Lec_08

# Lec_08 - Lecture 8: Time-Response Characteristics (I)...

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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 ) exists 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 T = 0 : 1 s + ¡ 1 ¡ e ¡ Ts 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 (cont.) C ( s ) 4 s +2 R ( s ) + ¡ E ( s ) E ¤ ( s ) T = 0 : 1 s 1 ¡ e ¡ Ts 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
3 Motivating Example (cont.) C ( s ) 4 s +2 R ( s ) + ¡ E ( s ) E ¤ ( s ) T = 0 : 1 s 1 ¡ e ¡ Ts s A plot of the sampled output: c ( kT ) = 0 : 667[1 ¡ (0 : 4562) k ] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Steady-state value 0.667 (why?) What happens in between sampling moments? Motivating Example (cont.) C ( s ) 4 s +2 R ( s ) E ( s ) + ¡ E ¤ ( s ) T = 0 : 1 s 1 ¡ e ¡ Ts s Further goal: find c ( t ) under unit step input r ( t )= u ( t ) C ( s ) = G ( s ) E ¤ ( s ) E ( z ) = R ( z ) ¡ C ( z ) = z ¡ 0 : 8187 z ¡ 0 : 4562 ¢ z z ¡ 1 Therefore, C ( s ) = 4 s ( s +2) ¢ e sT ¡ 0 : 8187 e sT ¡ 0 : 4562 = 4 s ( s +2) ¢ [1 ¡ 0 : 363 e ¡ sT ¡ 0 : 165 e ¡ 2 sT + ¢¢¢ ] (superposition of a sequence of delayed step responses of the open-loop system) In general, the continuous output is only computed through numerical simulation.

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4 C ( s ) 4 s +2 R ( s ) + ¡ E ( s ) Approximating Continuous-Time System Step response (vs. sampled system) 0 0.1 0.2 0.3
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## This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue.

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Lec_08 - Lecture 8: Time-Response Characteristics (I)...

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