In a feedback control system, if \(G(s)=\frac{4}{s(s+3) }\)and \(H(s)=\frac{1}{s}\), then the closed-loop system will be of type

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Answer (Detailed Solution Below)

Option 2 : 2
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Concept:

The type of the system is defined as the number of poles at the origin of the open-loop transfer function G(s) H(s).

For an open-loop transfer function as shown:

\(G\left( s \right)H\left( s \right) = \frac{{{b_m}{s_m} + {b_{m - 1}}{s^{m - 1}} + \ldots + {b_0}}}{{{s^c}({a_n}{s^n} + {a_{n - 1}}{s^{n - 1}} + \ldots + {a_0})}}\)

The above system is a type ‘c’ system with an order of n + c.

Calculation:

Given:

\(G(s)=\frac{4}{s(s+3) }\) \(H(s)=\frac{1}{s}\)

\(G(s)H(s) = \frac{4}{s(s+3)}\frac{1}{s}\)

\(G(s)H(s) = \frac{4}{s^2(s+3)}\)

Hence, the type of system is 2.

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