数字信号处理MATLAB上机作业M 2.21.题目The square wave and the sawtooth wave are two periodic sequences as sketched in figure ing the function stem. The input data specified by the user are: desired length L of the sequence, peak value A, and the period N. For the square wave sequence an additional user-specified parameter is the duty cycle, which is the percent of the period for which the signal is positive. Using this program generate the first 100 samples of each of the above sequences with a sampling rate of 20 kHz ,a peak value of 7, a period of 13 ,and a duty cycle of 60% for the square wave.2.程序% 用户定义各项参数参数A = input('The peak value =');L = input('Length of sequence =');N = input('The period of sequence =');FT = input('The desired sampling frequency =');DC = input('The square wave duty cycle = ');% 产生所需要的信号t = 0:L-1;T = 1/FT;x = A*sawtooth(2*pi*t/N);y = A*square(2*pi*(t/N),DC);% Plotsubplot(2,1,1)stem(t,x);ylabel('幅度');xlabel('n');subplot(2,1,2)stem(t,y);ylabel('幅度');xlabel('n');3.结果4.结果分析M 2.41.题目(a)Write a matlab program to generate a sinusoidal sequence x[n]= Acos(ω0 n+Ф) and plot thesequence using the stem function. The input data specified by the user are the desired length L, amplitude A, the angular frequency ω0 , and the phase Фwhere 0<ω0 <pi and 0<=Ф<=2pi. Using this program generate the sinusoidal sequences shown in figure 2.15. (b)Generate sinusoidal sequences with the angular frequencies given in Problem 2.22.Determine the period of each sequence from the plot and verify the result theoretically. 2.程序%用户定义的参数L = input('Desired length = ');A = input('Amplitude = ');omega = input('Angular frequency = ');phi = input('Phase = ');%信号产生n = 0:L-1;x = A*cos(omega*n + phi);stem(n,x);xlabel('n');ylabel('幅度');title(['\omega_{o} = ',num2str(omega)]);3.结果(a)ω0=0ω0=0.1πω0=0.8πω0=1.2π(b)ω0=0.14πω0=0.24πω0=0.34πω0=0.68πω0=0.75π4.结果分析M 2.51.题目Generate the sequences of problem 2.21(b) to 2.21(e) using matlab.2.程序(b)n = 0 : 99;x=sin(0.6*pi*n+0.6*pi);stem(n,x);xlabel('n');ylabel('幅度');(c)n = 0 : 99;x=2*cos(1.1*pi*n-0.5*pi)+2*sin(0.7*pi*n);stem(n,x);xlabel('n');ylabel('幅度');(d)n = 0 : 99;x=3*sin(1.3*pi*n-4*cos(0.3*pi*n+0.45*pi));stem(n,x);xlabel('n');ylabel('幅度');(e)n = 0 : 99;x=5*sin(1.2*pi*n+0.65*pi)+4*sin(0.8*pi*n)-cos(0.8*pi*n);stem(n,x);xlabel('n');ylabel('幅度');(f)n = 0 : 99;x=mod(n,6);stem(n,x);xlabel('n');ylabel('幅度');3.结果(b)(c)(d)(e)(f)4.结果分析M 2.61.题目Write a matlab program to plot a continuous-time sinusoidal signal and its sampled version and verify figure 2.19. You need to use the hold function to keep both plots.2.程序%用户定义的参数fo = input('Frequency of sinusoid in Hz = ');FT = input('Samplig frequency in Hz = ');%产生信号t = 0:0.001:1;g1 = cos(2*pi*fo*t);plot(t,g1,'-')xlabel('时间t');ylabel('幅度')holdn = 0:1:FT;gs = cos(2*pi*fo*n/FT);plot(n/FT,gs,'o');hold off3.结果4.结果分析M 3.11.题目Using program 3_1 determine and plot the real and imaginary parts and the magnitude and phase spectra of the following DTFT for various values of r and θ:G(e jω)=1, 0<r<1.1−2r(cosθ)e−jω+r2e−2jω2.程序%program 3_1%discrete-time fourier transform computatition%k=input('Number of frequency points = ');num=input('Numerator coefficients= ');den=input('Denominator coefficients= ');%computer the frequency responsew=0:pi/k:pi;h=freqz(num,den,w);%plot the frequency responsesubplot(221)plot(w/pi,real(h));gridtitle('real part')xlabel('\omega/\pi');ylabel('Amplitude') subplot(222)plot(w/pi,imag(h));gridtitle('imaginary part')xlabel('\omega/\pi');ylabel('Amplitude') subplot(223)plot(w/pi,abs(h));gridtitle('magnitude spectrum')xlabel('\omega/\pi');ylabel('magnitude') subplot(224)plot(w/pi,angle(h));gridtitle('phase spectrum')xlabel('\omega/\pi');ylabel('phase,radians')3.结果(a)r=0.8 θ=π/6(b)r=0.6 θ=π/34.结果分析M 3.41.题目Using matlab verify the following general properties of the DTFT as listed in Table 3.2:(a) Linearity, (b) time-shifting, (c) frequency-shifting, (d) differentiation-in-frequency, (e) convolution, (f) modulation, and (g) Parseval’s relation. Since all data in matlab have to be finite-length vectors, the sequences to be used to verify the properties are thus restricted to be of finite length.2.程序%先定义两个信号N = input('The length of the sequence = ');k = 0:N-1;%g为正弦信号g = 2*sin(2*pi*k/(N/2));%h为余弦信号h = 3*cos(2*pi*k/(N/2));[G,w] = freqz(g,1);[H,w] = freqz(h,1);%*************************************************************************%% 线性性质alpha = 0.5;beta = 0.25;y = alpha*g+beta*h;[Y,w] = freqz(y,1);figure(1);subplot(211),plot(w/pi,abs(Y));xlabel('\omega/\pi');ylabel('|Y(e^j^\omega)|');title('线性叠加后的频率特性');grid;% 画出Y 的频率特性subplot(212),plot(w/pi,alpha*abs(G)+beta*abs(H));xlabel('\omega/\pi');ylabel('\alpha|G(e^j^\omega)|+\beta|H(e^j^\omega)|');title('线性叠加前的频率特性');grid;% 画出alpha*G+beta*H 的频率特性%*************************************************************************% % 时移性质n0 = 10;%时移10个的单位y2 = [zeros([1,n0]) g];[Y2,w] = freqz(y2,1);G0 = exp(-j*w*n0).*G;figure(2);subplot(211),plot(w/pi,abs(G0));xlabel('\omega/\pi');ylabel('|G0(e^j^\omega)|');title('G0的频率特性');grid;% 画出G0的频率特性subplot(212),plot(w/pi,abs(Y2));xlabel('\omega/\pi');ylabel('|Y2(e^j^\omega)|');title('Y2的频率特性');grid;% 画出Y2 的频率特性%*************************************************************************% % 频移特性w0 = pi/2; % 频移pi/2r=256; %the value of w0 in terms of number of samplesk = 0:N-1;y3 = g.*exp(j*w0*k);[Y3,w] = freqz(y3,1);% 对采样的512个点分别进行减少pi/2,从而生成G(exp(w-w0))k = 0:511;w = -w0+pi*k/512;G1 = freqz(g,1,w);figure(3);subplot(211),plot(w/pi,abs(Y3));xlabel('\omega/\pi');ylabel('|Y3(e^j^\omega)|');title('Y3的频率特性');grid;% 画出Y3的频率特性subplot(212),plot(w/pi,abs(G1));xlabel('\omega/\pi');ylabel('|G1(e^j^\omega)|');title('G1的频率特性');grid;% 画出G1 的频率特性%*************************************************************************% % 频域微分k = 0:N-1;y4 = k.*g;[Y4,w] = freqz(y4,1);%在频域进行微分y0 = ((-1).^k).*g;G2 = [G(2:512)' sum(y0)]';delG = (G2-G)*512/pi;figure(4);subplot(211),plot(w/pi,abs(Y4));xlabel('\omega/\pi');ylabel('|Y4(e^j^\omega)|');title('Y4的频率特性');grid;% 画出Y4的频率特性subplot(212),plot(w/pi,abs(delG));xlabel('\omega/\pi');ylabel('|delG(e^j^\omega)|');title('delG的频率特性');grid;% 画出delG的频率特性%*************************************************************************% % 相乘性质y5 = conv(g,h);%时域卷积[Y5,w] = freqz(y5,1);figure(5);subplot(211),plot(w/pi,abs(Y5));xlabel('\omega/\pi');ylabel('|Y5(e^j^\omega)|');title('Y5的频率特性');grid;% 画出Y5的频率特性subplot(212),plot(w/pi,abs(G.*H));%频域乘积xlabel('\omega/\pi');ylabel('|G.*H(e^j^\omega)|');title('G.*H的频率特性');grid;% 画出G.*H的频率特性%*************************************************************************% % 帕斯瓦尔定理y6 = g.*h;%对于freqz函数，在0到2pi直接取样[Y6,w] = freqz(y6,1,512,'whole');[G0,w] = freqz(g,1,512,'whole');[H0,w] = freqz(h,1,512,'whole');% Evaluate the sample value at w = pi/2% and verify with Y6 at pi/2H1 = [fliplr(H0(1:129)') fliplr(H0(130:512)')]';val = 1/(512)*sum(G0.*H1);% Compare val with Y6(129) i.e sample at pi/2 % Can extend this to other points similarly% Parsevals theoremval1 = sum(g.*conj(h));val2 = sum(G0.*conj(H0))/512;% Comapre val1 with val23.结果(a)(b)(c)(d)(e)4.结果分析M 3.81.题目Using matlab compute the N-point DFTs of the length-N sequences of Problem 3.12 for N=3, 5, 7, and 10. Compare your results with that obtained by evaluating the DTFTs computed in Problem 3.12 at ω= 2pik/N, k=0, 1,……N-1.2.程序%用户定义N的长度N = input('The value of N = ');k = -N:N;y1 = ones([1,2*N+1]);w = 0:2*pi/255:2*pi;Y1 = freqz(y1, 1, w);%对y1做傅里叶变换Y1dft = fft(y1);k = 0:1:2*N;plot(w/pi,abs(Y1),k*2/(2*N+1),abs(Y1dft),'o');grid;xlabel('归一化频率');ylabel('幅度');(a)clf;N = input('The value of N = ');k = -N:N;y1 = ones([1,2*N+1]);w = 0:2*pi/255:2*pi;Y1 = freqz(y1, 1, w);Y1dft = fft(y1);k = 0:1:2*N;plot(w/pi,abs(Y1),k*2/(2*N+1),abs(Y1dft),'o');xlabel('Normalized frequency');ylabel('Amplitude');(b)%用户定义N的长度N = input('The value of N = ');k = -N:N;y1 = ones([1,2*N+1]);y2 = y1 - abs(k)/N;w = 0:2*pi/255:2*pi;Y2 = freqz(y2, 1, w);%对y1做傅里叶变换Y2dft = fft(y2);k = 0:1:2*N;plot(w/pi,abs(Y2),k*2/(2*N+1),abs(Y2dft),'o');grid;xlabel('归一化频率');ylabel('幅度');(c)%用户定义N的长度N = input('The value of N = ');k = -N:N;y3 =cos(pi*k/(2*N));w = 0:2*pi/255:2*pi;Y3 = freqz(y3, 1, w);%对y1做傅里叶变换Y3dft = fft(y3);k = 0:1:2*N;plot(w/pi,abs(Y3),k*2/(2*N+1),abs(Y3dft),'o');grid;xlabel('归一化频率');ylabel('幅度');3.结果(a)N=3N=5 N=7N=10 (b)N=3N=5 N=7N=10 (c)N=3N=5 N=7N=104.结果分析M 3.191.题目Using Program 3_10 determine the z-transform as a ratio of two polynomials in z-1 from each of the partial-fraction expansions listed below:(a)X1(z)=−2+104+z−1−82+z−1,|z|>0.5,(b)X2(z)=3.5−21−0.5z−1−3+z−11−0.25z−2,|z|>0.5,(c)X3(z)=5(3+2z−1)2−43+2z−1+31+0.81z−2,|z|>0.9,(d)X4(z)=4+105+2z−1+z−16+5z−1+z−2,|z|>0.5.2.程序% Program 3_10% Partical-Fraction Expansion to rational z-Transform %r = input('Type in the residues = ');p = input('Type in the poles = ');k = input('Type in the constants = ');[num, den] = residuez(r,p,k);disp('Numberator polynominal coefficients');disp(num) disp('Denominator polynomial coefficients'); disp(den)4.结果分析M 4.61.题目Plot the magnitude and phase responses of the causal IIR digital transfer functionH(z)=0.0534(1+z−1)(1−1.0166z−1+z−2) (1−0.683z−1)(1−1.4461z−1+0.7957z−2).What type of filter does this transfer function represent? Determine the difference equation representation of the above transfer function.2.程序b=[0.0534 -0.00088644 -0.00088644 0.0534];a=[1 -2.1291 1.7833863 -0.5434631];figure(1)freqz(b,a);figure(2)[H,w]=freqz(b,a);plot(w/pi,abs(H)),grid;xlabel('Normalized Frequency (\times\pi rad/sample)'),ylabel('Magnitude');幅度化成真值之后：4.结果分析H(z)=0.0534−0.00088644z−1−0.00088644z−2+0.0534z−31−2.1291z−1+1.7833863z−2−0.5434631z−3M 4.71.题目Plot the magnitude and phase responses of the causal IIR digital transfer functionH(z)=(1−z−1)4(1−1.499z−1+0.8482z−2)(1−1.5548z−1+0.6493z−2).2.程序b=[1 -4 6 -4 1];a=[1 -3.0538 3.8227 -2.2837 0.5472]; figure(1)freqz(b,a);figure(2)[H,w]=freqz(b,a);plot(w/pi,abs(H)),grid;xlabel('Normalized Frequency (\times\pi rad/sample)'), ylabel('Magnitude');3.结果4.结果分析。
