Why do we require soundness for fine structural ultrapowers? So for this question you may assume the background fine structure is that of "Fine structure and iteration trees".
I actually have two questions which are related to properties of the degree of some node in an iteration tree.
First is that when constructing the $n$th fine structural ultrapower of some ppm, we require it to be $n$-sound. But as far as I see it, if the critical point of the extender is below the $n$th projectum, then the extender measures all the $r\Sigma_n$ subsets of its critical point, so we can actually go ahead with the construction. So why do we require $n$-soundness?
Secondly, do all fine structural ultrapowers provide new information? i.e. if $\mathcal{M}$ is some ppm and $E$ is an extender over some $\kappa < \rho_{n+1}(\mathcal{M})$, is it possible that $\operatorname{Ult}_{n+1}(\mathcal{M}; E) = \operatorname{Ult}_{n}(\mathcal{M}; E)$? Does soundness play a role here?
 A: First question: I haven't personally considered situations where taking an ultrapower at degree higher than the soundness degree seemed to be relevant. Of course that certainly doesn't mean that it's not, and I think something like that is done in Jensen's $\Sigma^*$ fine structure. However, there is an example in one of the standard texts showing that at the level of infinitely many measurable cardinals, there is a $2$-sound mouse $M$,
with $M$ having a measurable cardinal $\mu<\rho_2^M$, such that given any real $x$, there is an linear iteration $M$, with critical points above $\rho_1^M$, such that $x$ is
boldface-$\Sigma_2$ definable over the eventual iterate (but is not over $M$, since $\omega<\mu<\rho_2^M$). So, as they say there, random information can be coded into $\Sigma_2$ over $0$-sound structures, so one should take the $1$-core before considering $\Sigma_2$ (or $r\Sigma_2$). As I say, though, I think that one forms larger ultrapowers via $\Sigma^*$ fine structure, but I don't know the details.
Second question: Assuming $M$ is $(n+1)$-sound and $\mathrm{crit}(E)<\rho_{n+1}^M$, it can certainly be that $\mathrm{Ult}_{n}(M,E)=\mathrm{Ult}_{n+1}(M,E)$, even with the natural factor map $\sigma:\mathrm{Ult}_n(M,E)\to\mathrm{Ult}_{n+1}(M,E)$ being the identity. Equivalently, it can be that all pairs $[a,f]$ used in forming the degree $n+1$ ultrapower are represented (mod measure 1) with pairs $[a,f]$ used in the degree $n$ ultrapower. For a simple example, this happens (for all extenders over $M$ and $n<\omega$ simultaneously) if $M$ models $\mathrm{ZF}^-$. But there are also more interesting examples (for more details on the following things see e.g. $\S$2 of A premouse inheriting strong cardinals from $V$ and $\S$3 of Fine structure from normal iterability): Suppose $\kappa=\mathrm{crit}(E)<\rho_{n+1}^M$ and there is no boldface-$\mathrm{r}\Sigma_{n+1}^M$-cofinal function from $\kappa$ to $\rho_n^M$. Then $U_n=\mathrm{Ult}_n(M,E)=\mathrm{Ult}_{n+1}(M,E)=U_{n+1}$ and the natural factor map $\sigma$ is the identity. To see this, use the following observations: (i) $\rho_n^{U_{n+1}}=\sup\sigma``\rho_n^{U_n}$, since both are the sups of the ultrapower maps applied to $\rho_n^M$  (for $U_n$, just because it is a degree $n$ ultrapower;
for $U_{n+1}$, by the cofinality assumption) and as the maps commute. (ii) $\sigma$ is an $n$-embedding (use (i) and commutativity for the $\mathrm{r}\Sigma_{n+1}$-elementarity). (iii) $U_n,U_{n+1}$ are both $(n+1)$-sound and $\rho_{n+1}^{U_{n+1}}=\rho_{n+1}^{U_n}\leq\mathrm{crit}(\sigma)$
and $\sigma$ preserves $\vec{p}_{n+1}$. This leads to (iv) $U_{n+1}\subseteq\mathrm{rg}(\sigma)$, which is enough.
