《实验四线性时不变离散时间系统的频域分析》由会员分享，可在线阅读，更多相关《实验四线性时不变离散时间系统的频域分析（42页珍藏版）》请在金锄头文库上搜索。
1、实验四线性时不变离散时间系 统的频域分析Name:Chen yifan 20112121006Section:Laboratory Exercise 4LINEAR, TIME-INVARIANT DISCRETE-TIME SYSTEMS:FREQUENCY-DOMAIN REPRESENTATIONS4.1 TRANSFER FUNCTION AND FREQUENCY RESPONSEProject 4.1 Transfer Function AnalysisAnswers:Q4.1 The modified Program P3_1 to compute and plot the ma
2、gnitude and phase spectra of amoving average filter of Eq. (2.13) for 0 w 2p is shown below:w=0:pi/511:2*pi;M=input(M= );num=ones(1,M)/M;h=freqz(num,1,w);subplot(2,1,1);plot(w/pi,abs(h);grid;title(H(ejomega)幅度谱 );xlabel(omega/pi);ylabel(振幅);subplot(2,1,2);plot(w/pi,angle(h);grid;title(相位谱 H(ejomega)
3、;xlabel(omega/pi);ylabel(以弧度为单位的相位);This program was run for the following three different values of:corresponding frequency responses are shown belowMand the plots of theThe types of symmetries exhibited by the magnitude and phase spectra are due to The type of filter represented by the moving aver
4、age filter is-M=3j wjwH(e ) 幅 度 谱幅振10.500 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2位相的位单为度弧以-M=10;420-2-4pw/相 位 谱 H(e0 0.2 0.4 0.6 0.8 1pw/j w)1.2 1.4 1.6 1.8 2H(e )幅 度 谱幅振10.500 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2w/p相 位 谱 H(ejw)4位相的位单为度弧以20-2-40 0.2 0.4 0.6 0.8 1pw/1.2 1.4 1.6 1.8 2jwM=20;H(e )幅 度 谱幅振1.5
5、10.500 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2w/p相 位 谱 H(ejw)4位相的位单为度弧以20-2-40 0.2 0.4 0.6 0.8 1pw/1.2 1.4 1.6 1.8 2The results of Question Q2.1 can now be explained as follows-By the graph, you can seethat it represents a low-pass filter.Q4.2 The plot of the frequency response of the causal LTI discret
6、e-time system of QuestionQ4.2 obtained using the modified program is given below :w=0:pi/511:pi;num=0.15 0 -0.15;den=1 -0.5 0.7;h=freqz(num,den,w);subplot(2,1,1);plot(w/pi,abs(h);grid; title(H(ejomega)幅度谱 ); xlabel(omega/pi);ylabel(振幅); subplot(2,1,2);plot(w/pi,angle(h);grid;title(相位谱 H(ejomega);jwx
7、xxxxx (omega/pi);yyyyyy (以弧度为单位的相位);H(e )幅 度 谱幅振10.500 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1w/p相 位 谱 H(ejw)2位相的位单为度弧以10-1-20 0.1 0.2 0.3 0.4 0.5pw/0.6 0.7 0.8 0.9 1The type of filter represented by this transfer function is-It saysbandpass filter is obtained by diagrams can be BPFQ4.3 The plot of th
8、e frequency response of the causal LTI discrete-time system of QuestionQ4.3 obtained using the modified program is given below :w=0:pi/511:pi;num=0.15 0 -0.15;den=0.7 -0.5 1h=freqz(num,den,w);subplot(2,1,1);jwplot(w/pi,abs(h);grid; title(H(ejomega)幅度谱 ); xlabel(omega/pi);ylabel(振幅); subplot(2,1,2);p
9、lot(w/pi,angle(h);grid;title(相位谱 H(ejomega); xlabel(omega/pi);ylabel(以弧度为单位的相位);H(e )幅 度 谱幅振10.500 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1w/p相 位 谱 H(ejw)位相的位单为度弧以420-2-40 0.1 0.2 0.3 0.4 0.5pw/0.6 0.7 0.8 0.9 1The type of filter represented by this transfer function is-O causal lineartime-invariant di
10、screte time system frequency response, determine the type filterThe difference between the two filters of Questions 4.2 and 4.3 is-More onthe topic of figure, the amplitude spectrum isthe same, the phase spectrum, on the topic that is continuous, and kinds of phase jump line, .I shall choose the fil
11、ter of Question Q4. 3 for the following reason-theamplitude spectrum is the same, the phase spectrum, on the topic that is continuous, and kinds of phase jump line .Q4.6 The pole-zero plots of the two filters of Questions 4.2 and 4.3 developed usingzplaneare shown beloww=0:pi/511:pi;num=0.15 0 -0.15;den=1 -0.5 0.7;h=zplane(num,den);w=0:pi/511:pi;num=0.15 0 -0.15;den=0.7 -0.5 1;h=zplane(num,den);:tari10.80.60.4r 0.2PyangamI0-0.2-0.4-0.6-0.8-1-1 -0.5 0Real Part0.5 1trriaPyangamI10.50
