Truncating signed integers

Assume that $x^{'}$ is the result of truncating $x$ , then:
- If $x = B 2 U_{w} (x)$ and $x^{'} = B 2 U_{k} (x^{'})$ , then $x^{'} = U 2 T_{k} (x mod 2^{k})$

Explanation of the formula

Here, $x mod 2^{k}$ will need to be first treated as an unsigned number.
Then, applying $U 2 T_{k}$ (unsigned-to-two’s complement, $k$ -bit word) on the above result, will convert the unsigned result to a signed number. This basically changes the weight of the highest bit (MSB) from $2^{k - 1}$ to $- 2^{k - 1}$ .