The error detector — the summing junction at the input of every closed loop — subtracts the feedback signal from the reference :

\(E(s)=R(s)-B(s)\qquad\text{where}\qquad B(s)=H(s)\,C(s)\)

which is option 3.

Two details separate the right answer from the near miss.

First, subtraction, not addition. Negative feedback is what makes a closed loop stable and self-correcting: if the output exceeds the reference, the error goes negative and the controller reduces its drive. Adding the two signals (options 1 and 2) would give positive feedback, and the loop would run away to a limit rather than settle.

Second, the feedback signal, not the output. This is why option 4 is wrong. The feedback path contains the sensor and its conditioning, described by \(H(s)\), so what returns to the junction is \(H(s)C(s)\) and not \(C(s)\) itself. The two coincide only in a unity-feedback system, where \(H=1\) — a special case, not the general one. Physically the distinction matters because the reference and the output are usually different kinds of quantity: a thermostat compares a set-point voltage with a thermocouple voltage, not with a temperature.

The closed-loop transfer function follows directly from that subtraction :

\(C=GE=G\left(R-HC\right)\quad\Rightarrow\quad \dfrac{C}{R}=\dfrac{G}{1+GH}\)

The \(1+GH\) in the denominator — the source of every benefit feedback brings — exists only because the junction subtracts. Had it added, the denominator would be \(1-GH\), which vanishes when \(GH=1\) and gives the Barkhausen condition for an oscillator.

Practical error detectors take many forms — a potentiometer bridge, a differential amplifier, a synchro pair for shaft angles, or simply a subtraction in software — but all perform the same operation, and the steady-state value of E is what the error constants K_p, K_v and K_a quantify.

Hence, the error detector gives the difference of the reference signal and the feedback signal.