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Unformatted text preview: ECE 220, Section 051 Problem Lab Number 7 Solution Chapter 8, Laplace Transforms Part I. Fundamental Properties of Laplace Transforms 1. Use the basic definition of the Laplace transform to evaluate: f ( t ) = e a ( t t 1 ) u ( t t 1 ) Check using Laplace transform properties. Solution:(Using Basic Definition) F ( s ) = ∞ f ( t ) e st dt = ∞ e a ( t t 1 ) u ( t t 1 ) e st dt ⇒ F ( s ) = e at 1 ∞ t 1 e t ( a + s ) dt = e at 1 ( s + a ) e t ( s + a ) ∞ t 1 ⇒ F ( s ) = e at 1 ( s + a ) e t 1 ( s + a ) = e at 1 st 1 at 1 s + a = e st 1 s + a Solution:(Using Laplace Transform Properties) f ( t ) = e a ( t t 1 ) u ( t t 1 ) = g ( t t 1 ) Now, g ( t ) = e at u ( t ) So, G ( s ) = 1 s + a As f ( t ) = g ( t t 1 ), F ( s ) = e st 1 s + a 2. Use the basic definition of the Laplace transform to show that: L [ f ( t a ) u ( t a )] = F ( s ) e as where F ( s ) = L [ f ( t )] Check using Laplace transform properties. Solution:(Using Basic Definition) H ( s ) = ∞ h ( t ) e st dt = ∞ f ( t a ) u ( t a ) e st dt ⇒ H ( s ) = ∞ a f ( t a ) e st e at e at dt = ∞ a f ( t a ) e st e as e as dt ⇒ H ( s ) = e as ∞ a f ( t a ) e s ( t a ) dt = e as ∞ f ( t ) e s ( t ) dt = e as F ( s ) 1 Solution:(Using Laplace Transform Properties) h ( t ) = f ( t a ) u ( t a ) = g ( t a ) Now, g ( t ) = f ( t ) u ( t ) So, H ( s ) = e as G ( s ) H ( s ) = e as F ( s ) 3. Use the basic definition of the Laplace transform to show that: L [ e at f ( t )] = F ( s + a ) where F ( s ) = L [ f ( t )] Solution:(Using Basic Definition) H ( s ) = ∞ h ( t ) e st dt = ∞ e at f ( t ) e st dt ⇒ H ( s ) = ∞ a f ( t ) e t ( s + a ) dt = ∞ f ( t ) e t ( s + a ) dt ⇒ H ( s ) = F ( s + a ) 4. Use the results from the above problems to evaluate the Laplace transform of: (a) e at cos( ωt ) Solution: s a ( s a ) 2 + ω 2 (b) e at sin( ωt ) Solution: ω ( s a ) 2 + ω 2 using the known transform pairs...
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 Summer '08
 NILSON
 Laplace Transform Properties, est dt

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