Control systems — unit step response at t = 0⁺: For the transfer function G(s) = (s + 1) / (2s + 1), what is the initial value (at t = 0⁺) of the unit step response?

Introduction / Context:
Evaluating the very first value of a time response is a common check in control engineering. The initial value theorem provides a fast way to compute y(0⁺) directly from the Laplace-domain representation without fully inverting transforms. This problem reinforces comfort with that theorem for a stable, proper transfer function driven by a unit step.

Given Data / Assumptions:

• Transfer function: G(s) = (s + 1) / (2s + 1).
• Input: unit step → U(s) = 1 / s.
• System is linear time-invariant; initial rest is assumed.

Concept / Approach:
For a unit step input, Y(s) = G(s) * 1/s. The initial value theorem states y(0⁺) = lim_{s→∞} s * Y(s) provided poles are in the left half-plane and the limit exists. This avoids full partial fractions and time-domain inversion for the initial value alone.

Step-by-Step Solution:

Verification / Alternative check:
One can also compute the full step response via partial fractions to see the exponential term vanish at t = 0⁺, leaving the same initial value of 0.5. Both approaches agree.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting to multiply Y(s) by s in the initial value theorem, or mistakenly applying the final value theorem for y(0⁺). Ensure stability assumptions hold when using these theorems.

Final Answer:
1/2