设一个周期性功率信号s(t)的周期为 T 0 T_0 T0,其频谱为:
C n = C ( n f 0 ) = 1 T 0 ∫ − T 0 2 T 0 2 s ( t ) e − j π f 0 t d t (1) C_n = C(nf_0) = \frac 1 {T_0}\int_{-\frac {T_0} 2}^{\frac {T_0} 2}s(t)e^{-j\pi f_0t}dt \tag{1} Cn=C(nf0)=T01∫−2T02T0s(t)e−jπf0tdt(1)
式中, f 0 = 1 T 0 f_0 = \frac 1 {T_0} f0=T01,n为整数, − ∞ < n < ∞ -\infty < n < \infty −∞<n<∞
s ( t ) = ∑ n = − ∞ ∞ C n e j 2 π n t / T 0 (2) s(t) = \sum_{n = -\infty}^\infty C_ne^{j2\pi nt/T_0}\tag{2} s(t)=n=−∞∑∞Cnej2πnt/T0(2)
C 0 = 1 T 0 ∫ − T 0 2 T 0 2 s ( t ) d t (3) C_0 = \frac 1 {T_0}\int_{-\frac {T_0} 2}^{\frac {T_0} 2}s(t)dt\tag{3} C0=T01∫−2T02T0s(t)dt(3)
C n = ∣ C n ∣ e j θ n -双边谱,复频谱 C_n = |C_n|e^{j\theta_n}{\text{-双边谱,复频谱}} Cn=∣Cn∣ejθn-双边谱,复频谱
C n -整幅 θ n -相位 C_n {\text{-整幅}} \theta_n {\text{-相位}} Cn-整幅θn-相位
性质
对于物理可实现的实信号,有
C − n = 1 T 0 ∫ − T 0 2 T 0 2 s ( t ) e j π f 0 t d t = [ 1 T 0 ∫ − T 0 2 T 0 2 s ( t ) e − j π f 0 t d t ] ∗ = C n ∗ (4) C_{-n} = \frac 1 {T_0}\int_{-\frac {T_0} 2}^{\frac {T_0} 2}s(t)e^{j\pi f_0t}dt = [\frac 1 {T_0}\int_{-\frac {T_0} 2}^{\frac {T_0} 2}s(t)e^{-j\pi f_0t}dt]^* = C_n^*\tag{4} C−n=T01∫−2T02T0s(t)ejπf0tdt=[T01∫−2T02T0s(t)e−jπf0tdt]∗=Cn∗(4)
正频率部分和负频率部分存在复数共轭关系,即 C n C_n Cn的模偶对称,相位奇对称。
将4式代入2式可得:
s ( t ) = ∑ n = − ∞ ∞ C n e j 2 π n t / T 0 = C 0 + ∑ n = 1 ∞ [ a n 2 + b n 2 c o s ( 2 π n t / T 0 + θ ) ] (5) s(t) = \sum_{n = -\infty}^\infty C_ne^{j2\pi nt/T_0} = C_0+\sum_{n=1}^\infty [\sqrt{a_n^2+b_n^2}cos(2\pi nt/T_0+\theta)] \tag{5} s(t)=n=−∞∑∞Cnej2πnt/T0=C0+n=1∑∞[an2+bn2cos(2πnt/T0+θ)](5)
表明:
信号功率则为:
P = lim T → ∞ 1 T ∫ − ∞ ∞ ∣ S T ( f ) ∣ 2 d f = ∫ − ∞ ∞ P ( f ) d f (12) P = \lim_{T\rightarrow \infty}\frac 1 T \int_{-\infty}^{\infty}|S_T(f)|^2df=\int_{-\infty}^\infty P(f)df \tag{12} P=T→∞limT1∫−∞∞∣ST(f)∣2df=∫−∞∞P(f)df(12)
若此功率信号具有周期性,且周期为T,则:
P = 1 T ∫ − T 0 2 T 0 2 s 2 ( t ) d t = ∑ n = − ∞ ∞ ∣ C n ∣ 2 (13) P = \frac 1 T \int_{- \frac {T_0} 2}^{\frac {T_0}2} s^2(t)dt = \sum_{n=-\infty}^\infty |C_n|^2 \tag{13} P=T1∫−2T02T0s2(t)dt=n=−∞∑∞∣Cn∣2(13)
C n C_n Cn为此周期信号的傅里叶级数的系数, ∣ C n ∣ 2 |C_n|^2 ∣Cn∣2为第n次谐波的功率。
利用 δ \delta δ函数可将上式表示为:
P = ∫ − ∞ ∞ ∣ C ( f ) ∣ 2 δ ( f − n f 0 ) d f (14) P = \int_{-\infty}^{\infty}|C(f)|^2\delta(f-nf_0)df \tag{14} P=∫−∞∞∣C(f)∣2δ(f−nf0)df(14)
式中: C ( f ) = { C n f = n f 0 0 其他处 C(f)=\left\{\begin{matrix}C_n &&& f=nf_0 \\0 &&&其他处\end{matrix}\right. C(f)={Cn0f=nf0其他处
上式的被积因子就是此信号的功率谱密度P(f)了,即
P ( f ) = ∑ n = − ∞ ∞ ∣ C ( f ) ∣ 2 δ ( f − n f 0 ) (15) P(f) = \sum_{n=-\infty}^\infty |C(f)|^2\delta(f-nf_0) \tag{15} P(f)=n=−∞∑∞∣C(f)∣2δ(f−nf0)(15)
R ( 0 ) = lim T → ∞ 1 T ∫ − T / 2 T / 2 s 2 ( t ) d t = P R(0) = \mathop {\lim }\limits_{T \to \infty } \frac{1}{T}\int_{ - T/2}^{T/2} {{s^2}(t)dt} = P R(0)=T→∞limT1∫−T/2T/2s2(t)dt=P
R ( τ ) = ∫ − ∞ ∞ P ( f ) e j 2 π f τ d f R(\tau ) = \int_{ - \infty }^\infty {P(f){e^{j2\pi f\tau }}df} R(τ)=∫−∞∞P(f)ej2πfτdf P ( f ) = ∫ − ∞ ∞ R ( τ ) e − j 2 π f τ d τ P(f) = \int_{ - \infty }^\infty {R(\tau ){e^{ - j2\pi f\tau }}d\tau } P(f)=∫−∞∞R(τ)e−j2πfτdτ
R 12 ( τ ) R_{12}(\tau) R12(τ)和时间 t 无关,只和时间差 τ \tau τ有关。
R 12 ( τ ) R_{12}(\tau) R12(τ)和两个信号相乘的前后次序有关:
【证】令 x = t + τ x = t + \tau x=t+τ,则
R 21 ( τ ) = ∫ − ∞ ∞ s 2 ( t ) s 1 ( t + τ ) d t = ∫ − ∞ ∞ s 2 ( x − τ ) s 1 ( x ) d x = ∫ − ∞ ∞ s 1 ( x ) s 2 [ x + ( − τ ) ] d x = R 12 ( − τ ) \begin{array}{l} {R_{21}}(\tau ) = \int_{ - \infty }^\infty {{s_2}(t){s_1}(t + \tau )dt = \int_{ - \infty }^\infty {{s_2}(x - \tau ){s_1}(x)dx} } \\ = \int_{ - \infty }^\infty {{s_1}(x){s_2}[x + ( - \tau )]dx = {R_{12}}( - \tau )} \end{array} R21(τ)=∫−∞∞s2(t)s1(t+τ)dt=∫−∞∞s2(x−τ)s1(x)dx=∫−∞∞s1(x)s2[x+(−τ)]dx=R12(−τ)
互相关函数 R 12 ( τ ) R_{12}(\tau) R12(τ)和互能量谱密度 S 1 2 ( f ) S_12(f) S12(f)是一对傅里叶变换
R 12 ( τ ) = ∫ − ∞ ∞ S 12 ( f ) e j 2 π f τ d f {R_{12}}(\tau) = \int_{ - \infty }^\infty {{S_{12}}(f){e^{j2\pi f\tau }}df} R12(τ)=∫−∞∞S12(f)ej2πfτdf S 12 ( f ) = ∫ − ∞ ∞ R 12 ( τ ) e − j 2 π f τ d τ {S_{12}}(f) = \int_{ - \infty }^\infty {{R_{12}}(\tau ){e^{ - j2\pi f\tau }}d\tau } S12(f)=∫−∞∞R12(τ)e−j2πfτdτ
定义:
性质: