Question

The open loop transfer function of a unity feedback system is $G(s) = \frac{10}{s(s+2)}$, what is the steady state error for a unit step input?

• A. Infinite
• B. $0$
• C. $0.1$
• D. $0.2$

Hint:

Remember the system tracking table: - Type 0 system $\rightarrow$ Finite error for step input. - Type 1 (or higher) system $\rightarrow$ Zero steady-state error ($e_{ss} = 0$) for step input because the open-loop integrator provides infinite DC gain.

Answer 1

The Correct Option is B

Concept: The steady-state error ($e_{ss}$) of a closed-loop control system depends on both the type of the open-loop transfer function $G(s)H(s)$ and the nature of the reference input. The general expression for steady-state error using the Final Value Theorem is: \[ e_{ss} = \lim_{s \rightarrow 0} \frac{s \cdot R(s)}{1 + G(s)H(s)} \]

Step 1: Identify system Type and Input configuration.
The given open-loop transfer function is: \[ G(s) = \frac{10}{s(s+2)} \] Since there is a single isolated integrator pole at the origin ($s^1$ in the denominator), this is a Type 1 system. The feedback system is a unity configurations, meaning $H(s) = 1$. The reference input is a unit step function, which in the Laplace domain is: \[ R(s) = \frac{1}{s} \]

Step 2: Calculate the Position Error Constant ($K_p$).
For a step input, the steady-state error is given by: \[ e_{ss} = \frac{1}{1 + K_p} \] where $K_p$ is the static position error constant defined as: \[ K_p = \lim_{s \rightarrow 0} G(s)H(s) = \lim_{s \rightarrow 0} \frac{10}{s(s+2)} \] Evaluating this limit as $s$ approaches 0: \[ K_p = \frac{10}{0 \cdot (0+2)} = \frac{10}{0} \rightarrow \infty \]

Step 3: Evaluate the final steady-state error.
Substitute the infinite value of $K_p$ into the steady-state error formula: \[ e_{ss} = \frac{1}{1 + \infty} = 0 \] This shows that a Type 1 system tracks a step input with zero steady-state error.