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The transfer function \(\frac{1}{2s+1}\) will have
1. dc gain 1 and high frequency gain 1
2. dc gain 0 and high frequency gain ∞ 
3. dc gain 1 and high frequency gain 0
4. dc gain 0 and high frequency gain 1

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Correct Answer - Option 3 : dc gain 1 and high frequency gain 0

Concept:

DC gain:

The DC gain is the ratio of the magnitude of the steady-state step response to the magnitude of step input.

DC Gain of a system is the gain at the steady-state which is at t tending to infinity i.e., s tending to zero.

DC gain is nothing but the error coefficients.

For type 0 system:

\({K_P} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)

For type 1 system: 

\({K_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)\)

For type 2 system: 

\({K_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)\)

High-frequency gain:

The high-frequency gain of a system is the gain at the steady-state which is at t tending to 0 i.e., s tending to infinity.

Calculation:

Given:

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

\(DC \ Gain = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)

\(DC \ Gain = \mathop {\lim }\limits_{s \to 0} \frac{1}{2s+1}\)

DC Gain = 1

High-frequency gain is calculated as:

\( = \mathop {\lim }\limits_{s \to ∞ } G\left( s \right)\)

\( = \mathop {\lim }\limits_{s \to ∞ } \frac{1}{2s+1}\)

High frequency gain = 0

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