Step Response Analysis. Frequency Response, Relation Between Model - PowerPoint PPT Presentation

Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 Gustav Nilsson 17 November 2016 Lund University, Department of Automatic Control

Content 1. Step Response Analysis 2. Frequency Response 3. Relation between Model Descriptions 2

Step Response Analysis

Steup Response From the last lecture, we know that if the input u ( t ) is a step, then the output in the Laplace domain is Y ( s ) = G ( s ) U ( s ) = G ( s )1 s It is possible to do an inverse transform of Y ( s ) to get y ( t ), but is it possible to claim things about y ( t ) by only studying Y ( s )? We will study how the poles affects the step response. (The zeros will be discussed later) 4

Initial and Final Value Theorem Let F ( s ) be the Laplace transformation of f ( t ), i.e., F ( s ) = L ( f ( t ))( s ). Given that the limits below exist, it holds that: Initial value theorem lim t → 0 f ( t ) = lim s → + ∞ sF ( s ) Final value theorem lim t → + ∞ f ( t ) = lim s → 0 sF ( s ) For a step response we have that: s → 0 sG ( s )1 t → + ∞ y ( t ) = lim lim s → 0 sY ( s ) = lim s = G (0) 5

First Order System Singularity Chart Step Response 1 1 0 . 5 T = 1 T = 2 T = 5 y ( t ) Im 0 0 . 5 − 0 . 5 − 1 0 − 1 . 5 − 1 − 0 . 5 0 0 . 5 0 5 10 15 t Re K G ( s ) = 1 + sT One pole in s = − 1 / T Step response: Y ( s ) = G ( s )1 K � 1 − e − t / T � L − 1 s = − − → y ( t ) = K s (1 + sT ) 6

First Order System Singularity Chart Step Response 1 1 0 . 5 T = 1 T = 2 T = 5 y ( t ) Im 0 0 . 5 − 0 . 5 − 1 0 − 1 . 5 − 1 − 0 . 5 0 0 . 5 0 5 10 15 t Re K G ( s ) = 1 + sT Final value: K t → + ∞ y ( t ) = lim lim s → 0 sY ( s ) = lim s (1 + sT ) = K s → 0 6

First Order System Singularity Chart Step Response 1 1 0 . 5 T = 1 T = 2 T = 5 y ( t ) Im 0 0 . 5 − 0 . 5 − 1 0 − 1 . 5 − 1 − 0 . 5 0 0 . 5 0 5 10 15 t Re K G ( s ) = 1 + sT T is called the time-constant: y ( T ) = K (1 − e − T / T ) = K (1 − e − 1 ) ≈ 0 . 63 K I.e., T is time it takes for the step response to reach 63% of its final value 6

First Order System Singularity Chart Step Response 1 1 0 . 5 T = 1 T = 2 T = 5 y ( t ) Im 0 0 . 5 − 0 . 5 − 1 0 − 1 . 5 − 1 − 0 . 5 0 0 . 5 0 5 10 15 t Re K G ( s ) = 1 + sT Derivative at zero: s 2 K s (1 + sT ) = K t → 0 ˙ lim y ( t ) = s → + ∞ s · sY ( s ) = lim lim T s → + ∞ 6

Second Order System With Real Poles Singularity Chart Step Response 1 1 0 . 5 T = 1 T = 2 y ( t ) Im 0 0 . 5 − 0 . 5 − 1 0 − 1 . 5 − 1 − 0 . 5 0 0 . 5 0 5 10 15 t Re K G ( s ) = (1 + sT 1 )(1 + sT 2 ) Poles in s = − 1 / T 1 and s = − 1 / T 2 . Step response:  � � 1 − T 1 e − t / T 1 − T 2 e − t / T 2  K T 1 � = T 2 T 1 − T 2 y ( t ) = � 1 − e − t / T − t T e − t / T �  K T 1 = T 2 = T 7

Second Order System With Real Poles Singularity Chart Step Response 1 1 0 . 5 T = 1 T = 2 y ( t ) Im 0 0 . 5 − 0 . 5 − 1 0 − 1 . 5 − 1 − 0 . 5 0 0 . 5 0 5 10 15 t Re K G ( s ) = (1 + sT 1 )(1 + sT 2 ) Final value: sK t → + ∞ = lim lim s → 0 sY ( s ) = lim s (1 + sT 1 )(1 + sT 2 ) = K s → 0 7

Second Order System With Real Poles Singularity Chart Step Response 1 1 0 . 5 T = 1 T = 2 y ( t ) Im 0 0 . 5 − 0 . 5 − 1 0 − 1 . 5 − 1 − 0 . 5 0 0 . 5 0 5 10 15 t Re K G ( s ) = (1 + sT 1 )(1 + sT 2 ) Derivative at zero: s 2 K t → 0 ˙ lim y ( t ) = s → + ∞ s · sY ( s ) = lim lim s (1 + sT 1 )(1 + sT 2 ) = 0 s → + ∞ 7

Second Order System With Complex Poles K ω 2 0 G ( s ) = , 0 < ζ < 1 s 2 + 2 ζω 0 s + ω 2 0 Relative damping ζ , related to the angle ϕ ζ = cos( ϕ ) Singularity Chart 1 0 . 5 ω 0 ϕ Im 0 − 0 . 5 − 1 − 1 0 1 Re 8

Second Order System With Complex Poles K ω 2 0 G ( s ) = , 0 < ζ < 1 s 2 + 2 ζω 0 s + ω 2 0 Inverse transformation yields: � �� 1 � � 1 − ζ 2 e − ζω 0 t sin 1 − ζ 2 t + arccos ζ y ( t ) = K 1 − ω 0 � 8

Second Order System With Complex Poles K ω 2 0 G ( s ) = , 0 < ζ < 1 s 2 + 2 ζω 0 s + ω 2 0 Singularity Chart Step Response ω 0 = 1 . 5 1 1 ω 0 = 1 ω 0 = 0 . 5 y ( t ) Im 0 0 . 5 − 1 0 − 1 0 1 0 5 10 15 t Re 8

Second Order System With Complex Poles K ω 2 0 G ( s ) = , 0 < ζ < 1 s 2 + 2 ζω 0 s + ω 2 0 Singularity Chart Step Response 1 . 5 1 ζ = 0 . 3 ζ = 0 . 7 1 ζ = 0 . 9 y ( t ) Im 0 0 . 5 − 1 0 − 1 0 1 0 5 10 15 t Re 8

Frequency Response

Sinusoidial Input Given a transfer function G ( s ), what happens if we let the input be u ( t ) = sin( ω t )? 1 0 . 5 y ( t ) 0 − 0 . 5 − 1 0 5 10 15 20 t 1 0 . 5 u ( t ) 0 − 0 . 5 − 1 0 5 10 15 20 10 t

Sinusoidial Input It can be shown that if the input is u ( t ) = sin( ω t ), the output will be y ( t ) = a sin( ω t + ϕ ) where a = | G ( i ω ) | ϕ = arg G ( i ω ) So if we determine a and ϕ for different frequencies, we have a description of the transfer function. 11

Bode Plot Idea: Plot | G ( i ω ) | and arg G ( i ω ) for different frequencies ω . 10 1 Magnitude (abs) 10 0 10 − 1 10 − 2 10 − 3 10 − 2 10 − 1 10 0 10 1 10 2 0 Phase (deg) − 45 − 90 − 135 − 180 10 − 2 10 − 1 10 0 10 1 10 2 Frequency (rad/s) 12

Bode Plot - Products of Transfer Functions Let G ( s ) = G 1 ( s ) G 2 ( s ) G 3 ( s ) then log | G ( i ω ) | = log | G 1 ( i ω ) | + log | G 2 ( i ω ) | + log | G 3 ( i ω ) | arg G ( i ω ) = arg G 1 ( i ω ) + arg G 2 ( i ω ) + arg G 3 ( i ω ) This means that we can construct Bode plots of transfer functions from simple ”building blocks” for which we know the Bode plots. 13

Bode Plot of G ( s ) = K If G ( s ) = K then log | G ( i ω ) | = log( K ) arg G ( i ω ) = 0 14

Bode Plot of G ( s ) = K 10 1 Magnitude (abs) K = 4 K = 1 10 0 K = 0 . 5 10 − 1 10 − 2 10 − 1 10 0 10 1 10 2 45 0 Phase (deg) − 45 − 90 − 135 − 180 10 − 2 10 − 1 10 0 10 1 10 2 Frequency (rad/s) 14

Bode Plot of G ( s ) = s n If G ( s ) = s n then log | G ( i ω ) | = n log( ω ) arg G ( i ω ) = n π 2 15

Bode Plot of G ( s ) = s n 10 2 Magnitude (abs) n = 1 10 1 10 0 n = − 1 10 − 1 10 − 2 n = − 2 10 − 3 10 − 4 10 − 1 10 0 10 1 10 2 90 45 Phase (deg) 0 − 45 − 90 − 135 − 180 10 − 2 10 − 1 10 0 10 1 10 2 Frequency (rad/s) 15

Bode Plot of G ( s ) = (1 + sT ) n If G ( s ) = (1 + sT ) n then � 1 + ω 2 T 2 ) log | G ( i ω ) | = n log( arg G ( i ω ) = n arg(1 + i ω T ) = n arctan ( ω T ) For small ω log | G ( i ω ) | → 0 arg G ( i ω ) → 0 For large ω log | G ( i ω ) | → n log( ω T ) arg G ( i ω ) → n π 2 16

Bode Plot of G ( s ) = (1 + sT ) n 10 1 Magnitude (abs) n = 1 10 0 n = − 1 10 − 1 1 T n = − 2 10 − 2 10 − 1 10 0 10 1 90 45 Phase (deg) 0 − 45 − 90 − 135 − 180 10 − 1 10 0 10 1 Frequency (rad/s) 16

Bode Plot of G ( s ) = (1 + 2 ζ s /ω 0 + ( s /ω 0 ) 2 ) n G ( s ) = (1 + 2 ζ s /ω 0 + ( s /ω 0 ) 2 ) n For small ω log | G ( i ω ) | → 0 arg( i ω ) → 0 For large ω � ω � log | G ( i ω ) | → 2 n log ω 0 arg G ( i ω ) → n π 17

Bode Plot of G ( s ) = (1 + 2 ζ s /ω 0 + ( s /ω 0 ) 2 ) n 10 2 Magnitude (abs) ζ = 0 . 2 10 1 ζ = 0 . 1 ζ = 0 . 05 10 0 10 − 1 10 − 2 10 − 1 10 0 10 1 0 Phase (deg) − 45 − 90 − 135 − 180 10 − 1 10 0 10 1 Frequency (rad/s) 17

Bode Plot of G ( s ) = e − sL G ( s ) = e − sL Describes a pure time delay with delay L , i.e, y ( t ) = u ( t − L ) log | G ( i ω ) | = 0 arg G ( i ω ) = − ω L 18

Bode Plot of G ( s ) = e − sL 10 1 Magnitude (abs) 10 0 10 − 1 10 − 1 10 0 10 1 10 2 0 L = 0 . 1 Phase (deg) − 200 L = 5 L = 0 . 5 − 400 − 600 10 − 1 10 0 10 1 10 2 Frequency (rad/s) 18

Bode Plot of Composite Transfer Function Example Draw the Bode plot of the transfer function G ( s ) = 100( s + 2) s ( s + 20) 2 First step, write it as product of sample transfer functions: G ( s ) = 100( s + 2) s ( s + 20) 2 = 0 . 5 · s − 1 · (1 + 0 . 5 s ) · (1 + 0 . 05 s ) − 2 Then determine the corner frequencies: w c 2 =20 w c 1 =2 � �� � G ( s ) = 100( s + 2) � �� � s ( s + 20) 2 = 0 . 5 · s − 1 · (1 + 0 . 05 s ) − 2 (1 + 0 . 5 s ) · 19

Bode Plot of Composite Transfer Function w c 2 =20 w c 1 =2 � �� � G ( s ) = 100( s + 2) � �� � s ( s + 20) 2 = 0 . 5 · s − 1 · (1 + 0 . 05 s ) − 2 (1 + 0 . 5 s ) · 10 1 10 0 Magnitude (abs) 10 − 1 10 − 2 10 − 3 10 − 4 10 − 1 10 0 10 1 10 2 10 3 Frequency (rad/s) 20

Bode Plot of Composite Transfer Function w c 2 =20 w c 1 =2 � �� � G ( s ) = 100( s + 2) � �� � s ( s + 20) 2 = 0 . 5 · s − 1 · (1 + 0 . 05 s ) − 2 (1 + 0 . 5 s ) · 10 1 − 1 10 0 Magnitude (abs) 10 − 1 10 − 2 10 − 3 10 − 4 10 − 1 10 0 10 1 10 2 10 3 Frequency (rad/s) 20

Bode Plot of Composite Transfer Function w c 2 =20 w c 1 =2 � �� � G ( s ) = 100( s + 2) � �� � s ( s + 20) 2 = 0 . 5 · s − 1 · (1 + 0 . 05 s ) − 2 (1 + 0 . 5 s ) · 10 1 − 1 10 0 0 Magnitude (abs) 10 − 1 10 − 2 10 − 3 10 − 4 10 − 1 10 0 10 1 10 2 10 3 Frequency (rad/s) 20