What is the standard form of a first-order plus dead-time model?

What is the standard form of a first-order plus dead-time model?

• A. G(s) ≈ K/(τ s + 1)
• B. G(s) ≈ e^{-Ls}/(τ s + 1)
• C. G(s) ≈ (K s)/(τ s + 1)
• D. G(s) ≈ K e^{-Ls}/(τ s + 1)

Answer: (D)

The key idea is combining a simple first-order dynamic with a pure time delay (dead time). The standard first-order plus dead-time model is G(s) = K e^{-Ls} / (τ s + 1). Here, the denominator 1/(τ s + 1) represents a first-order response with time constant τ, so the system gradually approaches its final value. The factor e^{-Ls} introduces a pure delay of L seconds, meaning the response doesn’t start until after that lag. The gain K scales the final steady-state value. This form is preferred because it captures both the immediate lag due to transport or processing delay and the simple exponential approach to the final value. Dropping the delay (no e^{-Ls}) ignores the lag, and dropping the gain (no K) or using a different numerator would misrepresent the system’s steady-state or dynamics.

The key idea is combining a simple first-order dynamic with a pure time delay (dead time). The standard first-order plus dead-time model is G(s) = K e^{-Ls} / (τ s + 1). Here, the denominator 1/(τ s + 1) represents a first-order response with time constant τ, so the system gradually approaches its final value. The factor e^{-Ls} introduces a pure delay of L seconds, meaning the response doesn’t start until after that lag. The gain K scales the final steady-state value.

This form is preferred because it captures both the immediate lag due to transport or processing delay and the simple exponential approach to the final value. Dropping the delay (no e^{-Ls}) ignores the lag, and dropping the gain (no K) or using a different numerator would misrepresent the system’s steady-state or dynamics.