \frac{1}{2\pi} \int_{0}^{\pi}e^{j\omega n} d\omega = \left. \frac{1}{2\pi j n} e^{j\omega n} \right\vert _0^\pi = \frac{e^{j\pi n}-1}{2\pi j n}\\ [10pt] &=& \frac{(-1)^n-1}{2\pi j n} = \left\{\begin{array}{ll} 0, & \hbox{$n$\ even}, n\neq 0 \\ [5pt] j\dfrac{1}{\pi n}, & \hbox{$n$\ odd} \\ \end{array} \right. \end{eqnarray*}$$

For $n=0$ , going back to the original integral gives

$$h(0)=\frac{1}{2\pi} \int_{0}^{\pi}e^{j\omega \cdot 0 } d\omega = \frac{1}{2}.$$

Thus, the real part of $h(n)$ is $\delta(n)/2$ , and the imaginary part is $1/(\pi n)$ for odd $n$ and zero otherwise (as a result of bandlimiting). There is no aliasing.