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Lesson_27

# Lesson_27 - Lesson27 Challenge26 systems Challenge27 Lesson...

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Lesson 27 Lesson 27 Challenge 26 Lesson 27  -  Laplace Transform and continuous-time signals and  systems Challenge 27

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Lesson 27 Challenge 26 ) ( = ) ( + ) ( 0 0 t v dt t dv RC t v ) ( 1 = ) ( / - t u e RC t h RC t RC=1 y(t=0.5)=0.4 ) ( ) - 1 ( = ) ( - t u e t y t
Lesson 27 Lesson 27 Again, convolution is the issue.   How do you normally perform this  convolution operation? Response:  Laplace transform ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 τ τ τ τ τ τ d h t x d t h x t x t h t y - - - = - = =

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Lesson 27 Lesson 27 Laplace assumed that the solution to an ODE was: and postulated a mapping: What would happen if x(t)=e st ? ( 29 = = n i t i s i e a t x 1 ( 29 ( 29 - - = st e t x s X
Lesson 27 Lesson 27 Example x(t)=e σ 0 t  u(t).   ( 29 ( 29 0 0 ) 0 ( 0 0 1 σ σ σ - = = = - - s dt e dt e e s X t s st t for Re(s)> σ 0 What happens when s= σ 0 ?

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Lesson 27 Lesson 27 Table 1: Common Laplace Transforms x(t) Laplace Transform δ (t) 1 u(t) 1/s tu(t) 1/s 2 t n u(t) n!/s n+1 e at u(t) 1/(s-a) t e at u(t) 1/(s-a) 2 t n e at u(t) n!/(s-a) n+1 sin( ϖ 0 t) u(t) ϖ 0 /(s 2 + ϖ 0 2 ) cos( ϖ 0 t) u(t) s/(s 2 + ϖ 0 2 ) e at sin( ϖ 0 t) u(t) ϖ 0 /((s-a) 2 + ϖ 0 2 ) e at cos( ϖ 0 t) u(t) (s-a)/((s-a) 2 + ϖ 0 2 )
Lesson 27 Lesson 27 Table 1: Laplace Properties x(t) Laplace Transform x(t) X(s) y(t) Y(s) ax(t) +by(t) aX(s) +bY(s) x(t-t 0 ) e -st o X(s) x(at) (1/a)X(s/a) e at x(t) X(s-a) t n x(t) (-1) n d n X(s)/ds n x(0) lim s →∞ sX(s) if x(t) is causal and has no impulses orhigher order singularites x( ) lim s 0 sX(s) if all the poles have a negative real part with the exception being s=0.

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Lesson_27 - Lesson27 Challenge26 systems Challenge27 Lesson...

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