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

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $25.99Unlock all

Prepare for the Instrumentation Controls Lab (EE2327L) Exam with our comprehensive resources. Study with interactive quizzes, detailed explanations, and practice questions. Master the fundamentals of instrumentation and controls to excel in your exam!

Multiple Choice

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

Explanation:
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.

More Multiple Choice Questions
Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy