Consider the following statements related to steady state error for a control system:

1. Steady-state error can be calculated from a system’s closed-loop transfer function for a unity feedback system.

2. Steady-state error can be calculated from a system’s open-loop transfer function for a unity feedback system.

3. Steady-state error is the difference between the input and the output for a prescribed test input as time tends to infinity.

4. Many steady-state errors in control systems can arise from non-linear sources.

Which of the above statements are correct?

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  1. 1 and 3 only
  2. 1, 3 and 4 only
  3. 2 and 4 only
  4. 1, 2, 3 and 4

Answer (Detailed Solution Below)

Option 2 : 1, 3 and 4 only
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Detailed Solution

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Concept:
Stead state Analysis:

E(s) = Error signal, R(s) = Reference i/p or desired o/p

C(s) = Actual output

Steady-state error is the deviation from the desired output.

S.S.E (ess) = \(\rm lt_{s \rightarrow 0} S.E(s) = lt_{t \rightarrow ∞} S.e (t) \)

or ess = \(\rm lt_{s \rightarrow 0} S.[R(s) - c(s)] = lt_{t \rightarrow ∞} S.[r(t) - c(t)]\)

Hence, ess is the difference between the input and output for a prescribed test input as time tends to infinity.

⇒ S.S.E. can be calculated from a system's closed-loop transfer function for a unity feedback system. In the calculation S.S.E we take help of some error constants which are derived from the open loop transfer function.

hence, with the help of the open-loop transfer function, we are finding S.S.E of the closed-loop transfer function.

From the transfer function,

\(\frac{E(s)}{N(s)} = \frac{14}{1+ G(s)} \Rightarrow E(s) = \frac{R(s)}{1 + G(s)}\)

for step input:

ess = \(\rm lt_{s \rightarrow 0} S. \frac{R(s)}{1 + G(s)} = lt_{s \rightarrow 0} s. \frac{\frac{A}{s}}{1 + G(s)} = lt_{s \rightarrow 0} \frac{A}{1+ G(s)}\)

\(= \frac{A}{ 1 + K_p}\)

Kp = \(\rm lt_{s \rightarrow 0} G(s) \) = static position error constant

For ramp input:

ess = \(\rm lt_{s \rightarrow 0} \ S. \frac{R(c)}{1 + G(s)} = lt_{s \rightarrow 0} \ s. \frac{\frac{A}{s^2}}{1 + G(s)} \)

\(\rm = lt_{s \rightarrow 0} \frac{A}{s[1+ G(s)]} = \frac{A}{K_v}\)

where Kv = \(\rm lt_{s \rightarrow 0} \) = static velocity error constant

For parabolic input:

ess = \(\rm lt_{s \rightarrow 0} \ S. \frac{R(s)}{1 + G(s)} = lt_{s \rightarrow 0} \ s. \frac{\frac{A}{s^3}}{1 + G(s)} \)

where ka = \(= \rm lt_{s \rightarrow 0}\) s2 G(s)\(= \rm lt_{s \rightarrow 0} \frac{A}{s^2 [ 1 + G(s)]} = \frac{A}{K_a}\)

= static acceleration error constant

Type

Steady State Error (ess)

 

Step = \(\frac{A}{s}\)

Ramp = \(\frac{A}{s^2}\)

Parabolic = \(\frac{A}{s^3}\)

0

\(\frac{A}{1 + K_p}\)

1

0

\(\frac{A}{ K_r}\)

∞ 

2

0

0

\(\frac{A}{ K_a}\)

 

Note: If the type of system increases the error decreases, and accuracy increases but the stability decreases.

Many steady-state errors in control systems arise from nonlinear sources, such as backlash in gears or a motor that will not move unless the input voltage exceeds a threshold.

Statements 1, 3 and 4 are correct. Therefore, the correct option is (b)

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