2019학년도 1학기

반도체 공학과 공학수학1

주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition

부교재: 이상구 외 4인, 최신공학수학 I,  1st Edition

강의시간:  공학수학1, (화09:00-10:15)   (목10:30-11:45)

담당교수:  김응기 박사

Week 9

9주차

6.1: Laplace Transform, Linearity, First Shifting Theorem

8.1 라플라스 변환의 도입

6.2: Transforms of Derivatives and Integrals

8.2 라플라스 변환의 정의와 특성

6.3: Unit Step Function, t-shifting

Chapter 6.  Laplace Transform

Laplace Transform Method

The Laplace transform method is a powerful method for solving linear ODEs and corresponding initial value problems, as well as system of ODEs arising in engineering.

The process of solution consists of three steps

Step 1. The given ODE is transformed into an algebraic equation(“subsidiary equation”).

Laplace transform

Step 2. The subsidiary equation is solved by purely algebraic manipulations.

Algebraic operator

Step 3. The solution in step 2 is transformed back, resulting in the solution of the given           problem. - Inverse Laplace transform.

Solving IVP by Laplace transforms

Thus solving an ODE is reduced to an algebraic problem. This switching from calculus to algebra is called operational calculus.

<Relation of Time domain and Domain>

6.1 Laplace Transform, Inverse Laplace Transform,

Linear, S-shifting.

Definition of Laplace transform

is a function defined for all .

(1)

is the Laplace transform of a function .

The result is an “integral transform

with “kernel.

If , then

(1*)

is an inverse Laplace transform of .

Here is the inverse Laplace transform operator.

,

Notation

Original function are denoted by lowercase letters.

The Laplace transform of the function are denoted by capital letter.

is the transform of .

is the transform of .

Example 1  Laplace Transform

Let when . Find .

Solution

Sage Coding

 var('s, t') f(t) = 1 print f(t).laplace(t, s)

Evaluate

1/s

Example

Let when . Find .

Solution

Sage Coding

 var('s, t') f(t) = t print f(t).laplace(t, s)

Evaluate

s^(-2)

Example

Let when . Find .

Solution

Sage Coding

 var('s, t, n') assume(n>0) f(t) = t^n print f(t).laplace(t, s)

Evaluate

s^(-n - 1)*gamma(n + 1)

Example

Using Gamma function let when . Find .

Solution

Let or . Then or

Example 2  Laplace Transform of the Exponential Function.

Let when , where is a constant. Find .

Let when , where is a constant. Find .

Solution

Sage Coding

 var('s, t, a')  f(t) = e^(a*t) print f(t).laplace(t, s)

Evaluate
-1/(a - s)

 var('s, t, a')  f(t) = e^(-a*t) print f(t).laplace(t, s)

Evaluate
1/(a + s)

The Laplace transform is a “linear operation.”

Theorem 1 Linearity of the Laplace Transform

The Laplace transform is a linear operation, that is, for any function and whose transform exist and any constants and the transform of exists and

Proof

Let and .

Then and are any constant.

.

Generalize

If are any constants, have Laplace transform , respectively, then

Theorem 1*  Linearity of the inverse Laplace Transform

If and are any constants while and are the Laplace transform of and respectively, then

.

Proof

If and , then and .

Then if and are any constant.

Generalize

If are any constants, are the Laplace transform of  respectively, then

Example 3  Laplace Transform of the Hyperbolic Function

Let when , where is a constant. Find .

Let when , where is a constant. Find .

Solution

and in Example 2, and Theorem 1

Sage Coding

 var('s, t, a')  f(t) = cosh(a*t) print f(t).laplace(t, s)

Evaluate
-s/(a^2 – s^2)

 var('s, t, a')  f(t) = sinh(a*t) print f(t).laplace(t, s)

Evaluate
-a/(a^2 - s^2)

Example 4  Laplace Transform of the Trigonometric Function

Let when , where is a constant. Find .

Let when , where is a constant. Find .

Solution

Sage Coding

 var('s, t, w')  f(t) = cos(w*t) print f(t).laplace(t, s)

Evaluate
s/(s^2 + w^2)

 var('s, t, w')  f(t) = sin(w*t) print f(t).laplace(t, s)

Evaluate
w/(s^2 + w^2)

Theorem 2  First shifting Theorem, -shifting

If , then .

Proof 1

Using .

.

If , then

Proof 2

By assumption , then

Example 5  -Shifting : Damped Vibrations. Completing the Square

Find (a)               (b) .

Solution 1

(a) Let . Then

(b) Let . Then

Solution 2

Vibrations in Example 5

Sage Coding

 var('s, t, a, w')  f(t) = e^(a*t) * cos(w*t) print f(t).laplace(t, s)

Evaluate
-(a - s)/(a^2 - 2*a*s + s^2 + w^2)

 var('s, t, a, w')  f(t) = e^(a*t) * sin(w*t) print f(t).laplace(t, s)

Evaluate
w/(a^2 - 2*a*s + s^2 + w^2)

Example

Find the inverse of the transform .

Solution

Sage Coding

 var('t, s')  f(t) = (3*s - 137)/(s^2 + 2*s + 401) print f(t).inverse_laplace(s, t)

Evaluate
-(7*sin(20*t) - 3*cos(20*t))*e^(-t)

Existence and Uniqueness of Laplace Transform

A function has a Laplace transform if it does not grow two fast, say, if for all and some constants and it satisfies the “growth restriction

(2)          .

is called “growth of exponential order.”

Piecewise continuous

need not be continuous. is piecewise continuity. is piecewise continuous on a finite interval where is defined, if this interval can be divided into finitely many subintervals in each of which is continuous and has finite limits as approaches either endpoint of such a subinterval from the interior. This then gives finite jumps.

Theorem 3  Existence Theorem for Laplace Transforms

If is defined and piecewise continuous on every finite interval on and satisfies for all and some constants and , then the Laplace transform exists for all .

or  .

Proof

Laplace transform of

is piecewise continuous in every finite interval on . is continuity.

Then is integrable.

or , , is of growth of exponential order for

()

is integrable. Hence and is exist for all .

Theorem

If is exist, then .

Proof

Theorem

If , then is not exist.

Laplace Transform

 function function Unit Step Function Trigonometric  Function Step function Hyperbolic  Function ( positive) Exponential  Function

Some Important Properties of Laplace Transforms

Change of scale property

If , then

Proof

Let or  , then

If , then

Proof

By assumption if , then

6.2 Transforms of Derivatives and Integrals

The Laplace transform is a method of solving ODEs and initial value problems.

Operations of calculus on functions are replaced by operations of algebra on transforms.

Differentiation of will correspond to multiplication of by and integration of to division of by .

The Laplace Transform of Derivative

Theorem 1  Laplace Transform of Derivative

The transforms of first and second derivatives of satisfy

(a)

(b)

Proof

(a) Using integration by parts, we have

.

using the fact that is of growth of exponential oder as , so that

for . is not continuous at .

(b) Using by

.

Theorem 2  Laplace Transform of the Derivative of Any Order

Let be continuous for all and satisfy the growth restriction. Furthermore, let be piecewise continuous on every finite interval on the semi-axis . Then the transform of satisfies

(3)

Multiplication by

If , then .

Proof

Inverse Laplace transform of derivatives

If  , then

Proof

By assumption if , then

Example 1  Transform of a Resonance Term

Let . Using , find .

Solution

and ,

Sage Coding

 var('s, t, w')  f(t) = t*sin(w*t) print f(t).laplace(t, s)

Evaluate
2*s*w/(s^2 + w^2)^2

Example 2  Using , find and .

Solution

.

.

Sage Coding

 var('s, t, w')  f(t) = cos(w*t) print f(t).laplace(t, s)

Evaluate
s/(s^2 + w^2)

 var('s, t, w')  f(t) = sin(w*t) print f(t).laplace(t, s)

Evaluate
w/(s^2 + w^2)

Laplace Transform of Integral of a Function

We expect integration of to correspond to division of by .

Theorem 3  Laplace Transform of Integral

Let denote the transform of a function which is piecewise continuous for and satisfies a growth restriction , Sec. 6.1. Then for   and .

(4)       ,  thus .

Proof

Inverse Laplace Transform of Integral

If , then

Proof

By assumption , then

Multiplication by

If and , then

Proof

By assumption if , then

If , then

Proof

By assumption if , then

Find inverse Laplace transform.

(1)

(2)

Solution

(1)

(2)

Sage Coding

 var('t, s')  f = 1/(s*(s+ 1)) g = f.partial_fraction(s) print g print g.inverse_laplace(s, t)

Evaluate
-1/(s + 1) + 1/s

-e^(-t) + 1

 var('t, s')  f = 1/(s*(s^2 + 1)) g = f.partial_fraction(s) print g print g.inverse_laplace(s, t)

Evaluate
-s/(s^2 + 1) + 1/s

-cos(t) + 1

Example 3  Application of Theorem 3

Using Theorem 3, find the inverse of and .

Solution

Sage Coding

 var('t, s, w')  assume(w > 0) f = 1/(s*(s^2 + w^2)) g = f.partial_fraction(s) print g print g.inverse_laplace(s, t)

Evaluate
-s/((s^2 + w^2)*w^2) + 1/(s*w^2)

-cos(t*w)/w^2 + 1/w^2

 var('t, s, w')  assume(w > 0) f = 1/(s^2 *(s^2 + w^2)) g = f.partial_fraction(s) print g print g.inverse_laplace(s, t)

Evaluate
-1/((s^2 + w^2)*w^2) + 1/(s^2*w^2)

t/w^2 - sin(t*w)/w^3

Differential Equation, Initial Value Problem

We consider an initial value problem

(5)

where and are constant. Here is the given input(driving force) and is output(response to the input) to be obtained.

Step 1. Setting up the subsidiary equation

,

Step 2. Solution of the subsidiary equation by algebra

We divide by and the so-called transfer function

(6)        .

This gives the solution

(7)        .

If , this is simply , hence

and this explains the name of . Note that depends neither on nor on the initial conditions (but only and )

Step 3. Inversion of to obtain

We reduce to a sum of terms whose inverse can be found from the tables so that we obtain the solution of .

: transfer function

If , this is simply , hence

Example  4  Initial Value Problem: The Basic Laplace Steps

Solve   , , .

Solution

, ,

Fig. 116. Steps of the Laplace transform method

Sage Coding

 # 라플라스 변환을 이용한 풀이 # 방정식 구하기 y, s, t = var('y', 's', 't') y = function('y')(t) f = t eqy = diff(y, t, 2) - y == f EQY = eqy.laplace(t, s) Y = var('Y') strEQY = str(EQY).replace('laplace(y(t), t, s)', 'Y') print strEQY #  초깃값 대입하여 해 구하기 strEQY1 = strEQY.replace('y(0)', '1').replace('D[0](y)(0)', '1') EQ2 = sage_eval(strEQY1, locals={'s':s, 'Y':Y}) #  연립방정식 풀이 soln = solve(EQ2, Y)   soly = soln[0].rhs() print soly print soly.inverse_laplace(s, t)

Evaluate
s^2*Y - s*y(0) - Y - D[0](y)(0) == s^(-2)

(s^3 + s^2 + 1)/(s^4 - s^2)

-t - 1/2*e^(-t) + 3/2*e^t

 # 직접 풀이 t = var('t') y = function('y')(t) eqy = diff(y, t, 2) - y == t desolve(eqy, [y, t], ics = [0, 1, 1])

Evaluate
-t - 1/2*e^(-t) + 3/2*e^t

Example 5  Comparison with the Usual Method

Solve the initial value problem

, , .

Solution

Sage Coding

 # 라플라스 변환을 이용한 풀이 # 방정식 구하기 y, s, t = var('y', 's', 't') y = function('y')(t) f = 0 eqy = diff(y, t, 2) + diff(y, t) + 9*y == f EQY = eqy.laplace(t, s) Y = var('Y') strEQY = str(EQY).replace('laplace(y(t), t, s)', 'Y') print strEQY   #  초깃값 대입하여 해 구하기 strEQY1 = strEQY.replace('y(0)', '0.16').replace('D[0](y)(0)', '0') EQ2 = sage_eval(strEQY1, locals={'s':s, 'Y':Y}) #  연립방정식 풀이 soln = solve(EQ2, Y)   soly = soln[0].rhs() print soly print soly.inverse_laplace(s, t)

Evaluate
s^2*Y + s*Y - s*y(0) + 9*Y - y(0) - D[0](y)(0) == 0

4/25*(s + 1)/(s^2 + s + 9)

4/875*(sqrt(35)*sin(1/2*sqrt(35)*t) + 35*cos(1/2*sqrt(35)*t))*e^(-1/2*t)

 # 직접 풀이 t = var('t') y = function('y', t) eqy = diff(y, t, 2) + diff(y, t) + 9*y == 0 desolve(eqy, [y, t], ics = [0, 0.16, 0])

Evaluate
4/875*(sqrt(35)*sin(1/2*sqrt(35)*t) + 35*cos(1/2*sqrt(35)*t))*e^(-1/2*t)

1. Solving a nonhomogeneous ODE does not require first solving the homogeneous ODE.

2. Initial value are automatically taken care of.

3. Complicated inputs can be handle very efficiently, as we show in the next sections.

Example 6  Shifted Data Problems

This means initial value problems with conditions given at some instead of . For such a problem set , so that gives and the Laplace transform can be applied. For instance, solve

.

Solution

We have and we set .

Then the problem is

,

Sage Coding

 # 라플라스 변환을 이용한 풀이 # 방정식 구하기 y, s, t = var('y', 's', 't') y = function('y')(t) f = 2*t + pi/2 eqy = diff(y, t, 2) + y == f EQY = eqy.laplace(t, s) Y = var('Y') strEQY = str(EQY).replace('laplace(y(t), t, s)', 'Y') print strEQY   #  초깃값 대입하여 해 구하기 strEQY1 = strEQY.replace('y(0)', 'pi/2').replace('D[0](y)(0)', '2 - sqrt(2)') EQ2 = sage_eval(strEQY1, locals={'s':s, 'Y':Y}) #  연립방정식 풀이 soln = solve(EQ2, Y)   soly = soln[0].rhs() print soly print soly.inverse_laplace(s, t)

Evaluate
s^2*Y - s*y(0) + Y - D[0](y)(0) == 1/2*pi/s + 2/s^2

1/2*(pi*s^3 + 2*(sqrt(2) + 2)*s^2 + pi*s + 4)/(s^4 + s^2)

1/2*pi + sqrt(2)*sin(t) + 2*t

*위의 t 대신에 t – pi/4 대입해야 함

 # 직접 풀이 t = var('t') y = function('y')(t) eqy = diff(y, t, 2) + y == 2*t desolve(eqy, [y, t], ics=[pi/4, pi/2, 2 - sqrt(2)])

Evaluate
2*t - sin(t) + cos(t)

6.3 Unit Step Function. t-Shifting

Unit-step function or Heaviside function

Unit Step Function-

Solution

Unit step function

Unit Step Function(Heaviside Function)

Solution

Unit step function

Let for all negative . Then with is shifted to the right by the amount .

Time Shifting -Shifting : Replacing by in

Theorem 1  Second Shifting ; Time Shifting

If has the transform , then the “shifted function

(3)

has the transform . That is, if , then

(4)

Or, if we take the inverse on both sides, we can write

(4*)       .

Proof

(If , then )

Hence .

Taking inverse Laplace transform

.

If and , then .

Proof

Let then

If , then

Proof

By assumption if , then

Example 1  Application of Theorem 1 use of Unit Step Function

Write the following function using unit step function and find its transform.

.

Solution

,   ,

In terms of unit step function

Sage Coding

 var('s, t') f1 = 2*(unit_step(t) - unit_step(t-1)) f2 = 1/2*t^2*(unit_step(t-1) - unit_step(t-pi/2)) f3 = cos(t)*unit_step(t-pi/2) f = f1 + f2 + f3 print f.laplace(t, s)

Evaluate
-1/8*pi^2*e^(-1/2*pi*s)/s - e^(-1/2*pi*s)/(s^2 + 1) - 1/2*pi*e^(-1/2*pi*s)/s^2 - 3/2*e^(-s)/s + 2/s + e^(-s)/s^2 - e^(-1/2*pi*s)/s^3 + e^(-s)/s^3

Example

Write the following function using unit step function and find its transform

.

Solution

Periodic Function

Then we have for .

A function is said to be periodic if, for some nonzero constant , we have  called a

for all in the domain. is called a period of the function.

Consider the Laplace transform of periodic functions.

Let have period .

If and , then .

Proof

Example 2  Application of Both Shifting Theorems. Inverse Transform

Find the inverse transform of

.

Solution

,

,

,

Fig. 123. in Example 2

Example 3  Response of an -Circuit to a Single Rectangular wave

Find the current in the -circuit if a single rectangular wave with voltage is applied. The circuit is assumed to be quiescent before the wave is applied.

Solution

The input is .

Hence the circuit is  modeled by the integro-differential equation

.

We obtain the subsidiary equation

.

Solving this equation algebraically for , we get

where    and  ,

Hence Theorem 1 yields the solution

that is, if ,  and

where,  .

Fig. 124. RC-circuit, electromotive force , and current in Example 3

Example 4  Response of an -Circuit to a Sinusoidal Input Acting Over a Time Interval

Find the response (the current) of the -circuit, where is sinusoidal, acting for a short time interval only, say,

if   and     if

and current and charge are initially zero.

Solution

The electromotive force can be represented by .

Hence the model for the current in the circuit is the integro-differential equation

, ,

We obtain the subsidiary equation for

Solving it algebraically and noting that , we obtain

The inverse of

The second term of differ from the first term by the factor .

Since and ,

the second shifting theorem gives the inverse

.

Hence in the cosine and sine terms cancel, and the current for is

.

It goes to zero very rapidly, practically with in .

Fig. 125. RLC-circuit in Example 4

[한빛 아카데미] Engineering Mathematics with Sage:

[저자] 이상 구, 김영 록, 박준 현, 김응 기, 이재 화

Contents

A. 공학수학 1 – 선형대수, 상미분방정식+ Lab

Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html

Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html

Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html

Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html

Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html

Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html

Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html

Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html

B. 공학수학 2 - 벡터미적분, 복소해석 + Lab

Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html

Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html

Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html

Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html

Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html

Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html

Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html

Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html

Made by Prof. Sang-Gu LEE  sglee at skku.edu

http://matrix.skku.ac.kr/sglee/   with Dr. Jae Hwa LEE