Consider the cascaded system as shown in the figure. Neglecting the faster component of the transient response, which one of the following options is a first - order pole - only approximation such that the steady – state values of the unit step response of the original and the approximated systems are same ?

The correct answer is:
$\frac{2}{s+1}$

Step 1: Determine the Overall Transfer Function of the Cascaded System
From the block diagram visible in the image, the system consists of two cascaded blocks with the following transfer functions:
First block: $G_1\left(s\right)=\frac{1}{s+1}$
Second block: $G_2\left(s\right)=\frac{s+40}{s+20}$
Since these blocks are connected in series (cascade), the overall system transfer function $T\left(s\right)$ is the product of the individual block transfer functions:
$T\left(s\right)=G_1\left(s\right)\cdot G_2\left(s\right)=\frac{s+40}{(s+1)(s+20)}$

Step 2: Identify the Dominant Pole
The poles of the system are found by setting the denominator of $T\left(s\right)$ to zero:
$(s+1)(s+20)=0\Rightarrow s=-1,-20$
The pole at $s=-20$ is located far to the left in the complex plane compared to the dominant pole at $s=-1$. The transient term associated with $s=-20$ decays rapidly (as $e^{-20t}$).
By neglecting this faster component, the system is approximated by a first-order pole-only model containing only the dominant pole at $s=-1$:
$G_a\left(s\right)=\frac{A}{s+1}$
where $A$ is a constant to be determined.

Step 3: Match the Steady-State Step Response (DC Gain)
For a unit step input, the steady-state response of a stable system is equal to its transfer function evaluated at $s=0$ (the DC gain).
For the original system: