Computing inverse Laplace Transforms.

Similar documents
Generalized sources (Sect. 6.5). The Dirac delta generalized function. Definition Consider the sequence of functions for n 1, Remarks:

The Laplace Transform and the IVP (Sect. 6.2).

(f g)(t) = Example 4.5.1: Find f g the convolution of the functions f(t) = e t and g(t) = sin(t). Solution: The definition of convolution is,

Chapter DEs with Discontinuous Force Functions

Math 308 Exam II Practice Problems

Lecture 29. Convolution Integrals and Their Applications

Find the Fourier series of the odd-periodic extension of the function f (x) = 1 for x ( 1, 0). Solution: The Fourier series is.

The Laplace transform

f(t)e st dt. (4.1) Note that the integral defining the Laplace transform converges for s s 0 provided f(t) Ke s 0t for some constant K.

Laplace Transform. Chapter 4

MATH 251 Examination II November 5, 2018 FORM A. Name: Student Number: Section:

+ + LAPLACE TRANSFORM. Differentiation & Integration of Transforms; Convolution; Partial Fraction Formulas; Systems of DEs; Periodic Functions.

MA 266 Review Topics - Exam # 2 (updated)

Chapter 6: The Laplace Transform 6.3 Step Functions and

Solution: Homework 3 Biomedical Signal, Systems and Control (BME )

Systems Analysis and Control

Laplace Theory Examples

Applied Differential Equation. October 22, 2012

Math 307 Lecture 19. Laplace Transforms of Discontinuous Functions. W.R. Casper. Department of Mathematics University of Washington.

The Laplace Transform (Sect. 4.1). The Laplace Transform (Sect. 4.1).

Math Shifting theorems

ENGIN 211, Engineering Math. Laplace Transforms

Lecture Notes for Math 251: ODE and PDE. Lecture 22: 6.5 Dirac Delta and Laplace Transforms

Third In-Class Exam Solutions Math 246, Professor David Levermore Thursday, 3 December 2009 (1) [6] Given that 2 is an eigenvalue of the matrix

Math 2C03 - Differential Equations. Slides shown in class - Winter Laplace Transforms. March 4, 5, 9, 11, 12, 16,

e st f (t) dt = e st tf(t) dt = L {t f(t)} s

MATH 251 Examination II April 4, 2016 FORM A. Name: Student Number: Section:

MATH 251 Examination II April 3, 2017 FORM A. Name: Student Number: Section:

HIGHER-ORDER LINEAR ORDINARY DIFFERENTIAL EQUATIONS IV: Laplace Transform Method David Levermore Department of Mathematics University of Maryland

20.5. The Convolution Theorem. Introduction. Prerequisites. Learning Outcomes

HIGHER-ORDER LINEAR ORDINARY DIFFERENTIAL EQUATIONS IV: Laplace Transform Method. David Levermore Department of Mathematics University of Maryland

Solutions of the Sample Problems for the Third In-Class Exam Math 246, Fall 2017, Professor David Levermore

{ sin(t), t [0, sketch the graph of this f(t) = L 1 {F(p)}.

MA 201, Mathematics III, July-November 2016, Laplace Transform

Ordinary Differential Equations. Session 7

(an improper integral)

Introduction. f(t) dt. lim. f(t) dt = lim. 1 (1+t)1 p. dt; here p>0. h = (1+h)1 p 1 p. dt = (1+t) p 1 p

The Laplace Transform

Math 3313: Differential Equations Laplace transforms

Section 6.4 DEs with Discontinuous Forcing Functions

we get y 2 5y = x + e x + C: From the initial condition y(0) = 1, we get 1 5 = 0+1+C; so that C = 5. Completing the square to solve y 2 5y = x + e x 5

Laplace Transforms and use in Automatic Control

Laplace Transform Part 1: Introduction (I&N Chap 13)

CHAPTER 8 Laplace Transforms

MathQuest: Differential Equations

MA 201, Mathematics III, July-November 2018, Laplace Transform (Contd.)

Honors Differential Equations

ECEN 420 LINEAR CONTROL SYSTEMS. Lecture 2 Laplace Transform I 1/52

Math 353 Lecture Notes Week 6 Laplace Transform: Fundamentals

Laplace Transform of Discontinuous Functions

MA26600 FINAL EXAM INSTRUCTIONS December 13, You must use a #2 pencil on the mark sense sheet (answer sheet).

MATH 251 Examination II April 7, 2014 FORM A. Name: Student Number: Section:

Ch 6.4: Differential Equations with Discontinuous Forcing Functions

Name: Solutions Exam 3

Math 341 Fall 2008 Friday December 12

APPM 2360: Midterm exam 3 April 19, 2017

SOLUTIONS FOR HOMEWORK SECTION 6.4 AND 6.5

REVIEW NOTES FOR MATH 266

Chapter 3 Convolution Representation

Chapter 31. The Laplace Transform The Laplace Transform. The Laplace transform of the function f(t) is defined. e st f(t) dt, L[f(t)] =

Section 6.5 Impulse Functions

Laplace Transform Theory - 1

Section Arclength and Curvature. (1) Arclength, (2) Parameterizing Curves by Arclength, (3) Curvature, (4) Osculating and Normal Planes.

= e t sin 2t. s 2 2s + 5 (s 1) Solution: Using the derivative of LT formula we have

x(t) = t[u(t 1) u(t 2)] + 1[u(t 2) u(t 3)]

EE 3054: Signals, Systems, and Transforms Summer It is observed of some continuous-time LTI system that the input signal.

LTI Systems (Continuous & Discrete) - Basics

Differential Equations Class Notes

2.161 Signal Processing: Continuous and Discrete Fall 2008

Practice Problems For Test 3

Practice Problems For Test 3

Ordinary Differential Equations

Module 4. Related web links and videos. 1. FT and ZT

Introduction & Laplace Transforms Lectures 1 & 2

9.5 The Transfer Function

Differential Equations

Math 221 Topics since the second exam

Ordinary differential equations

Faculty of Engineering, Mathematics and Science School of Mathematics

Review Sol. of More Long Answer Questions

Math 216 Second Midterm 19 March, 2018

ECE-202 FINAL April 30, 2018 CIRCLE YOUR DIVISION

Chapter 6: The Laplace Transform. Chih-Wei Liu

MAE143 A - Signals and Systems - Winter 11 Midterm, February 2nd

Solutions to Assignment 7

GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)

The Method of Laplace Transforms.

Short Solutions to Review Material for Test #2 MATH 3200

Exam 3 Review Sheet Math 2070

Higher Order Linear Equations

Math Exam 3 Solutions

Section 2.1 (First Order) Linear DEs; Method of Integrating Factors. General first order linear DEs Standard Form; y'(t) + p(t) y = g(t)

Math 216 Second Midterm 20 March, 2017

Matemáticas II: Segundo del Grado en Ingeniería Aeroespacial

Math 20D Final Exam 8 December has eigenvalues 3, 3, 0 and find the eigenvectors associated with 3. ( 2) det

E2.5 Signals & Linear Systems. Tutorial Sheet 1 Introduction to Signals & Systems (Lectures 1 & 2)

L2 gains and system approximation quality 1

Definition of the Laplace transform. 0 x(t)e st dt

Linear Filters. L[e iωt ] = 2π ĥ(ω)eiωt. Proof: Let L[e iωt ] = ẽ ω (t). Because L is time-invariant, we have that. L[e iω(t a) ] = ẽ ω (t a).

Time Response Analysis (Part II)

Transcription:

Review Exam 3. Sections 4.-4.5 in Lecture Notes. 60 minutes. 7 problems. 70 grade attempts. (0 attempts per problem. No partial grading. (Exceptions allowed, ask you TA. Integration table included. Complete Laplace Transform table included. Computing inverse Laplace Transforms. Find the inverse Laplace Transform of F (s = Solution: We start rewriting function F, F (s = e 2s e 2s s (s 3 2 + 25. s [(s 3 + 3 (s 3 2 = e 2s + 52 (s 3 2 + 5 2 F (s = e 2s (s 3 (s 3 2 + 5 2 + e 2s 5 5 (3 (s 3 2 + 5 2 F (s = e 2s L[cos(5t(s 3 + 3 5 e 2s L[sin(5t(s 3 Recall (4: L[f (t(s c = L[e ct f (t. F (s = e 2s L[e 3t cos(5t + 3 5 e 2s L[e 3t sin(5t.

Computing inverse Laplace Transforms. Find the inverse Laplace Transform of F (s = e 2s s (s 3 2 + 25. Solution: Recall: F (s = e 2s L[e 3t cos(5t + 3 5 e 2s L[e 3t sin(5t. Recall (3: e cs L[f (t = L[u(t c f (t c. F (s = L[u(t 2 e 3(t 2 cos(5(t 2 + 3 5 L[u(t 2 e3(t 2 sin(5(t 2. f (t = u(t 2 e 3(t 2 [ cos(5(t 2 + 3 5 sin(5(t 2. Computing Laplace Transforms of discontinuous functions. t 0 t 6, Find the LT of f (t = 2. 3 t 6. Solution: We need to rewrite the function f in terms of functions that apear in the LT table. We need a box function for the first part, and a step function for the second part. f (t = t 2[ u(t u(t 6 + 3 u(t 6. f (t = t ( 2 u(t + t 2 + 3 u(t 6 = t 2 u(t + ( t + 6 u(t 6. 2 f (t = 2[ t u(t (t 6 u(t 6.

Computing Laplace Transforms of discontinuous functions. t 0 t 6, Find the LT of f (t = 2. 3 t 6. Solution: Recall: f (t = 2[ t u(t (t 6 u(t 6. L[f (t = 2( L[t u(t L[(t 6 u(t 6, L[f (t = 2( L[t e 6s L[t, L[f (t = 2 ( s 2 e 6s s 2, We conclude that L[f (t = 2s 2 ( e 6s. Solving IVP with generalized sources. Use Laplace Transform to find y solution of y 2y + 2y = δ(t 2, y(0 =, y (0 = 3. Solution: Compute the LT of the equation, L[y 2 L[y + 2 L[y = L[δ(t 2 = e 2s L[y = s 2 L[y s y(0 y (0, L[y = s L[y y(0. (s 2 2s + 2 L[y s y(0 y (0 + 2 y(0 = e 2s L[y = (s 2 2s + 2 L[y s = e 2s (s + (s 2 2s + 2 + (s 2 2s + 2 e 2s.

Solving IVP with generalized sources. Use Laplace Transform to find y solution of y 2y + 2y = δ(t 2, y(0 =, y (0 = 3. Solution: Recall: L[y = (s + (s 2 2s + 2 + (s 2 2s + 2 e 2s. s 2 2s + 2 = 0 s ± = 2[ 2 ± 4 8, complex roots. s 2 2s + 2 = (s 2 2s + + 2 = (s 2 +. L[y = s + (s 2 + + (s 2 + e 2s L[y = (s + + (s 2 + + (s 2 + e 2s Solving IVP with generalized sources. Use Laplace Transform to find y solution of y 2y + 2y = δ(t 2, y(0 =, y (0 = 3. Solution: Recall: L[y = L[y = (s + 2 (s 2 + + (s 2 + e 2s, (s (s 2 + + 2 (s 2 + + e 2s (s 2 +, L[cos(at = s s 2 + a 2, L[sin(at = a s 2 + a 2, L[y = L[cos(t (s + 2 L[sin(t (s + e 2s L[sin(t (s.

Solving IVP with generalized sources. Use Laplace Transform to find y solution of y 2y + 2y = δ(t 2, y(0 =, y (0 = 3. Solution: Recall: L[y = L[cos(t (s + 2 L[sin(t (s + e 2s L[sin(t (s and L[f (t (s c = L[e ct f (t. Therefore, L[y = L[e t cos(t + 2 L[e t sin(t + e 2s L[e t sin(t. Also recall: e cs L[f (t = L[u c (t f (t c. Therefore, L[y = L[e t cos(t + 2 L[e t sin(t + L[u 2 (t e (t 2 sin(t 2. y(t = [ cos(t + 2 sin(t e t + u 2 (t sin(t 2 e (t 2. Solving IVP with discontinuous sources. 0, t < 2, y + 3y = g(t, y(0 = y (0 = 0, g(t = e (t 2, t 2. Solution: Express g using step functions, e t g ( t g(t = u 2 (t e (t 2. u ( t 2 L[u c (t f (t c = e cs L[f (t. Therefore, 2 t L[g(t = e 2s L[e t. We obtain: L[g(t = e 2s (s.

Solving IVP with discontinuous sources. 0, t < 2, y + 3y = g(t, y(0 = y (0 = 0, g(t = e (t 2, t 2. Solution: Recall: L[g(t = e 2s (s. (s 2 + 3 L[y = e 2s (s H(s = L[y + 3 L[y = L[g(t = e 2s (s. L[y = e 2s (s (s 2 + 3. (s (s 2 + 3 = a (s = a(s 2 + 3 + (bs + c(s + (bs + c (s 2 + 3 Solving IVP with discontinuous sources. 0, t < 2, y + 3y = g(t, y(0 = y (0 = 0, g(t = e (t 2, t 2. Solution: Recall: = a(s 2 + 3 + (bs + c(s. = as 2 + 3a + bs 2 + cs bs c = (a + b s 2 + (c b s + (3a c a + b = 0, c b = 0, 3a c =. a = b, c = b, 3b b = b = 4, a = 4, c = 4. H(s = 4 [ s s + s 2. + 3

Solving IVP with discontinuous sources. 0, t < 2, y + 3y = g(t, y(0 = y (0 = 0, g(t = e (t 2, t 2. Solution: Recall: H(s = [ 4 s s + s 2, L[y = e 2s H(s. + 3 H(s = [ 4 s s s 2 + 3 3 3 s 2, + 3 H(s = 4 [ L[e t L [ cos ( 3 t 3 L [ sin ( 3 t. [ H(s = L e 4( t cos ( 3 t 3 sin ( 3 t. Solving IVP with discontinuous sources. 0, t < 2, y + 3y = g(t, y(0 = y (0 = 0, g(t = e (t 2, t 2. [ Solution: Recall: H(s = L e 4( t cos ( 3 t 3 sin ( 3 t. h(t = 4 ( e t cos ( 3 t 3 sin ( 3 t, H(s = L[h(t. L[y = e 2s H(s = e 2s L[h(t = L[u 2 (t h(t 2. We conclude: y(t = u 2(t 4 y(t = u 2 (t h(t 2. Equivalently, [ e (t 2 cos ( 3 (t 2 3 sin ( 3 (t 2.

Laplace Transform and convolutions. Use convolutions to find f satisfying L[f (t = Solution: One way to solve this is with the splitting L[f (t = e 2s (s 2 + 3 (s = 3 e 2s 3 (s 2 + 3 L[f (t = e 2s 3 L[sin ( 3 t L[e t e 2s (s (s 2 + 3. L[f (t = 3 L[u 2 (t sin ( 3 (t 2 L[e t. (s, f (t = 3 t 0 u 2 (τ sin ( 3 (τ 2 e (t τ dτ. Definition of the Laplace Transform. Use the definition of the LT to find the LT of f (t = cosh(t. Solution: Recall that cosh(t = (e t + e t /2, and that L[cosh(t = 0 e st (et + e t 2 dt L[cosh(t = lim N 2 L[cosh(t = lim N 2 N 0 ( e (s t + e (s+t dt. [ e (s t (s e (s+t (s + L[cosh(t = [ 2 (s + = (s + 2 We conclude: L[cosh(t = s s 2. N 0. (s + + (s (s 2.

Solving IVP with impulsive forces. (Sect 6.5, Probl.7 Find the solution to the initial value problem y + y = δ(t π cos(t, y(0 = 0, y (0 = 0. Solution: Compute the Laplace Transform of the equation, L[y + L[y = L[δ(t π cos(t To compute the right-hand side above, we need the definition of the LT. Given any smooth function f and a constant c, holds L[δ(t cf (t = 0 e st f (t δ(t c dt = [ e st f (t t=c We have used that c+ɛ c ɛ δ(t c g(t dt = g(c. We obtain the formula: L[δ(t cf (t = f (c e cs. Solving IVP with impulsive forces. (Sect 6.5, Probl.7 Find the solution to the initial value problem y + y = δ(t π cos(t, y(0 = 0, y (0 = 0. Solution: Recall: L[δ(t cf (t = f (c e cs. Hence s 2 L[y + L[y = L[δ(t π cos(t = cos(πe πs = e πs L[y = e πs s 2 + = e πs L [ sin(t. Recall the property (3: e cs L[f (t = L [ u(t c f (t c ; L[y = L [ u(t π sin(t π y(t = u(t π sin(t π.

Laplace Transform and convolutions. Given any function g(t with Laplace transform G(s = L[g(t, find the function f satisfying L[f (t = e 2s (s 2 + 3 G(s. Solution: One way to solve this is with the splitting L[f (t = e 2s (s 2 + 3 G(s = e 2s 3 3 (s 2 + 3 G(s, L[f (t = e 2s 3 L[sin ( 3 t L[g(t L[f (t = 3 L[u 2 (t sin ( 3 (t 2 L[g(t. f (t = 3 t 0 u 2 (τ sin ( 3 (τ 2 g(t τ dτ. Solving IVP with step functions in the source. y 6y = g(t, y(0 = y (0 = 0, g(t = 0, t < π, sin(t π, t π. Solution: Express g using step functions, sin ( t u ( t pi g ( t g(t = u π (t sin(t π. L[u c (t f (t c = e cs L[f (t. pi t Therefore, L[g(t = e πs L[sin(t. We obtain: L[g(t = e πs s 2 +.

Solving IVP with step functions in the source. 0, t < π, y 6y = g(t, y(0 = y (0 = 0, g(t = sin(t π, t π. Solution: L[g(t = e πs s 2 +. (s 2 6 L[y = e πs s 2 + H(s = L[y 6 L[y = L[g(t = e πs s 2 +. L[y = e πs (s 2 + (s 2 6. (s 2 + (s 2 6 = (s 2 + (s + 6(s 6 H(s = a (s + 6 + b (s 6 + (cs + d (s 2 +. Solving IVP with step functions in the source. y 6y = g(t, y(0 = y (0 = 0, g(t = Solution: H(s = a (s + 6 + b (s 6 0, t < π, + (cs + d (s 2 +. (s 2 + (s + 6(s 6 = a (s + 6 + b (s 6 sin(t π, t π. + (cs + d (s 2 + = a(s 6(s 2 + + b(s + 6(s 2 + + (cs + d(s 2 6. The solution is: a = 4 6, b = 4 6, c = 0, d = 7.

Solving IVP with step functions in the source. y 6y = g(t, y(0 = y (0 = 0, g(t = Solution: H(s = [ 4 6 (s + 6 + 0, t < π, sin(t π, t π. (s 6 2 6 (s 2 + H(s = [ 4 L [ e 6 t + L [ e 6 t 2 6 L[sin(t 6 [ H(s = L 4 6 ( e 6 t + 6 t e 2 6 sin(t. h(t = [ 4 e 6 t + 6 t e 2 6 sin(t 6. H(s = L[h(t. Solving IVP with step functions in the source. y 6y = g(t, y(0 = y (0 = 0, g(t = Solution: Recall: 0, t < π, sin(t π, t π. L[y = e πs H(s, where H(s = L[h(t, and h(t = [ 4 e 6 t + 6 t e 2 6 sin(t. 6 L[y = e πs L[h(t = L[u π (t h(t π y(t = u π (t h(t π. Equivalently: y(t = u π(t 4 6 [ e 6 (t π + e 6 (t π 2 6 sin(t π.

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback