Solutions_F11_ECH157_HW4

# Solutions_F11_ECH157_HW4 - 1(1 K a K c Since KaKc 0 1 Hence...

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a) For s M s X = ) ( 1 1 ) 1 )( 1 ( ) ( + τ + - + = + τ - = s C s B s A s s s KM s Y KM s s KM A s = + τ - = = ) 1 )( 1 ( lim 0 1 lim ( 1) 1 s KM KM B s s = = = τ + τ + 2 1/ lim 1 1 (1 ) 1 1 s KM KM KM C s s →- τ - τ = = = - τ +  - +  τ τ  Then, + τ τ - + τ - = τ - / 1 1 1 1 ) ( t t e e KM t y For M =2 , K = 3, and τ = 3, the Simulink response is shown:

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2 4 6 8 10 12 14 16 18 20 -12 -10 -8 -6 -4 -2 0 x 10 8 Time Output Figure S6.14a. Unit step response for part a). b) If ) 1 )( 1 ( ) ( 2 2 + τ - = - s s Ke s G s then, ) 2 ( 1 1 1 ) ( / ) 2 ( 2 2 - + τ τ - + τ - = τ - - - t S e e KM t y t t Note presence of positive exponential term. c) Approximating G 2 ( s ) using a Padé function ) 1 )( 1 ( ) 1 )( 1 )( 1 ( ) 1 ( ) ( 2 + τ + = - + τ + - = s s K s s s s K s G Note that the two remaining poles are in the LHP. d)
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Solutions_F11_ECH157_HW4 - 1(1 K a K c Since KaKc 0 1 Hence...

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