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# 8 - Lecture 8 Time-Response Characteristics(I System...

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1 Lecture 8: Time-Response Characteristics (I) System Time-Response Characteristic Equations and Closed-Loop Poles Steady-State Error Reading: Chapter 6 (except 6.4) System Time-Response Problem : given a sampled-data closed-loop system, find its output c ( t ) under arbitrary input r ( t ) Challenge : the continuous-time output c ( t ) is difficult to compute Solution : Transfer function from C ( z ) to R ( z ) may exist when r ( t ) is sampled before feeding into any analog transfer function. Hence for any continuous-time input r ( t ), only its sampled data r ( kT ), k =0,1,… matter in determining the sampled output c ( kT ), k =0,1,… Modified Problem : find the sampled output given any sampled input.

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2 Problem : find the system response c ( kT ) to a unit step input r ( t )=1( t ) Step 1: find the transfer function from R ( z ) to C ( z ) C ( s ) 4 s +2 R ( s ) E ( s ) E ¤ ( s ) Motivating Example: I T = 0 : 1 s + ¡ 1 ¡ e ¡ T s s G ( s ) = 1 ¡ e ¡ Ts s 4 s +2 C ( z ) = G ( z ) 1+ G ( z ) R ( z ) G ( z ) = 0 : 3625 z ¡ 0 : 8187 C ( z ) = 0 : 3625 z ¡ 0 : 4562 R ( z ) (a closed-loop pole at z =0.4562) (an open-loop pole at z =0.8187) Motivating Example: II C ( s ) 4 s +2 R ( s ) + ¡ E ( s ) E ¤ ( s ) T = 0 : 1 s 1 ¡ e ¡ T s s C ( z ) = 0 : 3625 z ¡ 0 : 4562 R ( z ) Step 2: find the ( z -transform) of the sampled input: Step 3: find the sampled output: R ( z ) = Z [ R ( s )] = z z ¡ 1 C ( z ) = 0 : 3625 z ¡ 0 : 4562 z z ¡ 1 = 0 : 667 z z ¡ 1 ¡ 0 : 667 z z ¡ 0 : 4562