Laplace Transforms. Examples. Is this equation differential? y 2 2y + 1 = 0, y 2 2y + 1 = 0, (y ) 2 2y + 1 = cos x,
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1 Laplace Transforms Definiion. An ordinary differenial equaion is an equaion ha conains one or several derivaives of an unknown funcion which we call y and which we wan o deermine from he equaion. The equaion may conain y iself as well as given funcions and consans. Examples. Is his equaion differenial? y 2 2y + 1 =, y 2 2y + 1 =, (y ) 2 2y + 1 = cos x, (y ) 2 2y + 1 = cos y, yy 2y + 1 = xe y, y y + x2 y = ˆ x e y(u)2 du, y y + x2 y = ˆ x e u2 du 1
2 1 Laplace Transform. Inverse Transform The Laplace Transform is used o reduce a differenial equaion o an algebraic one. Definiion. Le f() be a given funcion defined for all F (s) = e s f() d is he Laplace ransform of f(). I is denoed by L (f) L (f) = F (s) The operaion is also called he Laplace ransform. The original funcion f() is called he inverse ransform or he inverse of F (s) and is denoed by L 1 (F ) f() = L 1 (F ) Example. f() = 1, L (f) = L (1) = 1 s, f() = e a, L (e a ) = 1 s a, s > Re(a) 2
3 Some Properies of Laplace Transform 1. Lineariy Example. L (cosh a) = L (cos ω) = 2. Firs Shifing Theorem L (af + bg) = a L (f) + b L (g) If f() has he ransform F (s), s > k, hen e a f() has F (s a), s a > k s s 2 a, L (sinh a) = a 2 s 2 a, 2 s s 2 + ω, L (sin ω) = ω 2 s 2 + ω 2 L (f()) = F (s), s > k, L (e a f()) = F (s a), s > k + a, Proof. F (s a) = e a f() = L 1 (F (s a)) e (s a) f() d = e s e a f() d = L (e a f()) Example. Damped vibraions L (e a s a cos ω) = (s a) 2 + ω, s > a, 2 L (e a sin ω) = ω (s a) 2 + ω 2, s > a 3
4 Transform of Derivaives 1. Laplace ransform of he firs derivaive of f Proof. L (f ) = L (f ) = s L (f) f() e s f () d = e s d(f()) = e s f() + s e s f() d = s L (f) f() 2. Laplace ransform of he second derivaive of f L (f ) = s 2 L (f) sf() f () Proof. f = (f ), hus L (f ) = s L (f ) f () = s 2 L (f) sf() f () 3. Laplace ransform of he hird derivaive of f L (f ) = s 3 L (f) s 2 f() sf () f () Proof. f = (f ), hus L (f ) = s L (f ) f () = s 3 L (f) s 2 f() sf () f () 4
5 Example. L ( 2 ) =? L (( 2 ) ) = L (2) = 2/s. Thus, L ( 2 ) = 2/s 3 Example. f() = cos ω, L (f) =? f = ω sin ω, f = ω 2 f Example. f() = y + ay + by L (f) = (s 2 + as + b)l (y) (s + a)y() y () Iniial Value Problem y + ay + by = r(), y() = k, y () = k 1, a, b are cons Here r() is he inpu (driving force) applied o he mechanical sysem, and y() is he oupu (response of he sysem). In Laplace s mehod we do 3 seps 1. Noaion: Y = L (y), R = L (r) Transform he equaion (s 2 + as + b)y (s + a)y() y () = R This is he subsidiary equaion. 2. Solve i for Y Y (s) = [(s + a)y() + y ()] Q(s) + R(s)Q(s) where 1 Q(s) = s 2 + as + b is called he ransfer funcion. In he simples case y() = y () = Y (s) = R(s)Q(s) Q = Y R = L (oupu) L (inpu) 5
6 3. Do he inverse Laplace ransform y() = L 1 (Y ) Example. y y =, y() = 1, y () = 1 Example. y + 2y + y = e, y() = 1, y () = 1 2 Uni Sep Funcion u() = { for < 1 for > A = i is undefined or can be defined o be, 1 or 1/2. The uni sep funcion is also called he Heaviside funcion and denoed by θ(). I can be hough of as an idealisaion of a smooh funcion which changes very quickly from o 1. For example ) 1 u() u ϵ () 1 u ɛ () = π an 1 ( ɛ Example. Wrie he funcion in a piece-wise form, and plo i 3 u( + 1) u( 2) + 2u( 3) Noe also ha u() + u( ) = 1 6
7 The uni sep funcion is used o model he effec of a driving force being on or off. If we wan o swich a driving force f() on a some ime = a we can muliply f() by u( a): f() f()u( a) To swich i off a = b we muliply i by u(b ) = 1 u( b): f() f() f()u( b) Thus, o swich on a = a and hen o swich off a = b we muliply i by u( a) u( b): f() f()(u( a) u( b)) f() f()u(-a) f()u(b-) f()(u(-a)-u(-b)) Example. Wrie he funcion in a piece-wise form, and plo i y() = cos 3 u( π) u( π) cos 3 + 2u( 2π) sin 3 Example. Wrie he funcion whose graph is shown in erms of he uni sep funcion Here f() = 4 sin π/2 7
8 Second shifing heorem; -shifing If f() has he ransform F (s), hen he shifed funcion { for < a f() = f( a)u( a) = f( a) for > a has he ransform e as F (s): Proof. L ( f( a)u( a) ) = e as F (s) e as F (s) = e as e sτ f(τ) dτ = = τ + a = = f( a)u( a) = L 1( e as F (s) ) ˆ = L (f( a)u( a)) Examples. f() = a e s f( a) d = e s(τ+a) f(τ) dτ f() = 1, L ( f) =? 2 for < < 2 4 for 2 < < 4 sin π for > 4 e s f( a)u( a) d, L (f) =? F (s) = 2 s 2 2e 2s s 2 4e 2s s + se πs, f() =? s
9 Example. Response of an RC-circui o a single square wave. Find he curren i() in he circui below if a single square wave wih volage V is applied. The circui is assumed o be quiescen before he square wave is applied. C v() v v() R The equaion of he circui is The soluion is i() = R i() + q() C = v() for < < a K 1 e RC for a < < b (K 1 K 2 )e RC for > b a b K 1 = V R e a RC, K2 = V R e b RC 9
10 3 Dirac s dela funcion Ofen we need o describe phenomena of an impulsive naure which involve he acion of very large forces or volages over very shor inervals of ime. We define he impulse of a driving force f() over a ime inerval a a + k o be he inegral of f() from a o a + k. We consider he case of a very shor ime inervals, i.e. very small k and is limi k where he force acs only for an insan bu we wan he impulse o say finie. Consider f k (-a) f k ( a) = { 1 k for a < < a + k for oherwise Is impulse I k is 1 for all k. Le us represen f k ( a) as 1/k Area = 1 f k ( a) = u( a) u( a k) k The limi of f k ( a) as k is denoed by δ( a) δ( a) = lim k f k ( a) = lim k u( a) u( a k) k a a+k = u ( a) and is called he Dirac dela funcion (or he uni impulse funcion) Dirac s dela funcion is no a funcion in he ordinary sense because δ( a) = { for = a for oherwise, bu I is a so-called disribuion or generalised funcion. 1 δ( a) d = 1
11 Neverheless, he Laplace ransform of Dirac s dela funcion is welldefined as is seen from he Laplace ransform of f k ( a) L (f k ( a)) = Taking he limi k, we ge L (u( a)) L (u( a k)) k = e as e (a+k)s ks L (δ( a)) = e as as 1 e ks = e ks Example. Response of a damped vibraing sysem o a single square wave and o a uni impulse y + 3y +2y = r(), y() =, y () = (A) r() = u( 1) u( 2) (B) r() = δ( 1) y
12 4 Convoluion Convoluion Theorem. The produc of he ransforms F (s) = L (f) and G(s) = L (g) is he ransform H(s) = L (h) of he convoluion h() of f() and g() which is denoed by (f g)() and defined by h() = (f g)() = ˆ f(τ)g( τ) dτ Example. H(s) = 1/(s 2 + 1) 2 Properies of he convoluion 1. Commuaive law: f g = g f 2. Disribuive law: f (g 1 + g 2 ) = f g 1 + f g 2 3. Associaive law: (f g) v = f (g v) 4. f = f = 5. In general, f 1 f 12
13 Differenial equaions and he convoluion y + ay + by = r(), y() = k, y () = k 1, a, b are cons 1 Inroduce Y = L (y), R = L (r), Q(s) =, and ge s 2 +as+b Y (s) = [(s + a)y() + y ()] Q(s) + R(s)Q(s) If y() = y () = Y (s) = R(s)Q(s) y() = ˆ q( τ)r(τ) dτ I is he inegral represenaion of he soluion. Example. Response of a damped vibraing sysem o a single square wave y + 3y +2y = u( 1) u( 2), y() =, y () = Inegral equaions and he convoluion Example y() = + ˆ sin( τ)y(τ) dτ 13
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