Lecture 4: Laplace Transforms

Transcription

1 Lur 4: Lapla Transforms Lapla and rlad ransformaions an b usd o solv diffrnial quaion and o rdu priodi nois in signals and imags. Basially, hy onvr h drivaiv opraions ino mulipliaion, diffrnial quaions ino algbrai quaions, and priodi funions ar assoiad wih hir frqunis. This lar propry is vry usful in imag nhanmn whr on wans o disard or filr unwand priodi omponns. For xampl, in h sixh lur h fas Fourir will b usd o filr h following fuzzy sin funion. 2 1 Fuzzy Sin Funion Fas Fourir Transform of Fuzzy Sin Funion: Idnifis Frqunis Th Lapla ransform is radiionally inrodud baus i is usful in solving iniial valu problms for ordinary diffrnial quaions. Th pross an b oulind in h following hr sps: us h Lapla ransform o onvr h iniial valu problm (ofn assoiad wih h im domain) o an algbrai problm (ofn assoiad wih frquny domain), solv h algbrai problm, and onvr bak o h im domain for h soluion of h iniial valu problm.

2 Dfiniion. Th Lapla ransform of f() is a funion of a paramr s >, F(s), s L( f) = F( s) f( ) d. Baus h impropr ingral is a linar opraion, L(f) also linar, ha is, L(f) = L(f) and L(f + g) = L(f) + L(g). Anohr asy propry o show is L(1) = 1/s: s L(1) = 1d = lim s d = = + 1/ s. s s lim /( s) /( s) Addiional propris ar summarizd as follows: Basi Propris of Lapla Transforms. Assum all funions ar suh ha h Lapla ransforms xis. 1. L(f) = L(f), 2. L(f + g) = L(f) + L(g), 3. L( n ) = n!/s n+1, 4. L(f ) = sl(f) f(), 5. L( a ) = 1/(s-a), 6. L(sin(b)) = b/(s 2 + b 2 ), 7. L(os(b)) = s/(s 2 + b 2 ) and 8. L(f*g) = L(f) L(g) whr ( f * g)( ) f( τ ) g( τ) dτ. Propry 4 is fundamnal as i onvrs h opraion of drivaiv ino a mulipliaiv opraion wih h paramr s. Th proof uss ingraion by pars wih v = -s and du = f ()d:

3 s L( f ') = lim f '( ) d s s s = lim f( ) f() ( s) f( ) d = f() + sl( f). Th proofs of propris 5-7 ar srai forward. Th las propry rfrs o h onvoluion of wo funions and rquirs a lil mor ffor. Th onvoluion propry and is analogus ar xrmly imporan. Appliaion o h soluion of iniial valu problms. Considr h u = (7 u) wih u() = 2. Firs, ompu h Lapla ransform of boh sids of h diffrnial quaion: Lu ( ') = L ( (7 u)) sl( u) u() = 7 L(1) L( u) sl( u) 2 = 7 / s L( u). Sond, solv h algbrai quaion for L(u): ( s+ ) L( u) = 7 / s Lu ( ) = ss ( + ) s = s s+ s+ = 7 L(1) 7 L( ) + 2 L( ) = L( ). Third, solv for u: u ( ) = Th onvoluion propry an b usd for mor ompliad problms. For xampl, onsidr h iniial valu problm

4 u + u = f() wih u() = givn. Apply h Lapla ransform o h diffrnial quaion and us h basi ruls o obain Lu ( ' + u) = L( f) sl( u) u() + L( u) = L( f ) ( s+ ) L( u) u() = L( f) 1 1 Lu ( ) = L( f) + u() s+ s+ = L( f) L( ) + u() L( ). ow apply h onvoluion propry wih g() = - o onlud u () = ( f* g)() + u() = f ( τ) g ( τ) dτ + u() ( τ ) ( τ) τ () 1 [ ( ) τ = f τ dτ + u ()]. = f d + u Malab has a symboli oolbox, whih has ommands for h Lapla ransform, lapla(), and is invrs, ilapla(). Th following india how asy i is o us hs. >> lapla() 1/s^2 >> lapla(sin(b*)) b/(s^2+b^2) >> lapla(xp(3*)*sin(b*))

5 b/((s-3)^2+b^2) >> lapla(xp(3*)*sin(1*)) 1/((s-3)^2+1) >> ilapla(ans) -1/4*(-4)^(1/2)*(xp((3+1/2*(-4)^(1/2))*) -xp((3-1/2*(-4)^(1/2))*)) >> simplify(ans) 1/2*i*(-xp((3+1*i)*)+xp((3-1*i)*)) Us Eulr's Formula o g xp(3*)*sin(1*) >> dsolv('dx=24 -(3/5)*x','x()=3','') 4-37*xp(-3/5*) This an also b solvd by h Lapla ransform, and h hird sp is >> ilapla(24/(s*(s+3/5)) + 3/(s+3/5)) 4-37*xp(-3/5*). Th Lapla ransform an idnify frqunis of priodi funions. This is indiad by h Lapla ransform of h sin and osin funions whr b = 2πf and f is a spifi frquny. An audio signal may hav a numbr of frqunis as wll as som frqunis assoiad wih bakground and quipmn nois. On an us ransform mhods o idnify and filr hs unwand pars of h signal. In ordr o dvlop ffiin mhods o do his, w will us a variaion of h Lapla ransform whr h paramr s is rplad by iw, i is h squar roo of -1 and w is a ral numbr.