This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 4 THE LAPLACE TRANSFORM Definition. Let f be a function defined for all t ≥ 0. The Laplace transform of f is the function F ( s ) defined by F ( s ) = L ( f ) = Z ∞ e st f ( t ) dt, (1) provided the improper integral on the right exists. The original function f ( t ) in (1) is called the inverse transform or inverse of F ( s ) and is denoted by L 1 ( F ); i.e., f ( t ) = L 1 ( F ( s )) . Notation: Original functions are denoted by lower case letters and their Laplace transforms by the same 1 letters in capitals. Thus F ( s ) = L ( f ( t )), Y ( s ) = L ( y ( t )) etc. Recall that, by definition, for any function h defined on [0 , ∞ ), Z ∞ h ( t ) dt = lim b →∞ Z b h ( t ) dt and the integral is said to converge if this limit exists. Because e st decreases so rapidly with t, the Laplace transform usually does exist [there are exceptions, however], and then we say that the function f HAS A WELLDEFINED LAPLACE TRANSFORM. Example 1. Let f ( t ) = e at , when t ≥ 0. Find F ( s ). 2 Solution. F ( s ) = L ( e at ) = Z ∞ e st e at dt = lim b →∞ Z b e st e at dt. Now Z b e ( a s ) t dt = ‰ b if s = a e b ( a s ) a s 1 a s if s 6 = a . If s < a , a s > 0 and e ( a s ) b → ∞ as b → ∞ . Thus when s ≤ a , R ∞ e ( a s ) t dt diverges. When s > a , a s < 0, and e ( a s ) b → 0 as b → ∞ , and then F ( s ) = L ( e at ) = 1 s a , s > a. (2) Example 2. Let f ( t ) = 1, t ≥ 0. Find F ( s ). 3 Solution. This function is the same as the one in Example 1 with a = 0, thus, L (1) = 1 s , s > . (3) The Laplace transform is a linear operation; i.e., the Laplace transform of a linear combination of func tions equals the same linear combination of their Laplace transforms. Thus Theorem L ( af ( t ) + bg ( t )) = aL ( f ) + bL ( g ) , (4) where a and b are constants. As a corollary, the inverse Laplace transform also satisfies the linearity property. L 1 ( aF ( s ) + bG ( s )) = aL 1 ( F ) + bL 1 ( g ) . (5) 4 Verification of (4) and (5) is easy. Example 3. Using (4), we obtain L (cosh at ) = L 1 2 ( e at + e at ) ¶ = 1 2 1 s a + 1 s + a ¶ = s s 2 a 2 , s > a ≥ . Example 4. If F ( s ) = 3 s + 5 s 7 , find f ( t ) = L 1 ( F ). Solution. Using (5), we have L 1 ( F ) = L 1 ( 3 s ) + 5 L 1 ( 1 s 7 ) = 3 · 1 + 5 · e 7 t = 3 + 5 e 7 t . 5 Example 5. Set a = iw in the formula (2) L ( e iwt ) = L (cos wt + i sin wt ) = L (cos wt ) + iL (sin wt ) = 1 s iw = s + iw s 2 + w 2 Equating real and imaginary parts, we get L (cos wt ) = s s 2 + w 2 , L (sin wt ) = w s 2 + w 2 . (6,7) Example 6. To show that L ( t n ) = n ! s n +1 , n = 0 , 1 , 2 ,... . Solution. L ( t n ) = Z ∞ e st t n dt = 1 s e st t n  ∞ + n s Z ∞ e st t n 1 dt....
View
Full
Document
 Spring '08
 McNines

Click to edit the document details