Closed-loop with dead time and first-order blocks (typical process + measurement + controller): which characteristic equation is consistent with two first-order lags ((τm s + 1), (τp s + 1)) in unity feedback and a controller gain Kc with overall process gain Kp and measurement gain Km including a pure dead time e^(−τd s)?
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Aτl s (τp s + 1) + Kc Kp (τl s + 1) e^(−τd s) = 0
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B(τm s + 1)(τp s + 1) + Km Kp e^(−τd s) = 0
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Cτl s (τm s + 1)(τp s + 1) + Kc Kp Km (τl s + 1) e^(−τd s) = 0
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D(τm s + 1)(τp s + 1) + Kc Kp Km e^(−τd s) = 0
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E1 + Kc Kp Km (τm s + 1)(τp s + 1) e^(−τd s) = 0
Answer
Correct Answer: (τm s + 1)(τp s + 1) + Kc Kp Km e^(−τd s) = 0
Explanation
Introduction / Context:Writing the characteristic equation from a block diagram is a core skill in control engineering. When a loop contains first-order dynamics for the process and the measuring element, a controller gain, and a pure dead time, the loop equation must reflect these factors in the denominator of the closed-loop transfer function.
Given Data / Assumptions:
- Unity feedback with measurement dynamics Km/(τm s + 1).
- Process dynamics Kp/(τp s + 1) with a dead time e^(−τd s).
- Controller gain Kc (no integral/derivative terms shown).
- No extra transport lag besides those stated.
Concept / Approach:For unity feedback, the characteristic equation is 1 + L(s) = 0 where L(s) is the loop transfer function. With L(s) = Kc * [Km/(τm s + 1)] * [Kp/(τp s + 1)] * e^(−τd s), multiply both sides by (τm s + 1)(τp s + 1) to obtain (τm s + 1)(τp s + 1) + Kc Kp Km e^(−τd s) = 0.
Step-by-Step Solution:Write L(s) = Kc Km Kp e^(−τd s)/[(τm s + 1)(τp s + 1)].Characteristic equation: 1 + L(s) = 0.Multiply through by (τm s + 1)(τp s + 1): (τm s + 1)(τp s + 1) + Kc Kp Km e^(−τd s) = 0.
Verification / Alternative check:Setting τd → 0 recovers the familiar polynomial (τm s + 1)(τp s + 1) + Kc Kp Km = 0, confirming consistency without dead time.
Why Other Options Are Wrong:Options with τl terms introduce an extra lag not stated.Options missing Kc (just Km Kp) omit the controller gain.Option with product multiplied by e^(−τd s) on the polynomial part misrepresents the structure of 1 + L(s) = 0.
Common Pitfalls:Placing e^(−τd s) on the wrong factor, forgetting to clear denominators, or misreading unity feedback versus sensor-in-loop placement.
Final Answer:(τm s + 1)(τp s + 1) + Kc Kp Km e^(−τd s) = 0