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EE3530-6

# EE3530-6 - Chapter 6 The Frequency-Response Design Method...

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Chapter 6 The Frequency-Response Design Method Frequency Response Let P ( s ) be a stable system and r ( t ) = sin ωt . Then the steady state response is y ss ( t ) = | P ( ) | sin( ωt + 6 P ( )) 6 P ( ) = tan - 1 Im P ( ) Re P ( ) P ( ) is called the frequency response . | P ( ) | is the magnitude frequency response 6 P ( ) is the phase frequency response Example P ( s ) = 10 ( s + 1)( s + 2) r ( t ) = sin ωt Y ( s ) = 10 ω/ (1 + ω 2 ) s + 1 - 10 ω/ (4 + ω 2 ) s + 2 + s Im { P ( ) } + ω Re { P ( ) } s 2 + ω 2 Then the transient response is y t ( t ) = 10 ω 1 + ω 2 e - t - 10 ω 4 + ω 2 e - 2 t the steady-state response is y ss ( t ) = Im P ( ) cos ωt + Re P ( ) sin ωt = | P ( ) | sin( ωt + 6 P ( )) 1

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0 2 4 6 8 10 -4 -3 -2 -1 0 1 2 3 4 Time t (sec) y ( t ) ω = 1 ω = 2 ω = 5 ω = 10 Figure 1: Time response to sinusoidal signals in Example ω 1 2 5 10 | P ( ) | 3.1623 1.5811 0.3642 0.0976 6 P ( ) -71.5651 -108.4349 -146.8887 -162.9795 However, if the system is not stable, for example, let P ( s ) = 1 s - 1 . Then y ( t ) = 10 ω 1 + ω 2 e t + | P ( ) | sin( ωt + 6 P ( )) It is impossible to observe the sinusoidal response, | P ( ) | sin( ωt + 6 P ( )), experimentally from the time response of y ( t ) since the e t term does not go away for all time. 2
Example Let f ( t ) be a square wave given by f ( t ) = 1 , 2 kT < t < (2 k + 1) T, - 1 , (2 k + 1) < t < 2( k + 1)

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