第三次作业参考答案

1.1 P135 题16

N?1X(k)???[x((N?1?n))NRN(n)WNn?0N?1?k(N?1?n)WNk(N?1)]???[x(n)NWNn?0N?1?kn]WNk(N?1)???[x(n)NWNn?0N?1(?k)n]?WN]?WNk(N?1)

???[x(n)NWNn?0(?k)nk(N?1)??(X(?k))NWNk(N?1)RN(k)

因为X(K)以N为周期延拓，所以X(?N2)?X(N?N2)。

1.2 P135 题17

17.设x1(n)?R5(n)。(1)求X1(ej?)?DTFT[x1(n)]，画出它的幅频特性和相频特性（标出主要坐标值）。 n(2)求X2(k)?DFT[x1((n))10R10(n)],并画出它的幅频特性.(3)求X3(k)?DFT[(?1)x1((n))10R10(n)],并画出它的幅频特性.(4)求x4(n)?IDFT[X2ep(k)].(5)求x5(n)?IDFT[Im[X2(k)]].(6)求x6(n)?IDFT[X2((N?1?k))NRN(k)].(7)求x7(n)?IDFT[W10X2(k)X2(k)]. 解：

??4?2k(1)X1(e?x(0)e?1?e?e

jw)?DTFT[x1(n)]??x(1)e?j2w?jw1?n???x(n)e?jwn??x(n)en?0?jwn?jw0?x(2)e?e?jw2?x(3)e1?e?jw?jw3?x(4)ee?j5w/2?jw/2?jw4j5w/2jw/2?j5w/2?j5w/2?jw/2?j5w?jw?e?e?j3w?j4w?1?e?(e(e?e?e)e?jw/2)?esin(5w/2)sin(w/2)e?j2wsin(5w/2)sin(w/2)幅频特性：

(1)X1(ejw)?sin(5w/2)sin(w/2)?sin(5w/2)/(5w/2)(5w/2)sin(w/2)/(w/2)(w/2)?5sinc(5w/2/?)sinc(w/2/?)，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

幅频响应54.543.53w = -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(); 幅度2.521.510.50-1-0.8-0.6-0.4-0.20ω/?0.20.40.60.81plot(w/pi,phi); xlabel('ω/\\pi'),ylabel('相位'); title('相频响应') e?j5w/2?jw/2e?e?j4w/2?e?j2w，??arg(e?2jw)

相频响应4321相位0-1-2-3-4-1-0.8-0.6-0.4-0.20ω/?0.20.40.60.81 9(2)X2(k)?DFT[x1((n))10R10(n)]??W10?W10?W10?W10?W10??1?WN5kk0kk2k3k4k?[x((n))1n?010R10(n)]W10kn1?WN?1?e?j2k?/N*5?j2?/N*k1?ee?e?j5k?/N?jk?/Nsin(5k?/N)sin(k?/N)e

sin(5k?/N)sin(k?/N)?j4k?/N,k?0,1,2,3,...9?5sinc(k/2)sinc(k/10)e?j2k?/5?5*sinc(5k/N)sinc(k/N)e?j4k?/N

N = 10; 543210k = 0:N-1 x1 = sinc(5*k/N); x2 = sinc(k/N); X = 5*abs(x1./x2); 0123456789% 5 3.2361 0 1.2361 0 1 0 1.2361 0 3.2361 420-2-4disp(X) subplot(2,1,1);stem(k,X); phi = -2*k*pi/5; 0123456789 subplot(2,1,2);plot(k,phi); (3)X3(k)?DFT[(?1)x1((n))10R10(n)]?W0k10n?Wk10?W2k10?W3k10?W4k10

N = 10; 54.543.532.521.510.500123456789k = 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 (4)x4(n)?IDFT[X2ep(k)]?x1((n))10R10(n)?[1 1 1 1 1 0 0 0 0 0]为实数?X2ep(k)?X2(k),即X2(k)圆周共轭对称?x4(n)?IDFT[X2ep(k)]?IDFT[X2(k)]?x1((n))10R10(n)?[1 1 1 1 1 0 0 0 0 0](5)IDFT[jIm[X2(k)]]对应x5(n)的奇对称分量1/2*(x10(n)R10(n)?x10(?n))R10(?n))?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]则x5(n)?IDFT[Im[X2(k)]]?IDFT[?j*jIm[X2(k)]]??j*[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]

(6)x6(n)?IDFT[X2((N?1?k))NRN(k)]???1N1N1NN?1?k?0N?1X2((N?1?k))NRN(k)WNX2((N?1?k))NRN(k)WNX2((N?1?k))NRN(k)WN?(N?1)n7?kn?k?0N?1(N?1?k)nWN?(N?1)n?k?00?(N?1?k)(?n)WN?(N?1)n

?x((N?n))NRNWN62?10?[1 0 0 0 0 0 1 1 1 1]WN892?102?10n?[W10 0 0 0 0 0 W10 W10 W10 W10 ]?[1 0 0 0 0 0 e-j6 e-j2?107 e-j8 e-j9 ](7)x7(n)?IDFT[W10X2(k)X2(k)]?IDFT[W10X2(k)]⑩IDFT[X2(k)]?x4(n?2)⑩x4(n)?[1 1 1 0 0 0 0 0 1 1]⑩[1 1 1 1 1 0 0 0 0 0 ]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 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 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?2k?2k

1.3 题18

N点实序列用，已知y(n)?x(N?1?n),u(n)?(?1)x(n),且x(n),u(n),y(n)都是以nX(k)表示Y(k),U(k),其中X(k)?DFT[x(n)],Y(K)?DFT[y(n)],U(k)?DFT[u(n)].解：

N?1N?1

Y(k)??n?0x(N?1?n)RN(n)W(N?1)knkN??n?0x(N?1?n)WN?(N?1?n)kWN(N?1)k

?X((?k))NRN(n)WNu(n)中(?1)?en?j?n?X(?k)WN(N?1)k?X((?k))NWNRN(k)?kN?WN2

n

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