In Field Orientation Control (FOC) of an induction motor the stator voltage and slip are controlled to orient the rotor current, $\tilde{i}'_{ar}$, in the negative $q$-axis orthogonal to the out-of-phase component of $\tilde{i}_{as}$, $i^e_{ds}$. Since the currents are controlled the electrical transients are minimized and the dynamics are the mechanical dynamics given by (10) on the Introduction tab. The stator voltage equation becomes $$\tilde{V}_{as} = \left(r_s + j\omega_e K_L L_{ms}\right)\tilde{I}_{as} - \omega_e\left(L_{ss}-K_L L_{ms}\right)I_s \mathrm{sin}\theta_{esi}(0)~~~~~~~~~~\text{(15)}$$
Example 8G. Torque Control. Let us consider the dynamics of an induction motor drive with field orientation control given in Example 8F. The drive is originally operating at Operating Point 1 of Fig. 8F-1. The load line is $$ T_L = K \omega^2_r ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{(8G-1)}$$ where for $T_e = 10 \text{ Nm}$ at Operating Point 1 with $\omega_r=367.4 \text{ rad/s}$ from Table 8F-1 $$ K = \frac{10}{(367.4)^2} = 7.41 \times 10^{-5} \text{ Nms}^2 ~~~~~~\text{(8G-2)}$$ This mode of operation is shown in the Figure 3, which is Fig. 8G-1 in the text.
The commanded torque is changed from 10 Nm to 5 Nm. The motor slows and after the mechanical dynamics subside the new Operating Point 2 is at 2 on Fig. 8G-1. At this point the new speed from (8G-1) is $$ T_L = K \omega^2_r ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$ 5 = 7.41 \times 10^{-5} \omega^2_r ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$ \omega_r = \left(\frac{5}{7.41 \times 10^{-5}}\right)^{1/2} = 259.8 \text{ rad/s} ~~~~~~~~~~~~~\text{(8G-3)}$$ We could liken this scenario to reducing the accelerator of an electric car by one half. Now let us return to Operating Point 1, the accelerator is depressed and we return to Operating Point 1 as shown in Fig. 8G-1.
