ADSP重点习题

第二章
- 例题2.1.5
- 习题2.12
- 习题2.13
第三章
- 习题3.2
- 习题3.7
- - 尤利-沃克方程：
  - PACS（部分自相关序列的计算）
- 习题3.11
- 习题3.21
- - 低阶极点模型(低阶AP模型)
  - 低阶零点模型(低阶MA模型)
- 习题3.23
- - 极点-零点模型

第二章

例题2.1.5

Consider the following harmonic process
x ( n ) = c o s ( 0.1 π n + φ 1 ) + 2 s i n ( 1.5 n + φ 2 ) . x(n)=cos(0.1\pi n+ \varphi_{1})+2sin(1.5n+\varphi_{2})\,. x(n)=cos(0.1πn+φ1)+2sin(1.5n+φ2).
where φ 1 \varphi_{1} φ1 and φ 2 \varphi_{2} φ2 are IID random variables uniformly distributed in the interval [ 0 , 2 π ] [0,2\pi] [0,2π]. The first component of x ( n ) x(n) x(n) is periodic with ω 1 = 0.1 π \omega_1=0.1\pi ω1=0.1π and period equal to 20 while the second component is almost periodic with ω 1 = 0.1 π = 1.5 \omega_1=0.1\pi=1.5 ω1=0.1π=1.5. Thus sequence x ( n ) x(n) x(n) is almost periodic.
The mean of x ( n ) x(n) x(n) is
μ x ( n ) = E { x ( n ) } = E { c o s ( 0.1 π n + φ 1 ) + 2 s i n ( 1.5 n + φ 2 ) } = 0 \mu_x(n)=E\{ x(n)\} \\ = E\{cos(0.1\pi n+ \varphi_{1})+2sin(1.5n+\varphi_{2})\} \\ = 0 μx(n)=E{x(n)}=E{cos(0.1πn+φ1)+2sin(1.5n+φ2)}=0
and the autocorrelation sequence(using mutual independence between φ 1 \varphi_{1} φ1 and φ 2 \varphi_{2} φ2) is:
r x ( n 1 , n 2 ) = E { x ( n 1 ) x ∗ ( n 2 ) } = E { c o s ( 0.1 π n 1 + φ 1 ) c o s ( 0.1 π n 2 + φ 1 ) } + E { 2 s i n ( 1.5 n 1 + φ 2 ) 2 c o s ( 1.5 n 2 + φ 2 ) } = 1 2 c o s [ 0.1 π ( n 1 − n 2 ) ] + 2 c o s [ 1.5 ( n 1 − n 2 ) ] r_{x}(n_{1},n_{2})=E\{ x(n_{1})x^{*}(n_{2})\} \\ =E\{cos(0.1\pi n_{1}+\varphi_{1})cos(0.1\pi n_{2}+\varphi_{1})\} \\ +E\{2sin(1.5n_{1}+\varphi_{2})2cos(1.5n_{2}+\varphi_{2})\} \\ =\frac{1}{2}cos[0.1\pi (n_{1}-n_{2})] + 2cos[1.5(n_{1}-n_{2})] rx(n1,n2)=E{x(n1)x∗(n2)}=E{cos(0.1πn1+φ1)cos(0.1πn2+φ1)}+E{2sin(1.5n1+φ2)2cos(1.5n2+φ2)}=21cos[0.1π(n1−n2)]+2cos[1.5(n1−n2)]
or
r x ( l ) = 1 2 c o s 0.1 π l + 2 c o s 1.5 l r_{x}(l)=\frac{1}{2}cos0.1\pi l + 2cos1.5l rx(l)=21cos0.1πl+2cos1.5l
The power spectrum R x ( e j w ) R_{x}(e^{jw}) Rx(ejw) is given by:
R x ( e j w ) = 2 π δ ( ω + 1.5 ) + π 2 δ ( ω + 0.1 π ) + 2 π δ ( ω − 1.5 ) + π 2 δ ( ω − 0.1 π ) R_{x}(e^{jw})=2\pi \delta(\omega+1.5)+\frac{\pi}{2}\delta(\omega+0.1 \pi) \\+2\pi \delta(\omega-1.5)+\frac{\pi}{2}\delta(\omega-0.1 \pi) Rx(ejw)=2πδ(ω+1.5)+2πδ(ω+0.1π)+2πδ(ω−1.5)+2πδ(ω−0.1π)
补充：
积化和差，和差化积公式：

习题2.12

A causal LTI system which is described by the difference equation
y ( n ) = 1 2 y ( n − 1 ) + x ( n ) + 1 3 x ( n − 1 ) y(n)=\frac{1}{2}y(n-1)+x(n)+\frac{1}{3}x(n-1) y(n)=21y(n−1)+x(n)+31x(n−1)
is driven by a zero-mean WSS process with autocorrelation r x ( l ) = 0. 5 ∣ l ∣ r_{x}(l)=0.5^{|l|} rx(l)=0.5∣l∣.
(a)Determine the PSD and the autocorrelation of the output sequence y ( n ) y(n) y(n).
易得： H ( z ) = 1 + 1 3 z − 1 1 − 1 2 z − 1 , ∣ z ∣ < 1 2 R x ( z ) = 3 4 5 4 − 1 2 ( z + z − 1 ) , 1 2 < ∣ z ∣ < 2 H(z)=\frac{1+\frac{1}{3}z^{-1}}{1-\frac{1}{2}z^{-1}}, |z|<\frac{1}{2} \\ R_{x}(z)=\frac{\frac{3}{4}}{\frac{5}{4}-\frac{1}{2}(z+z^{-1})}, \frac{1}{2}<|z|<2 H(z)=1−21z−11+31z−1,∣z∣<21Rx(z)=45−21(z+z−1)43,21<∣z∣<2

(b)Determine the cross-correlation r x y ( l ) r_{xy}(l) rxy(l) and cross-PSD R x y ( e j w ) R_{xy}(e^{jw}) Rxy(ejw) between the input and output signals.

知识点：
R x y ( z ) = H ∗ ( 1 z ∗ ) R x ( z ) R y x ( z ) = H ( z ) R x ( z ) R y ( z ) = H ( z ) H ∗ ( 1 z ∗ ) R x ( z ) R_{xy}(z)=H^{*}(\frac{1}{z^{*}})R_{x}(z) \\ R_{yx}(z)=H(z)R_{x}(z) \\ R_{y}(z)=H(z)H^{*}(\frac{1}{z^{*}})R_{x}(z) Rxy(z)=H∗(z∗1)Rx(z)Ryx(z)=H(z)Rx(z)Ry(z)=H(z)H∗(z∗1)Rx(z)

习题2.13

A WSS process with PSD R x ( e j w ) = 1 / ( 1.64 + 1.6 c o s w ) R_{x}(e^{jw})=1/(1.64+1.6cosw) Rx(ejw)=1/(1.64+1.6cosw) is applied to a causal system described by the following difference equation
y ( n ) = 0.6 y ( n − 1 ) + x ( n ) + 1.25 x ( n − 1 ) y(n)=0.6y(n-1)+x(n)+1.25x(n-1) y(n)=0.6y(n−1)+x(n)+1.25x(n−1)
Compute (a)the PSD of the output and (b)the cross-PSD R x y ( e j w ) R_{xy}(e^{jw}) Rxy(ejw) between input and output.

涉及知识点同上题

第三章

习题3.2

Consider a zero-mean random sequence x ( n ) x(n) x(n) with PSD
R x ( e j w ) = 5 + 3 c o s w 17 + 8 c o s w R_{x}(e^{jw})=\frac{5+3cosw}{17+8cosw} Rx(ejw)=17+8cosw5+3cosw
(a)Determine the innovations representation of the process x ( n ) x(n) x(n)

(b)Find the autocorrelation sequence r x ( l ) r_{x}(l) rx(l).

知识点：
R x ( z ) = σ w 2 H ( z ) H ∗ ( 1 z ∗ ) R_{x}(z)=\sigma_{w} ^{2}H(z)H^{*}(\frac{1}{z^{*}}) Rx(z)=σw2H(z)H∗(z∗1)

习题3.7

Use the Yule-Walker equations to determine the autocorrelation and partial autocorrelation coefficients of the following AR models,assuming that w ( n ) w(n) w(n)~ W N ( 0 , 1 ) WN(0,1) WN(0,1).
(a) x ( n ) = 0.5 x ( n − 1 ) + w ( n ) x(n)=0.5x(n-1)+w(n) x(n)=0.5x(n−1)+w(n)

(b) x ( n ) = 1.5 x ( n − 1 ) − 0.6 x ( n − 2 ) + w ( n ) x(n)=1.5x(n-1)-0.6x(n-2)+w(n) x(n)=1.5x(n−1)−0.6x(n−2)+w(n)

知识点：

尤利-沃克方程：

对于全极点模型，由(4.2.15)和(4.2.16):
∑ k = 0 P a k r h ( l − k ) = d 0 h ∗ ( − l ) , − ∞ < l < ∞ ∑ k = 0 P a k r h ( l − k ) = 0 , l > 0 \sum_{k=0}^{P}a_{k}r_{h}(l-k)=d_{0}h^{*}(-l),-\infty<l<\infty \\ \sum_{k=0}^{P}a_{k}r_{h}(l-k)=0,l>0 k=0∑Pakrh(l−k)=d0h∗(−l),−∞<l<∞k=0∑Pakrh(l−k)=0,l>0
写成矩阵形式，即尤利-沃克(Yule-Walker)方程：
[ r h ( 0 ) r h ( 1 ) ⋯ r h ( P ) r h ∗ ( 1 ) r h ( 0 ) ⋯ r h ( P − 1 ) ⋮ ⋮ ⋱ ⋮ r h ∗ ( P ) r h ∗ ( P − 1 ) ⋯ r h ( 0 ) ] [ 1 a 1 ⋮ a p ] = [ ∣ d 0 ∣ 2 0 ⋮ 0 ] \left[\begin{matrix} r_{h}(0) & r_{h}(1) & \cdots & r_{h}(P) \\ r^{*}_{h}(1) & r_{h}(0) & \cdots & r_{h}(P-1) \\ \vdots & \vdots & \ddots & \vdots \\ r_{h} ^{*}(P) & r_{h}^{*}(P-1) & \cdots & r_{h}(0) \end{matrix}\right] \left[ \begin{matrix} 1 \\ a_{1} \\ \vdots \\ a_{p} \end{matrix}\right]=\left[\begin{matrix}|d_0|^2 \\ 0 \\ \vdots \\ 0\end{matrix}\right] ⎣⎢⎢⎢⎡rh(0)rh∗(1)⋮rh∗(P)rh(1)rh(0)⋮rh∗(P−1)⋯⋯⋱⋯rh(P)rh(P−1)⋮rh(0)⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡1a1⋮ap⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡∣d0∣20⋮0⎦⎥⎥⎥⎤
即为：
R h a = − r h \mathbf{R_{h}a}=-\mathbf{r_{h}} Rha=−rh
同时，我们也可以根据输出过程 x ( n ) x(n) x(n)的自相关表示模型参数：
R x a = − r x \mathbf{R_{x}a}=-\mathbf{r_{x}} Rxa=−rx
其中 r x = σ w 2 r h \mathbf{r_{x}}=\sigma_{w}^2 \mathbf{r_{h}} rx=σw2rh，因为 r x ( l ) = σ w 2 r h ( l ) r_{x}(l)=\sigma_{w}^2r_{h}(l) rx(l)=σw2rh(l)

PACS（部分自相关序列的计算）

[ 1 ρ ( 1 ) ⋯ ρ ( m − 1 ) ρ ∗ ( 1 ) 1 ⋯ ⋮ ⋮ ⋮ ⋱ ⋮ ρ ∗ ( m − 1 ) ⋯ ρ ∗ ( 1 ) 1 ] [ a 1 ( m ) a 2 ( m ) ⋮ a m ( m ) ] = [ ρ ∗ ( 1 ) ρ ∗ ( 2 ) ⋮ ρ ∗ ( m ) ] \left[\begin{matrix} 1 & \rho(1) & \cdots & \rho(m-1) \\ \rho^{*}(1) & 1 & \cdots &\vdots \\ \vdots & \vdots & \ddots & \vdots \\ \rho ^{*}(m-1) & \cdots & \rho^{*}(1) & 1 \end{matrix}\right] \left[ \begin{matrix} a^{(m)}_1 \\ a^{(m)}_{2} \\ \vdots \\ a^{(m)}_{m} \end{matrix}\right]=\left[\begin{matrix}\rho^{*}(1) \\ \rho^{*}(2) \\ \vdots \\ \rho^{*}(m) \end{matrix}\right] ⎣⎢⎢⎢⎢⎡1ρ∗(1)⋮ρ∗(m−1)ρ(1)1⋮⋯⋯⋯⋱ρ∗(1)ρ(m−1)⋮⋮1⎦⎥⎥⎥⎥⎤⎣⎢⎢⎢⎢⎡a1(m)a2(m)⋮am(m)⎦⎥⎥⎥⎥⎤=⎣⎢⎢⎢⎡ρ∗(1)ρ∗(2)⋮ρ∗(m)⎦⎥⎥⎥⎤,对于 m > P , a m ( m ) = 0 m>P,a^{(m)}_{m}=0 m>P,am(m)=0，我们可以使用序列 a m ( m ) , m = 1 , 2 , ⋯ a^{(m)}_{m},m=1,2,\cdots am(m),m=1,2,⋯这一被称为部分自相关的序列去确定全极点模型的阶。 ρ ( l ) \rho(l) ρ(l)为归一化的自相关序列， ρ ( l ) = r ( l ) r ( 0 ) \rho(l)=\frac{r(l)}{r(0)} ρ(l)=r(0)r(l)

习题3.11

Consider the following AR(2) models:(i) x ( n ) = 0.6 x ( n − 1 ) + 0.3 x ( n − 2 ) + w ( n ) x(n)=0.6x(n-1)+0.3x(n-2)+w(n) x(n)=0.6x(n−1)+0.3x(n−2)+w(n) and (ii) x ( n ) = 0.8 x ( n − 1 ) − 0.5 x ( n − 2 ) + w ( n ) x(n)=0.8x(n-1)-0.5x(n-2)+w(n) x(n)=0.8x(n−1)−0.5x(n−2)+w(n),where w ( n ) w(n) w(n)~ W G N ( 0 , 1 ) WGN(0,1) WGN(0,1).
(a)Find the general expression for the normalized autocorrelation sequence ρ ( l ) \rho(l) ρ(l), and determine σ x 2 \sigma_{x}^2 σx2.

(b)Plot ρ ( l ) 0 15 {\rho(l)}^{15}_{0} ρ(l)015 and check if the models exhibit pseudoperiodic behavior.

From these plots, x ( n ) = 0.6 x ( n − 1 ) + 0.3 x ( n − 2 ) + w ( n ) x(n) = 0.6x(n − 1) + 0.3x(n − 2) + w(n) x(n)=0.6x(n−1)+0.3x(n−2)+w(n) does not show any pseudo-periodic behavior, while x ( n ) = 0.8 x ( n − 1 ) − 0.5 x ( n − 2 ) + w ( n ) x(n) = 0.8x(n − 1) − 0.5x(n − 2) + w(n) x(n)=0.8x(n−1)−0.5x(n−2)+w(n) does show pseudo-periodic behavior.
( c)Justify your answer in part(b) by ploting the PSD of the two models.
Clearly, the first model has no fundamental frequency.While the second system does indeed have a large no zero frequency component.
知识点：
二阶模型确定好 r ( 0 ) , r ( 1 ) r(0),r(1) r(0),r(1)的值和 r ( l ) r(l) r(l)与 r ( l − 1 ) , r ( l − 2 ) r(l-1),r(l-2) r(l−1),r(l−2)的关系式之后，就可以画出整个 r ( l ) , ρ ( l ) r(l),\rho(l) r(l),ρ(l)曲线。

习题3.21

Consider the MA(2) model x ( n ) = w ( n ) − 0.1 w ( n − 1 ) + 0.2 w ( n − 2 ) x(n)=w(n)-0.1w(n-1)+0.2w(n-2) x(n)=w(n)−0.1w(n−1)+0.2w(n−2).
(a) Is the process x ( n ) x(n) x(n) stationary,why?
The process x ( n ) x(n) x(n) is a linear combination of a stationary process w ( n ) w(n) w(n), therefore x ( n ) x(n) x(n) is stationary.
(b) Is the model minimum-phase,why?

( c) Determine the autocorrelation and the partial autocorrelation of the process.

知识点：

低阶极点模型(低阶AP模型)

一阶全极点模型：AP(1)
一个AP(1)模型有传递函数如下：
H ( z ) = d 0 1 + a z − 1 H(z)=\frac{d_0}{1+az^{-1}} H(z)=1+az−1d0,如果 − 1 < a < 1 -1<a<1 −1<a<1，则 H ( z ) H(z) H(z)是最小相位的。PACS长度为一： a 1 = − r ( 1 ) r ( 0 ) = − ρ ( 1 ) a_1=-\frac{r(1)}{r(0)}=-\rho(1) a1=−r(0)r(1)=−ρ(1)
滤波器冲激响应： h ( n ) = d 0 ( − a ) n u ( n ) h(n)=d_0(-a)^nu(n) h(n)=d0(−a)nu(n)
自相关序列： r ( l ) = r ( 0 ) ( − a ) ∣ l ∣ , r ( 0 ) = d 0 2 1 − a 2 r(l)=r(0)(-a)^{|l|},r(0)=\frac{d^2_0}{1-a^2} r(l)=r(0)(−a)∣l∣,r(0)=1−a2d02
自相关z变换： R ( z ) = d 0 2 ( 1 + a z − 1 ) ( 1 + a z ) R(z)=\frac{d^2_0}{(1+az^{-1})(1+az)} R(z)=(1+az−1)(1+az)d02
频谱： R ( e j w ) = d 0 2 1 + 2 a c o s w + a 2 R(e^{jw})=\frac{d^2_0}{1+2acosw+a^2} R(ejw)=1+2acosw+a2d02
二阶全极点模型：AP(2)
AP(2)模型系统函数：
H ( z ) = d 0 1 + a 1 z − 1 + a 2 z − 2 = d 0 ( 1 − p 1 z − 1 ) ( 1 − p 2 z − 1 ) H(z)=\frac{d_0}{1+a_1z^{-1}+a_2z^{-2}}=\frac{d_0}{(1-p_1z^{-1})(1-p_2z^{-1})} H(z)=1+a1z−1+a2z−2d0=(1−p1z−1)(1−p2z−1)d0
其中： a 1 = − ( p 1 + p 2 ) , a 2 = p 1 p 2 a_1=-(p_1+p_2),a_2=p_1p_2 a1=−(p1+p2),a2=p1p2
最小相位条件：
− 1 < a 2 < 1 a 2 − a 1 > − 1 a 2 + a 1 > − 1 -1<a_2<1 \\ a_2-a_1>-1 \\ a_2+a_1>-1 −1<a2<1a2−a1>−1a2+a1>−1

低阶零点模型(低阶MA模型)

一阶全零点模型：AZ(1)
H ( z ) = G ( 1 + d 1 z − 1 ) H(z)=G(1+d_1z^{-1}) H(z)=G(1+d1z−1) d 1 d_1 d1取任何值，模型稳定，当 − 1 < d 1 < 1 -1<d_1<1 −1<d1<1时相位最小。
R h ( z ) = H ( z ) H ( z − 1 ) = G 2 [ d 1 z + ( 1 + d 1 2 ) + d 1 z − 1 ] R_h(z)=H(z)H(z^{-1})=G^2[d_1z+(1+d_1^2)+d_1z^{-1}] Rh(z)=H(z)H(z−1)=G2[d1z+(1+d12)+d1z−1]
自相关函数为上式的z反变换，有 r h ( 0 ) = G 2 ( 1 + d 1 2 ) , r h ( 1 ) = r h ( − 1 ) = G 2 d 1 r_h(0)=G^2(1+d_1^2),r_h(1)=r_h(-1)=G^2d_1 rh(0)=G2(1+d12),rh(1)=rh(−1)=G2d1,其他情况时r_h(l)=0。归一化自相关函数：
ρ ( l ) = { 1 l = 0 d 1 1 + d 1 2 l = ± 1 0 ∣ l ∣ ≥ 2 \rho(l)=\left\{ \begin{array}{lr} 1 & l=0 \\ \frac{d_1}{1+d_1^2} & l=\pm1\\ 0 & |l|\ge2 \end{array} \right. ρ(l)=⎩⎨⎧11+d12d10l=0l=±1∣l∣≥2
二阶全零点模型：AZ(2)
系统函数为： H ( z ) = G ( 1 + d 1 z − 1 + d 2 z − 2 ) H(z)=G(1+d_1z^{-1}+d_2z^{-2}) H(z)=G(1+d1z−1+d2z−2)
对于 d 1 , d 2 d_1,d_2 d1,d2所有值，系统稳定，如果满足：
− 1 < d 2 < 1 d 2 − d 1 > − 1 d 2 + d 1 > − 1 -1<d_2<1 \\ d_2-d_1>-1 \\ d_2+d_1>-1 −1<d2<1d2−d1>−1d2+d1>−1
则系统为最小相位系统。归一化自相关函数：
ρ ( l ) = { 1 l = 0 d 1 ( 1 + d 2 ) 1 + d 1 2 + d 2 2 l = ± 1 d 2 1 + d 1 2 + d 2 2 l = ± 2 0 ∣ l ∣ ≥ 2 \rho(l)=\left\{ \begin{array}{lr} 1 & l=0 \\ \frac{d_1(1+d_2)}{1+d_1^2+d_2^2} & l=\pm1\\ \frac{d_2}{1+d_1^2+d_2^2} & l=\pm2 \\ 0 & |l|\ge2 \end{array} \right. ρ(l)=⎩⎪⎪⎪⎨⎪⎪⎪⎧11+d12+d22d1(1+d2)1+d12+d22d20l=0l=±1l=±2∣l∣≥2
频谱： R h ( e j w ) = G 2 [ ( 1 + d 1 2 + d 2 2 ) + 2 d 1 ( 1 + d 2 ) c o s w + 2 d 2 c o s 2 w ] R_h(e^{jw})=G^2[(1+d_1^2+d_2^2)+2d_1(1+d_2)cosw+2d_2cos2w] Rh(ejw)=G2[(1+d12+d22)+2d1(1+d2)cosw+2d2cos2w]

习题3.23

Determine the coefficients of a PZ(2,1) model with autocorrelation values r h ( 0 ) = 19 , r h ( 1 ) = 9 , r h ( 2 ) = − 5 , r h ( 3 ) = − 7 r_h(0)=19,r_h(1)=9,r_h(2)=-5,r_h(3)=-7 rh(0)=19,rh(1)=9,rh(2)=−5,rh(3)=−7
易得：
[ 9 19 − 5 9 ] [ a 1 a 2 ] = [ 5 7 ] \left[\begin{matrix} 9 & 19 \\ -5 & 9 \\ \end{matrix}\right] \left[\begin{matrix} a_1 \\ a_2 \end{matrix}\right]=\left[\begin{matrix}5 \\ 7\end{matrix}\right] [9−5199][a1a2]=[57]
所以， a 1 = − 1 / 2 , a 2 = 1 / 2 a_1=-1/2,a_2=1/2 a1=−1/2,a2=1/2。由
r a ( l ) = ∑ k = 0 P − ∣ l ∣ a k a k + ∣ l ∣ ∗ , − P ≤ l ≤ P r_a(l)=\sum^{P-|l|}_{k=0}a_ka^{*}_{k+|l|},-P\le l \le P ra(l)=k=0∑P−∣l∣akak+∣l∣∗,−P≤l≤P得： r a ( 0 ) = 3 / 2 , r a ( ± 1 ) = − 3 / 4 , r a ( ± 2 ) = 1 / 2 r_a(0)=3/2,r_a(\pm1)=-3/4,r_a(\pm2)=1/2 ra(0)=3/2,ra(±1)=−3/4,ra(±2)=1/2由：
r d ( l ) = ∑ k = − P P r a ( k ) r h ( l − k ) r_d(l)=\sum_{k=-P}^{P}r_a(k)r_h(l-k) rd(l)=k=−P∑Pra(k)rh(l−k)得： R d ( z ) = 4 z + 10 + 4 z − 1 = 4 ( 1 + 1 2 z − 1 ) ( z + 2 ) R_d(z)=4z+10+4z^{-1}=4(1+\frac{1}{2z^{-1}})(z+2) Rd(z)=4z+10+4z−1=4(1+2z−11)(z+2)取其因果部分得到 D ( z ) D(z) D(z)，即 D ( z ) = 2 [ 1 + 1 / ( 2 z − 1 ) ] D(z)=2[1+1/(2z^{-1})] D(z)=2[1+1/(2z−1)]和 d 1 = 1 / 2 d_1=1/2 d1=1/2。
知识点：

极点-零点模型

x ( n ) = − ∑ k = 1 P a k x ( n − k ) + ∑ k = 0 Q d k w ( n − k ) x(n)=-\sum^{P}_{k=1}a_kx(n-k)+\sum^{Q}_{k=0}d_kw(n-k) x(n)=−k=1∑Pakx(n−k)+k=0∑Qdkw(n−k)
其冲激响应可以写成递归形式：
h ( n ) = − ∑ k = 1 P a k h ( n − k ) + d n , n ≥ 0 h(n)=-\sum_{k=1}^{P}a_kh(n-k)+d_n,n\ge 0 h(n)=−k=1∑Pakh(n−k)+dn,n≥0
其中， d n = 0 , n > Q d_n=0,n>Q dn=0,n>Q
自相关：
H ( Z ) H(Z) H(Z)的复谱：
R h ( z ) = H ( z ) H ( 1 z ∗ ) = D ( z ) D ( 1 / z ∗ ) A ( z ) A ( 1 / z ∗ ) = R d ( z ) R a ( z ) R_h(z)=H(z)H(\frac{1}{z^{*}})=\frac{D(z)D(1/z^{*})}{A(z)A(1/z^{*})}=\frac{R_d(z)}{R_a(z)} Rh(z)=H(z)H(z∗1)=A(z)A(1/z∗)D(z)D(1/z∗)=Ra(z)Rd(z)
对于自相关序列 r h ( l ) r_h(l) rh(l)，有：
∑ k = 0 P a k r h ( l − k ) = 0 , l > Q \sum_{k=0}^{P}a_kr_h(l-k)=0,l>Q k=0∑Pakrh(l−k)=0,l>Q,写成矩阵形式为：
[ r h ( Q ) r h ( Q + 1 ) ⋯ r h ( Q + P − 1 ) r h ∗ ( Q − 1 ) r h ( Q ) ⋯ r h ( Q + P − 2 ) ⋮ ⋮ ⋱ ⋮ r h ∗ ( Q − P + 1 ) r h ∗ ( Q − P + 2 ) ⋯ r h ( Q ) ] [ a 1 a 2 ⋮ a p ] = [ ∣ r h ( Q − 1 ) r h ( Q − 2 ) ⋮ r h ( Q − P ) ] \left[\begin{matrix} r_{h}(Q) & r_{h}(Q+1) & \cdots & r_{h}(Q+P-1) \\ r^{*}_{h}(Q-1) & r_{h}(Q) & \cdots & r_{h}(Q+P-2) \\ \vdots & \vdots & \ddots & \vdots \\ r_{h} ^{*}(Q-P+1) & r_{h}^{*}(Q-P+2) & \cdots & r_{h}(Q) \end{matrix}\right] \left[ \begin{matrix} a_{1} \\ a_{2} \\ \vdots \\ a_{p} \end{matrix}\right]=\left[\begin{matrix}|r_h(Q-1) \\ r_h(Q-2) \\ \vdots \\ r_h(Q-P)\end{matrix}\right] ⎣⎢⎢⎢⎡rh(Q)rh∗(Q−1)⋮rh∗(Q−P+1)rh(Q+1)rh(Q)⋮rh∗(Q−P+2)⋯⋯⋱⋯rh(Q+P−1)rh(Q+P−2)⋮rh(Q)⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡a1a2⋮ap⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡∣rh(Q−1)rh(Q−2)⋮rh(Q−P)⎦⎥⎥⎥⎤
同时有：
r a ( l ) = ∑ k = 0 P − ∣ l ∣ a k a k + ∣ l ∣ ∗ , − P ≤ l ≤ P r_a(l)=\sum^{P-|l|}_{k=0}a_ka^{*}_{k+|l|},-P\le l \le P ra(l)=k=0∑P−∣l∣akak+∣l∣∗,−P≤l≤P因为 R d ( z ) = R a ( z ) R h ( z ) R_d(z)=R_a(z)R_h(z) Rd(z)=Ra(z)Rh(z),所以有：
r d ( l ) = ∑ k = − P P r a ( k ) r h ( l − k ) r_d(l)=\sum_{k=-P}^{P}r_a(k)r_h(l-k) rd(l)=k=−P∑Pra(k)rh(l−k)

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ADSP重点习题第二章-第三章（原版书第四章）