Steady State Error MCQ Quiz - Objective Question with Answer for Steady State Error - Download Free PDF

Explanation:

Feedback Control System of Type 2

Definition: A feedback control system of type 2 is a system where the open-loop transfer function has two integrators. This means that the system has two poles at the origin in the s-plane. The presence of these two integrators allows the system to handle certain types of input signals with zero steady-state error.

Steady-State Error: The steady-state error of a control system is the difference between the desired output and the actual output as time approaches infinity. This error is a critical parameter in evaluating the performance of control systems.

Ramp Input: A ramp input is a type of input signal that increases linearly with time. Mathematically, it can be represented as R(t) = At, where A is a constant. When analyzing control systems, it is crucial to understand how the system responds to different types of inputs, including step, ramp, and parabolic inputs.

Type 2 System and Ramp Input: For a feedback control system of type 2, the steady-state error for a ramp input can be determined using the final value theorem and the concept of system type. The system type is determined by the number of integrators in the open-loop transfer function.

Let's denote the open-loop transfer function as G(s)H(s). For a type 2 system:

G(s)H(s) = K / s2 * G'(s)

where K is a gain constant and G'(s) is a polynomial that does not have poles at the origin.

When a ramp input R(t) = At is applied, the Laplace transform of the ramp input is R(s) = A / s².

The steady-state error for a ramp input can be found using the final value theorem:

e_ss = lim (s → 0) s * E(s)

where E(s) is the Laplace transform of the error signal. The error signal E(s) is given by:

E(s) = R(s) / (1 + G(s)H(s))

Substitute R(s) and G(s)H(s):

E(s) = (A / s2) / (1 + K / s2 * G'(s))

For a type 2 system, the dominant term in the denominator for small values of s is K / s². Therefore:

E(s) ≈ (A / s2) / (K / s2) = A / K

Using the final value theorem:

e_ss = lim (s → 0) s * (A / K) = 0

Conclusion: The steady-state error for a ramp input in a type 2 feedback control system is zero. Hence, the correct answer is option 2: Zero.

Additional Information

To further understand the analysis, let's evaluate the other options:

Option 1: Infinite

This option suggests that the steady-state error for a ramp input is infinite. This would be true for a type 0 system (a system with no integrators in the open-loop transfer function). However, for a type 2 system, the steady-state error is zero, so this option is incorrect.

Option 3: Unity

This option suggests that the steady-state error for a ramp input is unity. This would be true for a type 1 system (a system with one integrator in the open-loop transfer function). However, for a type 2 system, the steady-state error is zero, so this option is incorrect.

Option 4: More than unity but not infinite

This option suggests that the steady-state error for a ramp input is more than unity but not infinite. This is not accurate for a type 2 system, where the steady-state error is zero for a ramp input. Therefore, this option is incorrect.

Conclusion:

Understanding the relationship between the type of a feedback control system and its steady-state error for different types of inputs is essential for evaluating system performance. For a type 2 system, the steady-state error for a ramp input is zero, making option 2 the correct choice. This analysis helps in designing and tuning control systems to achieve desired performance criteria based on the type of input signals they are expected to handle.

Concept

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

\(TF = \frac{L[Output]}{L[Input]}\)

\(TF = \frac{{C\left( s \right)}}{{U\left( s \right)}}\)

Calculation

Given, \(U(s) = {1\over s}\)

C(0) = 1

\(\frac{{C\left( s \right)}}{{U\left( s \right)}}={2\over s}\)

\(C(s)={2\over s}\times U(s)\)

\(C(s)={2\over s}\times {1\over s}\)

\(C(s)={2\over s^2}\)

If the initial condition is considered, then the inverse Laplace of C(s) is:

\(C(t)=C(0)+{2t}\)

\(C(t)=1+{2t}\)

The correct answer is "option 2"

Concept:

• Nonlinear control theory covers a wider class of systems that do not obey the superposition principle.
• It applies to more real-world systems, because all real control systems are nonlinear.
• These systems are often governed by nonlinear differential equations.
• Nonlinear systems are the ones that don’t stick to simple, predictable patterns throughout their whole operation.
• Nonlinear Differential Equations: Methods to solve and understand the systems governed by nonlinear differential equations.
• Application:
  • Nonlinear Control Strategies:
    • Developing control techniques specifically tailored for nonlinear systems, such as sliding mode control, adaptive control, and back stepping control.
  • Nonlinear Stability Analysis:
    • Evaluating stability in systems that does not adhere to linear stability concepts using Lyapunov stability theory & other nonlinear stability analysis methods.

Concept:

Steady-state error is defined as the difference between the desired value and the actual value of a system output in the limit as time goes to infinity (i.e. when the response of the control system has reached steady-state).

• Steady-state error is a property of the input/output response for a linear system.
• A good control system will have a low steady-state error.

For example, the above figure is the time response of the First order Transfer Function against step Input. It can be seen that the steady-state error is zero.

IMPORTANT POINTS:

Steady-state error is highest if the input is parabolic, is generally lower for ramp input, and is even lower for a step input.

It should be noted that steady-state error depends on input, while stability does not depend on input.

Concept:

The steady-state error for a system is defined as:

\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

R(s) = Input

G(s) = open loop transfer function

H(s) = feedback gain = 1 for unity feedback system

Calculation:

Given r(t) = α tu(t)

Applying Laplace Transform 

 R(s) = α /s² 

From the figure G(s)H(s) =\(\frac{k}{S(S+2)}\)

Steady-state error for a unit ramp input

\({e_{ss}} = \frac{\alpha}{{{K_V}}}\)

Where

KV = Velocity constant

\({K_V} = \mathop {\lim }\limits_{s \to 0} s\;G\left( s \right)H\left( s \right)\)

\({K_V} = \mathop {\lim }\limits_{s \to 0} s\frac{{k}}{{s(s + 2})}\) =\(\frac{k}{2}\)

Putting the value of K_V we get 

\({e_{ss}} = \frac{2\alpha}{{{K}}}\)

Hence the correct answer is 1

Concept:

KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady-state error for different inputs is given by

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input.

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

Steady-state error for the Unit step input is,

\({E_{ss}} = \frac{A}{{1 + {k_p}}}\)

\({k_p} = \mathop {\lim }\limits_{s \to 0} G(s)~H(s)\)

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

= \(\frac{1}{2}\)

\({E_{ss}} = \frac{1}{{1 + \frac{1}{2}}} = \frac{2}{3} = 0.67\)

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

Steady state error for different inputs is given by

Calculation:

Given:

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

It is a type 0 system, so:

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

K_p = 1 = DC Gain

Steady-state error for unit step input is given as:

\(e_{ss}=\frac{1}{1+K_p}\)

e_ss = 0.5

Concept:

KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

From the above table, it is clear that for type – 2 system, a system shows zero steady-state error for Ramp-input.

Concept:

Kp = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and \(\infty \) steady-state error for parabolic-input.

Concept:

The steady-state error for a system is defined as:

\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

R(s) = Input

G(s) = open loop transfer function

H(s) = feedback gain = 1 for unity feedback system

Analysis:

For a unit step input, we have:

\(R\left( s \right) = \frac{1}{s}\)

The steady-state error becomes:

\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{s\left( {\frac{1}{s}} \right)}}{{1 + G\left( s \right)}} = 0.1\)

1 + G(0) = 10

G(0) = 9

Again we have the input as:

r(t) = 10 [u(t) – u(t - 1)]

\(R\left( s \right) = 10\left[ {\frac{1}{s} - \frac{1}{s}{e^{ - s}}} \right]\)

= \(10\left( {\frac{{1 - {e^{ - s}}}}{s}} \right)\)

∴ The steady-state error for the input will be:

\(e_{ss}' = \mathop {\lim }\limits_{s \to 0} \frac{{s \times 10\frac{{\left( {1 - {e^{ - s}}} \right)}}{s}}}{{1 + G\left( s \right)}}\)

= \(\frac{{10\left( {1 - {e^0}} \right)}}{{10}}\)

\(e_{ss}' = 0\)

Important Points

The shifting in input signal does not affect the steady-state error.  

Concept:

KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input ∞  , and steady-state error for parabolic-input.

Application:

Given 

\(G\left( s \right) = \frac{{10\left( {s + 2} \right)}}{{{s^2}\left( {s + 5} \right)}}\)

r(t) = u(t) + 2t u(t)

r(t) = r1(t) + r2 (t)

It is a type 2 system. Therefore, the steady-state error for both step and ramp input is zero.

Note:

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

Acceleration error constant is a measure of the steady-state error of the system when the input is parabolic function.

Positional error constant is a measure of the steady-state error of the system when the input is unit step function.

Velocity error constant is a measure of the steady-state error of the system when the input is ramp function.

Key Points

Given,

\({G_p}(s) = \frac{{2.2}}{{(1 + 0.1s)(1 + 0.4s)(1 + 1.2s)}}\) ...(1)

\({G_c}(s) = K\left( {\frac{{1 + {T_1}s}}{{1 + {T_2}s}}} \right)\)  ....(2)

Maximum steady state error due to external input = 0.1

For step external disturbance input D(s) = 1/s  .....(3)

Assume R(s) = 0,

⇒ C(s) = - E(s)

Then the output expression is given as

C(s) = [E(s)Gc(s) + D(s)]Gp(s)

- E(s) = E(s)Gc(s)Gp(s) + D(s)Gp(s)

Error function \(E(s)=\frac{{-D(s)G_p(s)}}{{1+G_c(s)G_p(s)}}\)   ....(4)

Steady state error is given by

\(e_{ss}=\mathop {\lim }\limits_{s \to 0} s.E(s)\)

From equations (1), (2), (3)&(4)

\(|e_{ss}|=\mathop {\lim }\limits_{s \to 0} {\frac {{s.\frac {{1}}{{s}}G_p(s)}}{{1+G_c(s)G_p(s)}}}\)

\(e_{ss}≥\frac{{2.2}}{{1+2.2K}}\)

\(\frac{{2.2}}{{1+2.2K}}\le0.1\)

K ≥ 9.54

Minimum value of K is 9.54

Concept:

KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input, and ∞ steady-state error for parabolic-input.

Calculation:

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

Velocity error coefficient will be:

\({K_v} = \mathop {lim}\limits_{s \to 0} s\frac{2}{{s\left( {s + 1} \right)}} = 2\)

Steady-state error, e_ss = 1/2 = 0.5