  Given an expansive matrix $R\in M_d({\mathbb Z})$ and a finite set of digit
$B$ taken from $ {\mathbb Z}^d/R({\mathbb Z}^d)$. It was shown previously that
if we can find an $L$ such that $(R,B,L)$ forms a Hadamard triple, then the
associated fractal self-affine measure generated by $(R,B)$ admits an
exponential orthonormal basis of certain frequency set $\Lambda$, and hence it
is termed as a spectral measure. In this paper, we show that if #$B<|\det
(R)|$, not only it is spectral, we can also construct arbitrarily sparse
spectrum $\Lambda$ in the sense that its Beurling dimension is zero.
