  Normal bases and self-dual normal bases over finite fields have been found to
be very useful in many fast arithmetic computations. It is well-known that
there exists a self-dual normal basis of $\mathbb{F}_{2^n}$ over $\mathbb{F}_2$
if and only if $4\nmid n$. In this paper, we prove there exists a normal
element $\alpha$ of $\mathbb{F}_{2^n}$ over $\mathbb{F}_{2}$ corresponding to a
prescribed vector $a=(a_0,a_1,...,a_{n-1})\in \mathbb{F}_2^n$ such that
$a_i={Tr}_{2^n|2}(\alpha^{1+2^i})$ for $0\leq i\leq n-1$, where $n$ is a
2-power or odd, if and only if the given vector $a$ is symmetric ($a_i=a_{n-i}$
for all $i, 1\leq i\leq n-1$), and one of the following is true.
  1) $n=2^s\geq 4$, $a_0=1$, $a_{n/2}=0$, $\sum\limits_{1\leq i\leq n/2-1,
(i,2)=1}a_i=1$;
  2) $n$ is odd, $(\sum\limits_{0\leq i\leq n-1}a_ix^i,x^n-1)=1$.
  Furthermore we give an algorithm to obtain normal elements corresponding to
prescribed vectors in the above two cases. For a general positive integer $n$
with $4|n$, some necessary conditions for a vector to be the corresponding
vector of a normal element of $\mathbb{F}_{2^n}$ over $\mathbb{F}_{2}$ are
given. And for all $n$ with $4|n$, we prove that there exists a normal element
of $\mathbb{F}_{2^n}$ over $\mathbb{F}_2$ such that the Hamming weight of its
corresponding vector is 3, which is the lowest possible Hamming weight.
