Steady State Error and Error Constants Questions and Answers

Q: The order and type of the system shown in the following figure is :

The system is Type : 1, Order : 3.The open loop transfer function is the product of the forward and feedback blocks:\[ G(s)H(s) = \frac{k_{1}k_{2}}{s(s^{2} + 4s + 16)} \]The type is the number of open loop poles at the origin: one factor of s, so type 1. The order is the degree of the characteristic polynomial, s(s2 + 4s + 16) + k1k2, which is 3.Type 2 or 3 would need more integrators, and order 1 or 2 ignores either the quadratic forward block or the integrator in the feedback path.

Q: In a control system, what is the percentage error in measurement if the variable range is 4.20 mA and the measured value is 13 mA with set point of 10 mA :

The percentage error is 18.75.The variable range printed '4.20 mA' is the standard 4-20 mA signal range, a span of 16 mA. The error is expressed as a percentage of that span:\[ \begin{aligned} e &= \frac{13 - 10}{20 - 4} \times 100 \\ &= \frac{3}{16} \times 100 = 18.75\% \end{aligned} \]Dividing by the full 20 mA instead of the 16 mA span gives 15%, and dividing by the set point gives 30%; neither is the percentage of span used in process control.

Identify the type and order of the system \(G(s) = \dfrac{1}{s}\).

A.Type - 0; Order - 1
B.Type - 1; Order - 1
C.Type - 0; Order - 0
D.Type - 1; Order - 0

Exam Year(s)

DSSSB AE - 2023,

Answer: B

The type of a system is the number of poles at the origin in the open loop transfer function, and the order is the highest power of \(s\) in the denominator (the number of poles in all).

\[ G(s) = \frac{1}{s} \]

There is one pole at \(s = 0\), so the type is 1, and the denominator is of degree 1, so the order is 1. This is an integrator. The options with type 0 would need no pole at the origin, and those with order 0 would need a transfer function with no \(s\) in the denominator. The answer is Type 1, Order 1, option B.

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A feedback control system has the transfer function given by \(F(s) = \dfrac{8(s + 2)(s + 4)}{s^3(s + 1)(s^2 + 5s + 6)}\). Which of the following is a type of the system?

A.Type-6 system
B.Type-3 system
C.Type-2 system
D.Type-1 system

Exam Year(s)

DSSSB AE - 2023,

Answer: B

The type of a system is the number of poles of the open loop transfer function at the origin, that is, the power of \(s\) in the denominator when written in the standard form.

\[ F(s) = \frac{8(s+2)(s+4)}{s^3(s+1)(s^2+5s+6)} \]

The factor \(s^3\) gives three poles at \(s = 0\). The factor \(s^2 + 5s + 6 = (s+2)(s+3)\) has its roots at \(-2\) and \(-3\), and the factor \(s + 1\) has its root at \(-1\), so none of them is at the origin. The system is of type 3, option B. The total number of poles, 6, is the order, which is why Type-6 is wrong.

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The order and type of the system shown in the following figure is :

Block diagram: R(s) enters a summing point, forward block k1/(s^2 + 4s + 16) gives C(s), feedback block k2/s returns C(s) to the summing point

A.Type : 1, Order : 3
B.Type : 3, Order : 1
C.Type : 2, Order : 2
D.Type : 1, Order : 1

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The system is Type : 1, Order : 3.

The open loop transfer function is the product of the forward and feedback blocks:

\[ G(s)H(s) = \frac{k_{1}k_{2}}{s(s^{2} + 4s + 16)} \]

The type is the number of open loop poles at the origin: one factor of s, so type 1. The order is the degree of the characteristic polynomial, s(s² + 4s + 16) + k₁k₂, which is 3.

Type 2 or 3 would need more integrators, and order 1 or 2 ignores either the quadratic forward block or the integrator in the feedback path.

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In a control system, what is the percentage error in measurement if the variable range is 4.20 mA and the measured value is 13 mA with set point of 10 mA :

A.18.75
B.20.50
C.24.00
D.28.75

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The percentage error is 18.75.

The variable range printed '4.20 mA' is the standard 4-20 mA signal range, a span of 16 mA. The error is expressed as a percentage of that span:

\[ \begin{aligned} e &= \frac{13 - 10}{20 - 4} \times 100 \\ &= \frac{3}{16} \times 100 = 18.75\% \end{aligned} \]

Dividing by the full 20 mA instead of the 16 mA span gives 15%, and dividing by the set point gives 30%; neither is the percentage of span used in process control.

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