Signed and Unsigned Loads in a 32-bit Registers
A $n$-bit bitstring can be interpreted semantically both as an unsigned integer in the range $0$ to $2^n-1$, and as a signed integer in the range $-2^{n-1}$ to $2^{n-1}-1$. There is always a way to assign an $n_1$-bit bitstring to an $n_2$-bit bitstring so that both have the same semantics, but the way to do that depends on whether the semantics is unsigned or signed. In the unsigned case, we use zero extension: we affix $n_2-n_1$ zeroes on the left. In the signed case, we use zero extension: we affix $n_2-n_1$ copies of the sign bit on the left.

As an example, consider $x=1111$, and suppose that we wish to assign it into an 8-bit bitstring. In the unsigned case, the value of $x$ is $15$, and the extension is $00001111$. In the signed case, the value of $x$ is $-1$, and the extension is $11111111$.

When $n_2 = n_1$, there is nothing to extend, so the signed and unsigned cases are the same. That's why the point is moot in such cases.