Question

The open loop transfer function of a unity feedback system is: $G(s) = \frac{50}{(1 + 0.1s)(1 + 2s)}$. The Position, Velocity and Acceleration error constants are respectively

• A. 0, 0, 250
• B. 50, 0, 0
• C. 0, 250, $\infty$
• D. $\infty$, 50, 0

Hint:

For a Type 'N' system: - If $N=0$, $K_p$ is a constant, $K_v=0, K_a=0$. - If $N=1$, $K_p=\infty$, $K_v$ is a constant, $K_a=0$. - If $N=2$, $K_p=\infty$, $K_v=\infty$, $K_a$ is a constant.

Answer 1

The Correct Option is B

1. Identify System Type: The given transfer function is $G(s) = \frac{50}{(1 + 0.1s)(1 + 2s)}$. Since there are no $s$ terms in the denominator outside of the brackets (no poles at $s = 0$), this is a

Type 0 system.

2. Calculate Position Error Constant ($K_p$): $$K_p = \lim_{s \to 0} G(s) = \frac{50}{(1 + 0)(1 + 0)} = 50\lt strong\gt 3. Calculate Velocity Error Constant ($K_v$):\lt /strong\gt K_v = \lim_{s \to 0} sG(s) = \lim_{s \to 0} \frac{50s}{(1 + 0.1s)(1 + 2s)} = 0\lt strong\gt 4. Calculate Acceleration Error Constant ($K_a$):\lt /strong\gt K_a = \lim_{s \to 0} s^2G(s) = \lim_{s \to 0} \frac{50s^2}{(1 + 0.1s)(1 + 2s)} = 0$$

Result: The constants are $K_p = 50, K_v = 0, K_a = 0$.