1. 자동제어 3강: 라플라스 역변환 강의실 : 산107 담당교수 : 고경철(기계공학부) 사무실 : 산학협력관 105B
강의실 : 산107
담당교수 : 고경철(기계공학부)
사무실 : 산학협력관 105B
면담시간 : 수시
강의관련:
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.
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.

2. 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.

3. 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.

4. 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.

5. (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.

6. 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.

7. 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.

8. (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.

9. 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.

10. 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.

11. (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.

12. (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.

13. 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.

14. 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.

15. 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.

16. 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 =
4.0000 p=
k= 2
num=[ ]
den=[ ]
[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.

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

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