2.2.1. Steady-State

The steady-state analysis predicts the behavior of the full-bridge Buck inverter–DC motor system, given by (21)–(24), when its variables and input are in equilibrium. This is,

$$0 = -\overline{\upsilon} + E\overline{u}\_{av\prime} \tag{25}$$

$$0 = i - \frac{\overline{v}}{R} - \overline{\mathfrak{i}}\_{a\nu} \tag{26}$$

$$0 = \overline{\upsilon} - R\_a \overline{i}\_a - k\_c \overline{\omega},\tag{27}$$

$$0 = k\_m \overline{i}\_a - b\overline{\omega}\_\prime \tag{28}$$

where the overline means the nominal or constant value of such variables and input. After performing some algebraic manipulations, the equilibrium point (25)–(28) can be expressed in terms of the variable of interest *ω* as follows,

$$
\vec{d}\_a = \frac{b}{k\_m} \overline{\omega}\_\prime \tag{29}
$$

$$
\overline{\boldsymbol{w}} = \left(\frac{b\boldsymbol{R}\_d}{k\_{\text{nr}}} + k\_{\text{\textell}}\right) \overline{\boldsymbol{w}}\_{\text{\textell}} \tag{30}
$$

$$
\vec{I} = \left(\frac{bR\_d + k\_c k\_m + bR}{k\_m R}\right)\overline{\omega}\_\prime \tag{31}
$$

$$
\overline{u}\_{\rm av} = \left(\frac{bR\_{\rm af} + k\_{\rm ef}k\_{\rm mf}}{Ek\_{\rm mf}}\right)\overline{\omega}.\tag{32}
$$
