y [ n ] = ∑ l = − ∞ n x [ j ] = ∑ l = − ∞ n − 1 x [ j ] + x [ n ] = y [ n − 1 ] + x [ n ] y[n] = \sum_{l=-\infty}^n x[j] = \sum_{l=-\infty}^{n-1} x[j] + x[n] = y[n-1] + x[n] y[n]=l=−∞∑nx[j]=l=−∞∑n−1x[j]+x[n]=y[n−1]+x[n]
(2) 滑动平均滤波器
y [ n ] = 1 M ∑ l = 0 M − 1 x [ n − l ] y[n] = \frac 1 M \sum_{l = 0}^{M - 1}x[n-l] y[n]=M1l=0∑M−1x[n−l]
(1) 线性系统
线性系统遵循叠加原理:定义输入 x 1 [ n ] → x_1[n] \rightarrow x1[n]→输出 y 1 [ n ] y_1[n] y1[n],
x 2 [ n ] → y 2 [ n ] . . . x_2[n] \rightarrow y_2[n] ... x2[n]→y2[n]...,若系统输入为以上信号的线性组合: x [ n ] = α x 1 [ n ] + β x 2 [ n ] x[n] = \alpha x_1[n] + \beta x_2[n] x[n]=αx1[n]+βx2[n],则输出一定满足: y [ n ] = α y 1 [ n ] + β y 2 [ n ] y[n] = \alpha y_1[n] + \beta y_2[n] y[n]=αy1[n]+βy2[n]。
(2) 时不变系统
时不变系统输入的偏移将导致输出也有同样的偏移:即如果 x 1 [ n ] → y 1 [ n ] x_1[n] \rightarrow y_1[n] x1[n]→y1[n],则一定满足 x [ n ] = x 1 [ n − n 0 ] ⇉ y [ n ] = y 1 [ n − n 0 ] x[n] = x_1[n - n_0] \rightrightarrows y[n] = y_1[n - n_0] x[n]=x1[n−n0]⇉y[n]=y1[n−n0]
时不变系统的以上特性不依赖与n的取值(与输入施加到系统的时刻无关)
线性时不变系统
(3) 因果系统
如果系统输出仅依赖与过去和现在时刻的输入(与将来的输入无关),在满足这种条件的系统称为因果系统。
即若存在输入输出关系对: x 1 [ n ] → y 1 [ n ] x_1[n]\rightarrow y_1[n] x1[n]→y1[n]且 x 2 [ n ] → y 2 [ n ] x_2[n]\rightarrow y_2[n] x2[n]→y2[n],则因果系统满足: x 1 [ n ] = x 2 [ n ] ∀ n < N ⇔ y 1 [ n ] = y 2 [ n ] ∀ n < N x_1[n] = x_2[n] \forall n
(4) 稳定系统
BIBO系统:若x[n]的系统响应是y[n],且对于所有的n值:
∣ x [ n ] ∣ < B x |x[n]|
即输入为有界序列,则对于所有的n值,输出也为有界序列:
∣ y [ n ] ∣ < B y |y[n]|
(1)定义:
给定一个系统,当输入为冲激函数 δ [ n ] \delta[n] δ[n]时,系统输出为冲激响应h[n]
线性时不变系统完全由冲激响应h[n]确定
(2)输入输出关系:
任意一个序列可表示为基础序列与其延迟或超前版本的加权和:
x [ n ] = ∑ k = − ∞ ∞ x [ k ] δ [ n − k ] x[n] = \sum_{k = -\infty}^{\infty}x[k]\delta[n - k] x[n]=k=−∞∑∞x[k]δ[n−k]
换一种表示,上式可理解为任意序列x[n]可看作本省与 δ [ n ] \delta[n] δ[n]的卷积
因此对于LTI系统,由y[n] = x[n] * h[n]
(3)系统级联:
若两个系统分别具有冲激响应 h 1 [ n ] h_1[n] h1[n]和 h 2 [ n ] h_2[n] h2[n],则两者级联的冲激响应h[n]为 h [ n ] = h 1 [ n ] ∗ h 2 h[n] = h_1[n]*h_2 h[n]=h1[n]∗h2
(4)逆系统:
考虑:
z [ n ] = h 2 [ n ] ∗ y [ n ] = h 2 [ n ] ∗ h 1 [ n ] ∗ x [ n ] z[n] = h_2[n]*y[n] = h_2[n]*h_1[n]*x[n] z[n]=h2[n]∗y[n]=h2[n]∗h1[n]∗x[n]
如果: h 2 [ n ] = δ [ n ] h_2[n] = \delta [n] h2[n]=δ[n],则称 h 2 [ n ] h_2[n] h2[n]是 h 1 [ n ] h_1[n] h1[n]的逆系统
(5)系统并联:
两个系统的冲激响应相加就是其并联系统的冲激响应:
h [ n ] = h 1 [ n ] + h 2 [ n ] h[n] = h_1[n] + h_2[n] h[n]=h1[n]+h2[n]
对于LTI系统:
(1)稳定性条件
y [ 0 ] = ∑ k = − ∞ ∞ ∣ x [ − k ] ∣ ∣ h [ k ] ∣ = ∑ k = − ∞ ∞ ∣ h [ k ] ∣ = ∞ y[0] = \sum_{k = -\infty}^\infty |x[-k]||h[k]| = \sum_{k = -\infty}^\infty |h[k]| = \infty y[0]=k=−∞∑∞∣x[−k]∣∣h[k]∣=k=−∞∑∞∣h[k]∣=∞
其不满足BIBO稳定性条件
(2) 因果性条件
(1)定义
一个线性时不变系统对复指数信号的响应也是同样一个复指数信号,不同的只是幅度上的变化。
当输入为复指数时,即 x [ n ] = e j ω 0 n x[n] = e^{j\omega_0n} x[n]=ejω0n
其输出 y [ n ] = ∑ m h [ m ] e j ω 0 ( n − m ) = ∑ m h [ m ] e − j ω 0 m e j ω 0 n = H ( e j ω 0 ) x [ n ] = ∣ H ( e j ω 0 ) ∣ e j ( ω 0 n + θ ( ω 0 ) ) y[n] = \sum_m h[m]e^{j\omega_0(n-m)} = \sum_m h[m]e^{-j\omega_0m}e^{j\omega_0n} = H(e^{j\omega_0})x[n] = |H(e^{j\omega_0})|e^{j(\omega_0n + \theta(\omega_0))} y[n]=∑mh[m]ejω0(n−m)=∑mh[m]e−jω0mejω0n=H(ejω0)x[n]=∣H(ejω0)∣ej(ω0n+θ(ω0))
其中 H ( e j ω 0 ) H(e^{j\omega_0}) H(ejω0)被称作LTI离散时间系统的频率响应
(2)实正弦
一个频率为 ω 0 \omega_0 ω0的实正弦信号通过一个具有实冲激函数h[n]的LSI系统,具有增益放缩 ∣ H ( e j ω 0 ) ∣ |H(e^{j\omega_0})| ∣H(ejω0)∣向往平移 θ ( ω 0 ) \theta(\omega_0) θ(ω0)
A c o s ( ω 0 n + ϕ ) ∗ h [ n ] = A ∣ H ( e j ω 0 ) ∣ c o s ( ω 0 n + ϕ + θ ( ω 0 ) ) Acos(\omega_0n+\phi) * h[n] = A|H(e^{j\omega_0})|cos(\omega_0n+\phi+\theta(\omega_0)) Acos(ω0n+ϕ)∗h[n]=A∣H(ejω0)∣cos(ω0n+ϕ+θ(ω0))
(3)因果指数序列的暂态响应与稳态响应
y [ n ] = x [ n ] ∗ h [ n ] = ∑ m = − ∞ n h [ m ] e j ω 0 ( n − m ) − ∑ m = n + 1 ∞ h [ m ] e j ω 0 ( n − m ) = H ( e j ω 0 ) e j ω 0 n − ( ∑ m = n + 1 ∞ h [ m ] e − j ω 0 m ) e j ω 0 n y[n] = x[n] * h[n] = \sum_{m = -\infty}^n h[m]e^{j\omega_0(n-m)} - \sum_{m = n + 1}^\infty h[m]e^{j\omega_0(n-m)} = H(e^{j\omega_0})e^{j\omega_0n} - (\sum_{m = n + 1}^\infty h[m]e^{-j\omega_0m})e^{j\omega_0n} y[n]=x[n]∗h[n]=m=−∞∑nh[m]ejω0(n−m)−m=n+1∑∞h[m]ejω0(n−m)=H(ejω0)ejω0n−(m=n+1∑∞h[m]e−jω0m)ejω0n
暂态响应: H ( e j ω 0 ) e j ω 0 n H(e^{j\omega_0})e^{j\omega_0n} H(ejω0)ejω0n,与纯正弦输入相同
稳态响应: ( ∑ m = n + 1 ∞ h [ m ] e − j ω 0 m ) e j ω 0 n (\sum_{m = n + 1}^\infty h[m]e^{-j\omega_0m})e^{j\omega_0n} (∑m=n+1∞h[m]e−jω0m)ejω0n,门控的结果
三点滤波器: h [ n ] = α β α h[n] = {\alpha ~~~ \beta ~~~ \alpha} h[n]=α β α
频率响应: ∣ H ( e j ω 0 ) ∣ = ∣ β + 2 α c o s ( ω 0 ) ∣ |H(e^{j\omega_0})| = |\beta+2\alpha cos(\omega_0)| ∣H(ejω0)∣=∣β+2αcos(ω0)∣