자동제어 3강: 라플라스 역변환 강의실 : 산107 담당교수 : 고경철(기계공학부) 사무실 : 산학협력관 105B 강의실 : 산107 담당교수 : 고경철(기계공학부) 사무실 : 산학협력관 105B 면담시간 : 수시 Email : kckoh@sunmoon.ac.kr 강의관련: http://mecha.sunmoon.ac.kr 나의 소개: 내가 나를 소개한다는 것. 산업계에서 줄곳 산업용 제어기기 개발. 공장자동화 용 로봇제어기, PC-base이동로봇 제어시스템, 방전가공기, NC, 자동용접시스템, 지능제어. 배경: 기계->시스템->제어->로봇공학. 강단에 선 소감. 주로 이론과 실제를 조화한 교육이 되도록 노력. 2005. 3. 21 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

역라플라스변환 - Finding the time function f(t) from Laplace transform F(s) Inverse Laplace Transformation - Finding the time function f(t) from Laplace transform F(s) - Notation: Inversion integral P2 Where c is the abscissa of convergence. P1 Inversion integral process - Direct integral: “too complex” - Look-up table method: -> simpler! - background: Unique correspondence of a time function f(t) and its Laplace transform F(S) for any continuous time function Partial-fraction expansion method for finding inverse Laplace transforms - Frequently, then, c P3 Path of inversion integral: Pi’s are the singular points of the F(s) 제어시스템 자동제어 vs 시스템 교과서 선정배경: 기계시스템, 물리적 개념위주 역사, 수학적 배경, 이론, 설계, 시뮬레이션 진행방법: MATLAB Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

Frequency Shift Theorem Ex.2.2 Problem: Find the inverse Laplace transform of From 2.2(4) since therefore Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

General Partial Fraction When F(S) involves distinct poles only (2) Expending into a sum of simple partial fractions Where Ki (i=1,2,,….,n) are constants => called the residue at the pole s= - pi (3) Calculation of the residue (4) Inverse Laplace transform Since For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

(Case 1) Real and Distinct Root Case (Eq.2.7) Expending into a sum of simple partial fractions To find the inverse Laplace transform of By partial fraction, Where ak (k=1,2) are calculated by Thus, For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

연습1: 서로 다른실근 Practice 1 Expending into a sum of simple partial fractions Where ak (k=1,2) are calculated by Thus, For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

연습2: 서로다른실근 Ex. 2-5 For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

(Case2)Double root Case Expending into a sum of simple partial fractions To find the inverse Laplace transform of For t>=0 By partial fraction, Where ak (k=1,2,3) are calculated by differentiating Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

General Double root Partial Fraction When F(S) involves distinct poles only (2) Expending into a sum of simple partial fractions Where Ki (i=1,2,,….,n) are constants => called the residue at the pole s= - pi (3) Calculation of the residue For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

연습3:중근을 갖는 경우 When F(S) involves multiple poles (1) Partial-fraction expansion Considering (2) Partial fraction Expansion Where bk (k=1,2,,….,n) are determined as follows: (3) Calculation of the residue (4) Inverse Laplace transform For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

(Case 3) Imaginary Root Case Expending into a sum of simple partial fractions To find the inverse Laplace transform of By partial fraction, Where ak (k=1,2) are calculated by Thus, Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

(Case 3) Imaginary Root Case Expending into a sum of simple partial fractions From cosine and sine Laplacians Transform to similar form Thus, Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

Differential Equation Ex.2.3 Laplace transform solution of a differential equation Eq.(2.4) (1) Laplace Transform of Eq.(2.4) (2) Solving for the response Y(s) (3) Partial fraction expansion Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

Differential Equation Ex.2.3 Laplace transform solution of a differential equation (4) Calculation of coefficients Ki by Eq.(2.13) (5) Inverse Laplace transform For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

연습4: 미분방정식의 해 Solving linear, time-invariant differential equations Ex. 2-8 Find the solution x(t) of the differential equation Using, The given differential equation becomes Using, Thus, For t>=0 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

MATLAB을 이용한 부분분수분해 Partial-Fraction expansion with MATLAB Considering The command: [r,p,k]=residue(num,den) num=[b0 b1 ... Bn] den=[1 a1 ... An] [r,p,k]=residue(num,den) 연습5 r = -6.0000 -4.0000 4.0000 p= -3.0000 -2.0000 -1.0000 k= 2 num=[2 5 3 6] den=[1 6 11 6] [r,p,k]=residue(num,den) Thus, Cf. [num,den]=residue(r,p,k) Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

연습6: practice 6 Expand B(s)/A(s) into partial fraction with MATLAB Before use of MATLAB, ㅁnum=[0 1 2 3] num = 0 1 2 3 ㅁden=[1 3 3 1] den = 1 3 3 1 ㅁ[r,p,k]=residue(num,den) r = 1.0000 0.0000 2.0000 p = -1.0000 k = [] Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

HW#3 Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.