For the closed-loop system shown, the transfer function $\frac{E(s)}{R(s)}$ is

The correct answer is:
$\frac{1}{1+GH}$

Step-by-step Explanation:

By analyzing the provided block diagram, we can identify the following signals and system components:
- $R\left(s\right)$ is the input reference signal entering the summing junction with a positive (+) sign.
- $E\left(s\right)$ is the error (or actuating) signal output from the summing junction.
- $G$ is the forward path block transfer function.
- $C\left(s\right)$ is the system output signal.
- $H$ is the feedback path block transfer function.
- A negative (-) feedback sign is present at the summing junction where the feedback signal returns.

From the diagram, the error signal $E\left(s\right)$ is the difference between the input reference $R\left(s\right)$ and the feedback signal:
$E\left(s\right)=R\left(s\right)-B\left(s\right)$
where $B\left(s\right)$ represents the feedback signal after passing through block $H$.

The feedback signal $B\left(s\right)$ is calculated as:
$B\left(s\right)=C\left(s\right)·H$

The system output $C\left(s\right)$ is the result of the error signal passing through the block $G$:
$C\left(s\right)=E\left(s\right)·G$

Substituting the expression for $C\left(s\right)$ into the equation for $B\left(s\right)$:
$B\left(s\right)=E\left(s\right)·G·H$

Substitute this feedback expression back into the error signal equation:
$E\left(s\right)=R\left(s\right)-E\left(s\right)·G·H$

Rearrange the equation to isolate all terms with $E\left(s\right)$ on the left-hand side:
$E\left(s\right)+E\left(s\right)·G·H=R\left(s\right)$

Factor out $E\left(s\right)$:
$E\left(s\right)(1+GH)=R\left(s\right)$

Divide both sides by $R\left(s\right)$ and then by $(1+GH)$ to find the transfer function:
$\frac{E\left(s\right)}{R\left(s\right)}=\frac{1}{1+GH}$