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lecture 5

# lecture 5 - Summary of Lecture#5 p 584 Table 12.2 Basic...

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1 1 Summary of Lecture #5 9/01/2010 Basic properties of Laplace transform Physical meaning and Examples Use Laplace transform to solve differential equations , an example Use Laplace transform to solve circuit problems, two examples Partial fraction by MATLAB 2 p. 584 Table 12.2 3 Properties of Laplace transform F(s) = L (f(t)u(t)), f(t)u(t) = L –1 (F(s)) Linearity L [a 1 f 1 (t)u(t)+a 2 f 2 (t)u(t)] = a 1 F 1 (s)+a 2 F 2 (s) Time shift (right shift only) L [f(t - T) u(t - T)] = e - sT F (s) , T > 0 Multiplication by t : Multiplication by t n : dF(s) [ f (t)u(t) d t ] s = - L n n n n d F(s) [ f(t)u(t)] ( 1) s t d = - L 4 Time shift 0 1 2 3 t in s g(t) 4 g(t) = f(t - 1) 0 1 2 3 f(t) t in s Original function Time shifted Function (right shift) 5 Question: Use the properties of Laplace transform to find L (f(t)). (a) Given g 1 (t) = 2e - 4(t - 3) u(t - 5), find G 1 (s). (b) Given g 2 (t) = 2e - 4(t - 3) u(t + 5), find G 2 (s) . It looks simple. But be very careful. 6 Frequency shift L [e - at f (t)] = F (s + a) 1st- and 2nd- order time differentiation Higher-order time differentiation See Eq. 12.43 and 44 (p. 578) - = - - - - = - = 0 t 2 2 2 dt ) t ( df ) 0 ( sf ) s ( F s ] dt f d [ ) 0 ( f ) s ( sF ] dt df [ L L initial values initial derivative

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2 0 1 2 3 ϖ f( ϖ ) Original spectrum 0 1 2 3 ϖ g( ϖ ) 4 Frequency shift g( ϖ ) = f( ϖ- 1) 8
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lecture 5 - Summary of Lecture#5 p 584 Table 12.2 Basic...

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