GATE 2020 EE Q.34: solution and answer

GATE 2020 · EEQ.34 · 2 marks · MCQ

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

(A)
$\frac{10}{0.5 s + 1}$
(B)
$\frac{2}{0.5 s + 1}$
(C)
$\frac{2}{s + 0.5}$
(D)
$\frac{10}{s + 0.5}$

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Answer and solution

ANSWER

The motor transfer function is $\frac{Ω ( s )}{V ( s )} = \frac{2}{0.5 s + 1}$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2020 EE (IIT Delhi)

With armature inductance neglected, the electrical time constant drops out and the PMDC motor's speed-to-voltage transfer function is first order: $\frac{Ω ( s )}{V ( s )} = \frac{K}{τ s + 1}$ . A monotonic rise with no overshoot is what a first-order response looks like.
For a step input of $5 V$ , $V (s) = \frac{5}{s}$ , and the first-order step response is $ω (t) = 5 K (1 - e^{- t / τ})$ .
Steady state, using the final value theorem: $ω (\infty) = lim_{s \to 0} s \cdot \frac{K}{τ s + 1} \cdot \frac{5}{s} = 5 K$ . Setting this equal to $10 rad/s$ gives $K = \frac{10}{5} = 2 rad/s per V$ .
Time constant: $\frac{ω ( 0.5 )}{ω ( \infty )} = \frac{6.32}{10} = 0.632 = 1 - e^{- 1}$ . So $e^{- 0.5/ τ} = e^{- 1}$ , which means $t = 0.5 s$ is exactly one time constant: $τ = 0.5 s$ .
Therefore $\frac{Ω ( s )}{V ( s )} = \frac{2}{0.5 s + 1}$ which is option (B).
Checking the other options: (C) $\frac{2}{s + 0.5} = \frac{4}{2 s + 1}$ has DC gain $4$ and $τ = 2 s$ . (D) $\frac{10}{s + 0.5} = \frac{20}{2 s + 1}$ has DC gain $20$ and $τ = 2 s$ . (A) has the right time constant, but its DC gain is $10$ , so a $5 V$ step would settle at $50 rad/s$ , not $10 rad/s$ .

KEY IDEA

If armature inductance is negligible, a PMDC motor behaves as a first-order system $\frac{K}{τ s + 1}$ . Its DC gain is the final speed divided by the step voltage ( $10/5 = 2$ ). Its time constant is the time taken to reach $63.2%$ of the final value ( $0.5 s$ ).

WATCH OUT FOR

Using the final speed ( $10 rad/s$ ) as the gain and forgetting to divide by the $5 V$ input, which gives option (A). Another mistake is reading the pole form $\frac{K}{s + a}$ as if $a$ were the time constant. In fact $τ = 1/ a$ and the DC gain is $K / a$ , so $\frac{2}{s + 0.5}$ has $τ = 2 s$ and DC gain $4$ .

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GATE 2020 EE Q.34: solution and answer

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