数字信号处理上机作业.doc

DSP上机作业M2.22.30(b)n=0:1:40;x=sin(0.8*pi*n+0.8*pi);stem(n,x);xlabel(n);ylabel(x(n)(c) n=0:1:40;x=real(exp(i*pi*n*0.2)+imag(exp(i*pi*n*0.1);stem(n,x);xlabel(n);ylabel(x(n) (d)n=0:1:40;x=3*cos(1.3*pi*n)-4*sin(0.5*pi*n+0.5*pi);stem(n,x);xlabel(n);ylabel(x(n)(e)n=0:1:40;x=5*cos(1.5*pi*n+0.75*pi)+4*cos(0.6*pi*n)-sin(0.5*pi*n);stem(n,x);xlabel(n);ylabel(x(n)(分析：此题着重用到了stem函数，n的取值本应是一个基本周期，但为了仿真以及图形清晰易辨，n取140）M2.4t=0:0.001:1;f=input(Frequency of sinusoid in Hz=);FT=input(Sampling frequency in Hz=);g1=sin(2*pi*f*t);plot(t,g1,-);xlabel(time);ylabel(Amplitude);hold on n=0:1:FT;g2=sin(2*pi*f*n/FT);plot(n/FT,g2,o);hold offFrequency of sinusoid in Hz=10Sampling frequency in Hz=50（分析：此题表明了正弦信号与其采样信号的关系）M2.5t=0:0.001:0.5;g1=cos(3*pi*t);g2=cos(17*pi*t);g3=cos(23*pi*t);plot(t,g1,-,t,g2,-,t,g3,:);xlabel(time);ylabel(Amplitude);holdn=0:1:5;gs=cos(0.3*pi*n);plot(n/10,gs,o);hold off（分析：此题验证了公式（2.57），说明同样的采样信号有可能对应不止一个原信号，由此引出the Nyquist sampling theorem）M2.6Number of input samples = 5Number of input samples = 7Number of input samples = 9As the length of the moving average filter is increased,the output of the filter gets more smoother.However,the delay between the input and the output sequences also increases.(分析：长度越大，曲线越平滑，可是越远离原有曲线，即延时越大了。如取长度20，则得下图此时确实很平滑了，可是延时之大也可见一斑）M3.1（1）r=0.8,=0.5Number of frequency points = 513Numerator coefficients = 1 0 0Denominator coefficients = 1 -1.4042 0.64(2)r=0.9,=0.9Number of frequency points = 513Numerator coefficients = 1 0 0Denominator coefficients = 1 -1.1189 0.81(分析：（1）对应，（2） 对应，语句h = freqz(num, den, w)中num,den分别为传递函数分子和分母多项式系数向量，w为复频率响应计算点数，不同的r,值对应着不同的传递函数）M3.2(a)Number of frequency points = 513Numerator coefficients = 1 zeros(1,20) -1Denominator coefficients = zeros(1,10) 1 -1(b)Number of frequency points = 513Numerator coefficients = 1 zeros(1,10) -1Denominator coefficients = 1 -1 (c)Number of frequency points = 513Numerator coefficients = 1 zeros(1,9) -2 zeros(1,9) 1Denominator coefficients = zeros(1,9) 10 -20 10(d)Number of frequency points = 513Numerator coefficients = 100 -99 zeros(1,9) -2 zeros(1,9) -99 100Denominator coefficients = zeros(1,10) 10 -20 10(e)Number of frequency points = 513Numerator coefficients = 0.3129 zeros(1,19) 0.3129 0.3129Denominator coefficients = zeros(1,9) 2 -3.9508 1(分析：此题较为复杂，首先要把3.19题中的式子进行DTFT，然后化成与M3.1相同的形式，再运用Program3-1才可以得到答案，虽然思路很清楚，可是错综复杂，运算很繁琐。）M5.2(a) g=5 -2 2 0 4 3;h=3 1 -2 2 -4 4;Y=fft(g).*fft(h);y=ifft(Y)y =-6 9 -16 20 -4 45(b) x=2-i -1-3i 4-3i 1+2i 3+2i;v=-3 2+4i -1+4i 4+2i -3+i;Y=fft(x).*fft(v);y=ifft(Y)y = 11.0000 +25.0000i -9.0000 +48.0000i 3.0000 +17.0000i 29.0000 + 0.0000i -10.0000 +12.0000i(c) n=0:4;x=cos(pi*0.5*n);y=3.n;H=fft(x).*fft(y);h=ifft(H)h = -23.0000 -69.0000 35.0000 105.0000 73.0000(分析：用circonv程序验证：（a) x1=5 -2 2 0 4 3;x2=3 1 -2 2 -4 4; 3 4 -4 2 -2 1 1 3 4 -4 2 -2 -2 1 3 4 -4 2 2 -2 1 3 4 -4 -4 2 -2 1 3 4 4 -4 2 -2 1 3ans =-6 9 -16 20 -4 45(b)x1=2-i -1-3i 4-3i 1+2i 3+2i;x2=-3 2+4i -1+4i 4+2i -3+i; -3.0000 -3.0000 + 1.0000i 4.0000 + 2.0000i -1.0000 + 4.0000i 2.0000 + 4.0000i 2.0000 + 4.0000i -3.0000 -3.0000 + 1.0000i 4.0000 + 2.0000i -1.0000 + 4.0000i -1.0000 + 4.0000i 2.0000 + 4.0000i -3.0000 -3.0000 + 1.0000i 4.0000 + 2.0000i 4.0000 + 2.0000i -1.0000 + 4.0000i 2.0000 + 4.0000i -3.0000 -3.0000 + 1.0000i -3.0000 + 1.0000i 4.0000 + 2.0000i -1.0000 + 4.0000i 2.0000 + 4.0000i -3.0000 ans = 11.0000 +25.0000i -9.0000 +48.0000i 3.0000 +17.0000i 29.0000 -10.0000 +12.0000i(c)n=0:4;x1=cos(pi*0.5*n);x2=3.n; 1 81 27 9 3 3 1 81 27 9 9 3 1 81 27 27 9 3 1 81 81 27 9 3 1ans = -23.0000 -69.0000 35.0000 105.0000 73.0000通过核实与题目中计算结果一样，本题所作的是离散傅里叶变换（）的时域循环卷积定理，即时域循环卷积，频域相乘）M5.3 m=0:9;x=exp(-0.6*m);X=fft(x)X = Columns 1 through 5 2.2109 1.3423 - 0.7788i 0.8611 - 0.5412i 0.7112 - 0.3174i 0.6580 - 0.1470i Columns 6 through 10 0.6441 0.6580 + 0.1470i 0.7112 + 0.3174i 0.8611 + 0.5412i 1.3423 + 0.7788i X1=fft(conj(x);G1=conj(X(1) X(10:-1:2)G1 = Columns 1 through 5 2.2109 1.3423 - 0.7788i 0.8611 - 0.5412i 0.7112 - 0.3174i 0.6580 - 0.1470i Columns 6 through 10 0.6441 0.6580 + 0.1470i 0.7112 + 0.3174i 0.8611 + 0.5412i 1.3423 + 0.7788i X1X1 = Columns 1 through 5 2.2109 1.3423 - 0.7788i 0.8611 - 0.5412i 0.7112 - 0.3174i 0.6580 - 0.1470i Columns 6 through 10 0.6441 0.6580 + 0.1470i 0.7112 + 0.3174i 0.8611 + 0.5412i 1.3423 + 0.7788i(G1=X1说明） x2=conj(x(1) x(10:-1:2);X2=fft(x2)X2 = Columns 1 through 5 2.2109 1.3423 + 0.7788i 0.8611 + 0.5412i 0.7112 + 0.3174i 0.6580 + 0.1470i Columns 6 through 10 0.6441 0.6580 - 0.1470i 0.7112 - 0.3174i 0.8611 - 0.5412i 1.3423 - 0.7788i conj(X)ans = Columns 1 through 5 2.2109 1.3423 + 0.7788i 0.8611 + 0.5412i 0.7112 + 0.3174i 0.6580 + 0.1470i Columns 6 through 10 0.6441 0.6580 - 0.1470i 0.7112 - 0.3174i 0.8611 - 0.5412i 1.3423 - 0.7788i（X2=conj说明）x3=real(x);X3=fft(x3);G3=0.5*(X+conj(X(1) X(10:-1:2)G3 = Columns 1 through 5 2.2109 1.3423 - 0.7788i 0.8611 - 0.5412i 0.7112 - 0.3174i 0.6580 - 0.1470i Columns 6 through 10 0.6441 0.6580 + 0.1470i 0.7112 + 0.3174i 0.8611 + 0.5412i 1.3423 + 0.7788i X3X3 = Columns 1 through 5 2.2109 1.3423 - 0.7788i 0.8611 - 0.5412i 0.7112 - 0.3174i 0.6580 - 0.1470i Columns 6 through 10 0.6441 0.6580 + 0.1470i 0.7112 + 0.3174i 0.8611 + 0.5412i 1.3423 + 0.7788i（X3=G3说明） x4=j*imag(x);X4=fft(x4);G4=0.5*(X-conj(X(1) X(10:-1:2)G4 = 0 0 0 0 0 0 0 0 0 0 X4X4 = 0 0 0 0 0 0 0 0 0 0（X4=G4说明） x5=0.5*(x+conj(x(1) x(10:-1:2);X5=fft(x5)X5 = 2.2109 1.3423 0.8611 0.7112 0.6580 0.6441 0.6580 0.7112 0.8611 1.3423 real(X)ans = 2.2109 1.3423 0.8611 0.7112 0.6580 0.6441 0.6580 0.7112 0.8611 1.3423（X5=real(X)说明) x6=0.5*(x-conj(x(1) x(10:-1:2);X6=fft(x6)X6 = Columns 1 through 5 0 0 - 0.7788i 0 - 0.5412i 0 - 0.3174i 0 - 0.1470i Columns 6 through 10 0 0 + 0.1470i 0 + 0.3174i 0 + 0.5412i 0 + 0.7788i j*imag(X)ans = Columns 1 through 5 0 0 - 0.7788i 0 - 0.5412i 0 - 0.3174i 0 - 0.1470i Columns 6 through 10 0 0 + 0.1470i 0 + 0.3174i 0 + 0.5412i 0 + 0.7788i(X6=j*imag(X)说明）M6.1Type in the numerator coefficients = 2 -5 13.48 -7.78 9Type in the denominator coefficients = 4 7.2 20 -0.8 8Numfactors = 1.000000000000000 -2.100000000000000 5.000000000000004 1.000000000000000 -0.400000000000000 0.899999999999995Denfactors = 1.000000000000000 2.000000000000000 4.999999999999990 1.000000000000000 -0.199999999999999 0.400000000000009Numerator factors 1.000000000000000 -2.100000000000000 5.000000000000004 1.000000000000000 -0.400000000000000 0.899999999999995Denominator factors 1.000000000000000 2.000000000000000 4.999999999999990 1.000000000000000 -0.199999999999999 0.400000000000009Gain constant 0.500000000000000Thus,The pole-zero plot of it is given below:There are 3 ROCS associated with it:R1:|z|,R2:|z|.The inverse z-transform corresponding to the ROC R1 is a left-sided sequence,the inverse z-transform corresponding to the ROC R2 is a two-sided sequence,and the inverse z-transform corresponding to the ROC R3 is a right-sided sequence.(b)Type in the numerator coefficients = 5 3.5 21.5 -4.6 18Type in the denominator coefficients = 5 15.5 31.7 22.52 4.8Numfactors = 1.000000000000000 1.200000000000000 3.999999999999996 1.000000000000000 -0.500000000000000 0.900000000000004Denfactors = 1.000000000000000 2.100000000000003 4.000000000000012 1.000000000000000 0.600000000000037 0 1.000000000000000 0.399999999999960 0Numerator factors 1.000000000000000 1.200000000000000 3.999999999999996 1.000000000000000 -0.500000000000000 0.900000000000004Denominator factors 1.000000000000000 2.100000000000003 4.000000000000012 1.000000000000000 0.600000000000037 0 1.000000000000000 0.399999999999960 0Gain constant 1Thus,The pole-zero plot of it is given below:There are 4 ROCS associated with it:R1:|z|0.4,R2:0.4|z|0.6,R3:0.6|z|2.The inverse z-transform corresponding to the ROC R1 is a left-sided sequence,the inverse z-transform corresponding to the ROC R2 is a two-sided sequence,the inverse z-transform corresponding to the ROC R3is a two-sided sequence,and the inverse z-transform corresponding to the ROC R4 is a right-sided sequence.(分析：此题通过MATLAB实现了传递函数由多项式形式到因式分解形式之间的转换，且由因式分解形式可清楚地看出零极点，进而分析左序列，右序列，双边序列与收敛域的关系）M6.3(a)Type in the residues = 3 -0.8 -7/6Type in the poles = 0 -1/5 -1/6Type in the constants = 0Numerator polynomial coefficients 1.033333333333333 0.733333333333333 0.100000000000000 0Denominator polynomial coefficients 1.000000000000000 0.366666666666667 0.033333333333333 0Hence,(b)Type in the residues