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Solution_to_problem_5.6 - The poles of 1 2 1 2 1 2 1 1 K K...

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5.6 – 1 Solution to problem 5.6 a) The overall gain of G is G ( s =0) (recall the Final Value Theorem). 1 2 K K K = b) We just need to check the response at 1 2 5 3 8 t τ τ = + = + = : 8 8 5 3 (8) 5 3 1 0.599 0.632 5 3 y e e KM - - - = - = - Therefore, the equivalent time constant is not equal to 1 2 τ τ + . c) y ( t ) will show oscillations for a step input if the poles of the transfer function are complex.
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Unformatted text preview: The poles of ( )( ) 1 2 1 2 1 2 1 1 K K G G G s s = ⋅ = + + are 1 1-and 2 1-, which are real numbers, therefore the output cannot show oscillations for a step input....
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