1.

For the block diagram shown below :The characteristic equation is

A. τl S (τps + 1) + Kc . Kp (τlS + 1) e-τd s = 0
B. (τmS + 1) (τpS + 1) + Km . Kp e-τd S = 0
C. τl S(τm S + 1) (τp S + 1) +Kc . Kp . Km (τl s + 1) e-τdS = 0
D. (τms + 1) (τp s + 1) + Kc . Kp . Km . e-τdS = 0
Answer» D. (œÑms + 1) (œÑp s + 1) + Kc . Kp . Km . e-œÑdS = 0


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