Question

Position, velocity and acceleration errors of type 2 control system respectively, are

• A. \( 0, 0, \frac{1}{K_a} \)
• B. \( 0, \frac{1}{K_v}, \infty \)
• C. \( \frac{1}{1+K_p}, \infty, \infty \)
• D. \( \frac{1}{K_p}, \frac{1}{K_v}, \frac{1}{K_a} \)

Hint:

System Type = Number of open-loop poles at the origin \(s=0\).
Type 0: Finite \(K_p\), \(K_v=0, K_a=0\). Errors: \(1/(1+K_p)\) for step, \(\infty\) for ramp, \(\infty\) for parabola.
Type 1: \(K_p=\infty\), Finite \(K_v\), \(K_a=0\). Errors: \(0\) for step, \(1/K_v\) for ramp, \(\infty\) for parabola.
Type 2: \(K_p=\infty, K_v=\infty\), Finite \(K_a\). Errors: \(0\) for step, \(0\) for ramp, \(1/K_a\) for parabola.

Answer 1

The Correct Option is A

The type of a control system is defined by the number of poles of its open-loop transfer function \(G(s)H(s)\) located at the origin (\(s=0\)). A type 2 system has two poles at the origin. Steady-state errors for different types of systems and inputs:
Position Error Constant (\(K_p\)): \(K_p = \lim_{s \to 0} G(s)H(s)\) Steady-state error for unit step input: \(e_{ss,step} = \frac{1}{1+K_p}\)
Velocity Error Constant (\(K_v\)): \(K_v = \lim_{s \to 0} s G(s)H(s)\) Steady-state error for unit ramp input: \(e_{ss,ramp} = \frac{1}{K_v}\)
Acceleration Error Constant (\(K_a\)): \(K_a = \lim_{s \to 0} s^2 G(s)H(s)\) Steady-state error for unit parabolic input: \(e_{ss,parabolic} = \frac{1}{K_a}\) For a Type 2 system, \(G(s)H(s)\) has a factor of \(1/s^2\).
\(K_p = \lim_{s \to 0} G(s)H(s)\). Since there's \(1/s^2\), as \(s \to 0\), \(K_p \to \infty\). Position error \(e_{ss,step} = \frac{1}{1+K_p} = \frac{1}{1+\infty} = 0\).
\(K_v = \lim_{s \to 0} s G(s)H(s)\). Since there's \(1/s^2\), \(s G(s)H(s)\) still has \(1/s\). As \(s \to 0\), \(K_v \to \infty\). Velocity error \(e_{ss,ramp} = \frac{1}{K_v} = \frac{1}{\infty} = 0\).
\(K_a = \lim_{s \to 0} s^2 G(s)H(s)\). The \(s^2\) factor cancels the \(1/s^2\) term from the type 2 system. So, \(K_a\) will be a finite non-zero constant (assuming no zeros at origin in the rest of \(G(s)H(s)\)). Acceleration error \(e_{ss,parabolic} = \frac{1}{K_a}\) (which is finite and non-zero). So, for a type 2 system: Position error (for step input) = 0. Velocity error (for ramp input) = 0. Acceleration error (for parabolic input) = \(1/K_a\) (finite non-zero). This matches option (a). \[ \boxed{0, 0, \frac{1}{K_a}} \]