Question

For a Type-1 system, the steady-state error to a unit ramp input is:

• A. Zero
• B. 1/Kv
• C. Infinite
• D. 1/(1+Kp)

Hint:

Standard Steady-State Error Table:
Type 0: \( e_{ss}(\text{step}) = \frac{1}{1+K_p} \), \( e_{ss}(\text{ramp}) = \infty \)
Type 1: \( e_{ss}(\text{step}) = 0 \), \( e_{ss}(\text{ramp}) = \frac{1}{K_v} \)
Type 2: \( e_{ss}(\text{step}) = 0 \), \( e_{ss}(\text{ramp}) = 0 \), \( e_{ss}(\text{parabolic}) = \frac{1}{K_a} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the steady-state error of a Type-1 control system when subjected to a unit ramp input.
Steady-state error (\( e_{ss} \)) is a measure of the system's accuracy in tracking a specific input after the transient response has decayed to zero.

Step 2: Key Formula or Approach:
The Type of a control system is determined by the number of open-loop poles located at the origin of the s-plane (integrators).
For a unit ramp input \( r(t) = t \cdot u(t) \), the Laplace transform is \( R(s) = \frac{1}{s^2} \).
The steady-state error is given by:
\[ e_{ss} = \lim_{s \to 0} \frac{s R(s)}{1 + G(s)H(s)} = \lim_{s \to 0} \frac{1}{s + s G(s)H(s)} = \frac{1}{\lim_{s \to 0} s G(s)H(s)} \] We define the velocity error constant \( K_v \) as:
\[ K_v = \lim_{s \to 0} s G(s)H(s) \] Therefore, the steady-state error is:
\[ e_{ss} = \frac{1}{K_v} \]

Step 3: Detailed Explanation:
Let us analyze the behavior of different system types under a ramp input:

• Type-0 System:
- No poles at the origin. Therefore, \( K_v = \lim_{s \to 0} s G(s)H(s) = 0 \).
- The steady-state error is \( e_{ss} = \frac{1}{0} = \infty \). A Type-0 system cannot track a ramp input.

• Type-1 System:
- Exactly one pole at the origin, meaning we can write \( G(s)H(s) = \frac{K(1 + s T_a\dots)}{s(1 + s T_1\dots)} \).
- Evaluating the velocity error constant:
\[ K_v = \lim_{s \to 0} s \cdot \frac{K(1 + s T_a)}{s(1 + s T_1)} = K \] - Since \( K_v \) is a finite constant, the steady-state error is a finite value given by \( e_{ss} = \frac{1}{K_v} \).

• Type-2 System:
- Two poles at the origin. Thus, \( K_v = \infty \), which yields a steady-state error of \( e_{ss} = \frac{1}{\infty} = 0 \).

Step 4: Final Answer:
For a Type-1 system, the steady-state error to a unit ramp input is \( 1/K_v \).