Lecture (1). Laplace Transform. LAPLACE TRANSFORM (L.T.) Definition ofthe Laplace Transform: Suppose that f (t) is a piecewise continuous function. TheLapIace tmnsfonn of f (t) denoted by L defined as: L = i 0 whet? L is called the Laplace transfonn operator. From (1) it is clear that L is a function ofs, say F(s) and is found by the methods of ordinalY integrations and is denoted by (2).
Existence Conditions: For f (t) to have a Laplace transfonn, the followingtwo conditions should be satisfied: 1- Piecewise continuous(P.W.C.): A function f (t) is called piecewise continuous onan intetval O g , if the intelval can be subdivided into a finite number ofintercals on each of which f (t) is bounded and continuous. Here is a sketch of a piecewise continuous ftnction. 2- Exponential order: A function f (t) is of exponential order oc if there exists some number oc such that lim f (tk¯d = M, where cc, M are finite positive values..
Some Useful Theorems: Theorem 1: linearitv property: From the definition we can prove that: U{clfl (t) ± = clL{fl ± C2L{f2 where L{fi Fl(s) and F2(s) Example(l) From the find the Laplace transfonn of: (i)ce-at , and (ii)ct. wherea andcare constants. Solution (i) L {ce - -at Special cases: 1- Ata=o -4 Vlt=cf e dt=—c 2- At c= e (G7)0 - 1 - (7Z7) 1 3- Atc=l a=-b * L 1 4- At s.
—st (ii) L= = ctg— —€fe-stdt Similarly we can prove that L t —st also L t and so on Example (2): Find (i) and Solution @Using the definition of Using theorem (I) and the result of example (I ) ¯ 12 ¯ L— —L at e -af Similarly we can prove that: s.
Using the definition of sin(at) : sin(at)= [ Using theorem (1) and the result of example (1) 1 iat . L = 2.
The following table summarizes the Laplace Transforms of Some Elementary Functions Laplace Transforms of Some Elementary Functions: c = constant 7 —at cos(at) sin (at) cosh(at) sinh(at) s.
Example (3) Find the Laplace transforms of the given functions. (a) f(t)=6e +e3t +5t3 —9 (b) g(t)=4cos(4t)—9sin(4t)+2cos(IOt) Solution We'll do these examples in a little more detail than is typically used since this is the first time we're using the table. (a) f (t) = 6e + 5t3 _ 9 1 1 3! s 9 s 6 1 30 9 (b) g (t) = 2cos(IOt) 4 +2 36 2s 4s (c) h(t) 3sinh(2t)+3sin(2t) 2 2 e -(2)2 6 6 s2—4 s2 +4.
Example (4) Find the Laplace transforms ofthe given functions. (a) f(t) — sin2(5t) (b) f(t) = (c) e3t sinh(5t) (d) f(t) = e2t+5 (e) f (t) = sin (4t + 3) Solution 1 (a) — [1 —cos10t] 2 —I 1 (b) f(t) = cos(3t)sin(5t)= — [sin (5— 3)t + sin(5 + 3)tl 2 = —[sin(2t)+ sin(8t)] 1 2 s2 + (8)2.
(c) (d) f(t)— e3t sinh(5t) 1 e 5t —5t 2 2 2 (s-8XS+2) s2 -6s-16 2t+5 =e2t X F (S) = e5L f(t) = sin (4t + 3) = sin cos(4t)sin (3) F(s) = cos (3)L+ sin(3)L 4cos(3)+ s sin (3) 4 = cos (3) + sin(3) s2 + (4)2 s2 + (4)2.
Example (5) t Solution 1 Let g(t) = coss t = costcos2 t = —cost[cos2t+ 1] 2 1 1 1 [cos3t + cost] + —cost — = —costcos2t+ — cost = — — —cos3t + — cost 2 2 4 4 4 (14 cost+-co# 3 1 L s s Theorem 2: First Shifting Propertv: If = F(s) exists for s > oc , then L {etJ(t) - 00 F(s — a) Shift on s-axis.
Example (6) Find the Laplace transform ofthe following functions: (a) f(t)= est sinh (5t) (b) e3t + cos(6t)— e3f c046t) 5 Solution (a) Since s2 — 25 Applying the first shift property, using ct=3 then 5 L{e3f sinh (s -3)2-25 1 s (b) G(s) - Example (7) Find 5 sin3t)J Solution 3 s Since cos3t + 5sin 2 Applying the first shift property, using -1, then 2s +15 +1)+15 LI t(2cos3t+ 5sin (s +1)2+9 s + 2S+10.
Example (8) Find L cosh3tsin5t Solution f (t) = e2t cosh3tsin 5t = e 5 Since L{sin s +25 1 5 Then sin 5t = —e sin 5t + 2 sin 5t 2 1 5 2 2 — + 25 2 ( s +1)2+25 Theorem 3: Multiplication bv t: If F(s) exists for s > , then -dF(s)! Generally; dsn.
Example (9)Find L Solution b Since L{sin Applying theorem 5 , Example (I O) Find Solution First method 16 Since L —d Then b ds s —6 and L e ds s Hence L 3t + 4t = 3L ds (s — 2bs 1 (s -6)3 3 8.
Second method Since L t + 4t Then L 3t+4t (s-6)3 Example (11) Evaluate L? Solution Let Then L 2sin (-)2 (-)2— ds ds 12s -16 . L sin 2t —.