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1、数字信号处理第6章课后作业答案6.1（1）已知IIR数字滤波器的系统函数为 (1) 试写出滤波器的差分方程，并分别画出直接I型、直接型、转置直接型、级联型和并联型结构图。 解:经化解原式可得：直接I型：x(n)y(n)2-52-3/4+5/41/8直接型:x(n)y(n)5/4-3/41/82-52 级联型：注意，对于级联型，一定要化成负幂次，再写系数！经对原式进行分解得：x(n)y(n)1-1/2-2.511/42并联型：注意：系数b,a是的系数！b=0,0,16,-40,16;a=8,-10,6,-1,0;K,z,d=residue(b,a)KK1=K(1),K(2);zz1=z(1),z
2、(2);b2,a2=residue(KK1,zz1,0)经原式分解得: x(n)y(n)1-0.5-0/。50.254.811.2-5.6-166.4Matlab程序：clear; fp=5000; wp=2*pi*fp; fs=10000; ws=2*pi*fs; ap=3; as=30; N,wc=cheb1ord(wp,ws,ap,as,s); B,A=cheby1(N,ap,wc,s)freqs(B,A);系统函数：图：6.6试设计一个巴特沃斯型模拟带通滤波器，并用Matlab 验证结果，要求带宽为200Hz，中心频率为1000Hz，通带内衰减不大于3dB,在频率小于830Hz或大于1
3、200Hz处的衰减不小于25dB. 解：（1） 模拟带通滤波器的技术指标要求为： =400rad/s; =2210rad/s; =3dB; =25dB; 因为：=400；=40000002； 所以可得：=2210rad/s; =1810rad/s; =2400rad/s; =ll1810rad/s; (2) 归一化频率为： （3）归一化原型模拟低通滤波器的技术指标要求为： ；=1.83；（4）设计归一化原型低通滤波器： （5）查表得： （6） （7） Matlab程序：fp=905,1105;fs=830,1200;wp=2*pi*fp;ws=2*pi*fs;ap=3;as=25;N,wc=b
4、uttord(wp,ws,ap,as,s); %巴特沃斯型模拟带通滤波器B,A=butter(N,wc,s);f=500:1500;w=2*pi*f;H=freqs(B,A,w);subplot(2,1,1);plot(f,20*log10(abs(H);grid on;axis(500,1500,-80,5);xlabel(f/Hz);ylabel(幅度/dB);subplot(2,1,2);plot(f,angle(H);grid on;axis(500,1500,-5,5);xlabel(f/Hz);ylabel(相位/dB); 6.7解：;Matlab程序：clear;b=1,1;a=
5、1,5,6;Fs=10;B,A=impinvar(b,a,Fs);H,w=freqz(B,A,whole);plot(w/pi,20*log10(abs(H);68试用双线性变换法设计一个巴特沃斯型低通数字滤波器，并用matlab，验证结果，给定技术指标为采样频率为1000Hz。 解： (1) 数字低通滤波器的技术指标要求为 采用双线性变换法，相应的模拟低通滤波器的技术指标为：（2） 设计对应摸拟滤波器并求其指标 （3）（4）Matlab程序：clear;clc;Fs=1000;wp=2*pi*100/Fs;ws=2*pi*300/Fs;ap=3;as=20;Wap=2*Fs*tan(wp/2
6、);Was=2*Fs*tan(ws/2);N,wc=buttord(Wap,Was,ap,as,s)b,a=butter(N,wc,s);B,A=bilinear(b,a,Fs);H,w=freqz(B,A);plot(w*Fs/2/pi,abs(H);6.9解：本题选用双线性变换法设计巴特沃斯型数字高通滤波器。(1)数字高通滤波器的技术指标要求为 （2）采用双线性变换法，相应的模拟低通滤波器的技术指标为：（3） 设计对应摸拟滤波器并求其指标： （4）程序：Ap=3;As=20;Fs=1;Wp=0.8*pi;Ws=0.44*pi;Wap=2*Fs*tan(Wp/2);Was=2*Fs*tan(
7、Ws/2);N,Wac=buttord(Wap,Was,Ap,As,s)b,a=butter(N,Wac,high,s);B,A=bilinear(b,a,Fs);freqz(B,A);and performance test copies of the record. If necessary, review should be carried out; 4) for spring hangers (included simple spring, hangers and constant support hangers) it should also be recognized as set
8、ting and locking of loads. 5) check the surface quality, folded layering and without cracks, rust and other defects. 5) after completion of the test and control drawing number one by one, by series baled. Color alloy steel parts, the parts marking installation location and rotation about the directi
9、on you want. 7.3.14. hangers installation 7.3.14.1 hanger layout a. a clear design of hanger should be installed strictly in accordance with the drawings and designs shall not be installed wrong, missing, etc. B. own arrangement of piping support and hanger set and selection should be based on compr
10、ehensive analysis of general layout of piping systems; cold installation of steam pipe with particular attention reserved for compensation of thermal expansion displacement and orientation. C. support systems should be rational to withstand pipe loads, static load and incidental load; reasonable pip
11、ing displacement; guaranteed under various conditions, stress are within the allowed range. Strength, stiffness, and meet requirements to prevent vibration and soothing water, without affecting the adjacent equipment maintenance and other piping installation and expansion. D. equipment connected to
12、the interface to meet pipeline thrust (torque) limit requirements; increase the stability of piping systems to prevent pipeline . Tube wall thickness (mm) 2-3 4-6 7-10 weld form no slope mouth weld strengthening height h (mm) 1-1.5 1.5-2 weld width b (mm) 5-6 7-6 has slope mouth weld strengthening height h (mm) 1.5-2 2 weld width b (mm) cover had each edge slope mouth about 2 mm argon arc welding weld strengthening surface height and width tube wall thickness (mm) 2-3 3-4 5-6 weld form weld strengthening height h (mm) 1-1 .5 1.5-2 2-2.5 width b (mm)
