e j w 0 t e^{jw_0t} ejw0t
是否为周期信号? x ( t ) = x ( t + T ) x(t) = x(t+T) x(t)=x(t+T) 现假设 x ( t ) = e j w 0 t x(t) = e^{jw_0t} x(t)=ejw0t
e j w 0 t = e j w 0 ( t + T ) e^{jw_0t}=e^{jw_0(t+T)} ejw0t=ejw0(t+T)
当且仅当 w 0 T = 2 π → T = 2 π w 0 w_0T=2\pi\rightarrow T=\frac{2\pi}{w_0} w0T=2π→T=w02π
w 0 = 0 w_0=0 w0=0 无基波周期
w 0 ≠ 0 , T = 2 π w 0 w_0 \neq 0, T=\frac{2\pi}{w_0} w0=0,T=w02π
一组基波 w 0 w_0 w0的整数倍构成的信号:
ϕ k ( t ) = e j k w 0 t k = 0 , ± 1 , ± 2 ⋯ \phi_k(t)=e^{jkw_0t}\quad k=0,\pm1,\pm2\cdots ϕk(t)=ejkw0tk=0,±1,±2⋯
x ( t ) x(t) x(t)信号由若干的复指数信号构成,如
x ( t ) = ∑ k = − ∞ ∞ a k e j k w 0 t x(t) = \sum_{k=-\infty}^{\infty}a_ke^{jkw_0t} x(t)=k=−∞∑∞akejkw0t
我们可以得到傅立叶系数 FS
a k = 1 T ∫ T x ( t ) e − j k w 0 t d t a_k=\frac{1}{T}\int_Tx(t)e^{-jkw_0t}dt ak=T1∫Tx(t)e−jkw0tdt
证明过程:
例题:
傅立叶级数的性质
e j w 0 n e^{jw_0n} ejw0n
上面 n n n只能取整数。
e j ( w 0 + 2 π ) n = e j 2 π n e j w 0 n = e j w 0 n e^{j(w_0+2\pi)n} = e^{j2\pi n}e^{jw_0n} = e^{jw_0n} ej(w0+2π)n=ej2πnejw0n=ejw0n
上述式子表明 w 0 w_0 w0和 w 0 + 2 π w_0 + 2\pi w0+2π 一样,因此 w 0 w_0 w0的不断增加并不是和振荡增加对应
在 w 0 = π w_0 = \pi w0=π的时候,有
e j π n = ( e j π ) n = ( − 1 ) n e^{j\pi n} =(e^{j\pi})^n=(-1)^n ejπn=(ejπ)n=(−1)n
此时信号在每一点都符号都发生变化。
是否周期信号? x [ n ] = x [ n + N ] x[n]=x[n+N] x[n]=x[n+N],现 x [ n ] = e j w 0 n x[n]=e^{jw_0n} x[n]=ejw0n
e j w 0 ( n + N ) = e j w 0 n e^{jw_0(n+N)} = e^{jw_0n} ejw0(n+N)=ejw0n
则保证
e j w 0 N = 1 e^{jw_0N} = 1 ejw0N=1
则
w 0 N = 2 π m w_0 N = 2\pi m w0N=2πm
即
N = 2 π m w 0 N=\frac{2\pi m}{w_0} N=w02πm
要 N N N为整数,则 w 0 2 π = m N \frac{w_0}{2\pi}=\frac{m}{N} 2πw0=Nm为有理数,则该信号为周期。若 m m m和 N N N无公因子,则周期为 N N N
比较:
我们现在构建一组基波周期为 N N N的离散信号为
ϕ k [ n ] = e j k w 0 n = e j k 2 π N n k = 0 , ± 1 , ± 2 , ⋯ \phi_k[n] = e^{jkw_0n}=e^{jk\frac{2\pi}{N}n}\quad k=0,\pm1,\pm2,\cdots ϕk[n]=ejkw0n=ejkN2πnk=0,±1,±2,⋯
x [ n ] = ∑ k a k ϕ k [ n ] = ∑ k a k e j k w 0 n = ∑ k a k e j k 2 π N n x[n]=\sum_ka_k\phi_k[n]=\sum_ka_ke^{jkw_0n}=\sum_ka_ke^{jk\frac{2\pi}{N}n} x[n]=k∑akϕk[n]=k∑akejkw0n=k∑akejkN2πn
由于有 ϕ k [ n ] = ϕ k + r N [ n ] \phi_k[n]=\phi_{k+rN}[n] ϕk[n]=ϕk+rN[n],则 k k k只在 N N N个相继值区间频率是不相同的, k k k可取 0 , ⋯ , N − 1 0,\cdots,N-1 0,⋯,N−1,也可以取 3 , 4 , ⋯ , N + 2 3,4,\cdots,N+2 3,4,⋯,N+2
则
x [ n ] = ∑ k = 0 N − 1 a k ϕ k [ n ] = ∑ k = 0 N − 1 a k e j k w 0 n = ∑ k = 0 N − 1 a k e j k 2 π N n x[n] = \sum_{k=0}^{N-1}a_k\phi_k[n]=\sum_{k=0}^{N-1}a_ke^{jkw_0n}=\sum_{k=0}^{N-1}a_ke^{jk\frac{2\pi}{N}n} x[n]=k=0∑N−1akϕk[n]=k=0∑N−1akejkw0n=k=0∑N−1akejkN2πn
我们可以计算出离散时间傅立叶系数DFS:
a k = 1 N ∑ n = 0 N − 1 x [ n ] e − j k w 0 n − ∞ < k < ∞ a_k = \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-jkw_0n}\quad -\infty < k < \infty ak=N1n=0∑N−1x[n]e−jkw0n−∞<k<∞
证明过程:
例题:
离散时间傅立叶级数性质
周期矩形波:
a k = 1 T S a ( 2 k π w 0 T 1 ) a_k = \frac{1}{T}Sa(\frac{2k\pi w_0}{T_1}) ak=T1Sa(T12kπw0)
T a k = S a ( 2 π w T 1 ) Ta_k = Sa(\frac{2\pi w}{T_1}) Tak=Sa(T12πw)
我们可以将式(17)看成式(18)的包络函数的样本。
我们先把周期信号表示成
x ~ ( t ) = ∑ k = − ∞ ∞ a k e j k w 0 t \tilde{x}(t) = \sum_{k=-\infty}^{\infty}a_ke^{jkw_0t} x~(t)=k=−∞∑∞akejkw0t
a k = 1 T ∫ − T 2 T 2 x ~ ( t ) e j k w 0 t d t a_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\tilde{x}(t)e^{jkw_0t}dt ak=T1∫−2T2Tx~(t)ejkw0tdt
我们将其变化到非周期的,在 ∣ t ∣ < T 2 |t|<\frac{T}{2} ∣t∣<2T时, x ~ ( t ) = x ( t ) \tilde{x}(t)=x(t) x~(t)=x(t)
a k = 1 T ∫ − ∞ ∞ x ( t ) e − j k w 0 t d t a_k=\frac{1}{T}\int_{-\infty}^{\infty}x(t)e^{-jkw_0t}dt ak=T1∫−∞∞x(t)e−jkw0tdt
则,定义 T a k Ta_k Tak的包络 X ( j w ) X(jw) X(jw)为
X ( j w ) = ∫ − ∞ ∞ x ( t ) e − j w t d t X(jw)=\int_{-\infty}^{\infty}x(t)e^{-jwt}dt X(jw)=∫−∞∞x(t)e−jwtdt
a k = 1 T X ( j k w 0 ) a_k=\frac{1}{T}X(jkw_0) ak=T1X(jkw0)
其实周期信号的傅立叶变换也可以得到
x ~ ( t ) = ∑ k = − ∞ ∞ 1 T X ( j k w 0 ) e j k w 0 t \tilde{x}(t)=\sum_{k=-\infty}^{\infty}\frac{1}{T}X(jkw_0)e^{jkw_0t} x~(t)=k=−∞∑∞T1X(jkw0)ejkw0t
由 2 π T = w 0 \frac{2\pi}{T}=w_0 T2π=w0,则
x ~ ( t ) = 1 2 π ∑ k = − ∞ ∞ X ( j k w 0 ) e j k w 0 t w 0 \tilde{x}(t)=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}X(jkw_0)e^{jkw_0t}w_0 x~(t)=2π1k=−∞∑∞X(jkw0)ejkw0tw0
由于 T → ∞ T\rightarrow \infty T→∞, w 0 → 0 w_0\rightarrow0 w0→0, x ~ ( t ) → x ( t ) \tilde{x}(t)\rightarrow x(t) x~(t)→x(t),其实就是变成积分
x ( t ) = 1 2 π ∫ − ∞ ∞ X ( j w ) e j w t d w x(t)= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(jw)e^{jwt}dw x(t)=2π1∫−∞∞X(jw)ejwtdw
X ( j w ) = ∫ − ∞ ∞ x ( t ) e − j w t d t X(jw)=\int_{-\infty}^{\infty}x(t)e^{-jwt}dt X(jw)=∫−∞∞x(t)e−jwtdt
e j w 0 t ↔ 2 π δ ( w − w 0 ) e^{jw_0t}\leftrightarrow 2\pi\delta(w-w_0) ejw0t↔2πδ(w−w0)
周期信号:
X ( j w ) = ∑ k = − ∞ ∞ 2 π a k δ ( w − k w 0 ) X(jw) = \sum_{k=-\infty}^{\infty}2\pi a_k\delta(w-kw_0) X(jw)=k=−∞∑∞2πakδ(w−kw0)
逆变换:
x ( t ) = ∑ k = − ∞ ∞ a k e j k w 0 t x(t) = \sum_{k=-\infty}^{\infty}a_ke^{jkw_0t} x(t)=k=−∞∑∞akejkw0t
傅立叶变换性质
常用傅立叶变换
从离散时间周期信号 x ~ [ n ] \tilde{x}[n] x~[n]
x ~ [ n ] = ∑ k = 0 N − 1 a k e j k 2 π N n \tilde{x}[n]=\sum_{k=0}^{N-1}a_ke^{jk\frac{2\pi}{N}n} x~[n]=k=0∑N−1akejkN2πn
a k = 1 N ∑ n = 0 N − 1 x ~ [ n ] e − j k 2 π N n a_k=\frac{1}{N}\sum_{n=0}^{N-1}\tilde{x}[n]e^{-jk\frac{2\pi}{N}n} ak=N1n=0∑N−1x~[n]e−jkN2πn
变为非周期,即使 − N 1 ≤ n ≤ N 2 -N_1\leq n \leq N_2 −N1≤n≤N2的区间,使得 x [ n ] = x ~ [ n ] x[n] = \tilde{x}[n] x[n]=x~[n],这时候上下限可以变化
a k = 1 N ∑ n = − N 1 N 2 x [ n ] e − j k 2 π N n = 1 N ∑ n = − ∞ ∞ x [ n ] e − j k 2 π N n a_k=\frac{1}{N}\sum_{n=-N_1}^{N_2}x[n]e^{-jk\frac{2\pi}{N}n}=\frac{1}{N}\sum_{n=-\infty}^{\infty}x[n]e^{-jk\frac{2\pi}{N}n} ak=N1n=−N1∑N2x[n]e−jkN2πn=N1n=−∞∑∞x[n]e−jkN2πn
则定义为
X ( e j w ) = ∑ n = − ∞ ∞ x [ n ] e j w n X(e^{jw}) = \sum_{n=-\infty}^{\infty}x[n]e^{jwn} X(ejw)=n=−∞∑∞x[n]ejwn
则 a k a_k ak和 X ( e j w ) X(e^{jw}) X(ejw)之间的关系是:
a k = 1 N X ( e j k w 0 ) a_k = \frac{1}{N}X(e^{jkw_0}) ak=N1X(ejkw0)
代入
x ~ [ n ] = ∑ k = 0 N − 1 1 N X ( e j k w 0 ) e j k w 0 n \tilde{x}[n] = \sum_{k=0}^{N-1}\frac{1}{N}X(e^{jkw_0})e^{jkw_0n} x~[n]=k=0∑N−1N1X(ejkw0)ejkw0n
w 0 = 2 π N w_0=\frac{2\pi}{N} w0=N2π即 1 N = w 0 2 π \frac{1}{N}=\frac{w_0}{2\pi} N1=2πw0,则
x ~ [ n ] = 1 2 π ∑ n = 0 N − 1 X ( e j k w 0 ) e j k w 0 n w 0 \tilde{x}[n]=\frac{1}{2\pi}\sum_{n=0}^{N-1}X(e^{jkw_0})e^{jkw_0n}w_0 x~[n]=2π1n=0∑N−1X(ejkw0)ejkw0nw0
x [ n ] = 1 2 π ∫ 2 π X ( e j w ) e j w n d w x[n]=\frac{1}{2\pi}\int_{2\pi}X(e^{jw})e^{jwn}dw x[n]=2π1∫2πX(ejw)ejwndw
x [ n ] = e j w 0 n x[n] = e^{jw_0n} x[n]=ejw0n
X ( e j w ) = ∑ l = − ∞ ∞ 2 π δ ( w − w 0 − 2 π l ) X(e^{jw})=\sum_{l=-\infty}^{\infty}2\pi\delta(w-w_0-2\pi l) X(ejw)=l=−∞∑∞2πδ(w−w0−2πl)
验证:
周期信号:
x [ n ] = ∑ k = 0 N − 1 a k e j k 2 π N n x[n]=\sum_{k=0}^{N-1}a_ke^{jk\frac{2\pi}{N}n} x[n]=k=0∑N−1akejkN2πn
离散时间傅立叶变换
X ( e j w ) = ∑ k = − ∞ ∞ 2 π a k δ ( w − 2 π k N ) X(e^{jw})=\sum_{k=-\infty}^{\infty}2\pi a_k\delta(w-\frac{2\pi k}{N}) X(ejw)=k=−∞∑∞2πakδ(w−N2πk)
离散时间傅立叶变换性质
x p ( t ) = x ( t ) p ( t ) x_p(t)=x(t)p(t) xp(t)=x(t)p(t)
p ( t ) p(t) p(t)为冲激串函数, T T T为采样周期, p ( t ) p(t) p(t)的基波频率 w s = 2 π T w_s=\frac{2\pi}{T} ws=T2π,也叫采样频率
p ( t ) = ∑ n = − ∞ ∞ δ ( t − n T ) p(t)=\sum_{n=-\infty}^{\infty}\delta(t-nT) p(t)=n=−∞∑∞δ(t−nT)
x p ( t ) = ∑ n = = − ∞ ∞ x ( n T ) δ ( t − n T ) x_p(t)=\sum_{n==-\infty}^{\infty}x(nT)\delta(t-nT) xp(t)=n==−∞∑∞x(nT)δ(t−nT)
时域相乘特性:频域卷积
X p ( j w ) = 1 2 π ∫ − ∞ ∞ X ( j w ) P ( j ( w − θ ) ) d θ X_p(jw) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(jw)P(j(w-\theta))d\theta Xp(jw)=2π1∫−∞∞X(jw)P(j(w−θ))dθ
由于
P ( j w ) = 2 π T ∑ k = − ∞ ∞ δ ( w − k w s ) P(jw) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta(w-kw_s) P(jw)=T2πk=−∞∑∞δ(w−kws)
则 X ( j w ) ∗ δ ( w − w 0 ) = X ( j ( w − w 0 ) ) X(jw)*\delta(w-w_0) =X(j(w-w_0)) X(jw)∗δ(w−w0)=X(j(w−w0))
X p ( j w ) = 1 T ∑ k = − ∞ ∞ X ( j ( w − k w s ) ) X_p(jw)=\frac{1}{T}\sum_{k=-\infty}^{\infty}X(j(w-kw_s)) Xp(jw)=T1k=−∞∑∞X(j(w−kws))
如何从抽样信号恢复出原始信号 内插
*
若信号为
x ( t ) = c o s ( w 0 t + ϕ ) x(t) = cos(w_0t+\phi) x(t)=cos(w0t+ϕ)
x r ( t ) = c o s [ ( w s − w 0 ) t − ϕ ] x_r(t)=cos[(w_s-w_0)t-\phi] xr(t)=cos[(ws−w0)t−ϕ]
现在构建一下连续和离散之间的关系,我们把 x c ( t ) x_c(t) xc(t)和 y c ( t ) y_c(t) yc(t)的连续时间傅立叶变换用 X c ( j w ) X_c(jw) Xc(jw)和 Y c ( j w ) Y_c(jw) Yc(jw)表示,把 x d [ n ] x_d[n] xd[n]和 y d [ n ] y_d[n] yd[n]的离散时间傅立叶变换用 X d ( e j Ω ) X_d(e^{j\Omega}) Xd(ejΩ)和 Y d ( e j Ω ) Y_d(e^{j\Omega}) Yd(ejΩ)表示
x p ( t ) = ∑ n = − ∞ ∞ x c ( n T ) δ ( t − n T ) x_p(t)=\sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT) xp(t)=n=−∞∑∞xc(nT)δ(t−nT)
X p ( j w ) = ∑ n = − ∞ ∞ x c ( n T ) e − j w n T X_p(jw)=\sum_{n=-\infty}^{\infty}x_c(nT)e^{-jwnT} Xp(jw)=n=−∞∑∞xc(nT)e−jwnT
由于 x d [ n ] x_d[n] xd[n]的离散时间傅立叶变换为
X d ( e j Ω ) = ∑ n = − ∞ ∞ x d [ n ] e − j Ω n X_d(e^{j\Omega})=\sum_{n=-\infty}^{\infty}x_d[n]e^{-j\Omega n} Xd(ejΩ)=n=−∞∑∞xd[n]e−jΩn
我们可以得到
X d ( e j Ω ) = X p ( j Ω T ) X_d(e^{j\Omega})=X_p(\frac{j\Omega}{T}) Xd(ejΩ)=Xp(TjΩ)
之前我们得到
X p ( j w ) = 1 T ∑ k = − ∞ ∞ X c ( j ( w − w s ) ) X_p(jw)=\frac{1}{T}\sum_{k=-\infty}^{\infty}X_c(j(w-w_s)) Xp(jw)=T1k=−∞∑∞Xc(j(w−ws))
可以变为
X d ( e j Ω ) = 1 T ∑ k = − ∞ ∞ X c ( j ( Ω − 2 π k T ) ) X_d(e^{j\Omega})=\frac{1}{T}\sum_{k=-\infty}^{\infty}X_c(j(\Omega-\frac{2\pi k}{T})) Xd(ejΩ)=T1k=−∞∑∞Xc(j(Ω−T2πk))
其实就是个频率坐标转换,将 w = Ω T w=\Omega T w=ΩT,我们从下图得到 X ( e j Ω ) X(e^{j\Omega}) X(ejΩ)是 X p ( j w ) X_p(jw) Xp(jw)的重复,且是 Ω \Omega Ω的周期函数,周期为 2 π 2\pi 2π.
我们作出如下理解
我们通过对 x [ n ] x[n] x[n]进行采样得到新的序列 x p [ n ] x_p[n] xp[n],在采样周期为 N N N的时候有值,而采样点之间为0
我们根据上面可以写出
x p [ n ] = x [ n ] p [ n ] = ∑ k = − ∞ ∞ x [ k N ] δ [ n − k N ] x_p[n]=x[n]p[n]=\sum_{k=-\infty}^{\infty}x[kN]\delta[n-kN] xp[n]=x[n]p[n]=k=−∞∑∞x[kN]δ[n−kN]
X p ( e j w ) = 1 2 π ∫ 2 π P ( e j θ ) ( X ( e j ( w − θ ) ) d θ X_p(e^{jw})=\frac{1}{2\pi}\int_{2\pi}P(e^{j\theta})(X(e^{j(w-\theta}))d\theta Xp(ejw)=2π1∫2πP(ejθ)(X(ej(w−θ))dθ
P ( e j w ) = 2 π N ∑ k = − ∞ ∞ δ ( w − k w s ) P(e^{jw})=\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta(w-kw_s) P(ejw)=N2πk=−∞∑∞δ(w−kws)
X p ( e j w ) = 1 N ∑ k = 0 N − 1 X ( e j ( w − k w s ) ) X_p(e^{jw})=\frac{1}{N}\sum_{k=0}^{N-1}X(e^{j(w-kw_s)}) Xp(ejw)=N1k=0∑N−1X(ej(w−kws))
在采样周期的整数倍点上总是相等的
x r [ k N ] = x [ k N ] , k = 0 , ± 1 , ± 2 , ⋯ x_r[kN]=x[kN], \quad k=0,\pm1,\pm2,\cdots xr[kN]=x[kN],k=0,±1,±2,⋯
x p [ n ] x_p[n] xp[n]在采样点之间都是0,用一个新的序列 x b [ n ] x_b[n] xb[n]表示 x p [ n ] x_p[n] xp[n]每隔 N N N点的序列值,这一过程叫抽取,也叫减采样。
x b [ n ] = x p [ n N ] x_b[n]=x_p[nN] xb[n]=xp[nN]
X b ( e j w ) = ∑ k = − ∞ ∞ x b [ k ] e − j w k X_b(e^{jw})=\sum_{k=-\infty}^{\infty}x_b[k]e^{-jwk} Xb(ejw)=k=−∞∑∞xb[k]e−jwk
X b ( e j w ) = ∑ k = − ∞ ∞ x p [ k N ] e − j w k X_b(e^{jw})=\sum_{k=-\infty}^{\infty}x_p[kN]e^{-jwk} Xb(ejw)=k=−∞∑∞xp[kN]e−jwk
令 n = k N n=kN n=kN则
X b ( e j w ) = ∑ n 为 N 的 整 数 倍 x p [ n ] e − j w n N X_b(e^{jw})=\sum_{n为N的整数倍}x_p[n]e^{-\frac{jwn}{N}} Xb(ejw)=n为N的整数倍∑xp[n]e−Njwn
由于当 n n n不为 N N N的整数倍时, x p [ n ] = 0 x_p[n]=0 xp[n]=0,则
X b ( e j w ) = ∑ n = − ∞ ∞ x p [ n ] e − j w n N = X p ( e j w N ) X_b(e^{jw})=\sum_{n=-\infty}^{\infty}x_p[n]e^{-\frac{jwn}{N}}=X_p(e^{\frac{jw}{N}}) Xb(ejw)=n=−∞∑∞xp[n]e−Njwn=Xp(eNjw)
x ~ ( t ) = ∑ k = − ∞ ∞ a k e j k w 0 t \tilde{x}(t) = \sum_{k=-\infty}^{\infty}a_ke^{jkw_0t} x~(t)=k=−∞∑∞akejkw0t
a k = 1 T ∫ − T 2 T 2 x ~ ( t ) e − j k w 0 t d t a_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\tilde{x}(t)e^{-jkw_0t}dt ak=T1∫−2T2Tx~(t)e−jkw0tdt
x ( t ) = 1 2 π ∫ − ∞ ∞ X ( j w ) e j w t d w x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(jw)e^{jwt}dw x(t)=2π1∫−∞∞X(jw)ejwtdw
X ( j w ) = ∫ − ∞ ∞ x ( t ) e − j w t d t X(jw) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt X(jw)=∫−∞∞x(t)e−jwtdt
推导:我们先把周期信号表示成
x ~ ( t ) = ∑ k = − ∞ ∞ a k e j k w 0 t \tilde{x}(t) = \sum_{k=-\infty}^{\infty}a_ke^{jkw_0t} x~(t)=k=−∞∑∞akejkw0t
a k = 1 T ∫ − T 2 T 2 x ~ ( t ) e j k w 0 t d t a_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}\tilde{x}(t)e^{jkw_0t}dt ak=T1∫−2T2Tx~(t)ejkw0tdt
我们将其变化到非周期的,在 ∣ t ∣ < T 2 |t|<\frac{T}{2} ∣t∣<2T时, x ~ ( t ) = x ( t ) \tilde{x}(t)=x(t) x~(t)=x(t)
a k = 1 T ∫ − ∞ ∞ x ( t ) e − j k w 0 t d t a_k=\frac{1}{T}\int_{-\infty}^{\infty}x(t)e^{-jkw_0t}dt ak=T1∫−∞∞x(t)e−jkw0tdt
则,定义 T a k Ta_k Tak的包络 X ( j w ) X(jw) X(jw)为
X ( j w ) = ∫ − ∞ ∞ x ( t ) e − j w t d t X(jw)=\int_{-\infty}^{\infty}x(t)e^{-jwt}dt X(jw)=∫−∞∞x(t)e−jwtdt
a k = 1 T X ( j k w 0 ) a_k=\frac{1}{T}X(jkw_0) ak=T1X(jkw0)
其实周期信号的傅立叶变换也可以得到
x ~ ( t ) = ∑ k = − ∞ ∞ 1 T X ( j k w 0 ) e j k w 0 t \tilde{x}(t)=\sum_{k=-\infty}^{\infty}\frac{1}{T}X(jkw_0)e^{jkw_0t} x~(t)=k=−∞∑∞T1X(jkw0)ejkw0t
由 2 π T = w 0 \frac{2\pi}{T}=w_0 T2π=w0,则
x ~ ( t ) = 1 2 π ∑ k = − ∞ ∞ X ( j k w 0 ) e j k w 0 t w 0 \tilde{x}(t)=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}X(jkw_0)e^{jkw_0t}w_0 x~(t)=2π1k=−∞∑∞X(jkw0)ejkw0tw0
由于 T → ∞ T\rightarrow \infty T→∞, w 0 → 0 w_0\rightarrow0 w0→0, x ~ ( t ) → x ( t ) \tilde{x}(t)\rightarrow x(t) x~(t)→x(t),其实就是变成积分
x ( t ) = 1 2 π ∫ − ∞ ∞ X ( j w ) e j w t d w x(t)= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(jw)e^{jwt}dw x(t)=2π1∫−∞∞X(jw)ejwtdw
X ( j w ) = ∫ − ∞ ∞ x ( t ) e − j w t d t X(jw)=\int_{-\infty}^{\infty}x(t)e^{-jwt}dt X(jw)=∫−∞∞x(t)e−jwtdt
e j k w 0 t ↔ 2 π δ ( w − k w 0 ) e^{jkw_0t} \leftrightarrow 2\pi\delta(w-kw_0) ejkw0t↔2πδ(w−kw0)
x ~ ( t ) = ∑ k = − ∞ ∞ a k e − j k w 0 t \tilde{x}(t)= \sum_{k=-\infty}^{\infty}a_ke^{-jkw_0t} x~(t)=k=−∞∑∞ake−jkw0t
X ( j w ) = ∑ k = − ∞ ∞ 2 π a k δ ( w − k w 0 ) X(jw) = \sum_{k=-\infty}^{\infty}2\pi a_k\delta(w-kw_0) X(jw)=k=−∞∑∞2πakδ(w−kw0)
x ~ [ n ] = ∑ k = 0 N − 1 a k e j k w 0 n \tilde{x}[n] =\sum_{k=0}^{N-1}a_ke^{jkw_0n} x~[n]=k=0∑N−1akejkw0n
a k = 1 N ∑ n = 0 N − 1 x ~ [ n ] e − j k w 0 n a_k = \frac{1}{N}\sum_{n=0}^{N-1}\tilde{x}[n]e^{-jkw_0n} ak=N1n=0∑N−1x~[n]e−jkw0n
其中 w 0 = 2 π N w_0 = \frac{2\pi}{N} w0=N2π DFT就是DFS取得主值区间。
x [ n ] = 1 2 π ∫ 2 π X ( e j w ) e j w n d w x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{jw})e^{jwn}dw x[n]=2π1∫2πX(ejw)ejwndw
X ( e j w ) = ∑ n = − ∞ ∞ x [ n ] e − j w n X(e^{jw})=\sum_{n=-\infty}^{\infty}x[n]e^{-jwn} X(ejw)=n=−∞∑∞x[n]e−jwn
推导过程:从离散时间周期信号 x ~ [ n ] \tilde{x}[n] x~[n]
x ~ [ n ] = ∑ k = 0 N − 1 a k e j k 2 π N n \tilde{x}[n]=\sum_{k=0}^{N-1}a_ke^{jk\frac{2\pi}{N}n} x~[n]=k=0∑N−1akejkN2πn
a k = 1 N ∑ n = 0 N − 1 x ~ [ n ] e − j k 2 π N n a_k=\frac{1}{N}\sum_{n=0}^{N-1}\tilde{x}[n]e^{-jk\frac{2\pi}{N}n} ak=N1n=0∑N−1x~[n]e−jkN2πn
变为非周期,即使 − N 1 ≤ n ≤ N 2 -N_1\leq n \leq N_2 −N1≤n≤N2的区间,使得 x [ n ] = x ~ [ n ] x[n] = \tilde{x}[n] x[n]=x~[n],这时候上下限可以变化
a k = 1 N ∑ n = − N 1 N 2 x [ n ] e − j k 2 π N n = 1 N ∑ n = − ∞ ∞ x [ n ] e − j k 2 π N n a_k=\frac{1}{N}\sum_{n=-N_1}^{N_2}x[n]e^{-jk\frac{2\pi}{N}n}=\frac{1}{N}\sum_{n=-\infty}^{\infty}x[n]e^{-jk\frac{2\pi}{N}n} ak=N1n=−N1∑N2x[n]e−jkN2πn=N1n=−∞∑∞x[n]e−jkN2πn
则定义为
X ( e j w ) = ∑ n = − ∞ ∞ x [ n ] e j w n X(e^{jw}) = \sum_{n=-\infty}^{\infty}x[n]e^{jwn} X(ejw)=n=−∞∑∞x[n]ejwn
则 a k a_k ak和 X ( e j w ) X(e^{jw}) X(ejw)之间的关系是:
a k = 1 N X ( e j k w 0 ) a_k = \frac{1}{N}X(e^{jkw_0}) ak=N1X(ejkw0)
代入
x ~ [ n ] = ∑ k = 0 N − 1 1 N X ( e j k w 0 ) e j k w 0 n \tilde{x}[n] = \sum_{k=0}^{N-1}\frac{1}{N}X(e^{jkw_0})e^{jkw_0n} x~[n]=k=0∑N−1N1X(ejkw0)ejkw0n
w 0 = 2 π N w_0=\frac{2\pi}{N} w0=N2π即 1 N = w 0 2 π \frac{1}{N}=\frac{w_0}{2\pi} N1=2πw0,则
x ~ [ n ] = 1 2 π ∑ n = 0 N − 1 X ( e j k w 0 ) e j k w 0 n w 0 \tilde{x}[n]=\frac{1}{2\pi}\sum_{n=0}^{N-1}X(e^{jkw_0})e^{jkw_0n}w_0 x~[n]=2π1n=0∑N−1X(ejkw0)ejkw0nw0
x [ n ] = 1 2 π ∫ 2 π X ( e j w ) e j w n d w x[n]=\frac{1}{2\pi}\int_{2\pi}X(e^{jw})e^{jwn}dw x[n]=2π1∫2πX(ejw)ejwndw
e j k w 0 n ↔ ∑ l = − ∞ ∞ 2 π δ ( w − k w 0 − 2 π l ) e^{jkw_0n} \leftrightarrow \sum_{l=-\infty}^{\infty}2\pi\delta(w-kw_0-2\pi l) ejkw0n↔l=−∞∑∞2πδ(w−kw0−2πl)
x ~ [ n ] = ∑ k = 0 N − 1 a k e j k w 0 n \tilde{x}[n] = \sum_{k=0}^{N-1}a_ke^{jkw_0n} x~[n]=k=0∑N−1akejkw0n
x ~ [ n ] = ∑ k = − ∞ ∞ 2 π a k δ ( w − k w 0 ) \tilde{x}[n]=\sum_{k=-\infty}^{\infty}2\pi a_k\delta(w-kw_0) x~[n]=k=−∞∑∞2πakδ(w−kw0)
其中 w 0 = 2 π N w_0=\frac{2\pi}{N} w0=N2π