傅里叶级数与傅里叶变换_Part3_周期为2L的函数展开为傅里叶级数
傅里葉級(jí)數(shù)與傅里葉變換_Part3_周期為2L的函數(shù)展開(kāi)為傅里葉級(jí)數(shù)
參考鏈接：
DR_CAN老師的原視頻
0、復(fù)習(xí)Part2的內(nèi)容
參考鏈接：傅里葉級(jí)數(shù)與傅里葉變換_Part2_周期為2Π的函數(shù)展開(kāi)為傅里葉級(jí)數(shù)
對(duì)于周期為 T=2πT = 2\piT=2π周期函數(shù)，即f(x)=f(x+2π)f\left( x \right) = f\left( {x + 2\pi } \right)f(x)=f(x+2π) ，它的傅里葉級(jí)數(shù)展開(kāi)形式如下：
f(x)=a02+∑n=1∞ancos?nx+∑n=1∞bnsin?nxf\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nx} + \sum\limits_{n = 1}^\infty {{b_n}\sin nx}f(x)=2a0??+n=1∑∞?an?cosnx+n=1∑∞?bn?sinnx
其中，
a0=1π∫?ππf(x)dxan=1π∫?ππf(x)cos?nxdxbn=1π∫?ππf(x)sin?nxdx\begin{array}{l} {a_0} = \frac{1}{\pi }\int_{ - \pi }^\pi {f\left( x \right)dx} \\\\ {a_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {f\left( x \right)\cos nxdx} \\\\ {b_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {f\left( x \right)\sin nxdx} \end{array}a0?=π1?∫?ππ?f(x)dxan?=π1?∫?ππ?f(x)cosnxdxbn?=π1?∫?ππ?f(x)sinnxdx?
1、周期為2L的函數(shù)展開(kāi)為傅里葉級(jí)數(shù)
對(duì)于，f(t)=f(t+2L)f\left( t \right) = f\left( {t + 2L} \right)f(t)=f(t+2L) ，周期為T=2LT = 2LT=2L的函數(shù)
如果想用運(yùn)用Part2掌握的以T=2πT = 2\piT=2π為周期的f(t)=f(t+2π)f\left( t \right) = f\left( {t + 2\pi} \right)f(t)=f(t+2π)的傅里級(jí)數(shù)展開(kāi)公式，需要做一個(gè)換元。
令 x=πLt?t=Lπxx = \frac{\pi }{L}t \Rightarrow t = \frac{L}{\pi }xx=Lπ?t?t=πL?x
| 2L2L2L | x=πLt=πL?2L=2πx = \frac{\pi }{L}t = \frac{\pi }{L} \cdot 2L = 2\pix=Lπ?t=Lπ??2L=2π |
| 4L4L4L | 4π4\pi4π |
| 000 | 000 |
f(t)=f(πLx)?g(x)f\left( t \right) = f\left( {\frac{\pi }{L}x} \right) \triangleq g\left( x \right)f(t)=f(Lπ?x)?g(x)
通過(guò)換元，我們就把周期為T=2LT = 2LT=2L的函數(shù)，f(t)=f(t+2L)f\left( t \right) = f\left( {t + 2L} \right)f(t)=f(t+2L) 變成了周期為T=2πT = 2\piT=2π的函數(shù)g(t)=g(t+2π)g\left( t \right) =g\left( {t + 2\pi} \right)g(t)=g(t+2π) 。
根據(jù)Part2的內(nèi)容我們知道
g(x)=a02+∑n=1∞ancos?nx+∑n=1∞bnsin?nxg\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nx} + \sum\limits_{n = 1}^\infty {{b_n}\sin nx}g(x)=2a0??+n=1∑∞?an?cosnx+n=1∑∞?bn?sinnx
其中，
a0=1π∫?ππg(shù)(x)dxan=1π∫?ππg(shù)(x)cos?nxdxbn=1π∫?ππg(shù)(x)sin?nxdx\begin{array}{l} {a_0} = \frac{1}{\pi }\int_{ - \pi }^\pi {g\left( x \right)dx} \\\\ {a_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {g\left( x \right)\cos nxdx} \\\\ {b_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {g\left( x \right)\sin nxdx} \end{array}a0?=π1?∫?ππ?g(x)dxan?=π1?∫?ππ?g(x)cosnxdxbn?=π1?∫?ππ?g(x)sinnxdx?
👇
現(xiàn)在要做的，就是把x=πLtx = \frac{\pi }{L}tx=Lπ?t帶進(jìn)去
cos?nx=cos?nπLt,sin?nx=sin?nπLt\cos nx = \cos \frac{{n\pi }}{L}t,\sin nx = \sin \frac{{n\pi }}{L}tcosnx=cosLnπ?t,sinnx=sinLnπ?t
g(x)=f(t)g\left( x \right) = f\left( t \right)g(x)=f(t)
| ?π-\pi?π | t=Lπx=Lπ??π=?Lt = \frac{L}{\pi }x = \frac{L}{\pi } \cdot - \pi = - Lt=πL?x=πL???π=?L |
| π\(zhòng)piπ | LLL |
∫?ππdx=∫?LLdπLt=πL∫?LLdt\int_{ - \pi }^\pi {dx} = \int_{ - L}^L {d\frac{\pi }{L}t} = \frac{\pi }{L}\int_{ - L}^L {dt}∫?ππ?dx=∫?LL?dLπ?t=Lπ?∫?LL?dt
1π∫?ππdx=1π?(πL∫?LLdt)=1L∫?LLdt\frac{1}{\pi }\int_{ - \pi }^\pi {dx} = \frac{1}{\pi } \cdot \left( {\frac{\pi }{L}\int_{ - L}^L {dt} } \right) = \frac{1}{L}\int_{ - L}^L {dt}π1?∫?ππ?dx=π1??(Lπ?∫?LL?dt)=L1?∫?LL?dt
講上述算出的式子帶入到g(x)g\left( x\right)g(x)中。
f(t)=g(x)=a02+∑n=1∞ancos?nπLt+∑n=1∞bnsin?nπLtf\left( t \right) = g\left( x\right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos \pmb{\frac{{n\pi }}{L}t}} + \sum\limits_{n = 1}^\infty {{b_n}\sin \pmb{\frac{{n\pi }}{L}t}}f(t)=g(x)=2a0??+n=1∑∞?an?cosLnπ?tLnπ?t+n=1∑∞?bn?sinLnπ?tLnπ?t
其中
a0=1L∫?LLf(t)dtan=1L∫?LLf(t)cos?nπLtdtbn=1L∫?LLf(t)sin?nπLtdt\begin{array}{l} {a_0} =\pmb{ \frac{1}{L}}\int_{ - L}^L {f\left( t \right)dt} \\\\ {a_n} =\pmb{ \frac{1}{L}}\int_{ - L}^L {f\left( t \right)\cos \pmb{\frac{{n\pi }}{L}tdt}} \\\\ {b_n} = \pmb{\frac{1}{L}}\int_{ - L}^L {f\left( t \right)\sin \pmb{\frac{{n\pi }}{L}tdt}} \end{array}a0?=L1?L1?∫?LL?f(t)dtan?=L1?L1?∫?LL?f(t)cosLnπ?tdtLnπ?tdtbn?=L1?L1?∫?LL?f(t)sinLnπ?tdtLnπ?tdt?
在工程當(dāng)中，由于時(shí)間是t≥0t \ge 0t≥0的，所以ttt是從000開(kāi)始的，假設(shè)周期為 T=2LT=2LT=2L，ω?πL=2π2L=2πT\omega \triangleq \frac{\pi }{L} = \frac{{2\pi }}{{2L}} = \frac{{2\pi }}{T}ω?Lπ?=2L2π?=T2π? ，這個(gè)ω\omegaω本質(zhì)上就是角頻率。
再來(lái)看積分，因?yàn)?span id="ze8trgl8bvbq" class="katex--inline">?L-L?L ~ LLL是一個(gè)周期， 也000 ~ 2L2L2L是一個(gè)周期，因此 ∫?LLdt→∫02Ldt→∫0Tdt\int_{ - L}^L {dt} \to \int_0^{2L} {dt} \to \int_0^T {dt}∫?LL?dt→∫02L?dt→∫0T?dt，把上述的符號(hào)帶入到周期為T=2LT = 2LT=2L的傅里葉級(jí)數(shù)展開(kāi)公式當(dāng)中，就可以得到。
f(t)=a02+∑n=1∞ancos?nωt+∑n=1∞bnsin?nωtf\left( t \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos n\omega t} + \sum\limits_{n = 1}^\infty {{b_n}\sin n\omega t}f(t)=2a0??+n=1∑∞?an?cosnωt+n=1∑∞?bn?sinnωt
a0=2T∫0Tf(t)dtan=2T∫0Tf(t)cos?nωtdtbn=2T∫0Tf(t)sin?nωtdt\begin{array}{l} {a_0} = \frac{2}{T}\int_0^T {f\left( t \right)dt} \\\\ {a_n} = \frac{2}{T}\int_0^T {f\left( t \right)\cos n\omega tdt} \\\\ {b_n} = \frac{2}{T}\int_0^T {f\left( t \right)\sin n\omega tdt} \end{array}a0?=T2?∫0T?f(t)dtan?=T2?∫0T?f(t)cosnωtdtbn?=T2?∫0T?f(t)sinnωtdt?
伏筆： 考慮，當(dāng)T→∞T \to \inftyT→∞時(shí)，f(t)f\left( t \right)f(t) 不再為周期函數(shù)，那時(shí)候 f(t)f\left( t \right)f(t)該給如何展開(kāi)呢？ 這就是傅里葉變換啦。
2、例子
把下圖這個(gè)函數(shù)，傅里葉展開(kāi)一下
T=20,ω=2πT=2π20=110πT = 20,\omega = \frac{{2\pi }}{T} = \frac{{2\pi }}{{20}} = \frac{1}{{10}}\piT=20,ω=T2π?=202π?=101?π
a0=2T∫0Tf(t)dt=2T∫0T27dt+2T∫T2T3dt=7+3=10{a_0} = \frac{2}{T}\int_0^T {f\left( t \right)dt} = \frac{2}{T}\int_0^{\frac{T}{2}} {7dt} + \frac{2}{T}\int_{\frac{T}{2}}^T {3dt} = 7 + 3 = 10a0?=T2?∫0T?f(t)dt=T2?∫02T??7dt+T2?∫2T?T?3dt=7+3=10
an=2T∫0Tf(t)cos?nωtdt=2T∫0T27cos?nπ10tdt+2T∫T2T3cos?nπ10dt=110?(70nπsin?nπ10t∣010)+110?(30nπsin?nπ10t∣1020)=110?(0+0)=0\begin{aligned} {a_n} &= \frac{2}{T}\int_0^T {f\left( t \right)\cos n\omega tdt} = \frac{2}{T}\int_0^{\frac{T}{2}} {7\cos \frac{{n\pi }}{{10}}tdt} + \frac{2}{T}\int_{\frac{T}{2}}^T {3\cos \frac{{n\pi }}{{10}}dt} \\ & = \frac{1}{{10}} \cdot \left( {\frac{{70}}{{n\pi }}\sin \frac{{n\pi }}{{10}}t\left| {_0^{10}} \right.} \right) + \frac{1}{{10}} \cdot \left( {\frac{{30}}{{n\pi }}\sin \frac{{n\pi }}{{10}}t\left| {_{10}^{20}} \right.} \right) = \frac{1}{{10}} \cdot \left( {0 + 0} \right) = 0 \end{aligned}an??=T2?∫0T?f(t)cosnωtdt=T2?∫02T??7cos10nπ?tdt+T2?∫2T?T?3cos10nπ?dt=101??(nπ70?sin10nπ?t∣∣?010?)+101??(nπ30?sin10nπ?t∣∣?1020?)=101??(0+0)=0?
bn=2T∫0Tf(t)sin?nωtdt=2T∫0T27sin?nπ10tdt+2T∫T2T3sin?nπ10dt=110?(?70nπcos?nπ10t∣010)+110?(?30nπcos?nπ10t∣1020)=?7nπ(cos?nπ?1)?3nπ(cos?2nπ?cos?nπ)\begin{aligned} {b_n} &= \frac{2}{T}\int_0^T {f\left( t \right)\sin n\omega tdt} = \frac{2}{T}\int_0^{\frac{T}{2}} {7\sin \frac{{n\pi }}{{10}}tdt} + \frac{2}{T}\int_{\frac{T}{2}}^T {3\sin \frac{{n\pi }}{{10}}dt} \\ &= \frac{1}{{10}} \cdot \left( { - \frac{{70}}{{n\pi }}\cos \frac{{n\pi }}{{10}}t\left| {_0^{10}} \right.} \right) + \frac{1}{{10}} \cdot \left( { - \frac{{30}}{{n\pi }}\cos \frac{{n\pi }}{{10}}t\left| {_{10}^{20}} \right.} \right)\\ & = - \frac{7}{{n\pi }}\left( {\cos n\pi - 1} \right) - \frac{3}{{n\pi }}\left( {\cos 2n\pi - \cos n\pi } \right) \end{aligned}bn??=T2?∫0T?f(t)sinnωtdt=T2?∫02T??7sin10nπ?tdt+T2?∫2T?T?3sin10nπ?dt=101??(?nπ70?cos10nπ?t∣∣?010?)+101??(?nπ30?cos10nπ?t∣∣?1020?)=?nπ7?(cosnπ?1)?nπ3?(cos2nπ?cosnπ)?
當(dāng) nnn為偶數(shù)時(shí)，cos?nπ=cos?2nπ=1\cos n\pi = \cos 2n\pi = 1cosnπ=cos2nπ=1
bn=?7nπ(1?1)+?3nπ(1?1)=0{b_n} = - \frac{7}{{n\pi }}\left( {1 - 1} \right) + - \frac{3}{{n\pi }}\left( {1 - 1} \right) = 0bn?=?nπ7?(1?1)+?nπ3?(1?1)=0
當(dāng) nnn為奇數(shù)時(shí)，cos?nπ=?1,cos?2nπ=1\cos n\pi = - 1,\cos 2n\pi = 1cosnπ=?1,cos2nπ=1
bn=?7nπ(?1?1)?3nπ[1?(?1)]=8nπ{b_n} = - \frac{7}{{n\pi }}\left( { - 1 - 1} \right) - \frac{3}{{n\pi }}\left[ {1 - \left( { - 1} \right)} \right] = \frac{8}{{n\pi }}bn?=?nπ7?(?1?1)?nπ3?[1?(?1)]=nπ8?
所以
f(t)=102+∑n=1∞0?cos?π10nt+∑n=1∞bnsin?π10nt=5+∑n=1∞8nπ?sin?nπ10t,n=1,3,5,7,?\begin{aligned} f\left( t \right) &= \frac{{10}}{2} + \sum\limits_{n = 1}^\infty {0 \cdot \cos \frac{\pi }{{10}}nt} + \sum\limits_{n = 1}^\infty {{b_n}\sin \frac{\pi }{{10}}nt} \\ &= 5 + \sum\limits_{n = 1}^\infty {\frac{8}{{n\pi }} \cdot \sin \frac{{n\pi }}{{10}}} t,n = 1,3,5,7, \cdots \end{aligned}f(t)?=210?+n=1∑∞?0?cos10π?nt+n=1∑∞?bn?sin10π?nt=5+n=1∑∞?nπ8??sin10nπ?t,n=1,3,5,7,??
下面給出MatLAB仿真結(jié)果
MatLAB代碼如下:
系列學(xué)習(xí)鏈接：歡迎大家點(diǎn)贊、收藏、留言討論。
傅里葉級(jí)數(shù)與傅里葉變換_Part0_歐拉公式證明+三角函數(shù)和差公式證明
傅里葉級(jí)數(shù)與傅里葉變換_Part1_三角函數(shù)系的正交性
傅里葉級(jí)數(shù)與傅里葉變換_Part2_周期為2Π的函數(shù)展開(kāi)為傅里葉級(jí)數(shù)
傅里葉級(jí)數(shù)與傅里葉變換_Part3_周期為2L的函數(shù)展開(kāi)為傅里葉級(jí)數(shù)
傅里葉級(jí)數(shù)與傅里葉變換_Part4_傅里葉級(jí)數(shù)的復(fù)數(shù)形式
傅里葉級(jí)數(shù)與傅里葉變換_Part5_傅里葉級(jí)數(shù)推導(dǎo)傅里葉變換
傅里葉級(jí)數(shù)與傅里葉變換_Part6_離散傅里葉變換推導(dǎo)
傅里葉級(jí)數(shù)與傅里葉變換_Part7_離散傅里葉變換的性質(zhì)
總結(jié)
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