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:

Since the armature inductance of the permanent magnet DC (PMDC) motor is negligible, the transfer function relating the speed response
$\Omega \left(s\right)$
to the input voltage
$V\left(s\right)$
can be modeled as a standard first-order system:

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

where:
•
$K$
is the steady-state gain of the motor.
•
$\tau$
is the motor time constant.

1. Finding the Steady-State Gain (
$K$
):
The motor is initially at rest. At
$t=0$
, a constant DC voltage of 5 V is applied. This represents a step input of magnitude
$A=5$
V.

The speed of the motor eventually settles at a steady-state value of 10 rad/s. Under steady-state conditions, the relationship between the output speed and input voltage is given by:

${\omega }_{\text{ss}}=K·A$

Substituting the given values:

$10=K·5$

Solving for
$K$
:

$K=\frac{10}{5}=2$

2. Finding the Time Constant (
$\tau$
):
The speed response of a first-order system to a step input is given by the formula:

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

We are given that at
$t=0.5$
s, the speed
$\omega \left(0.5\right)=6.32$
rad/s. Substituting these values into the response equation:

$6.32=10(1-e^{-0.5/\tau })$

Divide both sides by 10:

$0.632=1-e^{-0.5/\tau }$

Rearranging the terms:

$e^{-0.5/\tau }=1-0.632=0.368$

Since
$e^{-1}\approx 0.368$
, we can equate the exponents:

$-\frac{0.5}{\tau }=-1$

Solving for
$\tau$
:

$\tau =0.5\text{s}$

Alternatively, the time constant is defined as the time required for the system response to reach 63.2% of its final steady-state value. Since
$10·63.2\%=6.32\text{rad/s}$
and this occurs at
$t=0.5\text{s}$
, the time constant is directly identified as
$\tau =0.5\text{s}$
.

3. Constructing the Transfer Function:
Substituting the steady-state gain
$K=2$
and the time constant
$\tau =0.5$
s into the transfer function model yields:

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