Let $h_1[n]$ and $h_2[n]$ be impulse responses of type I FIR linear-phase systems with lengths $M$ and $N$. What can be said about the phase response of the following filters?
$h_3[n] = h_1[n] + h_2[n]$
$h_3[n] = h_1[n]h_2[n]$
$h_3[n] = h_1[n] \star h_2[n]$
My try: We can easily show that $$H_1(e^{j\omega}) = e^{-j\omega M /2}(\sum_{k=0}^{M/2}a[k]\cos(k\omega)) , \ a[0] = h_1[M/2], a[k] = 2h_1[(M/2)-k] \ \\ k=1,\dots ,M/2$$ and similarly $$H_2(e^{j\omega}) = e^{-j\omega N /2}(\sum_{k=0}^{N/2}b[k]\cos(k\omega)) , \ b[0] = h_2[N/2], b[k] = 2h_2[(N/2)-k] \\ k=1,\dots ,N/2$$ So we can write $$H_1(e^{j\omega}) = e^{-j\omega M /2}A(\omega) \\ H_2(e^{j\omega}) = e^{-j\omega N /2}B(\omega)$$ where $A(\omega)$ and $B(\omega)$ are real-valued periodic functions. Since in the third case we have $$H_3(e^{j\omega}) = H_1(e^{j\omega})H_2(e^{j\omega}) = e^{-j\omega(M/2 + N/2)}A(\omega)B(\omega)$$ the phase is linear in this case but I don't know how to solve the other cases. Certainly the phase is linear in the first case if $M=N$ but what about $M\not = N$ and also the second case?
Edit: In the first case for $M\not = N$, we can choose $a[k]$ and $b[k]$ such that the $h_3[n]$ be of type I or II FIR system. For example let $$a[0] = b[0] = 1, a[k] = b[k] = 0 \ \ \ k \not = 0 \\ H_3(e^{j\omega}) = e^{-j\omega M/2} + e^{-j\omega N/2} = e^{-j\omega (N+M)/4}(e^{j\omega (M-N)/4} + e^{-j\omega (M-N)/4}) = \\ 2e^{-j\omega (N+M)/4}\cos((M-N)\omega/4)$$ If $M=6$ and $N=2$, we get type I system and if $M=4$ and $N=2$ the system is of type II. Certainly there are examples such that this factorization isn't possible. So in this case, what's the required condition for $h_1[n]$ and $h_2[n]$ to $h_3[n]$ be of linear phase?
For the second case, the analysis in the frequency domain doesn't seem possible. When $M=N$, it's easy to show $h_3[n]$ is of type I because of the symmetry around $n = M/2 = N/2$. I couldn't find an example such that $h_3[n] = h_1[n]h_2[n]$ be (anti)symmetric when $M \not = N$.
