第三次作业参考答案
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1.1 P135 题16 1(1)(1)01(1)01()(1)01()(1)0(1)()[((1))()][()][()][()](())()NkNnkNNNNNnNknkNNNNnNknkNNNNnNknkNNNNnkNNNNXkxNnRnWWxnWWxnWWxnWWXkWRk
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因为X(K)以N为周期延拓，所以()()22NNXXN。 1.2 P135 题17 1512110103110102217.()()(1)[()] (2)()[(())()],(3)()[(1)(())()],(4)()[()].(5)()[Im[nepxnRnXDTFTxnXkDFTxnRnXkDFTxnRnxnIDFTXkxnIDFTXj1
45
设。求(e)，画出它的幅频特性和相频特性（标出主要坐标值）。 求并画出它的幅频特性.求并画出它的幅频特性.求求221022()]].(6)()[((1))()].(7)()[()()].NNkkxnIDFTXNkRkxnIDFTWXkXk67求求
解： 41100123455/25/25/25/2234/2/2/2/(1)()[()]()()(0)(1)(2)(3)(4)1()11()jwjwnjwnnnjwjwjwjwjwjwjwjwjwjwjwjwjwjwjwjwjwjwjwXeDTFTxnxnexnexexexexexeeeeeeeeeeeeeee
2
2sin(5/2)sin(/2)sin(5/2)sin(/2)jwwwwew
幅频特性： 1sin(5/2)sin(5/2)/(5/2)(5/2)sin(5/2/)(1)()5sin(/2)sin(/2)/(/2)(/2)sin(/2/)jwwwwwcwXewwwwcw，注意下图对
横坐标进行归一化，w/pi=-1到1，则w=-pi到pi，也就是幅度谱的一个周期
-1-0.8-0.6-0.4-0.200.20.40.60.8100.511.522.533.544.55
ω/幅度幅频响应 w = -pi:0.01:pi;
x1 = sinc(5*w/2/pi); x2 = sinc(w/2/pi); Mag = 5*abs(x1./x2); plot(w/pi,Mag); xlabel('ω/\pi'),ylabel('幅度'); title('幅频响应')
phi = angle(exp(-2*j*w)); figure(); plot(w/pi,phi); xlabel('ω/\pi'),ylabel('相位'); title('相频响应')
5/24/22/2jwjwjwjweeee
，2arg()jwe
-1-0.8-0.6-0.4-0.200.20.40.60.81-4-3-2-101234
ω/相位相频响应
9211010110101000234101010101052/*55/2/*/4/(2)()[(())()][(())()]11sin(5/)11sin(/)sin(5/),0,1,2,3,...9sin(/)knnkkkkkkjkNjkNNkjNkjkNNjkNXkDFTxnRnxnRnWWWWWWWeekNWeekNkNekkN
4/2/5sin(5/)sin(/2)5*5sin(/)sin(/10)jkNjkckNckeeckNck 01234567890123450123456789-4-2024 N = 10;
k = 0:N-1 x1 = sinc(5*k/N); x2 = sinc(k/N); X = 5*abs(x1./x2); % 5 3.2361 0 1.2361 0 1 0 1.2361 0 3.2361 disp(X) subplot(2,1,1);stem(k,X);
phi = -2*k*pi/5; subplot(2,1,2);plot(k,phi);
31101002341010101010(3)()[(1)(())()]nkkkkk
XkDFTxnRnWWWWW
012345678900.511.522.533.544.55 N = 10; k = 0:N-1; for m = 1:10 X(m) =0; for n=1:5 X(m) = X(m) + exp(-j*2*pi/N*(n-1)*(m-1))*(-1).^(n-1); end end
disp(abs(X)) stem(k,abs(X)); ％ 1.0000 0 1.2361 0.0000 3.2361 5.0000 3.2361 0.0000 1.2361 0.0000
2110102222211010(4)()[()](())()[1 1 1 1 1 0 0 0 0 0]()(),()()[()][()](())()[1 1 1 1 1 0 0 0 0 0]epepepxnIDFTXkxnRnXkXkXkxnIDFTXkIDFTXkxnRnQ4
4为实数
即圆周共轭对称
210101010[Im[()]]()1/2*(()()())())0.5*([1 1 1 1 1 0 0 0 0 0 ][1 0 0 0 0 0 1 1 1 1])0.5*([0 1 1 1 1 0 -1 -1 -1 -1 ])[0 0.5 0.5 0.5 0.5 0 -0.5 -0.5 -0.5 -0.5]()IDFTjXkxnxnRnxnRnxnI5
5(5)对应的奇对称分量则22[Im[()]][*Im[()]]*[0 0.5 0.5 0.5 0.5 0 -0.5 -0.5 -0.5 -0.5][0 -0.5j -0.5j -0.5j -0.5j 0 0.5j 0.5j 0.5j 0.5j]DFTXkIDFTjjXkj 21201(1)(1)201(1)()(1)20(1)(6)()[((1))()]1((1))()1((1))()1((1))()(())[1 0 0 0 0 0 1 1 1NNNknNNNkNNknNnNNNNkNNknNnNNNNkNnNNNxnIDFTXNkRkXNkRkWNXNkRkWWNXNkRkWWNxNnRW6067891010101010
2222-j6-j7-j8-j9
10101010
1][ 0 0 0 0 0 ][1 0 0 0 0 0 e e e e ]nNWWWWWW
210222102244(7)()[()()][()][()](2)()[1 1 1 0 0 0 0 0 1 1][1 1 1 1 1 0 0 0 0 0 ]kkxnIDFTWXkXkIDFTWXkIDFTXkxnxn
7⑩⑩⑩ 1 1 1 0 0 0 0 0 1 1
1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 01 1 1 0 0 0 0 0 1 1
1 2 3 3 3 2 1 0 1 2 2 2 2 12 2 2 13 4 5 4 3 2 1 0 1 2
1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1
1.3 题18 已知点实序列，都是以且NnynunxnxnunNxnyn)(),(),(),()1()(),1()(用)].([)()],([)()],([)(),(),()(nuDFTkUnyDFTKYnxDFTkXkUkYkX其中表示解：
)())(()()())(()1()()1()()1()1()1()1(1010kRWkXWkXWnRkXWWnNxWnRnNxkYNkNNkNNkNNNNkNNknNNNnnkNNnN
nNNnjWenu2n)1()(中
