In a first-order process with a dead time, what does the time constant τ represent?

In a first-order process with a dead time, what does the time constant τ represent?

• A. Time to reach 63% of the final value after the dead time.
• B. Time to reach 50% of the final value.
• C. The dead time itself.
• D. The final value.

Answer: (A)

In a first-order process with dead time, the response after the delay follows an exponential rise toward the final value. The time constant τ sets how fast that rise occurs once the process actually starts changing. Mathematically, after a step input, the output is described as y(t) = K·u·[1 − e^{−(t−θ)/τ}] for t > θ, where θ is the dead time and K·u is the final value. At t = θ + τ, the exponential term becomes e^{−1}, so the output reaches about 0.632 (63%) of its final value. That means τ is the time, measured after the dead time, needed to get to 63% of the final value. The dead time itself is the pure delay before any response begins, not part of the rising time, and the final value is the steady-state level the system approaches. For reference, reaching 50% would occur at t = θ + τ·ln(2) ≈ θ + 0.693τ, not at τ.

In a first-order process with dead time, the response after the delay follows an exponential rise toward the final value. The time constant τ sets how fast that rise occurs once the process actually starts changing. Mathematically, after a step input, the output is described as y(t) = K·u·[1 − e^{−(t−θ)/τ}] for t > θ, where θ is the dead time and K·u is the final value.

At t = θ + τ, the exponential term becomes e^{−1}, so the output reaches about 0.632 (63%) of its final value. That means τ is the time, measured after the dead time, needed to get to 63% of the final value. The dead time itself is the pure delay before any response begins, not part of the rising time, and the final value is the steady-state level the system approaches.

For reference, reaching 50% would occur at t = θ + τ·ln(2) ≈ θ + 0.693τ, not at τ.