Let $\mathbf{a}^N = a_0 a_1\cdots a_{N-1}$ and let $\mathbf{a} = \mathbf{a}^N\mathbf{a}^N \cdots $ be an $N$-periodic sequence over the finite field $\mathbb{F}_q$ with $q$ elements, where $N \mid (q-1)$. Let $m \in \mathbb{N}$. If the linear complexity of $\mathbf{a}$ is $L$, what is the linear complexity of the sequence $\mathbf{a}_m = a_0^m a_1^m \cdots a_{N-1}^m a_0^m a_1^m \cdots$?
I know that for periodic sequences one may invoke for instance Blahut's theorem to obtain the linear complexity by looking at the Hamming weight of its DFT. However the DFT of powers of sequences, such as $\mathbf{a}_m$, is an $m$-th self-convolution, for which it is not clear to me its Hamming weight (i.e., the size of its support). Thus I wonder if there's some other way to obtain the linear complexity here?