M. J. Robers - 2/18/07 Web Appenix N - Derivaions of he Properies of he aplacetransform N.1 ineariy e z= x+ y where an are consans. Then = x+ y Zs an he lineariy propery is N.2 Time Shifing es = xe s + y 0 0 x+ y e z= g( 0 ), 0 0. Then Z s e = 0 =. Then = ze s 0 0 X+ s Y s e s. = g( 0 )e s. 0 = X+ s Y s = g Z s e s ( 0 ) = e s 0 g( )e s 0 0 = e s 0 G. s N.3 Complex-Frequency Shifing e s 0 be a complex consan. Then e s 0 g e s 0 g 0 e s 0 ge s ge ( ss 0 ) = G s s 0 0 N-1
M. J. Robers - 2/18/07 The complex-frequency-shifing propery of he aplace ransform is e s 0 g G s s 0 (N.1) N.4 Time Scaling e a be any posiive real consan. Then he aplace ransform of g( a) is e = a an = a g a 0 g( a)e s g a g( )e s / a = 1 a a g ( )e s / a = 1 a G s a, a > 0 0 g a 0 ( 1/ a)g s / a, a > 0 (N.2) N.5 Frequency Scaling e a be any posiive real consan. Then, using he ime-scaling propery of he aplace ransform g( a) ( 1/ a)g ( s / a), a > 0. e b = 1/ a. Then g( / b) bg ( bs), b > 0 or ( 1/b)g( / b) G ( bs), b > 0 an he frequencyscaling propery of he aplace ransform is ( 1/ a)g / a G ( as), a > 0 (N.3) N.6 Firs Time Derivaive The aplace ransform is efine by Evaluae he inegral by pars using Then = ge s G s. 0 uv = uv vu an le u = g an v = e s. N-2
M. J. Robers - 2/18/07 an 0 ge s u = ) an v = 1 s es = g 1 s es 0 + 1 s 0 )e s G = s 1 ( s g0 )+ 1 s 0 )e s (where i is unersoo ha Re= s is chosen o make G s exis). Then ) = 0 )e s = sg s g0 an he firs-ime-erivaive propery of he aplace ransform is ) sg s g0. (N.4) N.7 Nh Time Derivaive This propery can be proven using he previous propery for he firs ime erivaive an applying i o a firs ime erivaive o form a secon ime erivaive an hen generalizing he resul o he Nh ime erivaive. The secon ime erivaive of a funcion g is Therefore, using we ge 2 2 2 2 ) )= ) = s ) sg s g0 ) ) =0 2 2 ) = s { sg s g0 } ) =0 = s 2 G s sg 0 ) =0 The secon-ime-erivaive propery of he aplace ransform is N-3
M. J. Robers - 2/18/07 2 ) 2 s 2 G s sg 0 ). =0 Afer seeing he erivaion of his propery from he previous iffereniaion propery we can inucively generalize o he Nh erivaive. N g ( ) s N G s N N n1 s N n n=1 n1 ) =0 (N.5) N.8 Complex-Frequency Differeniaion Sar wih he efiniion of he aplace ransform Differeniaing wih respec o s = ge s G s. 0 ( s G s )= ge s s = ( s g e s ) = g 0 0 0 e s = ( g ) g ( s G s ) (N.6) N.9 Muliplicaion-Convoluion Dualiy The convoluion of g wih h is Since g is zero for ime < 0, g h g h h( ). = g h( ). = g From he efiniion of he aplace ransform, 0 N-4
M. J. Robers - 2/18/07 g g h h = 0 0 g ( )h( ) e s = g ( ) e s h( ). 0 0 Since h is zero for ime < 0, g h = 0 g e s h e = an =. Then g g g h h h = = 0 0 = H s g e 0 h s + es g( ) e s h 0 H s e s g( ) 0 =G s = G H s s The ime-omain convoluion propery of he aplace ransform is g h G H s s (N.7) The aplace ransform of a prouc of ime-omain funcions is g g h h = = 1 j2 0 0 + j j g h e s G ( w)e w w he s where is chosen o make G s an H s exis. Doing he inegraion firs, N-5
M. J. Robers - 2/18/07 If H s exiss hen an g h g = 1 j2 0 h + j j h G w s w he = 1 j2 + j 0 = H( s w) s w e ( ) w. G ( w)h( s w) w. j Therefore gh 1 j2 + j G ( w)h( s w) w (N.8) j N.10 Inegraion The inegraion propery is easy o prove, using he convoluion propery jus proven in he las secion an he fac ha g u u( ) = g( ) = g 0 Therefore g u 0 G U s = s G / s s g( ) 1 s G. s (N.9) N.11 Iniial Value Theorem Sar wih he firs-ime-erivaive propery of he aplace ransform e s. Then s 0 ) = 0 )e s )e s = s = sg s g0 ( ). sg s g0 N-6
M. J. Robers - 2/18/07 Case I. g is coninuous a = 0 0 )e s = s s sg s g0 If he aplace ransform of g, which is G, s exiss for Re= s > 0, he quaniy )e s approaches zero as s approaches infiniy an 0 = s sg s g0 g0 = an, since g is coninuous a = 0, g0 ( )= g0 + an Case II. g is isconinuous a = 0 g0 ( + )= sg. s s s. sg s a = 0 means ha he erivaive of g In his case, he isconinuiy of g has an impulse a = 0 an he srengh of he impulse is g0 ( + ) g0 ( ). Now he inegral )e s s becomes 0 s 0 0 + )e s = g0 ( + ) g0 s e s + s )e s. 0 0 + an, using he sampling propery of he impulse in he firs inegral, =0 Therefore, or s 0 )e s = s g0 ( + ) g0 ( )= s g0 ( + ) g0 sg s g0 = = ( g0+ ) g0 s sg s g0 g0 ( + )= sg s (N.10) s an he resul is he same as in Case I. N-7
M. J. Robers - 2/18/07 N.12 Final Value Theorem From he firs-ime-erivaive propery of he aplace ransform, 0 s0 0 s0 )e s )e s 0 ) g g0 = s0 = s0 = s0 = s0 sg s g0 sg s g0 sg s g0 sg s g0 Then, if he i g exiss, he final value heorem of he aplace ransform is g = s0 sg. s (N.11) N-8