Tiêu đề Laplace 1.pdf ✅

At the end of the session you will: 1. Define the Laplace transform. 2. Understand the linearity and first shifting properties of Laplace Transform. Laplace Transform Integral transform ODE Heaviside function Linearity Subsidiary Equation IVP First shifting LAPLACE TRANSFORMS LEARNING OBJECTIVES: KEY TERMS CORE CONTENT
DEFINITION. Let f(t) be a given function which is defined for all positive values of t. We multiply f(t) by �!"# and integrate with respect to t from zero to infinity. Then, if the resulting integral exists, it is a function of s, say F(s); where s is a parameter which may be real or complex. F(s)= ∫ � $ !"# ∘ f(t) dt The function F(s) is called the LAPLACE TRANSFORM of the original function f(t), and will be denoted by L(f). The symbol L, which transforms f(t) in F(s) is called the Laplace transformation operator. Thus, F(s)= L{f(t)}= ∫ � $ !"# ∘ f(t) dt The described operation of f(t) is called the Laplace transformation. Also, the Laplace transform F(s) is said to exists if the integral converge for some values of s; otherwise, it does not exist. Furthermore, the original function f(t) is called the INVERSE TRANSFORM or INVERSE of F(s) and will be denoted by L -1(F); the is, f(t)= L -1(F). Th original function shall be denoted by a lower case letter and transform by the same letter in capital. PROBLEM 1: [Derivation] Find the Laplace transform of f(t) = 1 when t > 0. SOLUTION: L {f(t)} = ∫ � $ !"# ∘ f(t) dt Hence, L {1} = ∫ � $ !"# ∘ (1) dt L {1} = ∫ � $ !"# ∘ dt Let u = −st du = −s dt Then, L {1} = − & " ∫ � $ ' ∘ du L {1} =− & " [�']∘ $ L {1} =− & " [�!"#]∘ $
L {1} =− & " ( & (!") ∘ $ L {1} = − & " ( & $ − & (#) L {1} = − & " (0 − 1) Therefore, L {1} = � � PROBLEM 2: [Derivation] Find the Laplace transform of f(t) =�+# SOLUTION: L {f(t)} = ∫ � $ !"# ∘ f(t) dt Hence, L {�+#} = ∫ � $ !"# ∘ (�+#) dt L{�+#} = ∫ � $ +#!"# ∘ dt L{�+#} = ∫ � $ !("!+)# ∘ dt let u = −(� − �)� du =−(� − �) dt So that, L {�+#} = − & ("!+) ∫ � $ ' ∘ du L {�+#} = − & ("!+) [�']∘ $ L {�+#} = − & ("!+) 3�!("!+)#4 ∘ $ L {�+#} = − & ("!+) ( & ((!%&)") ∘ $ L {�+#} = − & ("!+) ( & $ − & & ) L {�+#} = − & ("!+) (0 − 1) Then, L {�+#} = & ("!+) ; � − � > 0
PROBLEM 3: [Derivation] Find the Laplace transform of f(t) = sin bt SOLUTION: L {sin bt} = ∫ � $ !"# ∘ (sin bt) dt Using integration by parts, let u = �!"# dV = sin bt dt du = −��!"# dt V = − & . cos bt L{sin ��} = (− & . �!"# cos � �) ∘ $ − " . ∫ �!"#(cos ��) $ ∘ dt Using integration by parts again, let u = �!"# dV = cos bt dt du = −��!"# dt V = & . sin bt Hence, L {sin bt} = (− & . �!"# cos ��) ∘ $ − " . ( (%!" /01 .# . + " . ∫ �!"# sin �� $ ∘ �� = (− & . �!"# cos ��) ∘ $ − ( "(%!" /01 .# .( ) ∘ $ − "( .( ∫ �!"# sin �� $ ∘ �� ;∫ � $ !"# ∘ (sin ��) �� + "( .( ∫ � $ !"# ∘ sin �� �� = ( !.(%!" 23/ .#!"(%!" /01 .# .( ) ∘ $ (�4 + �4) ∫ � $ !"# ∘ (sin ��) �� = [−��!"# cos �� − ��!"# sin ��]∘ $ (�4 + �4) ∫ � $ !"# ∘ (sin ��) �� = − ( !. 23/ .# !" /01 .# (!" ) ∘ $ (�4 + �4) ∫ � $ !"# ∘ (sin ��) �� = − ( . 23/ .# 6 " /01 .# (!" ) ∘ $ (�4 + �4) ∫ � $ !"# ∘ (sin ��) �� = − ( . 23/ $ 6 " /01 $ () − . 23/ 7 6 " /01 7 (# ) (�4 + �4) ∫ � $ !"# ∘ (sin ��) �� = − (− . 23/ 76 " /01 7 (# ) (�4 + �4) ∫ � $ !"# ∘ (sin ��) �� = ? .67 & @ (�4 + �4) ∫ � $ !"# ∘ (sin ��) �� = �