数字信号处理米特拉第四版实验一答案

第十章上机实验数字信号处理是一门理论和实际密切结合的课程，为深入掌握课程内容，最好在学习理论的同时，做习题和上机实验。

上机实验不仅可以帮助读者深入的理解和消化基本理论，而且能锻炼初学者的独立解决问题的能力。

本章在第二版的基础上编写了六个实验，前五个实验属基础理论实验，第六个属应用综合实验。

实验一 系统响应及系统稳定性。

实验二 时域采样与频域采样。

实验三 用FFT 对信号作频谱分析。

实验四 IIR 数字滤波器设计及软件实现。

实验五 FIR 数字滤波器设计与软件实现实验六 应用实验——数字信号处理在双音多频拨号系统中的应用任课教师根据教学进度，安排学生上机进行实验。

建议自学的读者在学习完第一章后作实验一；在学习完第三、四章后作实验二和实验三；实验四IIR 数字滤波器设计及软件实现在。

学习完第六章进行；实验五在学习完第七章后进行。

实验六综合实验在学习完第七章或者再后些进行；实验六为综合实验，在学习完本课程后再进行。

10.1 实验一: 系统响应及系统稳定性1.实验目的（1）掌握 求系统响应的方法。

（2）掌握时域离散系统的时域特性。

（3）分析、观察及检验系统的稳定性。

2.实验原理与方法在时域中，描写系统特性的方法是差分方程和单位脉冲响应，在频域可以用系统函数描述系统特性。

已知输入信号可以由差分方程、单位脉冲响应或系统函数求出系统对于该输入信号的响应，本实验仅在时域求解。

在计算机上适合用递推法求差分方程的解，最简单的方法是采用MA TLAB 语言的工具箱函数filter 函数。

也可以用MATLAB 语言的工具箱函数conv 函数计算输入信号和系统的单位脉冲响应的线性卷积，求出系统的响应。

系统的时域特性指的是系统的线性时不变性质、因果性和稳定性。

重点分析实验系统的稳定性，包括观察系统的暂态响应和稳定响应。

系统的稳定性是指对任意有界的输入信号，系统都能得到有界的系统响应。

或者系统的单位脉冲响应满足绝对可和的条件。

数字信号处理实验答案 HEN system office room 【HEN16H-HENS2AHENS8Q8-HENH1688】实验一熟悉Matlab环境一、实验目的1.熟悉MATLAB的主要操作命令。

2.学会简单的矩阵输入和数据读写。

3.掌握简单的绘图命令。

4.用MATLAB编程并学会创建函数。

5.观察离散系统的频率响应。

二、实验内容认真阅读本章附录，在MATLAB环境下重新做一遍附录中的例子，体会各条命令的含义。

在熟悉了MATLAB基本命令的基础上，完成以下实验。

上机实验内容：（1）数组的加、减、乘、除和乘方运算。

输入A=[1 2 3 4]，B=[3 4 5 6]，求C=A+B，D=A-B，E=A.*B，F=A./B，G=A.^B并用stem语句画出A、B、C、D、E、F、G。

clear all;a=[1 2 3 4];b=[3 4 5 6];c=a+b;d=a-b;e=a.*b;f=a./b;g=a.^b;n=1:4;subplot(4,2,1);stem(n,a);xlabel('n');xlim([0 5]);ylabel('A');subplot(4,2,2);stem(n,b);xlabel('n');xlim([0 5]);ylabel('B');subplot(4,2,3);stem(n,c);xlabel('n');xlim([0 5]);ylabel('C');subplot(4,2,4);stem(n,d);xlabel('n');xlim([0 5]);ylabel('D');subplot(4,2,5);stem(n,e);xlabel('n');xlim([0 5]);ylabel('E');subplot(4,2,6);stem(n,f);xlabel('n');xlim([0 5]);ylabel('F');subplot(4,2,7);stem(n,g);xlabel('n');xlim([0 5]);ylabel('G');（2）用MATLAB实现下列序列：a) x(n)= 0≤n≤15b) x(n)=e+3j)n 0≤n≤15c) x(n)=3cosπn+π)+2sinπn+π) 0≤n≤15(n)=x(n+16),绘出四个周期。

7.2　课后习题详解7-1 用冲激响应不变法将以下Ha （s ）变换为H （z ），抽样周期为T 。

（1）H a （s ）＝（s ＋a ）/[（s ＋a ）2＋b 2]；（2）H a （s ）＝A/（s －s 0）n0，n 0为任意正整数。

解：（1）冲激响应不变法满足h （n ）＝h a （t ）|t ＝nT ＝h a （nT ），T 为抽样间隔。

这种变换法必须让H a （s ）先用部分分式展开。

由推出由冲激响应不变法可得（2）先引用拉氏变换的结论，可得按且可得可以递推求得7-2 设计一个模拟低通滤波器，要求其通带截止频率f p＝20Hz，其通带最大衰减为R p＝2dB，阻带截止频率为f st＝40Hz，阻带最小衰减为A s＝20dB，采用巴特沃思滤波器，画出滤波器的幅度响应。

解：巴特沃思模拟低通滤波器设计流程为：①利用教程（7-5-24）式求解滤波器阶次N；②利用教程（7-5-27a）式求解3dB截止频率Ωc；③查教程表7-2或表7-4获得归一化巴特沃思低通滤波器的系统函数H an（s）；④将H an（s）根据Ωc的值去归一化求得所需的系统函数H a（s）。

已知Ωp＝2π×20rad/s，Ωst＝2π×40rad/s，R p＝2dB，A s＝20dB。

（1）按给定的参数由教程（7-5-24）式可求得取N＝4。

（2）按教程（7-5-27a）式可求得巴特沃思滤波器3dB处的通带截止频率Ωc为（3）查教程表7-2可得N＝4时归一化巴特沃思低通滤波器H an（s）（4）去归一化，求得所需的H a（s）为滤波器的幅度响应如图7-1所示。

图7-17-3 设计一个模拟高通滤波器，要求其阻带截止频率f st＝30Hz，阻带最小衰减为A s＝25dB，通带截止频率为f p＝50Hz，通带最大衰减为R p＝1dB。

（1）采用巴特沃思滤波器；（2）采用切比雪夫滤波器；（3）利用MATLAB工具箱函数设计椭圆函数滤波器。

Name : SOLUTION Section :Laboratory Exercise 1DISCRETE-TIME SIGNALS: TIME-DOMAIN REPRESENTATION1.1GENERATION OF SEQUENCESProject 1.1Unit sample and unit step sequencesA copy of Program P1_1 is given below.% Program P1_1% Generation of a Unit Sample Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the unit sample sequence u = [zeros(1,10) 1 zeros(1,20)]; % Plot the unit sample sequence stem(n,u);xlabel('Time index n');ylabel('Amplitude'); title('Unit Sample Sequence'); axis([-10 20 0 1.2]); Answers : Q1.1The unit sample sequence u[n] generated by running Program P1_1 is shown below:Time index nA m p l i t u d eUnit Sample SequenceQ1.2The purpose of clf command is – clear the current figureThe purpose of axis command is – control axis scaling and appearanceThe purpose of title command is – add a title to a graph or an axis and specify textpropertiesThe purpose of xlabel command is – add a label to the x-axis and specify textpropertiesThe purpose of ylabel command is – add a label to the y-axis and specify the textpropertiesQ1.3The modified Program P1_1 to generate a delayed unit sample sequence ud[n] with a delay of 11 samples is given below along with the sequence generated by running this program .% Program P1_1, MODIFIED for Q1.3% Generation of a DELAYED Unit Sample Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the DELAYED unit sample sequence u = [zeros(1,21) 1 zeros(1,9)];% Plot the DELAYED unit sample sequence stem(n,u);xlabel('Time index n');ylabel('Amplitude'); title('DELAYED Unit Sample Sequence'); axis([-10 20 0 1.2]);Time index nA m p l i t u d eDELAYED Unit Sample SequenceQ1.4The modified Program P1_1 to generate a unit step sequence s[n] is given below along with the sequence generated by running this program .% Program Q1_4% Generation of a Unit Step Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the unit step sequence s = [zeros(1,10) ones(1,21)]; % Plot the unit step sequence stem(n,s);xlabel('Time index n');ylabel('Amplitude'); title('Unit Step Sequence'); axis([-10 20 0 1.2]);Time index nA m p l i t u d eUnit Step SequenceQ1.5The modified Program P1_1 to generate a unit step sequence sd[n] with an advance of 7 samples is given below along with the sequence generated by running this program .% Program Q1_5% Generation of an ADVANCED Unit Step Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the ADVANCED unit step sequence sd = [zeros(1,3) ones(1,28)];% Plot the ADVANCED unit step sequence stem(n,sd);xlabel('Time index n');ylabel('Amplitude'); title('ADVANCED Unit Step Sequence'); axis([-10 20 0 1.2]);Time index nA m p l i t u d eADVANCED Unit Step SequenceProject 1.2 Exponential signalsA copy of Programs P1_2 and P1_3 are given below .% Program P1_2% Generation of a complex exponential sequence clf;c = -(1/12)+(pi/6)*i; K = 2; n = 0:40;x = K*exp(c*n); subplot(2,1,1); stem(n,real(x));xlabel('Time index n');ylabel('Amplitude'); title('Real part'); subplot(2,1,2); stem(n,imag(x));xlabel('Time index n');ylabel('Amplitude'); title('Imaginary part');% Program P1_3% Generation of a real exponential sequence clf;n = 0:35; a = 1.2; K = 0.2; x = K*a.^n; stem(n,x);xlabel('Time index n');ylabel('Amplitude');Answers: Q1.6The complex-valued exponential sequence generated by running Program P1_2 is shown below :510152025303540Time index n A m p l i t u d eReal partTime index nA m p l i t u d eImaginary partQ1.7The parameter controlling the rate of growth or decay of this sequence is – the real part of c .The parameter controlling the amplitude of this sequence is - KQ1.8The result of changing the parameter c to (1/12)+(pi/6)*i is – since exp(-1/12) isless than one and exp(1/12) is greater than one, this change means that the envelope of the signal will grow with n instead of decay with n.Q1.9The purpose of the operator real is – to extract the real part of a Matlab vector.The purpose of the operator imag is – to extract the imaginary part of a Matlab vector. Q1.10The purpose of the command subplot is – to plot more than one graph in the sameMatlab figure.Q1.11The real-valued exponential sequence generated by running Program P1_3 is shown below :Time index nA m p l i t u d eQ1.12The parameter controlling the rate of growth or decay of this sequence is - aThe parameter controlling the amplitude of this sequence is - KQ1.13The difference between the arithmetic operators ^ and .^ is – “^” raises a square matrix toa power using matrix multiplication. “.^” raises the elements of a matrix or vector to a power; this is a “pointwise” operation.Q1.14The sequence generated by running Program P1_3 with the parameter a changed to 0.9 and the parameter K changed to 20 is shown below :5101520253035Time index nA m p l i t u d eQ1.15The length of this sequence is - 36It is controlled by the following MATLAB command line : n = 0:35;It can be changed to generate sequences with different lengths as follows (give an example command line and the corresponding length): n = 0:99; makes the length 100. Q1.16The energies of the real-valued exponential sequences x [n]generated in Q1.11 and Q1.14 and computed using the command sum are - 4.5673e+004 and 2.1042e+003.Project 1.3 Sinusoidal sequencesA copy of Program P1_4 is given below .% Program P1_4% Generation of a sinusoidal sequence n = 0:40;f = 0.1; phase = 0; A = 1.5; arg = 2*pi*f*n - phase; x = A*cos(arg);clf; % Clear old graphstem(n,x); % Plot the generated sequence axis([0 40 -2 2]); grid;title('Sinusoidal Sequence'); xlabel('Time index n'); ylabel('Amplitude'); axis; Answers:Q1.17 The sinusoidal sequence generated by running Program P1_4 is displayed below .0510152025303540Sinusoidal SequenceTime index nA m p l i t u d eQ1.18The frequency of this sequence is- f = 0.1 cycles/sample.It is controlled by the following MATLAB command line: f = 0.1;A sequence with new frequency 0.2 can be generated by the following command line:f = 0.2;The parameter controlling the phase of this sequence is- phaseThe parameter controlling the amplitude of this sequence is- AThe period of this sequence is - 2π/ω = 1/f = 10Q1.19 The length of this sequence is - 41It is controlled by the following MATLAB command line: n = 0:40;A sequence with new length __81_ can be generated by the following command line:n = 0:80;Q1.20 The average power of the generated sinusoidal sequence is–sum(x(1:10).*x(1:10))/10 = 1.1250Q1.21 The purpose of axis command is – to set the range of the x-axis to [0,40] and the range of the y-axis to [-2,2].The purpose of grid command is – to turn on the drawing of grid lines on the graph.Q1.22 The modified Program P1_4 to generate a sinusoidal sequence of frequency 0.9 is given below along with the sequence generated by running it.% Program Q1_22A% Generation of a sinusoidal sequencen = 0:40;f = 0.9;phase = 0;A = 1.5;arg = 2*pi*f*n - phase;x = A*cos(arg);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 40 -2 2]);grid;title('Sinusoidal Sequence');xlabel('Time index n');ylabel('Amplitude');axis;Sinusoidal SequenceTime index nA m p l i t u d eA comparison of this new sequence with the one generated in Question Q1.17 shows - thetwo graphs are identical. This is because, in the modified program, we have ω = 0.9*2π. This generates the same graph as a cosine with angular frequency ω - 2π = −0.1*2π. Because cosine is an even function, this is the same as a cosine with angular frequency +0.1*2π, which was the value used in P1_4.m in Q1.17.In terms of Hertzian frequency, we have for P1_4.m in Q1.17 that f = 0.1 Hz/sample. For the modified program in Q1.22, we have f = 0.9 Hz/sample, which generates the same graph as f = 0.9 – 1 = −0.1. Again because cosine is even, this makes a graph that is identical to the one we got in Q1.17 with f = +0.1 Hz/sample.A sinusoidal sequence of frequency 1.1 generated by modifying Program P1_4 is shown below.0510152025303540Sinusoidal SequenceTime index nA m p l i t u d eA comparison of this new sequence with the one generated in Question Q1.17 shows - thegraph here is again identical to the one in Q1.17. This is because a cosine of frequency f = 1.1 Hz/sample is identical to one with frequency f = 1.1 – 1 = 0.1 Hz/sample, which was the frequency used in Q1.17.Q1.23The sinusoidal sequence of length 50, frequency 0.08, amplitude 2.5, and phase shift of 90 degrees generated by modifying Program P1_4 is displayed below .NOTE: for this program, it is necessary to convert the phase of 90 deg to radians and account for the minus sign that appears in the statement “arg = 2*pi*f*n - phase;” as opposed to the plus sign shown in eq. (1.12) of the lab manual. The correct statement to generate the phase is “phase = -90/(2*pi);”. It is also necessary to modify the axis command to account for the new length and amplitude of the signal. The correct axis statement is “axis([0 50 -3 3]);”.05101520253035404550Sinusoidal SequenceTime index nA m p l i t u d eThe period of this sequence is - 2π/ω = 1/f = 1/0.08 = 1/(8/100) = 100/8 = 25/2.Therefore, the fundamental period is 25 and the graph has the “appearance” of going through 2 cycles of a cosine waveform during each period.Q1.24By replacing the stem command in Program P1_4 with the plot command, the plot obtained is as shown below :Sinusoidal SequenceTime index nA m p l i t u d eThe difference between the new plot and the one generated in Question Q1.17 is – instead ofdrawing stems from the x-axis to the points on the curve, the “plot ” command connects the points with straight line segments, which approximates the graph of a continuous-time cosine signal.Q1.25By replacing the stem command in Program P1_4 with the stairs command the plot obtained is as shown below :Sinusoidal SequenceTime index nA m p l i t u d eThe difference between the new plot and those generated in Questions Q1.17 and Q1.24 is– the “stairs” command produces a stairstep plot as opposed to the stem graph that was generated in Q1.17 and the straight-line interpolation plot that was generated in Q1.24.Project 1.4 Random signalsAnswers:Q1.26 The MATLAB program to generate and display a random signal of length 100 with elements uniformly distributed in the interval [–2, 2] is given below along with the plot of the random sequence generated by running the program:% Program Q1_26% Generation of a uniform random sequencen = 0:99;A = 2;rand('state',sum(100*clock)); % "seed" the generator% rand(1,100) is uniform in [0,1]% rand(1,100)-0.5 is uniform in [-0.5,0.5]% 4*(rand(1,100)-0.5) is uniform in [-2,2]x = 2*A*(rand(1,length(n))-0.5);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 length(n) -round(2*(A+0.5))/2 round(2*(A+0.5))/2]);grid;title('Uniform Random Sequence');xlabel('Time index n');ylabel('Amplitude');axis;102030405060708090100Uniform Random SequenceTime index nA m p l i t u d eQ1.27 The MATLAB program to generate and display a Gaussian random signal of length 75 withelements normally distributed with zero mean and a variance of 3 is given below along with the plot of the random sequence generated by running the program :% Program Q1_27% Generation of a Gaussian random sequence% NOTE: if X is a random variable with zero mean and % unity variance, then (aX + b) is a random variable % with mean b and variance a^2. This follows from % basic probability theory. n = 0:74;xmean = 0; % mean of xxstd = sqrt(3); % standard deviation of xrandn('state',sum(100*clock)); % "seed" the generator % generate the sequencex = xstd*randn(1,length(n)) + xmean; % setup the graph and plotclf; % Clear old graphstem(n,x); % Plot the generated sequence xmax = max(abs(x));Ylim = round(2*(xmax+0.5))/2; axis([0 length(n) -Ylim Ylim]); grid;title('Gaussian Random Sequence'); xlabel('Time index n'); ylabel('Amplitude'); axis;10203040506070Gaussian Random SequenceTime index nA m p l i t u d eQ1.28The MATLAB program to generate and display five sample sequences of a random sinusoidal signal of length 31{X[n]} = {A·cos(ωo n + φ)}where the amplitude A and the phase φ are statistically independent random variables with uniform probability distribution in the range 0≤A ≤4 for the amplitude and in the range 0 ≤ φ ≤ 2π for the phase is given below. Also shown are five sample sequences generated by running this program five different times .% Generates the "deterministic stochastic process" % called for in Q1.28. n = 0:30; f = 0.1; Amax = 4;phimax = 2*pi;rand('state',sum(100*clock)); % "seed" the generator A = Amax*rand;% NOTE: successive calls to rand without arguments % return a random sequence of scalars. Since this % random sequence is "white" (uncorrelated), it is % not necessary to re-seed the generator for phi. phi = phimax*rand;% generate the sequence arg = 2*pi*f*n + phi; x = A*cos(arg); % plotclf; % Clear old graphstem(n,x); % Plot the generated sequence Ylim = round(2*(Amax+0.5))/2; axis([0 length(n) -Ylim Ylim]); grid;title('Sinusoidal Sequence with Random Amplitude and Phase'); xlabel('Time index n'); ylabel('Amplitude'); axis;Sinusoidal Sequence with Random Amplitude and PhaseTime index nA m p l i t u d eSinusoidal Sequence with Random Amplitude and PhaseTime index nA m p l i t u d eTime index nA m p l i t u d eSinusoidal Sequence with Random Amplitude and PhaseTime index nA m p l i t u d eTime index nA m p l i t u d e1.2 SIMPLE OPERATIONS ON SEQUENCESProject 1.5Signal SmoothingA copy of Program P1_5 is given below .% Program P1_5% Signal Smoothing by Averaging clf; R = 51;d = 0.8*(rand(R,1) - 0.5); % Generate random noise m = 0:R-1;s = 2*m.*(0.9.^m); % Generate uncorrupted signal x = s + d'; % Generate noise corrupted signal subplot(2,1,1);plot(m,d','r-',m,s,'g--',m,x,'b-.');xlabel('Time index n');ylabel('Amplitude'); legend('d[n] ','s[n] ','x[n] ');x1 = [0 0 x];x2 = [0 x 0];x3 = [x 0 0]; y = (x1 + x2 + x3)/3; subplot(2,1,2);plot(m,y(2:R+1),'r-',m,s,'g--'); legend( 'y[n] ','s[n] ');xlabel('Time index n');ylabel('Amplitude'); Answers: Q1.29The signals generated by running Program P1_5 are displayed below :05101520253035404550-5510Time index nA m p l i t u d e2468Time index nA m p l i t u d eQ1.30The uncorrupted signal s[n]is - the product of a linear growth with a slowly decayingreal exponential.The additive noise d[n]is – a random sequence uniformly distributed between -0.4 and+0.4.Q1.31 The statement x = s + d CANNOT be used to generate the noise corrupted signalbecause – d is a column vector, whereas s is a row vector; it is necessary to transposeone of these vectors before adding them.Q1.32The relations between the signals x1, x2, and x3, and the signal x are – all three signalsx1, x2, and x3 are extended versions of x , with one additional sample appended at the left and one additional sample appended to the right. The signal x1 is a delayedversion of x , shifted one sample to the right with zero padding on the left. The signal x2 is equal to x , with equal zero padding on both the left and right to account for the extended length. Finally, x3 is a time advanced version of x , shifted one sample to the left with zero padding on the right.Q1.33The purpose of the legend command is – to create a legend for the graphs. In P1_5, the signals are plotted using different colors and line types; the legend providesinformation about which color and line type is associated with each signal.Project 1.6Generation of Complex SignalsA copy of Program P1_6 is given below.% Program P1_6% Generation of amplitude modulated sequenceclf;n = 0:100;m = 0.4;fH = 0.1; fL = 0.01;xH = sin(2*pi*fH*n);xL = sin(2*pi*fL*n);y = (1+m*xL).*xH;stem(n,y);grid;xlabel('Time index n');ylabel('Amplitude');Answers:Q1.34 The amplitude modulated signals y[n]generated by running Program P1_6 for various values of the frequencies of the carrier signal xH[n]and the modulating signal xL[n], andvarious values of the modulation index m are shown below:m=0.4; fH=0.1; fL=0.01:Time index nA m p l i t u d em=0.9; fH=0.1; fL=0.1:Time index nA m p l i t u d em=0.4; fH=0.1; fL=0.005:Time index nA m p l i t u d em=0.4; fH=0.25; fL=0.01:Time index nA m p l i t u d eQ1.35 The difference between the arithmetic operators * and .* is – “*” multiplies twoconformable matrices or vectors using matrix multiplication. “.*” takes the pointwise products of the elements of two matrices or vectors that have the same dimensions.A copy of Program P1_7 is given below .% Program P1_7% Generation of a swept frequency sinusoidal sequence n = 0:100; a = pi/2/100; b = 0;arg = a*n.*n + b*n; x = cos(arg);clf; stem(n, x);axis([0,100,-1.5,1.5]);title('Swept-Frequency Sinusoidal Signal'); xlabel('Time index n'); ylabel('Amplitude'); grid; axis; Answers:Q1.36 The swept-frequency sinusoidal sequence x[n] generated by running Program P1_7 isdisplayed below .Swept-Frequency Sinusoidal SignalTime index nA m p l i t u d eQ1.37The minimum and maximum frequencies of this signal are - The minimum occurs at n=0, where we have 2an+b = 0 rad/sample = 0 Hz/sample. The maximum occurs at n=100, where we have 2an+b = 200a = 200π(0.5)(0.01) = π rad/sample= 0.5 Hz/sample.Q1.38The Program 1_7 modified to generate a swept sinusoidal signal with a minimum frequency of0.1 and a maximum frequency of 0.3 is given below:Note: for a minimum frequency of 0.1 Hz/sample = π/5 rad/sample at n=0, we must have 2a(0) + b =π/5, which implies b=π/5. For a maximum frequency of 0.3 Hz/sample = 3π/5 rad/sample at n=100, we must have 2a(100) + π/5 = 3π/5, which implies a=π/500.% Program Q1_38% Generation of a swept frequency sinusoidal sequencen = 0:100;a = pi/500;b = pi/5;arg = a*n.*n + b*n;x = cos(arg);clf;stem(n, x);axis([0,100,-1.5,1.5]);title('Swept-Frequency Sinusoidal Signal');xlabel('Time index n');ylabel('Amplitude');grid; axis;INFORMATION1.3 WORKSPACEQ1.39The information displayed in the command window as a result of the who command is – a listing of the names of the variables defined in the current workspace.Q1.40The information displayed in the command window as a result of the whos command is – a long form listing of the variables defined in the current workspace, including the variable names, their dimensions (size), the number of bytes of storage required for each variable, and the datatype of each variable. The total number of bytes of storagefor the entire workspace is also displayed.1.4 OTHER TYPES OF SIGNALS (Optional)Project 1.8 Squarewave and Sawtooth SignalsAnswer:Q1.41MATLAB programs to generate the square-wave and the sawtooth wave sequences of the type shown in Figures 1.1 and 1.2 are given below along with the sequences generated by running these programs:% Program Q1_41A% Generation of the square wave in Fig. 1.1(a)n = 0:30;f = 0.1;phase = 0;duty=60;A = 2.5;arg = 2*pi*f*n + phase;x = A*square(arg,duty);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 30 -3 3]);grid;title('Square Wave Sequence of Fig. 1.1(a)');xlabel('Time index n');ylabel('Amplitude');axis;% Program Q1_41B% Generation of the square wave in Fig. 1.1(b)n = 0:30;f = 0.1;phase = 0;duty=30;A = 2.5;arg = 2*pi*f*n + phase;x = A*square(arg,duty);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 30 -3 3]);grid;title('Square Wave Sequence of Fig. 1.1(b)');xlabel('Time index n');ylabel('Amplitude');axis;% Program Q1_41C% Generation of the square wave in Fig. 1.2(a) n = 0:50;f = 0.05;phase = 0;peak = 1;A = 2.0;arg = 2*pi*f*n + phase;x = A*sawtooth(arg,peak);clf; % Clear old graphstem(n,x); % Plot the generated sequence axis([0 50 -2 2]);grid;title('Sawtooth Wave Sequence of Fig. 1.2(a)'); xlabel('Time index n');ylabel('Amplitude');axis;% Program Q1_41D% Generation of the square wave in Fig. 1.2(b) n = 0:50;f = 0.05;phase = 0;peak = 0.5;A = 2.0;arg = 2*pi*f*n + phase;x = A*sawtooth(arg,peak);clf; % Clear old graphstem(n,x); % Plot the generated sequence axis([0 50 -2 2]);grid;title('Sawtooth Wave Sequence of Fig. 1.2(b)'); xlabel('Time index n');ylabel('Amplitude');axis;051015202530Time index nA m p l i t u d eSquare Wave Sequence of Fig. 1.1(b)Time index nA m p l i t u d eTime index nA m p l i t u d eSawtooth Wave Sequence of Fig. 1.2(b)Time index nA m p l i t u d eDate: 14 September, 2006 Signature: HAVLICEK。

数字信号处理参考答案《解答题及分析题》一、解释下列名词：（1）DSP: 数字信号处理或者数字信号处理芯片；（2）MIPS: 每秒执行百万条指令 ；（3）MOPS: 每秒执行百万条操作 ；（4）FFT: 快速傅里叶变换 ；（5）MAC 时间: 完成一次乘法和一次加法的时间 ；（6）指令周期：执行一条指令所需要的时间，单位通常为（ns ）；（7）BOPS:每秒执行十亿次操作；（8）MFLOPS ：每秒执行百万次浮点操作；（9）TMS320C54X ：TI 公司的54系列定点DSP 芯片；（10）ADSP21XX:AD ：公司的21系列定点DSP 芯片；二、已知)()()]([n x n g n x T =判断系统是否为：① 因果系统；② 稳定系统；③ 线性系统；④ 移不变系统解：（1）求解系统的单位取样响应)(n h令)()(n n x δ=，则系统的单位取样响应)()()(n n g n h δ=① 当0<n 时，0)(=n h ，系统为因果系统；②0)(=∑+∞-∞=n n h ，是稳定系统； ③ 设)()()(),()()(2211n g n x n y n g n x n y ==由于)()()()([)(2121n by n ay n bx n ax T n y +=+=，④ 由于)()]([),()()(k n y k n X T k n g k n x k n y -≠---=-而， 因此，系统为移变系统。

其余几个题的判断方法与这个相同,略。

三、画方框图说明DSP 系统的设计步骤。

设计步骤：（1）根据实际问题的要求写出任务书确定设计目标；（2）算法研究并确定系统的性能指标；（3）选择DSP 芯片和外围芯片；（4）完成系统的硬件设计和软件设计；（5）完成系统的硬件仿真和软件调试；（6）系统集成和测试。

四、以TMS320C5402为例，说明一个典型的DSP 实时数字信号处理系统通常有哪些部分组成？画出系统组成的方框图。

Name: SOLUTIONSection:Laboratory Exercise 2DISCRETE-TIME SYSTEMS: TIME-DOMAIN REPRESENTATION 2.1 SIMULATION OF DISCRETE-TIME SYSTEMSProject 2.1The Moving Average SystemA copy of Program P2_1 is given below:% Program P2_1% Simulation of an M-point Moving Average Filter% Generate the input signaln = 0:100;s1 = cos(2*pi*0.05*n); % A low-frequency sinusoids2 = cos(2*pi*0.47*n); % A high frequency sinusoidx = s1+s2;% Implementation of the moving average filterM = input('Desired length of the filter = ');num = ones(1,M);y = filter(num,1,x)/M;% Display the input and output signalsclf;subplot(2,2,1);plot(n, s1);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude');title('Signal #1');subplot(2,2,2);plot(n, s2);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude');title('Signal #2');subplot(2,2,3);plot(n, x);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude');title('Input Signal');subplot(2,2,4);plot(n, y);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude');title('Output Signal');axis;Answers:Q2.1 The output sequence generated by running the above program for M = 2 with x[n] =s1[n]+s2[n] as the input is shown below .50100-2-1012Time index n A m p l i t u d eSignal #150100-2-1012Time index n A m p l i t u d eSignal #250100-2-1012Time index nA m p l i t u d eInput Signal50100-2-1012Time index nA m p l i t u d eOutput SignalThe component of the input x[n] suppressed by the discrete-time system simulated by this program is – Signal #2, the high frequency one (it is a low pass filter).Q2.2Program P2_1 is modified to simulate the LTI system y[n] = 0.5(x[n]–x[n–1]) and process the input x[n] = s1[n]+s2[n] resulting in the output sequence shown below :Note: the code is not required; however, it is included here to demonstrate a tricky way ofmaking the modification to P2_1.% Program Q2_2% Modification of P1_1 to convert it to a high pass filter % Generate the input signal n = 0:100;s1 = cos(2*pi*0.05*n); % A low-frequency sinusoid s2 = cos(2*pi*0.47*n); % A high frequency sinusoid x = s1+s2;% Implementation of high pass filterM = input('Desired length of the filter = ');% By comparing eq. (2.13) to (2.3), you can see that "num" % actually contains the impulse response (times the constant % M). What we are actually doing in Q2.2 is multiplying the % impulse response of the low pass filter in P2_1 by the % sequency (-1)^n. This shifts the low pass frequency% response up to be centered at f=0.25, making it a high % pass filter.num = (-1).^[0:M-1]; y = filter(num,1,x)/M;% Display the input and output signals clf;subplot(2,2,1); plot(n, s1);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude'); title('Signal #1'); subplot(2,2,2); plot(n, s2);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude'); title('Signal #2'); subplot(2,2,3); plot(n, x);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude'); title('Input Signal'); subplot(2,2,4); plot(n, y);axis([0, 100, -2, 2]);xlabel('Time index n'); ylabel('Amplitude'); title('Output Signal'); axis;Time index n A m p l i t u d eSignal #1-2-1012Time index n A m p l i t u d eSignal #2Time index nA m p l i t u d eInput Signal-2-1012Time index nA m p l i t u d eOutput SignalThe effect of changing the LTI system on the input is – The system is now a high pass filter.It passes the high-frequency input component s2 instead of the low frequency input component s1.Q2.3Program P2_1 is run for the following values of filter length M and following values of the fre-quencies of the sinusoidal signals s1[n] and s2[n]. The output generated for these different values of M and the frequencies are shown below.f1=0.05; f2=0.47; M=1550100-2-1012Time index n A m p l i t u d eSignal #150100Time index n A m p l i t u d eSignal #250100-2-1012Time index nA m p l i t u d eInput Signal50100Time index nA m p l i t u d eOutput SignalFrom these plots we make the following observations – with M=15, the low passcharacteristic is much more pronounced (the passband is now very narrow). s2 is still nearly eliminated in the output signal. s1 is still passed, but at an attenuated level.f1=0.30; f2=0.47; M=4-2-1012Time index n A m p l i t u d eSignal #1Time index n A m p l i t u d eSignal #2-2-1012Time index nA m p l i t u d eInput SignalTime index nA m p l i t u d eOutput SignalFrom these plots we make the following observations – with M=4, this filter performs moresmoothing than in the case M=2. Both s1 and s2 are high frequency in this case, and they are both substantially attenuated in the output.f1=0.05; f2=0.10; M=3-2-1012Time index n A m p l i t u d eSignal #1Time index n A m p l i t u d eSignal #2-2-1012Time index nA m p l i t u d eInput SignalTime index nA m p l i t u d eOutput SignalFrom these plots we make the following observations – here s1 and s2 are both low passand they are both visible in the filter output. However, s2, the higher frequency input,is attenuated slightly more than s1 in the system output.Q2.4 The required modifications to Program P2_1 by changing the input sequence to a swept-frequency sinusoidal signal (length 101, minimum frequency 0, and a maximum frequency 0.5) as the input signal (see Program P1_7) are listed below:% Program Q2_4% Modify P2_1 to use a swept frequency chirp input% Generate the input signaln = 0:100;a = pi/200;b = 0;arg = a*n.*n + b*n;x = cos(arg);% Implementation of the moving average filterM = input('Desired length of the filter = ');num = ones(1,M);y = filter(num,1,x)/M;% Display the input and output signalsclf;subplot(2,1,1);plot(n, x);axis([0, 100, -1.5, 1.5]);xlabel('Time index n'); ylabel('Amplitude');title('Input Signal');subplot(2,1,2);plot(n, y);axis([0, 100, -1.5, 1.5]);xlabel('Time index n'); ylabel('Amplitude');title('Output Signal');axis;The output signal generated by running this program is plotted below .102030405060708090100-101Time index n A m p l i t u d eInput Signal102030405060708090100-101Time index nA m p l i t u d eOutput SignalThe results of Questions Q2.1 and Q2.2 from the response of this system to the swept-frequency signal can be explained as follows : we see again that this system is a low passfilter. At the left of the graphs, the input signal is a low frequency sinusoid that is passed to the output without attenuation. As n increases, the frequency of the input rises, and increasing attenuation is seen at the output. In Q2.1, the input was a sum of two sinusoids s1 and s2 with f1=0.05 and f2=0.47. The swept frequency input of Q2.4 reaches a frequency of 0.05 at n=10, where there is virtually no attenuation in the output shown above. This “explains” why s1 was passed by the system in Q2.1. The swept frequency input of Q2.4 reaches a frequency of 0.47 at approximately n=94, where the attenuation of the system is substantial. This “explains” why s2 was almost completely suppressed in the output in Q2.1.There is no direct relationship between the result shown above for Q2.4 and the result obtained in Q2.2. However, using frequency domain concepts (Chapter 3) we can reason that, if the swept frequency signal was input to the system y[n] = 0.5(x[n] – x[n-1]), we would see a result opposite to what is shown above. Since the system would then be a high pass filter, there would be substantial attenuation of the output at the left side of the graph and virtually no attenuation at the right side of the graph. This “explains” why in Q2.2 the low frequency component s1 was suppressed in the system output, whereas the high frequency component s2 was passed.Project 2.2 (Optional) A Simple Nonlinear Discrete-Time SystemA copy of Program P2_2 is given below:% Program P2_2% Generate a sinusoidal input signalclf;n = 0:200;x = cos(2*pi*0.05*n);% Compute the output signalx1 = [x 0 0]; % x1[n] = x[n+1]x2 = [0 x 0]; % x2[n] = x[n]x3 = [0 0 x]; % x3[n] = x[n-1]y = x2.*x2-x1.*x3;y = y(2:202);% Plot the input and output signalssubplot(2,1,1)plot(n, x)xlabel('Time index n');ylabel('Amplitude');title('Input Signal')subplot(2,1,2)plot(n,y)xlabel('Time index n');ylabel('Amplitude');title('Output signal');Answers:Q2.5The sinusoidal signals with the following frequencies as the input signals were used to generate the output signals: f=0.05, f=0.1, f=0.25The output signals generated for each of the above input signals are displayed below:Time index nA m p l i t u d eTime index nA m p l i t u d eOutput SignalTime index nA m p l i t u d eThe output signals depend on the frequencies of the input signal according to the following rules : up to “edge effects” visible at the very left and right sides of the first graph, theoutput is given by y[n] = sin 2(2πf).This observation can be explained mathematically as follows : Let 2f ω=π. Then[]cos()x n n =ω and22[]cos ()x n n =ω (1.1)Now, [1]cos[(1)]cos()x n n n +=ω+=ω+ω and [1]cos[(1)]x n n −=ω−= cos()n ω−ω, so [1][1]cos()cos()x n x n n n +−=ω+ωω−ω. Applying the trigonometric identity 1122cos()cos()cos(2)cos(2)A B A B A B +−=+, this becomes11[1][1]cos(2)cos(2).22x n x n n +−=ω+ω (1.2) Applying the trigonometric identity 2cos(2)2cos ()1A A =− to the first term of (1.2), we obtain211[1][1]cos ()cos(2).22x n x n n +−=ω−+ω (1.3)Applying the identity 2cos(2)12sin ()A A =−to the second term of (1.3) gives us22[1][1]cos ()sin ().x n x n n +−=ω−ω (1.4)It follows immediately from (1.1) and (1.4) that222[][][1][1]sin ()sin (2).y n x n x n x n f =−+−=ω=πQ2.6The output signal generated by using sinusoidal signals of the form x[n] = sin(ωo n) +K as the input signal is shown below for the following values of ωo and K -ωo = 0.1; K = 0.5Time index nA m p l i t u d eOutput SignalThe dependence of the output signal y[n] on the DC value K can be explained as –Again let2.f ω=π Then []cos()x n n K =ω+ and222[]cos ()2cos().x n n K n K =ω+ω+ (1.5)We have [1]cos[(1)]cos()x n n K n K +=ω++=ω+ω+and[1]cos[(1)]cos().x n n K n K −=ω−+=ω−ω+ So2[1][1]cos()cos()cos()cos().x n x n n n K n K n K +−=ω+ωω−ω+ω+ω+ω−ω+(1.6)From (1.4), we have that 22cos()cos()cos ()sin (),n n n n ω+ωω−ω=ω−ωso (1.6) may be written as222[1][1]cos ()sin ()[cos()cos()].x n x n n n K n n K +−=ω−ω+ω+ω+ω−ω+(1.7)Applying the trigonometric identity cos()cos()2cos()cos()A B A B A B ++−=to the term in square brackets in (1.7), we have that22[1][1]cos ()sin ()2cos()cos()2.x n x n K n K +−=ω−ω+ωω+Combining the this result with (1.5), it follows immediately that2[]sin (2)2[1cos(2)]cos(2).y n f K f fn =π+−ππSo addition of the constant K to the input cosine signal has the effect of adding a ripple onto the result that was obtained before in Q2.5. This ripple is a cosine wave of frequency f and amplitude 2[1cos(2)],K f −πas can be seen plainly in the graph above.Project 2.3Linear and Nonlinear SystemsA copy of Program P2_3 is given below :% Program P2_3% Generate the input sequences clf;n = 0:40; a = 2;b = -3;x1 = cos(2*pi*0.1*n); x2 = cos(2*pi*0.4*n); x = a*x1 + b*x2;num = [2.2403 2.4908 2.2403]; den = [1 -0.4 0.75];ic = [0 0]; % Set zero initial conditionsy1 = filter(num,den,x1,ic); % Compute the output y1[n] y2 = filter(num,den,x2,ic); % Compute the output y2[n] y = filter(num,den,x,ic); % Compute the output y[n] yt = a*y1 + b*y2;d = y - yt; % Compute the difference output d[n] % Plot the outputs and the difference signal subplot(3,1,1) stem(n,y);ylabel('Amplitude');title('Output Due to Weighted Input: a \cdot x_{1}[n] + b \cdot x_{2}[n]'); subplot(3,1,2) stem(n,yt);ylabel('Amplitude');title('Weighted Output: a \cdot y_{1}[n] + b \cdot y_{2}[n]'); subplot(3,1,3) stem(n,d);xlabel('Time index n');ylabel('Amplitude'); title('Difference Signal');Answers: Q2.7The outputs y[n], obtained with weighted input, and yt[n], obtained by combining the two outputs y1[n] and y2[n] with the same weights, are shown below along with the difference between the two signals :0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]x 10-15Time index nA m p l i t u d eDifference SignalThe two sequences are – the same up to numerical roundoff. The system is – Linear.Q2.8 Program P2_3 was run for the following three different sets of values of the weightingconstants, a and b , and the following three different sets of input frequencies :1. a=1; b=-1; f1=0.05; f2=0.4;2. a=10; b=2; f1=0.10; f2=0.25;3. a=2; b=10; f1=0.15; f2=0.20;The plots generated for each of the above three cases are shown below :0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540x 10-15Time index nA m p l i t u d eDifference Signal0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540x 10-14Time index nA m p l i t u d eDifference SignalA m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540x 10-14Time index nA m p l i t u d eDifference SignalBased on these plots we can conclude that the system with different weights is – Linear.Q2.9 Program 2_3 was run with the following non-zero initial conditions – ic = [5 10];The plots generated are shown below -A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]Time index nA m p l i t u d eDifference SignalBased on these plots we can conclude that the system with nonzero initial conditions is –Nonlinear.Q2.10Program P2_3 was run with nonzero initial conditions and for the following three different sets of values of the weighting constants, a and b , and the following three different sets of input frequencies :1. a=1; b=-1; f1=0.05; f2=0.4;2. a=10; b=2; f1=0.10; f2=0.25;3. a=2; b=10; f1=0.15; f2=0.20;The plots generated for each of the above three cases are shown below :0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540Time index nA m p l i t u d eDifference Signal0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540Time index nA m p l i t u d eDifference Signal0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540Time index nA m p l i t u d eDifference SignalBased on these plots we can conclude that the system with nonzero initial conditions and different weights is – Nonlinear.Q2.11 Program P2_3 was modified to simulate the system :y[n] = x[n]x[n –1]The output sequences y1[n], y2[n],and y[n]of the above system generated by running the modified program are shown below :0510152025303540y 1[n]0510152025303540y 2[n]y t [n]0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]0510152025303540A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540Time index nA m p l i t u d eDifference SignalComparing y[n] with yt[n] we conclude that the two sequences are – Not the Same. This system is – Nonlinear.Project 2.4 Time-invariant and Time-varying SystemsA copy of Program P2_4 is given below:% Program P2_4% Generate the input sequencesclf;n = 0:40; D = 10;a = 3.0;b = -2;x = a*cos(2*pi*0.1*n) + b*cos(2*pi*0.4*n);xd = [zeros(1,D) x];num = [2.2403 2.4908 2.2403];den = [1 -0.4 0.75];ic = [0 0]; % Set initial conditions% Compute the output y[n]y = filter(num,den,x,ic);% Compute the output yd[n]yd = filter(num,den,xd,ic);% Compute the difference output d[n]d = y - yd(1+D:41+D);% Plot the outputssubplot(3,1,1)stem(n,y);ylabel('Amplitude');title('Output y[n]'); grid;subplot(3,1,2)stem(n,yd(1:41));ylabel('Amplitude');title(['Output due to Delayed Input x[n Ð', num2str(D),']']); grid; subplot(3,1,3)stem(n,d);xlabel('Time index n'); ylabel('Amplitude');title('Difference Signal'); grid;Answers: Q2.12The output sequences y[n] and yd[n] generated by running Program P2_4 are shown below -0510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]Time index nA m p l i t u d eDifference SignalThese two sequences are related as follows – y[n-10] = yd[n].The system is – Time Invariant.Q2.13 The output sequences y[n] and yd[n] generated by running Program P2_4 for thefollowing values of the delay variable D – 2; 6; 8.are shown below -0510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -2]0510152025303540Time index nA m p l i t u d eDifference Signal510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -6]0510152025303540Time index nA m p l i t u d eDifference Signal0510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -8]0510152025303540Time index nA m p l i t u d eDifference SignalIn each case, these two sequences are related as follows – y[n-D] = yd[n].The system is – Time Invariant.Q2.14 The output sequences y[n] and yd[n] generated by running Program P2_4 for thefollowing values of the input frequencies –1. f1=0.05; f2=0.40;2. f1=0.10; f2=0.25;3. f1=0.15; f2=0.20;are shown below –0510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference Signal510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference Signal0510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference SignalIn each case, these two sequences are related as follows – y[n-10] = yd[n].The system is – Time Invariant.Q2.15The output sequences y[n] and yd[n] generated by running Program P2_4 for non-zero initial conditions are shown below –0510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference SignalThese two sequences are related as follows – yd[n] is NOT equal to the shift of y[n].The system is – Time Varying.Q2.16The output sequences y[n] and yd[n] generated by running Program P2_4 for non-zero initial conditions and following values of the input frequencies –1. f1=0.05; f2=0.40;2. f1=0.10; f2=0.25;3. f1=0.15; f2=0.20;are shown below -0510152025303540A m p l i t u d eOutput y[n]A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference Signal0510152025303540A m p l i t u d eOutput y[n]510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference SignalA m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference SignalIn each case, these two sequences are related as follows – yd[n] is NOT given by theshift of y[n].The system is – Time Varying.Q2.17The modified Program 2_4 simulating the systemy[n] = n x[n] + x[n-1]is given below :% Program Q2_17% Modification of P2_4 to implement the system% given by (2.16).% Generate the input sequencesclf;n = 0:40; D = 10;a = 3.0;b = -2;x = a*cos(2*pi*0.1*n) + b*cos(2*pi*0.4*n);xd = [zeros(1,D) x];nd = 0:length(xd)-1;% Compute the output y[n]y = (n .* x) + [0 x(1:40)];% Compute the output yd[n]yd = (nd .* xd) + [0 xd(1:length(xd)-1)];% Compute the difference output d[n]d = y - yd(1+D:41+D);% Plot the outputssubplot(3,1,1)stem(n,y);ylabel('Amplitude');title('Output y[n]'); grid;subplot(3,1,2)stem(n,yd(1:41));ylabel('Amplitude');title(['Output due to Delayed Input x[n -', num2str(D),']']); grid;subplot(3,1,3)stem(n,d);xlabel('Time index n'); ylabel('Amplitude');title('Difference Signal'); grid;The output sequences y[n]and yd[n] generated by running modified Program P2_4 are shown below -0510152025303540A m p l i t u d eOutput y[n]0510152025303540A m p l i t u d eOutput due to Delayed Input x[n -10]0510152025303540Time index nA m p l i t u d eDifference SignalThese two sequences are related as follows – yd[n] is NOT the shifted version of y[n]. The system is – Time Varying.Q2.18 (optional) The modified Program P2_3 to test the linearity of the system of (2.16) is shown below :% Program Q2_18% Modify P2_3 for Q2.18.% Generate the input sequences clf;n = 0:40; a = 2;b = -3;x1 = cos(2*pi*0.1*n); x2 = cos(2*pi*0.4*n); x = a*x1 + b*x2;y1 = (n .* x1) + [0 x1(1:40)]; % Compute the output y1[n] y2 = (n .* x2) + [0 x2(1:40)]; % Compute the output y2[n] y = (n .* x) + [0 x(1:40)]; % Compute the output y[n] yt = a*y1 + b*y2;d = y - yt; % Compute the difference output d[n] % Plot the outputs and the difference signal subplot(3,1,1) stem(n,y);ylabel('Amplitude');title('Output Due to Weighted Input: a \cdot x_{1}[n] + b \cdot x_{2}[n]'); subplot(3,1,2)stem(n,yt);ylabel('Amplitude');title('Weighted Output: a \cdot y_{1}[n] + b \cdot y_{2}[n]'); subplot(3,1,3) stem(n,d);xlabel('Time index n');ylabel('Amplitude'); title('Difference Signal');The outputs y[n]and yt[n] obtained by running the modified program P2_3 are shown below :0510152025303540A m p l i t u d eOutput Due to Weighted Input: a ⋅ x 1[n] + b ⋅ x 2[n]A m p l i t u d eWeighted Output: a ⋅ y 1[n] + b ⋅ y 2[n]0510152025303540x 10-14Time index nA m p l i t u d eDifference SignalThe two sequences are – The same up to numerical roundoff.The system is – Linear.2.5 LINEAR TIME-INVARIANT DISCRETE-TIME SYSTEMS Project 2.5Computation of Impulse Responses of LTI SystemsA copy of Program P2_5 is shown below :% Program P2_5% Compute the impulse response y clf; N = 40;num = [2.2403 2.4908 2.2403]; den = [1 -0.4 0.75]; y = impz(num,den,N);% Plot the impulse response stem(y);xlabel('Time index n'); ylabel('Amplitude'); title('Impulse Response'); grid;Answers:Q2.19 The first 41 samples of the impulse response of the discrete-time system of Project 2.3generated by running Program P2_5 is given below:-3-2-11234Time index nA m p l i t u d eImpulse ResponseQ2.20 The required modifications to Program P2_5 to generate the impulse response of the followingcausal LTI system :y[n] + 0.71y[n-1] – 0.46y[n-2] – 0.62y[n-3]= 0.9x[n] – 0.45x[n-1] + 0.35x[n-2] + 0.002x[n-3]are given below :% Program Q2_20% Compute the impulse response y clf; N = 45;num = [0.9 -0.45 0.35 0.002]; den = [1.0 0.71 -0.46 -0.62]; y = impz(num,den,N);% Plot the impulse response stem(y);xlabel('Time index n'); ylabel('Amplitude'); title('Impulse Response'); grid;The first 45 samples of the impulse response of this discrete-time system generated by running the modified is given below :051015202530354045-1.5-1-0.50.511.52Time index nA m p l i t u d eImpulse Responsethe filter command is indicated below :% Program Q2_21% Compute the impulse response y clf; N = 40;num = [0.9 -0.45 0.35 0.002]; den = [1.0 0.71 -0.46 -0.62]; % input: unit pulse x = [1 zeros(1,N-1)]; % outputy = filter(num,den,x);% Plot the impulse response% NOTE: the time axis will be WRONG; h[0] will % be plotted at n=1; but this will agree with % the INCORRECT plotting that was also done % by program P2_5. stem(y);xlabel('Time index n'); ylabel('Amplitude'); title('Impulse Response'); grid;The first 40 samples of the impulse response generated by this program are shown below :0510152025303540-1.5-1-0.50.511.52Time index nA m p l i t u d eImpulse ResponseComparing the above response with that obtained in Question Q2.20 we conclude - They are the SAME.indicated below :% Program Q2_22% Compute the step response s clf; N = 40; n = 0:N-1;num = [2.2403 2.4908 2.2403]; den = [1.0 -0.4 0.75]; % input: unit step x = [ones(1,N)]; % outputy = filter(num,den,x); % Plot the step response stem(n,y);xlabel('Time index n'); ylabel('Amplitude'); title('Step Response'); grid;The first 40 samples of the step response of the LTI system of Project 2.3 are shown below :0510152025303540Time index nA m p l i t u d eStep ResponseProject 2.6 Cascade of LTI SystemsA copy of Program P2_6 is given below:% Program P2_6% Cascade Realizationclf;x = [1 zeros(1,40)]; % Generate the inputn = 0:40;% Coefficients of 4th order systemden = [1 1.6 2.28 1.325 0.68];num = [0.06 -0.19 0.27 -0.26 0.12];% Compute the output of 4th order systemy = filter(num,den,x);% Coefficients of the two 2nd order systemsnum1 = [0.3 -0.2 0.4];den1 = [1 0.9 0.8];num2 = [0.2 -0.5 0.3];den2 = [1 0.7 0.85];% Output y1[n] of the first stage in the cascadey1 = filter(num1,den1,x);% Output y2[n] of the second stage in the cascadey2 = filter(num2,den2,y1);% Difference between y[n] and y2[n]d = y - y2;% Plot output and difference signalssubplot(3,1,1);stem(n,y);ylabel('Amplitude');title('Output of 4th order Realization'); grid;subplot(3,1,2);stem(n,y2)ylabel('Amplitude');title('Output of Cascade Realization'); grid;subplot(3,1,3);stem(n,d)xlabel('Time index n');ylabel('Amplitude');title('Difference Signal'); grid;Answers:Q2.23The output sequences y[n], y2[n], and the difference signal d[n]generated by running Program P2_6 are indicated below:。

第十章上机实验第十章上机实验数字信号处理是一门理论和实际密切结合的课程，为深入掌握课程内容，最好在学习理论的同时，做习题和上机实验。

上机实验不仅可以帮助读者深入的理解和消化基本理论，而且能锻炼初学者的独立解决问题的能力。

本章在第二版的基础上编写了六个实验，前五个实验属基础理论实验，第六个属应用综合实验。

实验一系统响应及系统稳定性。

实验二时域采样与频域采样。

实验三用FFT对信号作频谱分析。

实验四IIR 数字滤波器设计及软件实现。

实验五FIR 数字滤波器设计与软件实现实验六应用实验——数字信号处理在双音多频拨号系统中的应用任课教师根据教学进度，安排学生上机进行实验。

建议自学的读者在学习完第一章后作实验一；在学习完第三、四章后作实验二和实验三；实验四IIR 数字滤波器设计及软件实现在。

学习完第六章进行；实验五在学习完第七章后进行。

实验六综合实验在学习完第七章或者再后些进行；实验六为综合实验，在学习完本课程后再进行。

10.1实验一:系统响应及系统稳定性1.实验目的（1）掌握求系统响应的方法。

（2）掌握时域离散系统的时域特性。

（3）分析、观察及检验系统的稳定性。

2.实验原理与方法在时域中，描写系统特性的方法是差分方程和单位脉冲响应，在频域可以用系统函数描述系统特性。

已知输入信号可以由差分方程、单位脉冲响应或系统函数求出系统对于该输入信号的响应，本实验仅在时域求解。

在计算机上适合用递推法求差分方程的解，最简单的方法是采用MA TLAB 语言的工具箱函数filter 函数。

也可以用 MATLAB语言的工具箱函数conv 函数计算输入信号和系统的单位脉冲响应的线性卷积，求出系统的响应。

系统的时域特性指的是系统的线性时不变性质、因果性和稳定性。

重点分析实验系统的稳定性，包括观察系统的暂态响应和稳定响应。

系统的稳定性是指对任意有界的输入信号，系统都能得到有界的系统响应。

或者系统的单位脉冲响应满足绝对可和的条件。

系统的稳定性由其差分方程的系数决定。

《数字信号处理(第四版)》部分课后习题解答一、简答题1. 什么是数字信号处理？数字信号处理（DSP）是指对数字信号进行处理和分析的一种技术。

它使用数学和算法处理模拟信号，从而实现信号的采样、量化、编码、存储和重构等过程。

DSP广泛应用于通信、音频处理、图像处理和控制系统中。

2. 数字信号处理的主要特点有哪些？•数字信号处理能够处理和分析具有广泛频谱范围的信号。

•数字信号处理能够实现高精度的信号处理和复杂的算法运算。

•数字信号处理能够实现信号的存储、传输和复原等功能。

•数字信号处理可以利用计算机等处理硬件进行实时处理和系统集成。

3. 数字信号处理的基本原理是什么？数字信号处理的基本原理是将连续时间的模拟信号转换成离散时间的数字信号，然后通过一系列的算法对数字信号进行处理和分析。

该过程主要涉及信号的采样、量化和编码等环节。

4. 什么是离散时间信号？离散时间信号是指信号的取样点在时间上呈现离散的情况。

在离散时间信号中，只能在离散时间点上获取信号的取样值，而无法观测到连续时间上的信号变化。

5. 描述离散时间信号的功率和能量的计算方法。

对于离散时间信号，其功率和能量的计算方法如下：•功率：对于离散时间信号x(n)，其功率可以通过求平方和的平均值来计算，即功率P = lim(T->∞) [1/T *∑|x(n)|^2]，其中T表示信号x(n)的观测时间。

•能量：对于离散时间信号x(n)，其能量可以通过求平方和来计算，即能量E = ∑|x(n)|^2。

二、计算题1. 设有一个离散时间周期序列x(n) = [2, 3, -1, 4, 0, -2]，求其周期N。

由于x(n)是一个周期序列，我们可以通过观察序列来确定其周期。

根据观察x(n)的取值，我们可以发现序列在n=1和n=5两个位置上取得了相同的数值。

因此，序列x(n)的周期为N = 5 - 1 = 4。

2. 设有一个信号x(t) = 2sin(3t + π/4)，请将其离散化为离散时间信号x(n)。

Name : SOLUTION Section :Laboratory Exercise 1DISCRETE-TIME SIGNALS: TIME-DOMAIN REPRESENTATION1.1GENERATION OF SEQUENCESProject 1.1Unit sample and unit step sequencesA copy of Program P1_1 is given below.% Program P1_1% Generation of a Unit Sample Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the unit sample sequence u = [zeros(1,10) 1 zeros(1,20)]; % Plot the unit sample sequence stem(n,u);xlabel('Time index n');ylabel('Amplitude'); title('Unit Sample Sequence'); axis([-10 20 0 1.2]); Answers : Q1.1The unit sample sequence u[n] generated by running Program P1_1 is shown below:Time index nA m p l i t u d eUnit Sample SequenceQ1.2The purpose of clf command is – clear the current figureThe purpose of axis command is – control axis scaling and appearanceThe purpose of title command is – add a title to a graph or an axis and specify textpropertiesThe purpose of xlabel command is – add a label to the x-axis and specify textpropertiesThe purpose of ylabel command is – add a label to the y-axis and specify the textpropertiesQ1.3The modified Program P1_1 to generate a delayed unit sample sequence ud[n] with a delay of 11 samples is given below along with the sequence generated by running this program .% Program P1_1, MODIFIED for Q1.3% Generation of a DELAYED Unit Sample Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the DELAYED unit sample sequence u = [zeros(1,21) 1 zeros(1,9)];% Plot the DELAYED unit sample sequence stem(n,u);xlabel('Time index n');ylabel('Amplitude'); title('DELAYED Unit Sample Sequence'); axis([-10 20 0 1.2]);Time index nA m p l i t u d eDELAYED Unit Sample SequenceQ1.4The modified Program P1_1 to generate a unit step sequence s[n] is given below along with the sequence generated by running this program .% Program Q1_4% Generation of a Unit Step Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the unit step sequence s = [zeros(1,10) ones(1,21)]; % Plot the unit step sequence stem(n,s);xlabel('Time index n');ylabel('Amplitude'); title('Unit Step Sequence'); axis([-10 20 0 1.2]);Time index nA m p l i t u d eUnit Step SequenceQ1.5The modified Program P1_1 to generate a unit step sequence sd[n] with an advance of 7 samples is given below along with the sequence generated by running this program .% Program Q1_5% Generation of an ADVANCED Unit Step Sequence clf;% Generate a vector from -10 to 20 n = -10:20;% Generate the ADVANCED unit step sequence sd = [zeros(1,3) ones(1,28)];% Plot the ADVANCED unit step sequence stem(n,sd);xlabel('Time index n');ylabel('Amplitude'); title('ADVANCED Unit Step Sequence'); axis([-10 20 0 1.2]);Time index nA m p l i t u d eADVANCED Unit Step SequenceProject 1.2 Exponential signalsA copy of Programs P1_2 and P1_3 are given below .% Program P1_2% Generation of a complex exponential sequence clf;c = -(1/12)+(pi/6)*i; K = 2; n = 0:40;x = K*exp(c*n); subplot(2,1,1); stem(n,real(x));xlabel('Time index n');ylabel('Amplitude'); title('Real part'); subplot(2,1,2); stem(n,imag(x));xlabel('Time index n');ylabel('Amplitude'); title('Imaginary part');% Program P1_3% Generation of a real exponential sequence clf;n = 0:35; a = 1.2; K = 0.2; x = K*a.^n; stem(n,x);xlabel('Time index n');ylabel('Amplitude');Answers: Q1.6The complex-valued exponential sequence generated by running Program P1_2 is shown below :510152025303540Time index n A m p l i t u d eReal partTime index nA m p l i t u d eImaginary partQ1.7The parameter controlling the rate of growth or decay of this sequence is – the real part of c .The parameter controlling the amplitude of this sequence is - KQ1.8The result of changing the parameter c to (1/12)+(pi/6)*i is – since exp(-1/12) isless than one and exp(1/12) is greater than one, this change means that the envelope of the signal will grow with n instead of decay with n.Q1.9The purpose of the operator real is – to extract the real part of a Matlab vector.The purpose of the operator imag is – to extract the imaginary part of a Matlab vector. Q1.10The purpose of the command subplot is – to plot more than one graph in the sameMatlab figure.Q1.11The real-valued exponential sequence generated by running Program P1_3 is shown below :Time index nA m p l i t u d eQ1.12The parameter controlling the rate of growth or decay of this sequence is - aThe parameter controlling the amplitude of this sequence is - KQ1.13The difference between the arithmetic operators ^ and .^ is – “^” raises a square matrix toa power using matrix multiplication. “.^” raises the elements of a matrix or vector to a power; this is a “pointwise” operation.Q1.14The sequence generated by running Program P1_3 with the parameter a changed to 0.9 and the parameter K changed to 20 is shown below :5101520253035Time index nA m p l i t u d eQ1.15The length of this sequence is - 36It is controlled by the following MATLAB command line : n = 0:35;It can be changed to generate sequences with different lengths as follows (give an example command line and the corresponding length): n = 0:99; makes the length 100. Q1.16The energies of the real-valued exponential sequences x [n]generated in Q1.11 and Q1.14 and computed using the command sum are - 4.5673e+004 and 2.1042e+003.Project 1.3 Sinusoidal sequencesA copy of Program P1_4 is given below .% Program P1_4% Generation of a sinusoidal sequence n = 0:40;f = 0.1; phase = 0; A = 1.5; arg = 2*pi*f*n - phase; x = A*cos(arg);clf; % Clear old graphstem(n,x); % Plot the generated sequence axis([0 40 -2 2]); grid;title('Sinusoidal Sequence'); xlabel('Time index n'); ylabel('Amplitude'); axis; Answers:Q1.17 The sinusoidal sequence generated by running Program P1_4 is displayed below .0510152025303540Sinusoidal SequenceTime index nA m p l i t u d eQ1.18The frequency of this sequence is- f = 0.1 cycles/sample.It is controlled by the following MATLAB command line: f = 0.1;A sequence with new frequency 0.2 can be generated by the following command line:f = 0.2;The parameter controlling the phase of this sequence is- phaseThe parameter controlling the amplitude of this sequence is- AThe period of this sequence is - 2π/ω = 1/f = 10Q1.19 The length of this sequence is - 41It is controlled by the following MATLAB command line: n = 0:40;A sequence with new length __81_ can be generated by the following command line:n = 0:80;Q1.20 The average power of the generated sinusoidal sequence is–sum(x(1:10).*x(1:10))/10 = 1.1250Q1.21 The purpose of axis command is – to set the range of the x-axis to [0,40] and the range of the y-axis to [-2,2].The purpose of grid command is – to turn on the drawing of grid lines on the graph.Q1.22 The modified Program P1_4 to generate a sinusoidal sequence of frequency 0.9 is given below along with the sequence generated by running it.% Program Q1_22A% Generation of a sinusoidal sequencen = 0:40;f = 0.9;phase = 0;A = 1.5;arg = 2*pi*f*n - phase;x = A*cos(arg);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 40 -2 2]);grid;title('Sinusoidal Sequence');xlabel('Time index n');ylabel('Amplitude');axis;Sinusoidal SequenceTime index nA m p l i t u d eA comparison of this new sequence with the one generated in Question Q1.17 shows - thetwo graphs are identical. This is because, in the modified program, we have ω = 0.9*2π. This generates the same graph as a cosine with angular frequency ω - 2π = −0.1*2π. Because cosine is an even function, this is the same as a cosine with angular frequency +0.1*2π, which was the value used in P1_4.m in Q1.17.In terms of Hertzian frequency, we have for P1_4.m in Q1.17 that f = 0.1 Hz/sample. For the modified program in Q1.22, we have f = 0.9 Hz/sample, which generates the same graph as f = 0.9 – 1 = −0.1. Again because cosine is even, this makes a graph that is identical to the one we got in Q1.17 with f = +0.1 Hz/sample.A sinusoidal sequence of frequency 1.1 generated by modifying Program P1_4 is shown below.0510152025303540Sinusoidal SequenceTime index nA m p l i t u d eA comparison of this new sequence with the one generated in Question Q1.17 shows - thegraph here is again identical to the one in Q1.17. This is because a cosine of frequency f = 1.1 Hz/sample is identical to one with frequency f = 1.1 – 1 = 0.1 Hz/sample, which was the frequency used in Q1.17.Q1.23The sinusoidal sequence of length 50, frequency 0.08, amplitude 2.5, and phase shift of 90 degrees generated by modifying Program P1_4 is displayed below .NOTE: for this program, it is necessary to convert the phase of 90 deg to radians and account for the minus sign that appears in the statement “arg = 2*pi*f*n - phase;” as opposed to the plus sign shown in eq. (1.12) of the lab manual. The correct statement to generate the phase is “phase = -90/(2*pi);”. It is also necessary to modify the axis command to account for the new length and amplitude of the signal. The correct axis statement is “axis([0 50 -3 3]);”.05101520253035404550Sinusoidal SequenceTime index nA m p l i t u d eThe period of this sequence is - 2π/ω = 1/f = 1/0.08 = 1/(8/100) = 100/8 = 25/2.Therefore, the fundamental period is 25 and the graph has the “appearance” of going through 2 cycles of a cosine waveform during each period.Q1.24By replacing the stem command in Program P1_4 with the plot command, the plot obtained is as shown below :Sinusoidal SequenceTime index nA m p l i t u d eThe difference between the new plot and the one generated in Question Q1.17 is – instead ofdrawing stems from the x-axis to the points on the curve, the “plot ” command connects the points with straight line segments, which approximates the graph of a continuous-time cosine signal.Q1.25By replacing the stem command in Program P1_4 with the stairs command the plot obtained is as shown below :Sinusoidal SequenceTime index nA m p l i t u d eThe difference between the new plot and those generated in Questions Q1.17 and Q1.24 is– the “stairs” command produces a stairstep plot as opposed to the stem graph that was generated in Q1.17 and the straight-line interpolation plot that was generated in Q1.24.Project 1.4 Random signalsAnswers:Q1.26 The MATLAB program to generate and display a random signal of length 100 with elements uniformly distributed in the interval [–2, 2] is given below along with the plot of the random sequence generated by running the program:% Program Q1_26% Generation of a uniform random sequencen = 0:99;A = 2;rand('state',sum(100*clock)); % "seed" the generator% rand(1,100) is uniform in [0,1]% rand(1,100)-0.5 is uniform in [-0.5,0.5]% 4*(rand(1,100)-0.5) is uniform in [-2,2]x = 2*A*(rand(1,length(n))-0.5);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 length(n) -round(2*(A+0.5))/2 round(2*(A+0.5))/2]);grid;title('Uniform Random Sequence');xlabel('Time index n');ylabel('Amplitude');axis;102030405060708090100Uniform Random SequenceTime index nA m p l i t u d eQ1.27 The MATLAB program to generate and display a Gaussian random signal of length 75 withelements normally distributed with zero mean and a variance of 3 is given below along with the plot of the random sequence generated by running the program :% Program Q1_27% Generation of a Gaussian random sequence% NOTE: if X is a random variable with zero mean and % unity variance, then (aX + b) is a random variable % with mean b and variance a^2. This follows from % basic probability theory. n = 0:74;xmean = 0; % mean of xxstd = sqrt(3); % standard deviation of xrandn('state',sum(100*clock)); % "seed" the generator % generate the sequencex = xstd*randn(1,length(n)) + xmean; % setup the graph and plotclf; % Clear old graphstem(n,x); % Plot the generated sequence xmax = max(abs(x));Ylim = round(2*(xmax+0.5))/2; axis([0 length(n) -Ylim Ylim]); grid;title('Gaussian Random Sequence'); xlabel('Time index n'); ylabel('Amplitude'); axis;10203040506070Gaussian Random SequenceTime index nA m p l i t u d eQ1.28The MATLAB program to generate and display five sample sequences of a random sinusoidal signal of length 31{X[n]} = {A·cos(ωo n + φ)}where the amplitude A and the phase φ are statistically independent random variables with uniform probability distribution in the range 0≤A ≤4 for the amplitude and in the range 0 ≤ φ ≤ 2π for the phase is given below. Also shown are five sample sequences generated by running this program five different times .% Generates the "deterministic stochastic process" % called for in Q1.28. n = 0:30; f = 0.1; Amax = 4;phimax = 2*pi;rand('state',sum(100*clock)); % "seed" the generator A = Amax*rand;% NOTE: successive calls to rand without arguments % return a random sequence of scalars. Since this % random sequence is "white" (uncorrelated), it is % not necessary to re-seed the generator for phi. phi = phimax*rand;% generate the sequence arg = 2*pi*f*n + phi; x = A*cos(arg); % plotclf; % Clear old graphstem(n,x); % Plot the generated sequence Ylim = round(2*(Amax+0.5))/2; axis([0 length(n) -Ylim Ylim]); grid;title('Sinusoidal Sequence with Random Amplitude and Phase'); xlabel('Time index n'); ylabel('Amplitude'); axis;Sinusoidal Sequence with Random Amplitude and PhaseTime index nA m p l i t u d eSinusoidal Sequence with Random Amplitude and PhaseTime index nA m p l i t u d eTime index nA m p l i t u d eSinusoidal Sequence with Random Amplitude and PhaseTime index nA m p l i t u d eTime index nA m p l i t u d e1.2 SIMPLE OPERATIONS ON SEQUENCESProject 1.5Signal SmoothingA copy of Program P1_5 is given below .% Program P1_5% Signal Smoothing by Averaging clf; R = 51;d = 0.8*(rand(R,1) - 0.5); % Generate random noise m = 0:R-1;s = 2*m.*(0.9.^m); % Generate uncorrupted signal x = s + d'; % Generate noise corrupted signal subplot(2,1,1);plot(m,d','r-',m,s,'g--',m,x,'b-.');xlabel('Time index n');ylabel('Amplitude'); legend('d[n] ','s[n] ','x[n] ');x1 = [0 0 x];x2 = [0 x 0];x3 = [x 0 0]; y = (x1 + x2 + x3)/3; subplot(2,1,2);plot(m,y(2:R+1),'r-',m,s,'g--'); legend( 'y[n] ','s[n] ');xlabel('Time index n');ylabel('Amplitude'); Answers: Q1.29The signals generated by running Program P1_5 are displayed below :05101520253035404550-5510Time index nA m p l i t u d e2468Time index nA m p l i t u d eQ1.30The uncorrupted signal s[n]is - the product of a linear growth with a slowly decayingreal exponential.The additive noise d[n]is – a random sequence uniformly distributed between -0.4 and+0.4.Q1.31 The statement x = s + d CANNOT be used to generate the noise corrupted signalbecause – d is a column vector, whereas s is a row vector; it is necessary to transposeone of these vectors before adding them.Q1.32The relations between the signals x1, x2, and x3, and the signal x are – all three signalsx1, x2, and x3 are extended versions of x , with one additional sample appended at the left and one additional sample appended to the right. The signal x1 is a delayedversion of x , shifted one sample to the right with zero padding on the left. The signal x2 is equal to x , with equal zero padding on both the left and right to account for the extended length. Finally, x3 is a time advanced version of x , shifted one sample to the left with zero padding on the right.Q1.33The purpose of the legend command is – to create a legend for the graphs. In P1_5, the signals are plotted using different colors and line types; the legend providesinformation about which color and line type is associated with each signal.Project 1.6Generation of Complex SignalsA copy of Program P1_6 is given below.% Program P1_6% Generation of amplitude modulated sequenceclf;n = 0:100;m = 0.4;fH = 0.1; fL = 0.01;xH = sin(2*pi*fH*n);xL = sin(2*pi*fL*n);y = (1+m*xL).*xH;stem(n,y);grid;xlabel('Time index n');ylabel('Amplitude');Answers:Q1.34 The amplitude modulated signals y[n]generated by running Program P1_6 for various values of the frequencies of the carrier signal xH[n]and the modulating signal xL[n], andvarious values of the modulation index m are shown below:m=0.4; fH=0.1; fL=0.01:Time index nA m p l i t u d em=0.9; fH=0.1; fL=0.1:Time index nA m p l i t u d em=0.4; fH=0.1; fL=0.005:Time index nA m p l i t u d em=0.4; fH=0.25; fL=0.01:Time index nA m p l i t u d eQ1.35 The difference between the arithmetic operators * and .* is – “*” multiplies twoconformable matrices or vectors using matrix multiplication. “.*” takes the pointwise products of the elements of two matrices or vectors that have the same dimensions.A copy of Program P1_7 is given below .% Program P1_7% Generation of a swept frequency sinusoidal sequence n = 0:100; a = pi/2/100; b = 0;arg = a*n.*n + b*n; x = cos(arg);clf; stem(n, x);axis([0,100,-1.5,1.5]);title('Swept-Frequency Sinusoidal Signal'); xlabel('Time index n'); ylabel('Amplitude'); grid; axis; Answers:Q1.36 The swept-frequency sinusoidal sequence x[n] generated by running Program P1_7 isdisplayed below .Swept-Frequency Sinusoidal SignalTime index nA m p l i t u d eQ1.37The minimum and maximum frequencies of this signal are - The minimum occurs at n=0, where we have 2an+b = 0 rad/sample = 0 Hz/sample. The maximum occurs at n=100, where we have 2an+b = 200a = 200π(0.5)(0.01) = π rad/sample= 0.5 Hz/sample.Q1.38The Program 1_7 modified to generate a swept sinusoidal signal with a minimum frequency of0.1 and a maximum frequency of 0.3 is given below:Note: for a minimum frequency of 0.1 Hz/sample = π/5 rad/sample at n=0, we must have 2a(0) + b =π/5, which implies b=π/5. For a maximum frequency of 0.3 Hz/sample = 3π/5 rad/sample at n=100, we must have 2a(100) + π/5 = 3π/5, which implies a=π/500.% Program Q1_38% Generation of a swept frequency sinusoidal sequencen = 0:100;a = pi/500;b = pi/5;arg = a*n.*n + b*n;x = cos(arg);clf;stem(n, x);axis([0,100,-1.5,1.5]);title('Swept-Frequency Sinusoidal Signal');xlabel('Time index n');ylabel('Amplitude');grid; axis;INFORMATION1.3 WORKSPACEQ1.39The information displayed in the command window as a result of the who command is – a listing of the names of the variables defined in the current workspace.Q1.40The information displayed in the command window as a result of the whos command is – a long form listing of the variables defined in the current workspace, including the variable names, their dimensions (size), the number of bytes of storage required for each variable, and the datatype of each variable. The total number of bytes of storagefor the entire workspace is also displayed.1.4 OTHER TYPES OF SIGNALS (Optional)Project 1.8 Squarewave and Sawtooth SignalsAnswer:Q1.41MATLAB programs to generate the square-wave and the sawtooth wave sequences of the type shown in Figures 1.1 and 1.2 are given below along with the sequences generated by running these programs:% Program Q1_41A% Generation of the square wave in Fig. 1.1(a)n = 0:30;f = 0.1;phase = 0;duty=60;A = 2.5;arg = 2*pi*f*n + phase;x = A*square(arg,duty);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 30 -3 3]);grid;title('Square Wave Sequence of Fig. 1.1(a)');xlabel('Time index n');ylabel('Amplitude');axis;% Program Q1_41B% Generation of the square wave in Fig. 1.1(b)n = 0:30;f = 0.1;phase = 0;duty=30;A = 2.5;arg = 2*pi*f*n + phase;x = A*square(arg,duty);clf; % Clear old graphstem(n,x); % Plot the generated sequenceaxis([0 30 -3 3]);grid;title('Square Wave Sequence of Fig. 1.1(b)');xlabel('Time index n');ylabel('Amplitude');axis;% Program Q1_41C% Generation of the square wave in Fig. 1.2(a) n = 0:50;f = 0.05;phase = 0;peak = 1;A = 2.0;arg = 2*pi*f*n + phase;x = A*sawtooth(arg,peak);clf; % Clear old graphstem(n,x); % Plot the generated sequence axis([0 50 -2 2]);grid;title('Sawtooth Wave Sequence of Fig. 1.2(a)'); xlabel('Time index n');ylabel('Amplitude');axis;% Program Q1_41D% Generation of the square wave in Fig. 1.2(b) n = 0:50;f = 0.05;phase = 0;peak = 0.5;A = 2.0;arg = 2*pi*f*n + phase;x = A*sawtooth(arg,peak);clf; % Clear old graphstem(n,x); % Plot the generated sequence axis([0 50 -2 2]);grid;title('Sawtooth Wave Sequence of Fig. 1.2(b)'); xlabel('Time index n');ylabel('Amplitude');axis;051015202530Time index nA m p l i t u d eSquare Wave Sequence of Fig. 1.1(b)Time index nA m p l i t u d eTime index nA m p l i t u d eSawtooth Wave Sequence of Fig. 1.2(b)Time index nA m p l i t u d eDate: 14 September, 2006 Signature: HAVLICEK。

·1·第1章 时域离散信号和系统1.1 引 言本章内容是全书的基础。

学生从学习模拟信号分析与处理到学习数字信号处理，要建立许多新的概念，数字信号和数字系统与原来的模拟信号和模拟系统不同，尤其是处理方法上有本质的区别。

模拟系统用许多模拟器件完成，数字系统用运算方法完成。

如果对本章中关于数字信号与系统的若干基本概念不清楚，那么在学习数字滤波器时，会感到不好掌握，因此学好本章是很重要的。

1.2 本章学习要点(1) 关于信号● 模拟信号、时域离散信号、数字信号三者之间的区别。

● 如何由模拟信号产生时域离散信号。

● 常用的时域离散信号。

● 如何判断信号是周期性的，其周期如何计算。

(2) 关于系统● 什么是系统的线性、时不变性，以及因果性、稳定性；如何判断。

● 线性、时不变系统输入和输出之间的关系；求解线性卷积的图解法、列表法、解析法，以及用MA TLAB 工具箱函数求解。

● 线性常系数差分方程的递推解法。

● 用MA TLAB 求解差分方程。

● 什么是滑动平均滤波器，它的单位脉冲响应是什么。

1.3 习题与上机题解答1.1 用单位脉冲序列及其加权和表示图P1.1所示的序列。

解：()(2)(1)2()(1)2(2)3(3)(4)2(6)x n n n n n n n n n δδδδδδδδ=+-+++-+-+-+-+-1.2 给定信号24,4≤≤1()4,0≤≤40,n n x n n +--⎧⎪=⎨⎪⎩其他 (1) 画出x (n )的波形，标上各序列值；(2) 试用延迟的单位脉冲序列及其加权和表示x (n )序列； (3) 令1()2(2)x n x n =-，画出1()x n 的波形； (4) 令2()(2)x n x n =-，画出2()x n 的波形。

·2·解：(1) 画出x (n )的波形，如图S1.2.1所示。

图P1.1 图S1.2.1(2) ()4(4)2(3)2(1)4()4(1)4(2)4(3)4(4)x n n n n n n n n n δδδδδδδδ=+-+++++-+-+-+--。