Chapter 6, Problem 6.36P

To determine

The modified model of the system shown in figure 1 for the gear pair attached between the motor shaft and the load. Also, compute the transfer functions $\frac{{\Omega }_L\left(s\right)}{V_f\left(s\right)}$ and $\frac{{\Omega }_L\left(s\right)}{T_L\left(s\right)}$.

Answer to Problem 6.36P

The system’s modified model comprises the following equations as shown:

$\begin{array}{l}{v}_f=R_f{i}_f+L_f\frac{d{i}_f}{dt}\\ N\cdot \left(I_m+I_L\cdot N^2\right)\frac{d{\omega }_L}{dt}=K_T\cdot i_f-\left(c_m+c_L\cdot N^2\right)\cdot N\cdot {\omega }_L-T_L\end{array}$

Also, the transfer functions are as follows:

$\frac{{\Omega }_L\left(s\right)}{V_f\left(s\right)}=\frac{K_T}{N}\cdot \frac{1}{\left(R_f+L_{f}s\right)\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)}$

$\frac{{\Omega }_L\left(s\right)}{T_L\left(s\right)}=-\frac{1}{N\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)}$.

Explanation of Solution

Given:

The speed control system for the field-controlled motor has been given as shown in figure 1:

Figure 1

This speed-control system is equipped with a gear train of ratio N such that:

$\frac{{\omega }_m}{{\omega }_L}=N$

Concept Used:

1. The system model equations for this speed-control system are as follows:

For field-circuit, $v_f=R_f{i}_f+L_f\frac{d{i}_f}{dt}$

For rotational coupling, $I\frac{d\omega }{dt}=T-c\omega -T_L=K_T{i}_f-c\omega -T_L$

Here, I and c are the effective inertia and effective damping of the system respectively.

2. The effective inertia and the effective damping for the rotational part equipped with a gear train of ratio N are as follows:

$I_{eff}=I_m+I_L\cdot N^2$

$c_{eff}=c_m+c_L\cdot N^2$

Calculation:

The model equations for the system are:

$\Rightarrow v_f=R_f{i}_f+L_f\frac{d{i}_f}{dt}\left(1\right)$

For the inertia $I$,

$I\frac{d\omega }{dt}=T-c\omega -T_L=K_T{i}_f-c\omega -T_L\left(2\right)$

Also, on considering the effect of gearbox in a rotational system, we get

Effective inertia, $I_{eff}=I_m+I_L\cdot N^2\left(3\right)$

Effective damping, $c_{eff}=c_m+c_L\cdot N^2\left(4\right)$

Thus, on using equations (2), (3) and (4), we have for effective inertia $I_{eff}$

$I\frac{d\omega }{dt}=T-c\omega -T_L=K_T{i}_f-c\omega -T_L$

$\begin{array}{l}\Rightarrow I_{eff}\frac{d{\omega }_m}{dt}=K_T{i}_f-c_{eff}{\omega }_m-T_L\because \frac{{\omega }_m}{{\omega }_L}=N\\ \Rightarrow N\cdot \left(I_m+I_L\cdot N^2\right)\frac{d{\omega }_L}{dt}=K_T\cdot i_f-\left(c_m+c_L\cdot N^2\right)\cdot N\cdot {\omega }_L-T_L\left(5\right)\end{array}$

Taking Laplace transform of equation (1) and (5) while keeping zero initial conditions, we have

$\begin{array}{l}{v}_f=R_f{i}_f+L_f\frac{d{i}_f}{dt}\\ \Rightarrow V_f\left(s\right)=R_f{I}_f\left(s\right)+L_{f}s{I}_f\left(s\right)\\ \Rightarrow I_f\left(s\right)=\frac{1}{\left(R_f+L_{f}s\right)}{V}_f\left(s\right)\left(6\right)\end{array}$

$\begin{array}{l}N\cdot \left(I_m+I_L\cdot N^2\right)\frac{d{\omega }_L}{dt}=K_T\cdot i_f-\left(c_m+c_L\cdot N^2\right)\cdot N\cdot {\omega }_L-T_L\\ \Rightarrow N\cdot \left(I_m+I_L\cdot N^2\right)s\cdot {\Omega }_L\left(s\right)=K_T\cdot I_f\left(s\right)-\left(c_m+c_L\cdot N^2\right)\cdot N\cdot {\Omega }_L\left(s\right)-T_L\left(s\right)\\ \Rightarrow \left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)\cdot N\cdot {\Omega }_L\left(s\right)=K_T\cdot I_f\left(s\right)-T_L\left(s\right)\left(7\right)\end{array}$

From equations (6) and (7)

$\begin{array}{l}\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)\cdot N\cdot {\Omega }_L\left(s\right)=K_T\cdot I_f\left(s\right)-T_L\left(s\right)\\ \Rightarrow \left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)\cdot N\cdot {\Omega }_L\left(s\right)=K_T\cdot \frac{1}{\left(R_f+L_{f}s\right)}{V}_f\left(s\right)-T_L\left(s\right)\left(8\right)\end{array}$

In case, for the multiple inputs system, for finding the transfer function for one system other inputs are temporarily set to zero that is,

For $\frac{{\Omega }_L\left(s\right)}{V_f\left(s\right)}$ setting $T_L\left(s\right)=0$, therefore from equation (8)

$\begin{array}{l}\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)\cdot N\cdot {\Omega }_L\left(s\right)=K_T\cdot \frac{1}{\left(R_f+L_{f}s\right)}{V}_f\left(s\right)-T_L\left(s\right)\\ \Rightarrow \left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)\cdot N\cdot {\Omega }_L\left(s\right)=K_T\cdot \frac{1}{\left(R_f+L_{f}s\right)}{V}_f\left(s\right)\\ \Rightarrow \frac{{\Omega }_L\left(s\right)}{V_f\left(s\right)}=\frac{K_T}{N}\cdot \frac{1}{\left(R_f+L_{f}s\right)\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)}\end{array}$

Similarly, for $\frac{{\Omega }_L\left(s\right)}{T_L\left(s\right)}$ setting $V_f\left(s\right)=0$, therefore from equation (8)

$\begin{array}{l}\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)\cdot N\cdot {\Omega }_L\left(s\right)=K_T\cdot \frac{1}{\left(R_f+L_{f}s\right)}{V}_f\left(s\right)-T_L\left(s\right)\\ \Rightarrow \left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)\cdot N\cdot {\Omega }_L\left(s\right)=-T_L\left(s\right)\\ \Rightarrow \frac{{\Omega }_L\left(s\right)}{T_L\left(s\right)}=-\frac{1}{N\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)}\end{array}$.

Conclusion:

The system’s modified model comprises the following equations as shown:

$\begin{array}{l}{v}_f=R_f{i}_f+L_f\frac{d{i}_f}{dt}\\ N\cdot \left(I_m+I_L\cdot N^2\right)\frac{d{\omega }_L}{dt}=K_T\cdot i_f-\left(c_m+c_L\cdot N^2\right)\cdot N\cdot {\omega }_L-T_L\end{array}$

Also, the transfer functions are as follows:

$\frac{{\Omega }_L\left(s\right)}{V_f\left(s\right)}=\frac{K_T}{N}\cdot \frac{1}{\left(R_f+L_{f}s\right)\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)}$

$\frac{{\Omega }_L\left(s\right)}{T_L\left(s\right)}=-\frac{1}{N\left(\left(I_m+I_L\cdot N^2\right)s+\left(c_m+c_L\cdot N^2\right)\right)}$.