Question

Consider a system with transfer-function \( G(s) = \frac{2}{s+1} \). A unit step function \( \mu(t) \) is applied to the system, which results in an output \( y(t) \). If \( e(t) = y(t) - \mu(t) \), then \( \lim_{t \to \infty} e(t) \) is \(\underline{\hspace{2cm}}\).

Hint:

For a step input, the error signal in a system with a transfer function \( G(s) \) is given by the difference between the output and the step input. The steady-state error can be computed as \( y_{\text{ss}} - 1 \).

Answer 1

Correct Answer: 1

The transfer function \( G(s) = \frac{2}{s + 1} \) represents a system with a DC gain of 2. Applying a unit step input \( \mu(t) \), the output will approach a constant value as \( t \to \infty \). The steady-state output for a step input is given by: \[ y_{\text{ss}} = \lim_{s \to 0} G(s) \cdot \frac{1}{s} = \frac{2}{1} = 2 \] Since \( e(t) = y(t) - \mu(t) \) and \( \mu(t) = 1 \), the error signal \( e(t) \) will approach: \[ \lim_{t \to \infty} e(t) = 2 - 1 = 1 \] Thus, \( \lim_{t \to \infty} e(t) = 1 \).