Consider a permanent magnet dc (PMDC) motor which is initially at rest. At t = 0, a dc voltage of 5V is applied to the motor. Its speed monotonically increases from 0 rad/s to 6.32 rad/s in 0.5s and finally settles at 10 rad/s. Assuming that the armature inductance of the motor is negligible, the transfer function for the motor is

The correct option is 2/0.5s+1.

Step-by-Step Explanation:

1. Understanding the System Model:

A permanent magnet DC (PMDC) motor with negligible armature inductance can be modeled as a first-order dynamic system.

The relationship between the output angular speed $\Omega \left(s\right)$ and the input armature voltage $V\left(s\right)$ in the Laplace domain is given by the transfer function $G\left(s\right)$:

$G\left(s\right)=\frac{\Omega \left(s\right)}{V\left(s\right)}=\frac{K}{\tau s+1}$

where:

• $K$ is the steady-state gain of the motor.

• $\tau$ is the time constant of the system.

2. Finding the Steady-State Gain ($K$):

A constant DC voltage of $V=5\text{V}$ is applied at $t=0$ (a step input of magnitude 5).

The speed eventually settles at a final steady-state value of ${\omega }_{ss}=10\text{rad/s}$.

The steady-state gain $K$ is defined as the ratio of the steady-state output to the step input value:

$K=\frac{{\omega }_{ss}}{V}=\frac{10}{5}=2\text{rad/(V·s)}$

3. Determining the Time Constant ($\tau$):

For a first-order step response starting from rest ($\omega \left(0\right)=0$), the speed response as a function of time $t$ is:

$\omega \left(t\right)={\omega }_{ss}\left(1-e^{-t/\tau }\right)$

At one time constant ($t=\tau$), the output reaches approximately $63.2\%$ of its final steady-state value:

$\omega \left(\tau \right)=10\times (1-e^{-1})\approx 10\times 0.632=6.32\text{rad/s}$

From the given problem data, the motor reaches $6.32\text{rad/s}$ at time $t=0.5\text{s}$. Therefore, the time constant of the motor is:

$\tau =0.5\text{s}$

4. Writing the Final Transfer Function:

Substituting the values of steady-state gain $K=2$ and time constant $\tau =0.5$ into the general transfer function form:

$G\left(s\right)=\frac{2}{0.5s+1}$