frequencyresp

frequencyresp - Outline Definition FR Plots Transfer...

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Outline Definition FR Plots Transfer Function FR Characteristics TF from Bode Summary EE5108: Instrumentation and Sensors Topic: Frequency Response Arthur TAY ECE, NUS 6–1 / 48

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Outline Definition FR Plots Transfer Function FR Characteristics TF from Bode Summary Outline Definition Frequency Response Plots Bode plots Frequency Response Plots of Common Transfer Functions Poles/zeros at the origin Poles/zeros on the real-axis Complex conjugate poles/zeros Transport lag Frequency Response Characteristics Low frequency characteristics High frequency characteristics Examples Transfer Function from Bode Diagram Summary Summary Practice Problems 6–2 / 48
Outline Definition FR Plots Transfer Function FR Characteristics TF from Bode Summary Definition I Frequency response – the response of a linear system to sinusoidal inputs. I Consider the following system G ( s ) = Y ( s ) U ( s ) where the input u ( t ) is a sine wave with amplitude A : u ( t ) = A sin( ω 0 t )1( t ) Its Laplace transform is U ( s ) = 0 s 2 + ω 2 0 with zero i.c., the output is given by Y ( s ) = G ( s ) 0 s 2 + ω 2 0 (1) 6–3 / 48

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Outline Definition FR Plots Transfer Function FR Characteristics TF from Bode Summary I Assuming that the poles of G ( s ) are distinct, a partial fraction expansion of Equation 1 gives Y ( s ) = a 1 s + p 1 + a 2 s + p 2 + . . . + a n s + p n + c 1 s + 0 + c 2 s - 0 Inverse LT gives y ( t ) = a 1 e - p 1 t + a 2 e - p 2 t + · · · + a n e - p n t | {z } transient + c 1 e - 0 t + c 2 e 0 t | {z } steady-state The output of (stable) system at steady-state is then given by y ss ( t ) = c 1 e - 0 t + c 2 e 0 t (2) where (via partial fraction expansion) c 1 = ( s + 0 ) G ( s ) 0 ( s + 0 )( s - 0 ) s = - 0 = - A 2 j G ( - 0 ) c 2 = ( s - 0 ) G ( s ) 0 ( s + 0 )( s - 0 ) s = 0 = A 2 j G ( 0 ) 6–4 / 48
Outline Definition FR Plots Transfer Function FR Characteristics TF from Bode Summary Substitute c 1 , c 2 into Equation 2, we have y ss ( t ) = c 1 e - 0 t + c 2 e 0 t = A - 1 2 j G ( - 0 ) e - 0 t + 1 2 j G ( 0 ) e 0 t = A - ( R{ G ( 0 ) } - j I{ G ( 0 ) } ) e - 0 t 2 j +( R{ G ( 0 ) } + j I{ G ( 0 ) } ) e 0 t 2 j = A R{ G ( 0 ) } e 0 t - e - 0 t 2 j + j I{ G ( 0 ) } e 0 t + e - 0 t 2 j = A [ R{ G ( 0 ) } sin( ω 0 t ) + I{ G ( 0 ) } cos( ω 0 t )] = B sin( ω 0 t + φ ) where B = A p ( R{ G ( 0 ) } ) 2 + ( I{ G ( 0 ) } ) 2 = A | G ( 0 ) | and φ = tan - 1 I{ G ( 0 ) } R{ G ( 0 ) } = G ( 0 ) 6–5 / 48

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Outline Definition FR Plots Transfer Function FR Characteristics TF from Bode Summary I For a LTI (stable) system, G ( s ) , with input, u ( t ) = A sin( ω 0 t ) , the output at steady-state is given by y ( t ) = B sin( ω 0 t + φ ) i.e. The output at steady-state is also a sinusoidal oscillating at the same frequency with a change in magnitude and phase given by B A = | G ( s ) | s = 0 and φ = G ( s ) | s = 0 6–6 / 48
Outline Definition FR Plots Transfer Function FR Characteristics TF from Bode Summary I Example: Given G ( s ) = 1 s +1 and u ( t ) = sin 10 t , find y ( t ) . Y ( s ) = G ( s ) U ( s ) = 1 s + 1 10 s 2 + 100 = 10 101 1 s + 1 + 10 101 - s + 1 s 2 + 100 Inverse LT gives y ( t ) = 10 101 e - t + 1 101 sin(10 t + φ ) = y 1 ( t ) + y 2 ( t ) where φ = tan - 1 ( - 10) = - 1 . 47 rad or - 84 . 2 o .

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