5. It's known that the continuous time 1-D signal $x_c(t) = e^{-at^2}$ has Fourier transform $X_c(\Omega) = \sqrt{\frac{\pi}{\alpha}}e^{-\frac{\Omega^2}{4\alpha}}$, where $\alpha > 0$. Next, we sample $x(t)$ at sampling interval T=1 to obtain a discrete-time signal $x[n] = x_c(nT) = e^{-\alpha n^2}$.
(a) Write the Fourier Transform $X(\omega)$ of $x[n]$ in terms of the parameter $\alpha$. This should be an aliased sum involving $X_c$. Use sum index $k$.
(b) Find an upper bound on $\alpha$ so that the $k = \pm 1$ aliased terms in $X(\omega)|_{k=0}$ are no larger than $10^{-3}\sqrt{\frac{\pi}{\alpha}}$. Note: assume $X(\omega, k) = 0$, if $\omega \notin [\omega - 2\pi(k+1), \omega - 2\pi(k-1)]$
(c) Consider the 2-D signal $x[n_1, n_2] = e^{-\alpha(n_1^2 + n_2^2)}$, and repeat part (a), find now $X(\omega_1, \omega_2)$. Use aliased sum index $(k_1, k_2)$.
(d) Find an upper bound on $\alpha$ so that the $(k_1, k_2) = (\pm 1, 0)$ and $(0, \pm 1)$ aliased terms in $X(\omega_1, \omega_2)|_{k_1=0, k_2=0}$ are no larger than $10^{-3}\sqrt{\frac{\pi}{\alpha}}$. Note: assume $X(\omega_1, \omega_2, k_1, k_2) = 0$, when $[\omega_1, \omega_2] \notin [\omega_1 - 2\pi(k_1 + 1), \omega_1 - 2\pi(k_1 - 1)] \times [\omega_2 - 2\pi(k_2 + 1), \omega_2 - 2\pi(k_2 - 1)]$. $\times$ here refers to the range for $\omega_1$ and $\omega_2$.
