文章目录

现代数字信号处理课后作业【第六章】
- 6-2 用双线性变换法及冲激响应不变法将下列模拟系统函数Ha(s)H_a(s)Ha(s)转变成数字系统函数H(z)H(z)H(z)
- - (1)Ha(s)=3(s+1)(s+3)T=0.5(1)H_a(s)=\dfrac{3}{(s+1)(s+3)} \ \ \ \ \ \ \ T=0.5(1)Ha(s)=(s+1)(s+3)3 T=0.5
  - (2)Ha(s)=1s2+s+1T=2(2)H_a(s)=\dfrac{1}{s^2+s+1} \ \ \ \ \ \ \ T=2(2)Ha(s)=s2+s+11 T=2
  - (3)Ha(s)=3s+22s2+3s+1T=0.1(3)H_a(s)=\dfrac{3s+2}{2s^2+3s+1} \ \ \ \ \ \ \ T=0.1(3)Ha(s)=2s2+3s+13s+2 T=0.1
- 6-3 用冲激响应不变法设计一个三阶巴特沃兹数字低通滤波器，截止频率为1kHz，抽样频率为6.28318kHz。
- 6-4 用双线性变换法设计一个三阶巴特沃兹数字低通滤波器，截止频率为fc=400Hzf_c=400Hzfc=400Hz，抽样频率为fs=2000Hzf_s=2000Hzfs=2000Hz。
- 6-6 用数字频带变换法设计一个二阶数字高通滤波器截止频率fc=300Hzf_c=300Hzfc=300Hz，抽样频率为fs=2000Hzf_s=2000Hzfs=2000Hz。

现代数字信号处理课后作业【第六章】

6-2 用双线性变换法及冲激响应不变法将下列模拟系统函数Ha(s)H_a(s)Ha(s)转变成数字系统函数H(z)H(z)H(z)

冲激响应不变法：
Ha(s)=∑i=1NAis−si⇒H(z)=∑i=1NAi1−esiTz−1H_a(s)=\sum\limits_{i=1}^{N}\dfrac{A_i}{s-s_i}\ \ \ \ \ \Rightarrow \ \ \ \ \ H(z)=\sum\limits_{i=1}^{N}\dfrac{A_i}{1-e^{s_iT}{z^{-1}}}Ha(s)=i=1∑Ns−siAi ⇒ H(z)=i=1∑N1−esiTz−1Ai

双线性变换法：
s=2T1−z−11+z−1⇒H(z)=H(s)∣s=2T1−z−11+z−1s=\dfrac{2}{T}\dfrac{1-z^{-1}}{1+z^{-1}}\ \ \ \ \ \Rightarrow \ \ \ \ \ H(z)=H(s)\bigg|_{s=\dfrac{2}{T}\dfrac{1-z^{-1}}{1+z^{-1}}}s=T21+z−11−z−1 ⇒ H(z)=H(s)∣∣∣∣s=T21+z−11−z−1

(1)Ha(s)=3(s+1)(s+3)T=0.5(1)H_a(s)=\dfrac{3}{(s+1)(s+3)} \ \ \ \ \ \ \ T=0.5(1)Ha(s)=(s+1)(s+3)3 T=0.5

冲激响应不变法：

Ha(s)=A1s+1+A2s+3=32⋅1s+1−32⋅1s+3H_a(s)=\dfrac{A_1}{s+1}+\dfrac{A_2}{s+3}=\dfrac{3}{2}\cdot\dfrac{1}{s+1}-\dfrac{3}{2}\cdot\dfrac{1}{s+3}Ha(s)=s+1A1+s+3A2=23⋅s+11−23⋅s+31

H(z)=32⋅11−e−Tz−1−32⋅11−e−3Tz−1\ \ \ H(z)=\dfrac{3}{2}\cdot\dfrac{1}{1-e^{-T}z^{-1}}-\dfrac{3}{2}\cdot\dfrac{1}{1-e^{-3T}z^{-1}} H(z)=23⋅1−e−Tz−11−23⋅1−e−3Tz−11

=32⋅11−e−0.5z−1−32⋅11−e−1.5z−1\ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{3}{2}\cdot\dfrac{1}{1-e^{-0.5}z^{-1}}-\dfrac{3}{2}\cdot\dfrac{1}{1-e^{-1.5}z^{-1}} =23⋅1−e−0.5z−11−23⋅1−e−1.5z−11

双线性变换法：

H(z)=Ha(s)∣s=2T1−z−11+z−1=3(s+1)(s+3)∣s=4⋅1−z−11+z−1=3(1+z−1)2(5−3z−1)(7−z−1)H(z)=H_a(s)\bigg|_{s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}}=\dfrac{3}{(s+1)(s+3)}\bigg|_{s={4}\cdot\frac{1-z^{-1}}{1+z^{-1}}}=\dfrac{3(1+z^{-1})^2}{(5-3z^{-1})(7-z^{-1})}H(z)=Ha(s)∣∣∣∣s=T21+z−11−z−1=(s+1)(s+3)3∣∣∣∣s=4⋅1+z−11−z−1=(5−3z−1)(7−z−1)3(1+z−1)2

=3+6z−1+3z−235−26z−1+3z−2\ \ \ \ \ \ \ \ \ \ =\dfrac{3+6z^{-1}+3z^{-2}}{35-26z^{-1}+3z^{-2}} =35−26z−1+3z−23+6z−1+3z−2

(2)Ha(s)=1s2+s+1T=2(2)H_a(s)=\dfrac{1}{s^2+s+1} \ \ \ \ \ \ \ T=2(2)Ha(s)=s2+s+11 T=2

冲激响应不变法：

Ha(s)=1(s+12−32j)(s+12+32j)=A1s+12−32j+A2s+12+32jH_a(s)=\dfrac{1}{(s+\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}j)(s+\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}j)}=\dfrac{A_1}{s+\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}j}+\dfrac{A_2}{s+\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}j}Ha(s)=(s+21−23j)(s+21+23j)1=s+21−23jA1+s+21+23jA2

A1=−j3A2=j3\ \ \ \ A_1=-\dfrac{j}{\sqrt{3}} \ \ \ \ \ \ A_2=\dfrac{j}{\sqrt{3}} A1=−3j A2=3j

H(z)=−j3⋅11−e(−12+32j)Tz−1+j3⋅11−e(−12−32j)Tz−1\ \ \ H(z)=-\dfrac{j}{\sqrt{3}}\cdot\dfrac{1}{1-e^{(-\frac{1}{2}+\frac{\sqrt{3}}{2}j)T}z^{-1}}+\dfrac{j}{\sqrt{3}}\cdot\dfrac{1}{1-e^{(-\frac{1}{2}-\frac{\sqrt{3}}{2}j)T}z^{-1}} H(z)=−3j⋅1−e(−21+23j)Tz−11+3j⋅1−e(−21−23j)Tz−11

=−j3⋅11−e−1+3jz−1+j3⋅11−e−1−3jz−1\ \ \ \ \ \ \ \ \ \ \ \ \ =-\dfrac{j}{\sqrt{3}}\cdot\dfrac{1}{1-e^{-1+\sqrt{3}j}z^{-1}}+\dfrac{j}{\sqrt{3}}\cdot\dfrac{1}{1-e^{-1-\sqrt{3}j}z^{-1}} =−3j⋅1−e−1+3jz−11+3j⋅1−e−1−3jz−11

双线性变换法：

H(z)=Ha(s)∣s=2T1−z−11+z−1=1s2+s+1∣s=1−z−11+z−1=1+2z−1+z−23+z−2H(z)=H_a(s)\bigg|_{s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}}=\dfrac{1}{s^2+s+1}\bigg|_{s=\frac{1-z^{-1}}{1+z^{-1}}}=\dfrac{1+2z^{-1}+z^{-2}}{3+z^{-2}}H(z)=Ha(s)∣∣∣∣s=T21+z−11−z−1=s2+s+11∣∣∣∣s=1+z−11−z−1=3+z−21+2z−1+z−2

(3)Ha(s)=3s+22s2+3s+1T=0.1(3)H_a(s)=\dfrac{3s+2}{2s^2+3s+1} \ \ \ \ \ \ \ T=0.1(3)Ha(s)=2s2+3s+13s+2 T=0.1

冲激响应不变法：

Ha(s)=3s+2(2s+1)(s+1)=A12s+1+A2s+1H_a(s)=\dfrac{3s+2}{(2s+1)(s+1)}=\dfrac{A_1}{2s+1}+\dfrac{A_2}{s+1}Ha(s)=(2s+1)(s+1)3s+2=2s+1A1+s+1A2

A1=1A2=1\ \ \ \ A_1=1 \ \ \ \ \ \ A_2=1 A1=1 A2=1

Ha(s)=12⋅1s+12+1s+1\ \ \ H_a(s)=\dfrac{1}{2}\cdot\dfrac{1}{s+\frac{1}{2}}+\dfrac{1}{s+1} Ha(s)=21⋅s+211+s+11

H(z)=12⋅11−e−12Tz−1+11−e−Tz−1=12⋅11−e−0.05z−1+11−e−0.1z−1\ \ \ H(z)=\dfrac{1}{2}\cdot\dfrac{1}{1-e^{-\frac{1}{2}T}z^{-1}}+\dfrac{1}{1-e^{-T}z^{-1}}=\dfrac{1}{2}\cdot\dfrac{1}{1-e^{-0.05}z^{-1}}+\dfrac{1}{1-e^{-0.1}z^{-1}} H(z)=21⋅1−e−21Tz−11+1−e−Tz−11=21⋅1−e−0.05z−11+1−e−0.1z−11

双线性变换法：

H(z)=Ha(s)∣s=2T1−z−11+z−1=3s+22s2+3s+1∣s=20⋅1−z−11+z−1=62+4z−1−58z−2861−1598z−1+741z−2H(z)=H_a(s)\bigg|_{s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}}=\dfrac{3s+2}{2s^2+3s+1}\bigg|_{s=20\cdot\frac{1-z^{-1}}{1+z^{-1}}}=\dfrac{62+4z^{-1}-58z^{-2}}{861-1598z^{-1}+741z^{-2}}H(z)=Ha(s)∣∣∣∣s=T21+z−11−z−1=2s2+3s+13s+2∣∣∣∣s=20⋅1+z−11−z−1=861−1598z−1+741z−262+4z−1−58z−2

6-3 用冲激响应不变法设计一个三阶巴特沃兹数字低通滤波器，截止频率为1kHz，抽样频率为6.28318kHz。

已知：三阶巴特沃兹低通滤波器N=3，fc=1kHz，fs=6.28318kHz已知：三阶巴特沃兹低通滤波器 N=3，f_c=1kHz，f_s=6.28318kHz已知：三阶巴特沃兹低通滤波器N=3，fc=1kHz，fs=6.28318kHz

T=1fs=16.28318⋅103≈1ΩcT=\dfrac{1}{f_s}=\dfrac{1}{6.28318\cdot10^3}\approx\dfrac{1}{Ω_c}T=fs1=6.28318⋅1031≈Ωc1

Ωc=2πfc=2000π，sk=Ωcejπ(12+2k−12N)k=1,2,...,2NΩ_c=2\pi f_c=2000\pi，s_k=Ω_ce^{j\pi (\frac{1}{2}+\frac{2k-1}{2N})}\ \ \ \ \ k=1,2,...,2NΩc=2πfc=2000π，sk=Ωcejπ(21+2N2k−1) k=1,2,...,2N

s1=Ωc(−12+32j)s2=−Ωcs3=Ωc(−12−32j)s_1=Ω_c(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}j)\ \ \ \ s_2=-Ω_c\ \ \ \ s_3=Ω_c(-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}j)s1=Ωc(−21+23j) s2=−Ωc s3=Ωc(−21−23j)

s4=Ωc(12−32j)s5=Ωcs6=Ωc(12+32j)s_4=Ω_c(\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}j)\ \ \ \ s_5=Ω_c\ \ \ \ s_6=Ω_c(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}j)s4=Ωc(21−23j) s5=Ωc s6=Ωc(21+23j)

为了使系统稳定，巴特沃兹系统函数由S左半平面极点构成：为了使系统稳定，巴特沃兹系统函数由S左半平面极点构成：为了使系统稳定，巴特沃兹系统函数由S左半平面极点构成：

Ha(s)=ΩcN∏k=1N(s−sk)∣N=3=Ωc3(s−s1)(s−s2)(s−s3)=A1s−s1+A2s−s2+A3s−s3H_a(s)=\dfrac{Ω_c^N}{\prod\limits_{k=1}^{N}(s-s_k)}\bigg|_{N=3}=\dfrac{Ω_c^3}{(s-s_1)(s-s_2)(s-s_3)}=\dfrac{A_1}{s-s_1}+\dfrac{A_2}{s-s_2}+\dfrac{A_3}{s-s_3}Ha(s)=k=1∏N(s−sk)ΩcN∣∣∣∣N=3=(s−s1)(s−s2)(s−s3)Ωc3=s−s1A1+s−s2A2+s−s3A3

A1=Ωc32j−32A2=ΩcA3=Ωc−32j−32A_1=\dfrac{Ω_c}{\dfrac{\sqrt{3}}{2}j-\dfrac{3}{2}}\ \ \ \ A_2=Ω_c\ \ \ \ A_3=\dfrac{Ω_c}{-\dfrac{\sqrt{3}}{2}j-\dfrac{3}{2}}A1=23j−23Ωc A2=Ωc A3=−23j−23Ωc

H(z)=∑i=1NAi1−esiTz−1∣T=1fs=1ΩcH(z)=\sum\limits_{i=1}^{N}\dfrac{A_i}{1-e^{s_iT}z^{-1}}\bigg|_{T=\frac{1}{f_s}=\frac{1}{Ω_c}}H(z)=i=1∑N1−esiTz−1Ai∣∣∣∣T=fs1=Ωc1

=Ωc32j−32⋅11−e−12+j32z−1+Ωc⋅11−e−1z−1+Ωc−32j−32⋅11−e−12−j32z−1\ \ \ \ \ \ \ \ \ \ =\dfrac{Ω_c}{\dfrac{\sqrt{3}}{2}j-\dfrac{3}{2}}\cdot\dfrac{1}{1-e^{-\frac{1}{2}+j\frac{\sqrt{3}}{2}}z^{-1}}+Ω_c\cdot\dfrac{1}{1-e^{-1}z^{-1}}+\dfrac{Ω_c}{-\dfrac{\sqrt{3}}{2}j-\dfrac{3}{2}}\cdot\dfrac{1}{1-e^{-\frac{1}{2}-j\frac{\sqrt{3}}{2}}z^{-1}} =23j−23Ωc⋅1−e−21+j23z−11+Ωc⋅1−e−1z−11+−23j−23Ωc⋅1−e−21−j23z−11

6-4 用双线性变换法设计一个三阶巴特沃兹数字低通滤波器，截止频率为fc=400Hzf_c=400Hzfc=400Hz，抽样频率为fs=2000Hzf_s=2000Hzfs=2000Hz。

注：模拟频率fcf_cfc：每秒经历多少个周期，单位Hz，即1/s；
模拟角频率ΩcΩ_cΩc：每秒经历多少弧度，单位rad/s；
数字频率www：每个采样点间隔之间的弧度，单位rad。
数字频率与模拟频率相互转化：w=2πfcfsw=\dfrac{2\pi f_c}{f_s}w=fs2πfc，fsf_sfs为抽样频率。

Ωc=2Ttanw2=2Ttan(2π⋅4002⋅2000)=2Ttan(0.2π)Ω_c=\dfrac{2}{T}tan\dfrac{w}{2}=\dfrac{2}{T}tan(\dfrac{2\pi \cdot400}{2\cdot2000})=\dfrac{2}{T}tan(0.2\pi)Ωc=T2tan2w=T2tan(2⋅20002π⋅400)=T2tan(0.2π)

H(z)=Ha(s)=1[(sΩc)2+sΩc+1](sΩc+1)∣s=2T⋅1−z−11+z−1H(z)=H_a(s)=\dfrac{1}{[(\dfrac{s}{Ω_c})^2+\dfrac{s}{Ω_c}+1](\dfrac{s}{Ω_c}+1)}\bigg|_{s=\frac{2}{T}\cdot\frac{1-z^{-1}}{1+z^{-1}}}H(z)=Ha(s)=[(Ωcs)2+Ωcs+1](Ωcs+1)1∣∣∣∣s=T2⋅1+z−11−z−1

sΩc∣s=2T⋅1−z−11+z−1，Ωc=2Ttan(0.2π)=1tan(0.2π)⋅1−z−11+z−1=α⋅1−z−11+z−1\dfrac{s}{Ω_c}\bigg|_{s=\frac{2}{T}\cdot\frac{1-z^{-1}}{1+z^{-1}}，Ω_c=\frac{2}{T}tan(0.2\pi)}=\dfrac{1}{tan(0.2\pi)}\cdot\dfrac{1-z^{-1}}{1+z^{-1}}=\alpha\cdot\dfrac{1-z^{-1}}{1+z^{-1}}Ωcs∣∣∣∣s=T2⋅1+z−11−z−1，Ωc=T2tan(0.2π)=tan(0.2π)1⋅1+z−11−z−1=α⋅1+z−11−z−1

其中α=1tan(0.2π)≈1.376其中\alpha=\dfrac{1}{tan(0.2\pi)}\approx1.376其中α=tan(0.2π)1≈1.376

H(z)=1+3z−1+3z−2+z−3[(α2−α+1)z−2+(−2α2+2)z−1+α2+α+1][1+α+(1−α)z−1]H(z)=\dfrac{1+3z^{-1}+3z^{-2}+z^{-3}}{[(\alpha^2-\alpha+1)z^{-2}+(-2\alpha^2+2)z^{-1}+\alpha^2+\alpha+1][1+\alpha+(1-\alpha)z^{-1}]}H(z)=[(α2−α+1)z−2+(−2α2+2)z−1+α2+α+1][1+α+(1−α)z−1]1+3z−1+3z−2+z−3

H(z)=1+3z−1+3z−2+z−3(1.517z−2−1.787z−1+4.269)(2.376−0.376z−1)H(z)=\dfrac{1+3z^{-1}+3z^{-2}+z^{-3}}{(1.517z^{-2}-1.787z^{-1}+4.269)(2.376-0.376z^{-1})}H(z)=(1.517z−2−1.787z−1+4.269)(2.376−0.376z−1)1+3z−1+3z−2+z−3

=1+3z−1+3z−2+z−310.143−5.851z−1+4.276z−2−0.57z−3\ \ \ \ \ \ \ \ \ \ =\dfrac{1+3z^{-1}+3z^{-2}+z^{-3}}{10.143-5.851z^{-1}+4.276z^{-2}-0.57z^{-3}} =10.143−5.851z−1+4.276z−2−0.57z−31+3z−1+3z−2+z−3

6-6 用数字频带变换法设计一个二阶数字高通滤波器截止频率fc=300Hzf_c=300Hzfc=300Hz，抽样频率为fs=2000Hzf_s=2000Hzfs=2000Hz。

双线性变换法求解二阶低通滤波器：

设低通滤波器截止频率700Hz,抽样频率2000Hz设低通滤波器截止频率700Hz,抽样频率2000Hz设低通滤波器截止频率700Hz,抽样频率2000Hz

Ωc=2Ttanw2=2Ttan2π⋅7002⋅2000=2Ttan7π20Ω_c=\dfrac{2}{T}tan\dfrac{w}{2}=\dfrac{2}{T}tan\dfrac{2\pi \cdot700}{2\cdot2000}=\dfrac{2}{T}tan\dfrac{7\pi}{20}Ωc=T2tan2w=T2tan2⋅20002π⋅700=T2tan207π

二阶巴特沃兹低通滤波器Ha(s)=1(sΩc)2+1.4142⋅sΩc+1二阶巴特沃兹低通滤波器H_a(s)=\dfrac{1}{(\dfrac{s}{Ω_c})^2+1.4142\cdot\dfrac{s}{Ω_c}+1}二阶巴特沃兹低通滤波器Ha(s)=(Ωcs)2+1.4142⋅Ωcs+11

Hl(z)=1(sΩc)2+1.4142⋅sΩc+1∣s=2T⋅1−z−11+z−1H_l(z)=\dfrac{1}{(\dfrac{s}{Ω_c})^2+1.4142\cdot\dfrac{s}{Ω_c}+1}\bigg|_{s=\frac{2}{T}\cdot\frac{1-z^{-1}}{1+z^{-1}}}Hl(z)=(Ωcs)2+1.4142⋅Ωcs+11∣∣∣∣s=T2⋅1+z−11−z−1

sΩc∣s=2T⋅1−z−11+z−1，Ωc=2Ttan7π20=1tan(7π20)⋅1−z−11+z−1=α⋅1−z−11+z−1\dfrac{s}{Ω_c}\bigg|_{s=\frac{2}{T}\cdot\frac{1-z^{-1}}{1+z^{-1}}，Ω_c=\frac{2}{T}tan\frac{7\pi}{20}}=\dfrac{1}{tan(\frac{7\pi}{20})}\cdot\dfrac{1-z^{-1}}{1+z^{-1}}=\alpha\cdot\dfrac{1-z^{-1}}{1+z^{-1}}Ωcs∣∣∣∣s=T2⋅1+z−11−z−1，Ωc=T2tan207π=tan(207π)1⋅1+z−11−z−1=α⋅1+z−11−z−1

其中α=1tan7π20≈0.51其中\alpha=\dfrac{1}{tan\dfrac{7\pi}{20}}\approx0.51其中α=tan207π1≈0.51

所以二阶数字低通滤波器函数为：所以二阶数字低通滤波器函数为：所以二阶数字低通滤波器函数为：

Hl(z)=1α2(1−z−11+z−1)2+1.4142⋅α1−z−11+z−1+1H_l(z)=\dfrac{1}{\alpha^2(\dfrac{1-z^{-1}}{1+z^{-1}})^2+1.4142\cdot\alpha\dfrac{1-z^{-1}}{1+z^{-1}}+1}Hl(z)=α2(1+z−11−z−1)2+1.4142⋅α1+z−11−z−1+11

=(1+z−1)2α2(1−z−1)2+1.4142⋅α(1−z−1)(1+z−1)+(1+z−1)2\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{(1+z^{-1})^2}{\alpha^2(1-z^{-1})^2+1.4142\cdot\alpha(1-z^{-1})(1+z^{-1})+(1+z^{-1})^2} =α2(1−z−1)2+1.4142⋅α(1−z−1)(1+z−1)+(1+z−1)2(1+z−1)2

数字频带变换法将低通转成高通：

β=−cos(wlc+wc)2cos(wlc−wc)2=−cos[(700+300)⋅2π2000⋅2]cos[(700−300)⋅2π2000⋅2]=0\beta=-\dfrac{cos\dfrac{(w_{lc}+w_c)}{2}}{cos\dfrac{(w_{lc}-w_c)}{2}}=-\dfrac{cos\big[\dfrac{(700+300)\cdot2\pi}{2000\cdot2}\big]}{cos\big[\dfrac{(700-300)\cdot2\pi}{2000\cdot2}\big]}=0β=−cos2(wlc−wc)cos2(wlc+wc)=−cos[2000⋅2(700−300)⋅2π]cos[2000⋅2(700+300)⋅2π]=0

zl−1⇒−z−1+β1+βz−1=−z−1z_l^{-1}\Rightarrow -\dfrac{z^{-1}+\beta}{1+\beta z^{-1}}=-z^{-1}zl−1⇒−1+βz−1z−1+β=−z−1

二阶数字高通滤波器函数H(z)=Hl(z)∣zl−1=−z−1二阶数字高通滤波器函数H(z)=H_l(z)\bigg|_{z_l^{-1}=-z^{-1}}二阶数字高通滤波器函数H(z)=Hl(z)∣∣∣∣zl−1=−z−1

H(z)=1−2z−1+z−2α2(1+z−1)2+1.4142⋅α(1+z−1)(1−z−1)+(1−z−1)2H(z)=\dfrac{1-2z^{-1}+z^{-2}}{\alpha^2(1+z^{-1})^2+1.4142\cdot\alpha(1+z^{-1})(1-z^{-1})+(1-z^{-1})^2}H(z)=α2(1+z−1)2+1.4142⋅α(1+z−1)(1−z−1)+(1−z−1)21−2z−1+z−2

=1−2z−1+z−2α2+1+1.4142α+(2α2−2)z−1+(α2−1.4142α+1)z−2其中α=1tan7π20≈0.51\ \ \ \ \ \ \ \ \ \ =\dfrac{1-2z^{-1}+z^{-2}}{\alpha^2+1+1.4142\alpha+(2\alpha^2-2)z^{-1}+(\alpha^2-1.4142\alpha+1)z^{-2}} \ \ \ \ 其中\alpha=\dfrac{1}{tan\dfrac{7\pi}{20}}\approx0.51 =α2+1+1.4142α+(2α2−2)z−1+(α2−1.4142α+1)z−21−2z−1+z−2 其中α=tan207π1≈0.51

H(z)=1−2z−1+z−21.981−1.479z−1+0.5389z−2H(z)=\dfrac{1-2z^{-1}+z^{-2}}{1.981-1.479z^{-1}+0.5389z^{-2}}H(z)=1.981−1.479z−1+0.5389z−21−2z−1+z−2