The steady-state error due to unit step input to a type-1 system is:
Zero
Steady-State Error of Type-1 System with Unit Step Input
Understanding the steady-state error is crucial in control systems analysis. The steady-state error ($e_{ss}$) is the difference between the desired output and the actual output as time approaches infinity. It depends on the type of the system and the type of input signal.
For a linear time-invariant system, the steady-state error due to a unit step input is given by the formula:
\(e_{ss} = \frac{A}{1 + K_p}\)
Where:
- \(A\) is the magnitude of the step input. For a unit step input, \(A = 1\).
- \(K_p\) is the position error constant.
The position error constant \(K_p\) is defined as:
\(K_p = \lim_{s \to 0} G(s)H(s)\)
Here, \(G(s)H(s)\) is the open-loop transfer function of the system.
Analyzing a Type-1 System
A system's type is determined by the number of poles at the origin (\(s=0\)) in its open-loop transfer function \(G(s)H(s)\). A type-1 system has exactly one pole at \(s=0\).
The open-loop transfer function of a type-1 system can generally be written in the form:
\(G(s)H(s) = \frac{K(s+z_1)(s+z_2)...}{s(s+p_1)(s+p_2)...}\)
Where K is a constant, and \(z_i\) and \(p_i\) are the non-zero poles and zeros.
Calculating \(K_p\) for a Type-1 System
Let's calculate the position error constant \(K_p\) for a type-1 system:
\(K_p = \lim_{s \to 0} G(s)H(s) = \lim_{s \to 0} \frac{K(s+z_1)(s+z_2)...}{s(s+p_1)(s+p_2)...}\)
As \(s\) approaches 0, the term \(s\) in the denominator goes to zero, while the numerator and the other terms in the denominator approach finite non-zero values (assuming none of the \(z_i\) or \(p_i\) are zero).
Therefore, the limit becomes:
\(K_p = \frac{K(0+z_1)(0+z_2)...}{0 \cdot (0+p_1)(0+p_2)...} = \frac{\text{finite non-zero value}}{0}\)
This results in \(K_p\) approaching infinity:
\(K_p = \infty\)
Calculating Steady-State Error for Unit Step Input
Now, substitute the value of \(K_p\) into the steady-state error formula for a unit step input (\(A=1\)):
\(e_{ss} = \frac{1}{1 + K_p} = \frac{1}{1 + \infty} = \frac{1}{\infty}\)
As a quantity divided by infinity approaches zero, the steady-state error is:
\(e_{ss} = 0\)
Thus, the steady-state error due to unit step input to a type-1 system is zero.
Important Questions from Steady State Error
With reference to the error analysis of the control systems, the term 'acceleration error constant' stands for:
Which one of the following coefficient is associated with Unit Ramp function?
If the output of the system at steady state does not agree with the input, then the system is said to have _________ which determines the _________ of the system.
A unity negative feedback closed loop system has a plant with the transfer function \(G(s) = \dfrac{1}{s^2 + 2s + 2}\) and a controller Ge(s) in the feedforward path. For a unit step input, the transfer function of the controller that gives minimum steady slate error is
The closed loop transfer function of a system is \(T\left( s \right) = \frac{4}{{\left( {{s^2} + 0.4s + 4} \right)}}\). The steady state error due to unit step input is ________.
