The complex Fourier series has an important limiting form when the period approaches infinity, i.e., T 0. 0 since it is proportional to 1/L, but

Transcription

1 Fourier Transforms The complex Fourier series has an imporan limiing form when he period approaches infiniy, i.e., T or L. Suppose ha in his limi () k = nπ L remains large (ranging from o ) and (2) c n since i is proporional o /L, bu hen we have g(k) = lim L f (x) = c n n= L π c = n c n e ikx = lim L c n f (x)e ikx dx = finie n= π L g(k)eikx where k = nπ L. The sum over n is in seps of Δn =. Thus, we can wrie using Δk = π L Δn which becomes infiniesimally small when L becomes large, as a sum over k, which becomes an inegral in he limi πδn f (x) = lim L c n L g(k)eikx = lim Δk Δkg(k)e ikx n= k = = g(k)e ikx dk We call g(k) he Fourier Transform of f(x) g(k) = f (x)e ikx dx = F( f ) and he las equaion is he so-called Fourier inversion formula. We can now obain an inegral represenaion of he delafuncion. This corresponds o he orhogonaliy condiion for he complex exponenial Fourier series. We subsiue he definiion of g(k) ino he inversion formula o ge Page

2 where Properies f (x) = g(k)e ikx dk = dke ikx = dx' f (x') δ(x x') = dke ik(x x ') dke ik(x x ') = dx' e ikx ' f (x') dx' f (x') δ(x x') The evaluaion of he inegrals involved in many Fourier ransforms involves complex inegraion, which we shall learn laer. We will jus sae some properies Examples: The Fourier ransform of he box funcion is f (x) = x a x a F( f ) = dx' e ikx ' f (x') = dx' e ikx ' = α α e ikx ' α ik α = 2sin kα k The Fourier ransform of he derivaive of a funcion is F df dx = df (x') dx' e ikx ' = dx' f (x')e ikx ' ( ik) dx' e ikx ' f (x') = ik dx' e ikx ' f (x') = ikf( f ) where we have assumed ha f (x) as x. This generalizes o F d n f dx n = (ik)n F( f ) Oher useful properies of he Fourier ransform are: F( f (x)) = g(k), F( f (x a)) = e ika g(k), F( f (x)e ax ) = g(k + ia) A shor able of Fourier Transforms is shown below: Page 2

3 Convoluions f (x) δ(x) x> e λx x< e cx 2 2 g(k) a + ik c e k 2 2c π + x 2 2 e k In general, we define he convoluion inegral by h() = f ( τ )g(τ ) dτ f g convoluion inegral The Fourier ransform of he produc of wo funcions can be given in erms of he Fourier ransforms of he individual funcion. FT [ f ()g()] = = de iω de iω FT [ f ()g()] = = = 3 3 dω ' dω ' f ()g() dω 'e iω ' G(ω ') dω '' G(ω ')F(ω '') dω ''e [ ] d e iω '' F(ω '') i(ω '+ω '' ω ) dω ''[ G(ω ')F(ω '')]δ(ω '+ ω '' ω) dω ' G(ω ')F(ω ω ') = G(ω) F(ω) This jus he convoluion of he Fourier ransforms G(ω) and F(ω). I is he fundamenal consrucion needed o solve ODEs using Green s funcions laer, as we shall see laer. Clearly, i gives a measure of he overlap of wo signals as a funcion of. The symmery of he Fourier ransform and is inverse operaion gives he resuls Page 3

4 InvFT [ F(ω)G(ω) ] = f () g() I[ f () g() ] = F(ω)G(ω) An example will help us visualize he real complexiy of he convoluion operaion. We consider he convoluion of wo funcions h() = f () g() = dτ f (τ )g( τ ) Where we choose f() = square pulse and g() = riangular pulse as shown below. Now h() = area under produc inegrand f (τ )g( τ ) as a funcion of. Procedure(mus be done carefully): () plo f (τ ) and g( τ ) versus τ ; h() is given by his process for all possible values. (2), say =-2; In his case i is clear ha here is never any overlap of he funcions and herefore heir produc is zero and hus h() = for. Page 4

5 Also from he picure below i is clear ha h() = for (say.5) and also for 7 < (say = 8) as shown below; Page 5

6 (4) < < 2, say =.5; Here he funcions only overlap in he range < < 2 as shown below. The red curve is f(au)*g(au-) for =.5. The area under he red curve is h(). In his overlap region, Page 6

7 h() = dτ f (τ )g( τ ) f (τ ) = 2 2 ( τ ) < ( τ ) < 5 g( τ ) = 5 oherwise The range < ( τ ) < 5 is equivalen o ( 5) < τ <. Over his range we have g( τ ) = 2 5 ( τ ) and h() = f () g() = dτ 4 5 ( τ ) < < 2 = < < 2 (5) 2 < < 6, say = 4 Here he square pulse is enirely inside he riangular pulse as shown below. Page 7

8 We have 2 h() = f () g() = dτ 4 5 ( τ ) 2 < < 6 = < < 6 (6) Final region 6 < < 7, say = 6.5; Triangular pulse has moved o he righ of he square pulse as shown below. We have We finally ge 2 h() = f () g() = dτ 4 5 ( τ ) 6 < < 7 5 = < < < h() = < < < < < 6 < < 7 Page 8

9 which looks like A very ricky and edious process o comprehend!! Correlaion The correlaion process gives a measure of he similariy of wo signals. One of is mos imporan applicaions is o pick a known signal ou of a sea of noise. The cross-correlaion beween f() and g() is defined as ψ fg () = dτ f (τ )g(τ ) The cross-correlaion of a funcion wih iself ψ ff () = dτ f (τ ) f (τ ) is called an auocorrelaion. This operaion is similar o he convoluion operaion excep ha he second funcion is no invered. I is jus as ricky as he convoluion operaion. We can wrie FT dτ f (τ )g(τ ) = F(ω)G( ω) Page 9

10 Examples: () The Square Pulse - Consider he funcion f () = T / 2 < < T / 2 oherwise f() is absoluely inegrable so i has a valid Fourier ransform. I is given by F(ω) = T /2 de iω = 2 sin ωt 2 π ω T /2 which looks like (for T=) In he limi T we have (T = 5, in fac, here) Page

11 We ge a sharp spike, bu he area remains consan. This implies ha as T F(ω) δ(ω) Formally, we have F(ω) = lim T T /2 T /2 de iω = de iω = δ(ω) (2) Transform of a Dela-Funcion - Consider he funcion f () = δ(). The ransform is F(ω) = de iω δ() = The inverse ransform is de iω = δ() Now FT df d = iωft ( f ) = iωf(ω) Therefore for he square pulse we have df = δ( + T / 2) δ( T / 2) d I df d Bu Thus, FT df d F(ω) = = I( δ( + T / 2) δ( T / 2) ) FT ( f ( )) = e iω FT ( f ()) iω (T /2) ( )FT δ() = e iω ( T /2) e 2 sin ωt 2 π ω ( ) = i 2 π sin ωt 2 for he square pulse ( as before) Remember his only makes sense inside an inegral. (3) Transform of a Gaussian - Consider he funcion = iωf(ω) f () = α π e α 2 2 = normalized Gaussian pulse Page

12 We choose α =. The peak is a α / π. The /2 maximum poins are separaed by Δ = / α. The area under he curve is =. The Fourier ransform is F(ω) = d α e α 2 2 e iω = α d π π 2 e (α 2 2 +iω ) We complee he square o evaluae he inegral. We have We hus have Le ( ) 2 γ α iω = α iω + γ γ = α + β 2αβ = iω β = iω 2α γ = β 2 = ω 2 F(ω) = π α 4α 2 ω 2 2 e 4α 2 de (α + x = α + iω dx = αd 2α F(ω) = π ω 2 2 e 4α 2 which is a differen Gaussian. An imporan feaure is f () Δ α F(ω) Δω 2α ΔωΔ 2 iω 2α )2 dxe x2 = e In general for any f() we have ΔωΔ c = consan. In he wave heory of quanum mechanics, his corresponds o he Heisenberg Uncerainy Principle. The Laplace Transform Anoher imporan inegral ransform is he Laplace ransform. For a funcion f(), we define he Laplace ransform by F(s) = de s f () = L( f ()) ω 2 4α 2 Page 2

13 The Laplace ransform is a linear operaor so ha L(af () + bg()) = al( f ()) + bl(g()) The Laplace ransform has a firs shifing propery expressed as if L( f ()) = F(s), hen L(e a f ()) = F(s a) The Laplace ransform has a second shifing propery expressed as f ( a) >a if L( f ()) = F(s), and g() = hen L(g()) = e as F(s) <a If we le = τ / a, hen and if aσ = s we ge Examples: F(s) = de s f () = sτ /a dτe a f (τ / a) sτ /a af(aσ ) = dτe f (τ / a) () Heaviside uni sep funcion L(H ()) = de s H () = de s = s = F(s) Since inegraion is only beween, his also says ha L() = de s = de s = s (2) Exponenial funcion f () = e a L(e a ) = de (a s) = = F(s a) s a or by firs shifing L(e a f ()) = F(s a) L() = s L(e a ) = L(e a ) = s a (3) Shifed sep funcion Page 3

14 H a () = >a <a By second shifing L(H a ()) = L(H ( a)) = e as L(H ()) = e as (4) Euler funcion e iθ = cosθ + isinθ. From (2) we have (5) Power funcion k (6) Power series L(e iω ) = = L(cosω + isinω) = L(cosω) + il(sinω) s iω s + iω = s iω s + iω = s s 2 + ω + i ω 2 s 2 + ω 2 s L(cosω) = s 2 + ω 2 ω L(sinω) = s 2 + ω 2 L( k ) = k e s d = s k + = k! Γ(k + ) = k + s s k + f () = n= a n n x k e x dx using x = s F(s) = L( f ()) = a n L( n ) = a n! n n= (7) Bessel Funcion - we will see laer ha he Bessel funcion can be wrien as ( ) k J () = 2 2k (k!) 2k = Bessel funcion of order 2 We have Now so k = ( ) k 2k L(J ()) = L( ) = 2 2k 2 (k!) k = k = n= s n+ ( ) k (2k)! 2 2k (k!) 2 s 2k + (2k)! = k = k (2k ) = 2 k k! (2k ) s Page 4

15 ( ) k (2k)! L(J ()) = = 2 2k (k!) 2 s 2k + k = s + ( ) k (2k ) 2 2k k! s 2k k = Now using he binomial heorem (+ x) n = + n n(n ) n(n )...(n k + ) x + x x k +...! 2! k! + /2 ( ) 2 ( ) ( ) k 3 2k s 2 = +! s 2 + 2! s ! or Wow! (7) Dirac Dela Funcion L(J ()) = s + s 2 L δ( ) /2 = [ ] = δ( )e s d s 2 + = e s There is a ables of Laplace ransforms. More examples - ypical elecric circui inpu funcions): () Consider he square pulse for <a f () = A for a<<b for >b We rewrie his funcion as f () = AH a () AH b () and hen using lineariy and (3) we have L( f ()) = A e as A e bs = A s s s e as e bs (2) Consider he ime-dependen pulse for << f () = 2 for <<2 for 2< We rewrie his funcion as f () = H () 2 H 2 () 2 s 2 k Page 5

16 g() = 2 is no in he proper form o use he second shif propery. We fix i by he following algebra f () = H () ( ) 2 + 2( ) + H 2 () ( 2) 2 + 4( 2) + 4 Now using he second shif propery and lineariy we have ( ) L( H 2 () ( 2) 2 + 4( 2) + 4 ) L( f ()) = L H () ( ) 2 + 2( ) + = ( ) 2 + 2( ) + e s d ( 2) 2 + 4( 2) + 4 e s d = e s x 2 + 2x + e sx dx e 2s x 2 + 4x + 4 e sx dx 2 = e s s s + 2 s 2 e 2s s s s Finally, le us use he able o find he inverse Laplace ransform. () Consider Now L(cosω) = (2) Consider Now F(s) = F(s) = 2s s s s 2 + ω 2 L(2cos2) = F(s) = 2 2s s s s 2 + 4s + 3 s f () = 2cos2 = L s 2 + ω 2 6s s 2 + 4s + 3 = 6s 6(s + 2) 2 6(s + 2) = = (s + 2) (s + 2) (s + 2) (s + 2) which is a form where we can use he firs shif propery. We have s L(e a f ()) = F(s a) and L(cosω) = s 2 + ω 2 or (s + 2) L(e 2 cos 3) = L(cos 3( + 2)) = (s + 2) and or ω L(e a f ()) = F(s a) and L(sinω) = s 2 + ω 2 Page 6

17 Therefore, or (3) Consider Now F(s) = 3 L(e 2 sin 3) = L(sin 3( + 2)) = (s + 2) (s + 2) (s + 2) (s + 2) = 6L(e 2 cos 3) 4L(e 2 sin 3) = L(6e 2 cos 3 4e 2 sin 3) 6s f () = L s 2 + 4s + 3 = 2e 2 (3cos 3 2sin 3) F(s) = 4e 2s s 2 6 = 4 e 2s s L(sinh 4) = s L s 2 6 = sinh 4 The second shif propery hen gives 4 L(sinh 4( 2)) = e 2s L(sinh 4) = e 2s s 2 6 or 4e 2s sinh 4( 2) for >2 f () = L s 2 6 = for <2 Oher Properies of he Laplace Transform Laplace Transform of Derivaives and Inegrals I will jus quoe some resuls here wihou proof. Examples: L( f ') = sl( f ) f () L( f (n) ) = s n L( f ) s n f () s n 2 f '()... f (n ) () L f (τ )dτ = s L( f ) () Consider f () = sin. We have Now f '() = cos sin and f ''() = sin + 2cos Page 7

18 (2) If or L( f '' ) = s 2 L( f ) sf () f '() = s 2 L( f ) ( ) = L( 2cos) L( sin ) s 2 L( sin) = L 2cos sin 2 s + s 2 L( sin) = s L 2cos 2 2s L( sin) = (s 2 + ) 2 2s L( sin2) = (s 2 + 4) 2 8 F(s) = (s 2 + 4) 2 L( f ) = F(s) = L 8 (s 2 + 4) 2 = s f (τ )dτ = s L( f ) L ( ) = 2 s L( cos) = 2 s 2 s 2 8s (s 2 + 4) 2 s 2 + τ sin2τdτ = s L( sin2) = 2s s (s 2 + 4) 2 L( f ) = 4L τ sin2τdτ f () = 4 τ sin2τdτ = 2 cos2 + sin2 Derivaives and Inegrals of he Laplace Transform L( n f ()) = ( ) n F (n) (s) L f () = F(σ )dσ Laplace Transforms of Periodic Funcions s We now consider funcions ha are periodic wih period a, i.e., The ransform is f () = f ( + na) n=,2,3,... Page 8

19 L( f ()) = f ()e s d = f ()e s d + f ()e s d + f ()e s d +... a a 2a 3a a 2a = f ()e s d + f ( + a)e s( +a) d + f ( + 2a)e s( +2a) d +... a a = f ()e s d + f ()e s( +a) d + f ()e s( +2a) d +... = + e sa + e 2sa +... a ( ) f ()e s d a a = e sa a a f ()e s d Inverse Laplace Transforms Parial Fracions () Consider wo polynomials P(s) and Q(s) such ha degree(q) > degree(p) and Q(s) = (s a )(s a 2 )(s a 3 )...(s a n ) wih all he roos a i disinc.we saw earlier ha we can hen wrie F(s) = P(s) Q(s) = A (s a ) + A 2 (s a 2 ) + A 3 (s a 3 ) A n (s a n ) We hen have lim s a k Example: P(s) Q(s) (s a k ) = A k Suppose we know (s a = lim k ) s ak Q(s) = P(a k ) lim s ak Q'(s) = P(a ) k Q'(a k ) F(s) = (s a P(s) = P(a k ) lim k ) = P(a k ) lim s ak Q(s) s ak s 3 + 3s 2 2s + 4 s(s )(s 2)(s 2 + 4s + 3) d ds (s a k ) d ds Q(s) hen wha is f() or how do we find he inverse Laplace ransform? We have Page 9

20 Using we ge so ha Using we have F(s) = s 3 + 3s 2 2s + 4 s(s )(s 2)(s 2 + 4s + 3) = s 3 + 3s 2 2s + 4 s(s )(s 2)(s + )(s + 3) = A s + A 2 (s ) + A 3 (s 2) + A 4 (s + ) + A 5 (s + 3) P(a A k = k ) [Q(s) / (s a k )] s=ak A = 2 / 3, A 2 = 3 / 4, A 3 = 2 / 3, A 4 = 2 / 3, A 5 = /2 F(s) = 2 / 3 s 3 / 4 (s ) + 2 / 3 (s 2) 2 / 3 (s + ) + /2 (s + 3) L(e a ) = s a f () = e e2 2 3 e + 2 e 3 When he roos of Q(s) are no all disinc we ge a differen resul. Suppose here is a repeaed facor (s a ) m, i.e., Q(s) = (s a ) m (s a 2 )(s a 3 )... We hen ge F(s) = P(s) Q(s) = B m (s a ) + B m m (s a ) B 2 m (s a ) + B 2 (s a ) + A 2 (s a 2 ) + A 3 (s a 3 ) +... where and Two final cases are: P(a A k = k ) [Q(s) / (s a k )] s=ak B i = (m )! B m = d (m i) ds (m i) P(a ) Q(s) / (s a ) m s=a P(s) Q(s) / (s a ) m s=a Q(s) = ((s a) 2 + b 2 )(s a )(s a 2 )(s a 3 )... Page 2

21 F(s) = P(s) Q(s) = B s + B 2 (s a) 2 + b 2 + A (s a ) + A 2 (s a 2 ) +... P(a A k = k ) [Q(s) / (s a k )] s=ak and P(s) B (a + ib) + B 2 = Q(s) / (s a) 2 + b 2 s=a+ib Q(s) = ((s a) 2 + b 2 ) 2 (s a )(s a 2 )(s a 3 )... F(s) = P(s) Q(s) = C s + C 2 ((s a) 2 + b 2 ) 2 + B s + B 2 (s a) 2 + b 2 + A (s a ) + A 2 (s a 2 ) +... P(a A k = k ) [Q(s) / (s a k )] s=ak A Convoluion Theorem P(s) c (a + ib) + c 2 = Q(s) / (s a) 2 + b 2 2 s=a+ib B (a + ib) + B 2 = d P(s) ds Q(s) / (s a) 2 + b 2 2 Suppose ha we have wo funcions f() and g(). We define he inegral If hen we have s=a+ib h() = f ( τ )g(τ ) dτ f g convoluion inegral L( f ()) = F(s) and L(g()) = G(s) H (s) = de s h() = F(s)G(s) = L( f g) The Error Funcion as an Example of Convoluion As we defined earlier, he error funcion is defined by From he able erf (x) = 2 π e 2 d x Page 2

22 L e = π s + L s + = e π Now he convoluion heorem says ha L e τ s s + = f g = f ( τ )g(τ )dτ = () πτ dτ where f () =, g() = e π F(s) = s, G(s) = s + Changing variables wih x = τ we ge L s s + = e τ πτ dτ = 2 e x2 π dx = erf ( ) Page 22