# Absolute 2D pose from previous pose and measurement

I have a pose $$\boldsymbol{x}_0 = (x, y, \theta)$$ in an absolute frame, right-handed and centered in (0, 0, 0). I obtain a measurement $$\boldsymbol{m} = (\Delta x, \Delta y, \Delta \theta)$$ in the previous pose coordinate system.

What is the new pose $$\boldsymbol{x}_1$$ in the absolute frame? If the previous pose had a covariance $$\Sigma_{x_0}$$, and the measurement has a covariance $$\Sigma_{m}$$, what is the covariance of the new pose $$\Sigma_{x_1}$$?

$$\bigg( \begin{matrix} x'\\ y' \\ \theta' \end{matrix}\bigg) = \bigg( \begin{matrix} cos(\theta) & -sin(\theta) & x0 \\ sin(\theta) & cos(\theta) & y0 \\ 0 & 0 & 1 \end{matrix}\bigg) \bigg( \begin{matrix} x \\ y \\ \theta \end{matrix}\bigg)$$
For the second question I think it's ok to also take into account the covariance matrices in the previous eq. $$\bigg( \begin{matrix} x'\\ y' \\ \theta' \end{matrix}\bigg) = \bigg( \begin{matrix} cos(\theta) & -sin(\theta) & x0 \\ sin(\theta) & cos(\theta) & y0 \\ 0 & 0 & 1 \end{matrix}\bigg) Q_{m} \bigg( \begin{matrix} x \\ y \\ \theta \end{matrix}\bigg)Q_{x1}$$