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:

Consider the closed-loop feedback control system shown in the block diagram below:

Let us analyze the relationships between the signals at the summing junction and the blocks in the system:
1. The reference input is $R\left(s\right)$, and the output is $C\left(s\right)$.
2. The feedback path contains the block $H$. The input to $H$ is the output $C\left(s\right)$, so the feedback signal fed to the summing junction is:
$\mathrm{Feedback\; Signal}=C\left(s\right)H$
3. The summing junction subtracts the feedback signal from the input $R\left(s\right)$ to produce the error signal $E\left(s\right)$:
$E\left(s\right)=R\left(s\right)-C\left(s\right)H$
4. The error signal $E\left(s\right)$ is the input to the forward path transfer function block $G$, which produces the output $C\left(s\right)$:
$C\left(s\right)=E\left(s\right)G$

Now, substitute the expression for $C\left(s\right)$ into the summing junction equation:
$E\left(s\right)=R\left(s\right)-\left[E\right(s\left)G\right]H$

Simplifying the equation:
$E\left(s\right)=R\left(s\right)-E\left(s\right)GH$

Move the term with $E\left(s\right)$ to the left-hand side:
$E\left(s\right)+E\left(s\right)GH=R\left(s\right)$

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

Finally, solve for the ratio of the error signal to the reference input, which is the transfer function $\frac{E\left(s\right)}{R\left(s\right)}$:
$\frac{E\left(s\right)}{R\left(s\right)}=\frac{1}{1+GH}$