ISRO EC 2023 Q.57: solution and answer

ISRO 2023 · ECQ.57 · 1 mark · MCQ

Which of the following is the error transfer function for negative feedback?

(A)
$\frac{1}{1 + G H}$
(B)
$\frac{G}{1 + G H}$
(C)
$\frac{G H}{1 + G H}$
(D)
$\frac{- 1}{1 + G H}$

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Answer and solution

ANSWER

The error transfer function is $E / R = \frac{1}{1 + G H}$ , which is option (a).

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

Take a negative feedback loop with forward path $G$ and feedback path $H$ . The reference is $R$ , the output is $C$ , and the error signal is $E = R - H C$ .
The forward path gives $C = GE$ . Substituting into the error equation gives $E = R - G H E$ .
Collect the terms in $E$ : $E (1 + G H) = R$ .
So the error transfer function is $\frac{E}{R} = \frac{1}{1 + G H}$
Option (b), $\frac{G}{1 + G H}$ , is $C / R$ : from $C = GE$ and $E = \frac{R}{1 + G H}$ we get $C / R = \frac{G}{1 + G H}$ . This is the closed-loop transfer function, not the error one.
Option (c), $\frac{G H}{1 + G H}$ , is the ratio of the fed-back signal $H C$ to $R$ . It equals $1 - \frac{E}{R}$ , since $1 - \frac{1}{1 + G H} = \frac{G H}{1 + G H}$ , so it is not $E / R$ itself.
Option (d) has a stray minus sign: $E / R$ is positive, because $E = R$ when $G H = 0$ (no feedback action), which gives $E / R = 1$ , not $- 1$ .

KEY IDEA

Write the error as $E = R - H C$ and eliminate $C$ using $C = GE$ ; this gives $E / R = 1/ (1 + G H)$ . The same denominator $1 + G H$ appears in every loop transfer function, and the numerator tells you which signal you are measuring.

WATCH OUT FOR

Choosing $G / (1 + G H)$ because it is the familiar closed-loop expression. That is $C / R$ , the output over the reference, not the error over the reference.

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Block diagrams & SFG questions, with notes