To solve this problem, we need to find the first-order pole-only approximation of the given cascaded transfer function and ensure that it has the same steady-state value for the unit step response as the original system.

The given cascaded system has two transfer functions:

• \(\frac{1}{s+1}\)
• \(\frac{s+40}{s+20}\)

The overall transfer function of the cascaded system is:

\(G(s) = \frac{1}{s+1} \cdot \frac{s+40}{s+20} = \frac{s+40}{(s+1)(s+20)}\)

Neglecting the faster component of the transient response, the pole at the higher frequency should be neglected since its transient effects decay faster. The system is dominated by the slower response, which corresponds to a lower frequency pole.

To find the steady-state response to a unit step input, evaluate the zero-frequency (s=0) gain of the original system:

\(G(s=0) = \frac{0+40}{(0+1)(0+20)} = \frac{40}{20} = 2\)

We need a first-order system with the same steady-state value, \(2\). This can be achieved by a first-order system:

\(G_{\text{approx}}(s) = \frac{K}{s+a}\)

Given options are:

• \(\frac{1}{s+1}\)
• \(\frac{2}{s+1}\)
• \(\frac{1}{s+20}\)
• \(\frac{2}{s+20}\)

To have the same steady-state gain, \(\frac{2}{s+1}\) is the correct approximation. This gives a steady-state value of:

\(G_{\text{approx}}(s=0) = \frac{2}{0+1} = 2\)

Thus, the approximation \(\frac{2}{s+1}\) yields the same steady-state gain.

Therefore, the correct option is:

\(\frac{2}{s+1}\)