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 10

10주차

6.4: Short Impulses, Dirac’s Delta Function, Partial Fractions

6.5: Convolution. Integral Equations

6.6: Differentiation and Integration of Transforms. ODEs with Variable Coefficients

8.3 라플라스 변환을 이용한 미분방정식의 풀이

8.4, 8.5 라플라스 변환의 응용

6.4 Short Impulses. Dirac’s Delta Function.

Partial Fractions

Short Impulses

(1)

where is positive and small. is define.

In mechanics, the integral of a force acting over a time interval is called the impulse of the forces. For electromotive force acting on circuits.

The impulse of in is

(2)          .

Fig. 132. The function in (1)

, we take the limit of as . This limit is denoted by , that is,

is called the Dirac delta function or the unit impulse function.

is called generalized function. The impulse of is . From and by taking the limit as we obtain

(3)             and

In particular, for a continuous function one uses the property

(4)

which is plausible by . is the shifting property of .

To obtain the Laplace transform of , we write

and take the transform

Defining the transform of by this limit, that is

Example 1  Mass-Spring System under a Square Wave

Determine the response of the damped mass-spring system under a square wave, modeled by

, , .

Solution

Inverse of

,

,

Fig. 133. Square wave and response in Example 1

Sage Coding

 t = var('t') y = function('y')(t) eqy = diff(y, t, 2) + 3*diff(y, t) + 2*y == unit_step(t - 1) - unit_step(t - 2) desolve(eqy, [y, t], ics = [0, 0, 0])

Evaluate
-1/4*(2*(e - e^2)*e^t - (2*e^(t + 2) - e^(2*t) - e^4)*sgn(t - 2) + (2*e^(t + 1) - e^(2*t) - e^2)*sgn(t - 1) - e^2 + e^4)*e^(-2*t)

Example 2  Hammerblow Response of a Mass-Spring System

Find the response of the system in Example with the square wave replaced by a unit impulse at time .

, .

Solution

,

The inverse of

Fig. 134. Response to a hammerblow in Example 2

Example 3  For-Terminal RLC-Network

Find the output voltage response if , , the input (a unit impulse at time ), and current and charge are zero at time .

Solution

The current and charge are related by , we obtain the model

.

We obtain the subsidiary equation for

.

Solution

By the first shifting theorem,

we obtain from damped oscillations for and rounding ,

we get

and   .

Fig. 135. Network and output voltage in Example 3

Example 4  Unrepeated Complex Factors. Damped Forced Vibrations

Solve the initial value problem for a damped mass-spring system acted upon by a sinusoidal force for some time interval,

.

Solution

By the second shifting theorem, we obtain the subsidiary equation

By the first shifting theorem (-shifting)

The sum of this and (9) is the solution for ,

(10)                   if

Figure shows (9) (for ) and (10) (for ), a beginning vibration, which goes to zero rapidly because of the damping and the absence of a driving force after .

Fig. 136. Example 4

Sage Coding

 t = var('t') y = function('y')(t) eqy = diff(y, t, 2) + 2*diff(y, t) + 2*y == 10*sin(2*t)*(unit_step(t) - unit_step(t - pi)) s = desolve(eqy, [y, t], ics = [0, 1, -5]) print s plot(s, (t, 0, 4*pi))

Evaluate
-(4*sin(t) - cos(t))*e^(-t) - 1/2*(e^t*sin(2*t)*sgn(t) + 2*e^t*cos(2*t)*sgn(t) - 2*(e^pi + 1)*cos(t) - 4*(e^pi + sgn(t) + 1)*sin(t) - (4*e^pi*sin(t) + 2*e^pi*cos(t) + e^t*sin(2*t) + 2*e^t*cos(2*t))*sgn(-pi + t) - 2*cos(t)*sgn(t))*e^(-t)

6.5 Convolution, Integral Equation

Convolution has to do with the multiplication of transforms.

.

The transform of a product is general different from the product of the transforms of the factors,

.

For example, take and . Then  ,

,

but and gives

Hence .

If , then is transform of the convolution of and , denoted by the standard notation * and defined by the integral

(1)         *.

Theorem 1  Convolution Theorem

If , then

(2)         *

Proof

Thus

, then inverse transform of

.

**. From (1)

*

Example 1  Convolution

Let . Find .

Solution

With and we thus obtain from   the answer

*

*

Calculate

Sage Coding

 var('t, s, a') H = 1/(s*(s - a))      # H(s)  h = H.inverse_laplace(s, t) print h

Evaluate
e^(a*t)/a - 1/a

Example 2  Convolution

Let . Find .

Solution

*

*

*

Sage Coding

 var('t, s, w') assume(w > 0) H = 1/(s^2 + w^2)^2      # H(s)  h = H.inverse_laplace(s, t) print h

Evaluate
-1/2*t*cos(t*w)/w^2 + 1/2*sin(t*w)/w^3

Theorem  Arbitrary function . Convolution has the properties as follows.

(a) Commutative **

*

then

*

(b) Distribution ***

*

**

(c) Associative ****

*****

*

****

(d) **

Example 3  Unusual Properties of Convolution

* in general. For instance,

Convolution of

**

Convolution of

***

*

Example 4  Repeated Complex Factors. Resonance

In an undamped mass-spring system, resonance occur if the frequency of the driving force equals the natural frequency of the system. Then the model is

where is the spring constant, and is the mass of the body attached to the spring.

Solution

The subsidiary equation is

as

The solution of our problem is

*

*

*

The first term grows without bound. In the case of resonance such a term must occur.

Sage Coding

 var('t, w0, K') assume(w0 > 0) y = function('y')(t) eqy = diff(y, t, 2) + w0^2*y == K*sin(w0*t) desolve(eqy, [y, t], ics = [0, 0, 0])

Evaluate
-1/2*K*t*cos(t*w0)/w0 + 1/2*K*sin(t*w0)/w0^2

Application to Nonhomogeneous Linear ODEs

The subsidiary equation of the ODE

(2)           ( constant)

has the solution

.

with and the transfer function.

The first term is positive, zero or negative, its inverse will be a linear combination of two exponential function, or of the from , or damped oscillation. The interesting is because can have various forms of practical importance, as we shall see. If , then , and the convolution theorem gives the solution

(3)         .

Example 5  Response of a Damped Vibrating System to a Single Square Wave

Using Convolution, determine the response of the damped mass-spring system model by

.

Solution

*

*

Hence

if only. Hence if

if . Hence if only.

Sage Coding

 t = var('t') y = function('y')(t) eqy = diff(y, t, 2) + 3*diff(y, t) + 2*y == unit_step(t - 1) - unit_step(t - 2) desolve(eqy, [y, t], ics = [0, 0, 0])

Evaluate
-1/4*(2*(e - e^2)*e^t - (2*e^(t + 2) - e^(2*t) - e^4)*sgn(t - 2) + (2*e^(t + 1) - e^(2*t) - e^2)*sgn(t - 1) - e^2 + e^4)*e^(-2*t)

Integral Equation

Example 6  A Volterra Integral Equation of the Second Kind

Solve the Volterra integral equation of the second kind

.

Solution

By we can written *.

Writing and applying the convolution theorem, we obtain

*

The solution is

The inverse transforms is

.

Example 7  Another Volterra Integral Equation of the Second Kind

Solve the Volterra integral equation

Solution

By we can written *.

Writing and applying the convolution theorem, we obtain

*

The solution is

The inverse transforms is

6.6 Differentiation and  Integration of Transforms.

ODEs variable Coefficients

Differentiation of Transforms

If , then .

If then

(1)            hence  .

hence

Differentiation of the transform of a function corresponds to the multiplication of the function by .

Example 1  Differentiation of Transforms

Solution

*

Integration of Transforms

If is exist, then for ,

(6)            hence  .

If , then  .

Proof

Division by

If , then .

Proof

By assumption if , then

Division by

If , then .

Proof

Let , then

or

If , then

Proof

By assumption if , then .

Integration of the transforms corresponds to the division of by .

Example 2  Differentiation and  Integration of Transforms

Find the inverse transform of .

Solution

By derivative

.

Taking the inverse transform and using , we obtain

.

Hence, the inverse of is . Alternatively, if we let

, then  .

From this and we get, in agreement with the answer just obtained,

.

Special Linear ODEs with variable coefficients

Let . Then. Hence by ,

(7)         .

Similarly, and by

(8)

Example 3  Laguerre’s Equation. Laguerre Polynomials

Laguerre’s ODE is

(9)          .

(10*)    thus .

We write and prove Rodrigues’s formula

(16)     .

They are Laguerre’s polynomials and are usually denoted by . We prove .

By the first shifting theorem (shifting),

Because the derivatives up to the order are zero at .

.

Now make another shift and divide by to get

.

[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