EEM 3117 Laplace Transformation Dr. Sezai Taskin Department of - PowerPoint PPT Presentation

Automatic Control EEM 3117 Laplace Transformation Dr. Sezai Taskin Department of Electrical&Electronics Engineering Faculty of Engineering, Manisa Celal Bayar University

Laplace Transformation The Laplace transform of a function f ( t ), defined for all real numbers t ≥ 0, is the function F ( s ), defined by: L      st L [ ( )] f t F s ( ) f t e ( ) dt 0 F(s) Slides 2 2

Why s-domain? • We can transform an ordinary differential equation into an algebraic equation which is easy to solve. • It is also easy to analyze and design interconnected (series, feedback etc.) systems. Slides 2 3

• Unit step function     L st [1] e dt 0     st e       s   0 1 t 0    ( ) ( ) f t u t 0 1     0 t 0   s s 1 1   L [1] s s Slides 2 4

Exponential function  at f t ( ) e          L at at st ( s a t ) [ e ] e e dt e dt 0 0      ( s a t ) e        ( s a ) 0 0 1       ( s a ) ( s a ) 1 1   L at [ e ]   s a s a Slides 2 5

Frequency shift 1 1  L   L at [1] [ e 1]  s s a   L at [ ( )] ( ) e f t F s a Slides 2 6

Sine and cosine functions  s     L L [sin t ] [cos t ]     2 2 2 2 s s Slides 2 7

Impulse function   f t ( )          st st L [ ( )] t ( ) t e dt e 1   0 t 0 Slides 2 8

Unit ramp   t t 0    ( ) ( ) f t u t   0 t 0     (integration by parts) udv uv vdu     L st [ ] t e t dt similarly 0         st st st e t e e 1   n !        t d dt L n [ ] t    2   s  s s s s n 1 0 0 0 Slides 2 9

Differentiated function  L [ ( )] ( ) f t F s    ( ) df t  L ( ) (0) sF s f     dt    n d f t ( )   n n 1 L   s F s ( ) s f (0) n   dt      n 2 ( n 1) (0)..... (0) s f f Slides 2 10

Integrated function    F s ( )  t L f t dt ( )     s 0 2 nd shifting theorem      L as f t ( a ) e F s ( ) Slides 2 11

Solution of differential equations using Laplace Transformation Transformed Differential Laplace Transformation Equation Equation Algebraic Particular Complementary Manipulation Integral function Transformed Solution Solution Inverse Laplace Slides 2 12

Example dx   2 x 5 dt 5    sx s ( ) x (0) 2 ( ) x s Transformed Equation s 5  x s ( ) Transformed Solution  s s ( 2)   5  t t      2 t 2 t x t ( ) 5 e dt e   2 0 0    2 t 2.5(1 e ) Slides 2 13

Convolution integral     F s ( ) F s F s ( ). ( ) f t ( ) f t ( ) f t ( ) 1 2 1 2 Ex: 1 1 1    F s ( )     2 s 3 s 2 ( s 1) ( s 2)      t 2 t f t ( ) e e  t      f ( ) f t ( ) d 1 2 0 t t              2( t ) 2 t e e d e e d 0 0        2 t t t 2 t e ( e 1) e e Slides 2 14