When dealing with series, n is always a positive integer. Remember at every, sine has a value of zero, which means

Transcription

1 Fourier Series Some Prelimiar Ideas: Odd/Eve Fuctios: Sie is odd, which meas si ( ) si Cosie is eve, which meas cos ( ) cos Secial values of siie a cosie at Whe dealig with series, is alwas a ositive iteger. Remember at ever, sie has a value of zero, which meas si Cosie, o the otherhad, alterates betwee ad. So at odd values of, cos ad at eve values of, cos ; which meas cos ( ) What is a Fourier series? The Fourier series are useful for describig eriodic heomea. The advatage that the Fourier series has over Talor series is that the fuctio itself does ot eed to be cotiuous. Take for eamle a square wave de ed b oe eriod as < < f() < <.... Could this easil be aroimated usig a olomial, like we did usig Talor series? Probabl ot ver well. Sice this a eriodic fuctio (ol oe eriod show), it might be more useful to use eriodic fuctios such as sie ad cosie. This is eactl what the Fourier series does. The Fourier series is de ed as f() a + X a cos + b si for a fuctio de ed o the iterval ( ; ) where a a b f()d f() cos d f() si d Looks like fu, right? What we are goig to cosider are two secial cases. The rst is whe f() is a costat fuctio, ad the secod is whe f() is a liear fuctio.

2 Case f() is a costat. Cosider f() k a < < b, where a ad b are umbers i [ ; ]; a b, ad k is a real umber: NOTE: The reaso I am usig a ad b for the bouds is that the fuctio might be broke ito idividual ieces withi a iecewise de ed fuctio, ad ou would take the itegrals idividuall. You will see this i the eamles.. Fidig a Fidig a : a b a f()d Substitutig i f() k ou get b a kd (k)b a k (b a). Fidig a Fidig a : a f() cos d For ow I am goig to igore the bouds ad cocetrate o the itegral itself: f() cos d Pluggig i f() k we get k cos d The we ca do u-substitutio: Substitutig i k cos u du Now lug i the bouds: k si jb z k u k (si u) k si si b si a! du d d du So all that ou eed to do ow is lug i the values for a; b; k; ad :. Fidig b Fidig b : b f() si d Similar to above, I am goig to igore the bouds for ow ad lug them back i at the ed. b f() si d

3 Agai, luggig i f() ou get k si d The we ca do u-substitutio: Substitutig i k si u k du k cos a cos b si udu k ( u k cos u)! du d d du cos b cos a The ish b substitutig back i for a; b; k; ad :. Summar: For a costat fuctio de ed b f() k o ad iterval (a; b) ( series ca be determied b ; ) ; the coe ciets of the Fourier a k (b a) a k si b b k cos a si a cos b. Eamles Where f() is a costat Now let s look at some eamles, startig with the oe listed at the begiig. < < Eamle f() < <.... First ote that (the etire legth of the eriod is to, ad is alwas half of that) The fuctio is also broke u oto arts: from to, f(), ad from to, f(). This meas that each sectio of this fuctio will be its ow searate itegral. So for a, we will have: a d + d

4 But we ca take advatage of the formulas give for this, we just eed to do it for each iterval the add them together: a k (b a) Iterval : ( ; ) Iterval : (; ) a ; b ; k ; a ; b ; k ; ( ( )) ( ) So to d a add the two values together: a + It is similar for dig a ad b. Calculate each searatel the add them together. a k si b si a Iterval : ( ; ) Iterval : (; ) a ; b ; k ; a ; b ; k ; si si si si (si si ( )) (si si ) ( si ( )) si So that meas a si ( ) + si() However...ote that sice is alwas i iteger, there will alwas be a whole value of iside each value of sie, ad sice si ; the a si () ; therefore a () + () b k cos a cos b Iterval : ( ; ) Iterval : (; ) a ; b ; k ; a ; b ; k ; cos cos cos (cos ( ) cos ()) (cos () (cos ( ) ) Which meas that b (cos ( ) ) + ( ( cos ) cos ) cos cos )

5 Now let s do some algebra. () Remember that cosie is eve, so cos( ) cos, so the egative i the rst eressio disaears. b (cos ) + ( cos ) () Distribute the egative i the rst eressio to reverse the isides: b ( cos ) + ( () Factor out cos ) b ( cos + cos ) ( () Factor out a b ( cos ) cos ) () Trick art: Remember that cosie alterates betwee ad at ever other, so cos ( ) b ( ( ) ) Which gives: a a b ( ( ) ) FINAL STEP: Plug coe ciets ito the Fourier series: f() a + X a cos + b si f() + X X () cos + ( ( ) ) ( () ) si si....

6 < < Eamle f() < < So this meas that : Agai, this is broke ito itervals: ( ; ) ; k (; ) ; k But sice k o the iterval ( ; ) ; all of the itegrals o that iterval will be zero. (Sice all terms are multilied b k, zero times athig is zero) So we ol eed to a attetio to the iterval (; ) ; where a ; b ; k ; Fidig a k (b a) a ( ) Fidig a k si b a si But remember, si so a Fid b k cos a b cos But remember, cos ( so b ( ( ) ) f() + X si a si (si ) cos b cos ( ) cos ) So luggig ito the Fourier series f() a + X () cos f() + X + ( ( ) ) ( ( ) ) si a cos si + b si ou get 6

7 8 >< Eamle f() >: < < < < < < < < Now the eriod has bee broke ito itervals ad (half the legth of the eriod): Iterval Iterval Iterval Iterval ( ; ) ; k ( ; ) ; k (; ) ; k (; ) ; k We ca igore itervals ad, sice k (so all of the itegrals will be zero). a k (b a) Iterval : Iterval : a ; b ; k ; a ; b ; k ; ( ( )) So a + a k si b si a ( ) Iterval : Iterval : a ; b ; k ; a ; b ; k ; si si si si si si si si si si si si *Note i the last ste, for iterval, sice sie is a odd fuctio, si ( ) si, so si si, ad for iterval, remember that si a si + si a 9 si 7

8 b k cos a cos b Iterval : Iterval : a ; b ; k ; a ; b ; k ; cos cos cos cos cos cos ( ) *Note for the last ste, i iterval, cosie is eve, so cos ( ) cos, ad for iterval, cos ( ) : b cos () Factor out cos + cos + cos () Distribute coe ciets cos + + cos () Combie like terms: b cos + ( ) So... a a 9 si ( ) ( ) b ( ) So luggig ito the Fourier series f() a + X f() X 9 + si cos + cos + ( ) a cos + b si ou get cos + ( ) si 8

9 Case : f() is liear f() k + m Now we will cosider the case were f() is liear. I the geeral case, we will sa f() k + m o ( ; ) : Agai, we will cosider a geeric eamle to derive "easier" to use formulas. Ad agai, sice the fuctio ma be broke u withi the eriod, we will derive the formulas usig the iterval (a; b) :. Calculatig a So a o b a b f()d. Sice we are usig f() k + m, we get a (k + m) d a a k(b a) + m b a k + m b a. Calculatig a As before, I am goig to igore the bouds for ow, as well as the, which I will ut i at the ed. a b a f() cos d Substitutig i f() k + m we get (as well as igorig the bouds ad ) (k + m) cos d Note the miture of a algebraic with a trigoometric fuctio. This meas itegratio b arts: u k + m dv cos du md & v **u-sub for itegratig dv si (k + m) si (k + m) (k + m) (k + m) si si si m m Now let s multil b the (k + m) si (m) si d si d [to do this itegratio, use u-sub as we have doe before] cos + m cos Ad ow evaluate from a to b: a k + mb si b we took o at the begiig ad distribute. + m cos k + m si k + ma si a + m cos b + m cos cos a 9

10 . Calculatig b b b a f() si d I am goig to follow the same rocedure as above with (k + m) si d solve b itegratio b arts: u k + m dv si & du md v cos (k + m) (k + m) (k + m) (k + m) cos cos cos cos Multil through b (k + m) cos k + m cos + m + m (m) cos d si + m si + m si + m si Ad lastl evaluate from a to b: k + m cos b cos a k + ma cos a k + mb + m cos b cos d si b + m si b si a si a. Summar If f() is liear i the form f() k + m, the coe ciets of the Fourier series ca be calculated as a k(b a) + m b a a k + mb si b k + ma si a + m cos b cos a b k + ma cos a k + mb cos b + m si b si a (OK, ot as eas as the revious kid of roblems...)

11 . Eamles Eamle f() < < < < 6 Not that o the rst iterval ( ; ) that f(), so we ol eed to cocetrate o the secod. So i the iterval (; ) ; ; k ; m ; a ; b Fidig a : a a k(b a) + m b a ( + ( + (9)) Fidig a : a k + mb + () si b si k + ma + () si + 6 (cos ) + 6 (( ) ) 6 (( ) ) Fidig b : b k + ma + () cos a cos + () k + mb 6 6 cos + (si ) 6 ( ) + 6 ( ) si a + m si + ()() cos b cos + () + m cos b cos si b si So...a a 6 (( ) ) b 6 ( ) cos a cos si si a

12 Pluggig the coe ciets i the Fourier series f() a + X f() + X 6 (( ) ) cos 6 + ( ) a cos si + b si ou get 6 Eamle f() + < < < < Now we have two itervals to worr about: Iterval ( ; ) : ; k ; m ; a ; b Iterval : (; ) : ; k ; m ; a ; b Fidig a : Iterval : Iterval : So a + k(b a) + m b a (( ( )) + ( ) + ( ) ( ) ( ) Fidig a : k + mb si b k + ma si a + m cos b cos a Iterval : Iterval : + () si + ( ( ( ) ) () si + ( ) cos )) () si + () cos si + () cos cos cos

13 (cos ) (( ) ) ( ( ) ) Addig them together gives a ( ( ) ) + ( ( ) ) a ( ( ) ) Fidig b : k + ma cos a k + mb cos b + m si b si a Iterval : Iterval : + ( ) () cos ( ) cos So b + () + () cos + () si si cos + () si The we have a a ( ( ) ) b luggig the coe ciets i the Fourier series f() a + X f() + X ( ( ) ) cos + () si X f() + ( ( ) ) cos si a cos + b si ou get

14 < < Eamle 6 f() < < Note this oe that it is a miture of both costat ad liear fuctios. Therefore, o the lower iterval, we ca use the formulas for a costat fuctio, ad for the liear, the secod set of formulas: Iterval : ( ; ) : ; k ; a ; b Iterval : (; ) : ; k ; m ; a ; b Fidig a : Iterval : k (b a) ( ( )) Iterval : k(b a) + m b a + So a Fidig a : Iterval : k si b si si Iterval : + () k + mb si si b si a ( si ( )) + () si + (cos ) (( ) ) k + ma si a si + () cos + m cos b cos cos a So a + (( ) ) (( ) )

15 Fidig b : Iterval : cos k cos a (cos ( ) ) (( ) ) Iterval : k + ma + () cos cos + cos cos a + () cos b k + mb cos cos b + () si + m si b si si a ( ) So b (( ) ) ( ) Which gives: a + a (( ) ) b (( ) ) luggig the coe ciets i the Fourier series f() a + X X f() + (( ) ) cos + f() + + X (( ) ) cos + ( ) a cos + b si ou get (( ) ) ( ) si ( ) si (( ) ) Covergece at Poits of Discotiuit Oe advatage that a Fourier series gives is that it uses a cotiuous fuctio to describe a fuctio that might have discotiuities. Sometimes it might be useful to d the value to which the Fourier series coverges at oits where there is a jum discotiuit withi oe eriod. This ca easil be foud b takig the average value of the fuctio at both sides of the oit of discotiuit.

16 If a fuctio has a discotiuit at, the value to which the Fourier series coverges at that oit is F f( ) + f(+ ) where f( ) is the value of f() o the left side, ad f( + ) is the value of f() o the right side. Let s look at the revious eamles: Eamle f() < < < <.... This fuctio is discotiuous at : O the left side, it has a value of f( ) side f( + ) F +, ad o the right So the Fourier series coverges to ( ; F ) (; ) Eamle f() < < < < The oit of discotiuit occurs at, with f( ) ad f( + ) F + So at the oit of discotiuit, the Fourier series coverges to ( ; F ) ; 8 >< Eamle f() >: < < < < < < < < This fuctio has oits of discotiuit ad ;, ad : At : f( ) ad f( + )! F + ( ) At : f( ) ad f( + )! F + 6

17 At : f( ) ad f( + )! F + So the oits of covergece ( ; F ) would be ; ; ; ad (; ) Eamle f() < < < < 6 I this fuctio, there are o jum discotiuities withi the grah of oe eriod. Therefore, there would be o oits of covergece for discotiuities. Eamle f() + < < < < Like the revious eamle, this fuctio has o jum discotiuities. < < Eamle 6 f() < < The fuctio is discotiuous at, with f( ) ad f( + ), so F + Givig ( ; F ) (; ) 7