§5.1. FOURIER级数 1/67回 上式可以表示为更简洁的形式 f(r) C ikon 51-13) 复指数函数系(5.1-12)的正交归一性 T/2 jkoxe-inoxdr k≠n T/2 IT=20,k Fourier展开系数为 f() d 20 显然有 (5.1-15) ●Fist●Prev·Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 11/67 þª±L«{'/ª f(x) = X ∞ k=−∞ Cke ikωx . (5.1-13) Eê¼êX(5.1-12)85 Z T/2 −T/2 e ikωx e −inωxdx = ( 0 , k , n, T = 2`, k = n, Fourier ÐmXê Ck = 1 2` Z ` −` f(ξ) h e i kπξ ` i∗ dξ. (5.1-14) w,k C ∗ k = C−k. (5.1-15)
§5.1. FOURIER级数 12/67回 5.1.6例 例1(P.91,1)交流电压E0 sin ot经过全波振流,成为 E(t)= Eol sin ot|.试将其展为 Fourier级数 解:交流电压 Eo sin ot在区间-π≤ot≤π上是一个周期,令 则经过振流后成为 E(x)=a0+>(ak cos kx +bk sin kx) 其中系数 E 00-2π f(r)d sinx dx E 丌2E ( cos r) f(r)cos kx dx ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 12/67 5.1.6 ~ ✿ ~¬(P.91§1)6>Ø E0 sin ωt ²L Å6§¤ E(t) = E0|sin ωt|©ÁòÙÐ Fourier ?ê© )µ6>Ø E0 sin ωt 3«m −π ≤ ωt ≤ π þ´±Ï§- ωt = x§K²L6¤µ E(x) = a0 + X ∞ k=1 (ak cos kx + bk sin kx) , Ù¥Xê a0 = 1 2π Z π −π f(x) dx = E0 π Z π 0 sin x dx = E0 π (− cos x) π 0 = 2E0 π ak = 1 π Z π −π f(x) cos kx dx
§5.1. FOURIER级数 13/67回 E(sin x)cos kx dx Eo sin x cos kx dx Eo sin x cos kx dx sin(kx + x)-sin(kx -x)]dx Eo cos(k 1)x cos(k-1)x k+1 k-1 0 (当k为奇数时,但k≠1) x=x,(当k为偶数时 当k=1时, Eo sin x cos x dx sin 2x dx= 0, 又令k=2n时,则 4E0 k 丌(1-4n2 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 13/67 = 1 π Z 0 −π E(− sin x) cos kx dx + 1 π Z π 0 E0 sin x cos kx dx = 2 π Z π 0 E0 sin x cos kx dx = 2 π Z π 0 E0 2 [sin(kx + x) − sin(kx − x)] dx = − E0 π cos(k + 1)x k + 1 − cos(k − 1)x k − 1 π 0 = ( 0 , (kÛê, k , 1). 4E0 π(1−k 2) , (kóê). k = 1§ a1 = 2 π Z π 0 E0 sin x cos x dx = E0 π Z π 0 sin2 2x dx = 0, q-k = 2n§K ak = a2n = 4E0 π(1 − 4n 2 ) , n = 1, 2, 3, · · ·
§5.1. FOURIER级数 14/67回 同理,可以计算的bk f(r)sin kx dx Eo sin x sin kx dx =0 所以 E(t)=a0+>an cos 2nx =a0+>an cos 2not 2E0,4E0、c0s2not 1-4i n=1 例2(P2,2)将锯齿波展为 Fourier级数.在(0,T)这个周期上, 该锯齿波可表为f(x)=x/3 解:锯齿波之周期为T.令 2L=T ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 14/67 Ón§±Obk bk = 1 π Z π −π f(x) sin kx dx = 2 π Z π 0 E0 sin x sin kx dx = 0. ¤± E(t) = a0 + X ∞ n=1 an cos 2nx = a0 + X ∞ n=1 an cos 2nωt = 2E0 π + 4E0 π X ∞ n=1 cos 2nωt 1 − 4n 2 . ✿ ~(P.92§2)òç¸ÅÐ Fourier ?ê©3 ( 0, T ) ù±Ïþ§ Tç¸ÅL f(x) = x/3© )µç¸Å±Ï T©- 2l = T
§5.1. FOURIER级数 15/67回 得 将l代入以2l为周期之 Fourier级数和 Fourier系数表达式即可得适 合本题 Fourier级数和 Fourier系数表达式: 2n丌 2n丌 f(x)=a0+ T t+b sin Fourier系数的计算如下: f(t)dt dx 2n丌 an=t f( cos t tdt ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §5.1. FOURIER ?ê 15/67 l = T 2 , ò l \± 2l ±Ï Fourier ?êÚ Fourier XêLª=· Ü K Fourier ?êÚ Fourier XêLªµ f(x) = a0 + X ∞ n=1 an cos 2nπ T t + bn sin 2nπ T t . Fourier XêOXeµ a0 = 1 T Z T 0 f(t) dt = 1 T Z T 0 1 3 x · dx = 1 3T · 1 2 x 2 T 0 = T 6 , an = 2 T Z T 0 f(t) cos 2nπ T t dt