作业要求链接： 信号与系统2022春季作业-第十三次次作业 : https://zhuoqing.blog.csdn.net/article/details/124978410
§13 参考答案
1.1 系统频率特性
1.1.1 确定系统的幅频特性
（1） 系统函数具有两个极点，分别位于 p 1 = − 2 , p 2 = − 3 p_1 = - 2,p_2 = - 3 p1=−2,p2=−3 。改系统可以看成两个低通滤波器串联。
H 1 ( s ) = 1 ( s + 2 ) ( s + 3 ) , R e [ s ] > − 2 H_1 \left( s \right) = {1 \over {\left( {s + 2} \right)\left( {s + 3} \right)}},\,\,\,{\mathop{\rm Re}\nolimits} \left[ s \right] > - 2 H1(s)=(s+2)(s+3)1,Re[s]>−2
▲ 图1.1.1 系统的零极点分布
下面使用Python中的系统仿真软件包给出系统的频率特性。
▲ 图1.1.2 系统的幅频特性和相频特性
from headm import *
from scipy.signal import bode
from scipy.signal import TransferFunction
sys = TransferFunction([1],[1,5,6])
w,mag,phase = bode(sys)
plt.subplot(2,1,1)
plt.semilogx(w, mag)
plt.xlabel("Frequency")
plt.ylabel("Magnitude")
plt.grid(True)
plt.tight_layout()
plt.subplot(2,1,2)
plt.semilogx(w, phase)
plt.xlabel("Frequency")
plt.ylabel("Phase")
plt.grid(True)
plt.tight_layout()
plt.show()
（2） 系统具有两个零点和两个极点。具体分布如下图所示：
H 2 ( s ) = s 2 s 2 + 2 s + 1 , R e [ s ] > − 1 H_2 \left( s \right) = { {s^2 } \over {s^2 + 2s + 1}},\,\,\,{\mathop{\rm Re}\nolimits} \left[ s \right] > - 1 H2(s)=s2+2s+1s2,Re[s]>−1
▲ 图1.1.3 系统的零极点分布
这个系统可以看成两个高通滤波串联。 下面给出系统的幅频特性和相频特性。
▲ 图1.1.4 系统的频率特性
sys = TransferFunction([1,0,0],[1,2,1])
w,mag,phase = bode(sys)
（3） 系统具有左右对称的零点和极点。 符合全通特性 的定义。
H 3 ( s ) = s 2 − s + 1 s 2 + s + 1 , R e [ s ] > − 1 2 H_3 \left( s \right) = { {s^2 - s + 1} \over {s^2 + s + 1}},\,\,{\mathop{\rm Re}\nolimits} \left[ s \right] > - {1 \over 2} H3(s)=s2+s+1s2−s+1,Re[s]>−21
▲ 图1.1.5 系统的零极点分布
▲ 图1.1.6 系统的频率特性
from scipy.signal import bode
from scipy.signal import TransferFunction
sys = TransferFunction([1,-1,1],[1,1,1])
w,mag,phase = bode(sys)
plt.subplot(2,1,1)
plt.semilogx(w, mag)
plt.xlabel("Frequency")
plt.ylabel("Magnitude")
plt.axis([min(w), max(w), -1, 1])
plt.grid(True)
plt.tight_layout()
plt.subplot(2,1,2)
plt.semilogx(w, phase)
plt.xlabel("Frequency")
plt.ylabel("Phase")
plt.grid(True)
plt.tight_layout()
plt.show()
1.1.2 确定系统的频率特性
（1） H ( z ) = 2 z z − 0.6 H\left( z \right) = { {2z} \over {z - 0.6}} H(z)=z−0.62z
系统函数具有一个位于原点的零点，以及位于0.6处的极点。当 z = e j ω z = e^{j\omega } z=ejω 在单位圆上从 1 到 -1 移动时，对应的极点复矢量逐步增加，造成幅频特性下降。 因此对应的是低通滤波器特性。该系统的零点与极点都位于单位圆内，所以相片呈现为最小相位系统， 根据复矢量几何特性可以分析处相频特性先减小、后增加。
▲ 图1.1.7 离散时间系统的零极点分布
▲ 图1.1.8 系统的幅频特性和相频特性
from headm import *
def hs(w):
z = exp(1j*w)
return 2*z/(z-0.6)
w = linspace(0, 2*pi, 500)
hw = hs(w)
hwabs = abs(hw)
hwangle = angle(hw)
plt.subplot(2,1,1)
plt.plot(w, hwabs, label='Amplitude')
plt.xlabel("Frequency")
plt.ylabel("Amplitude")
plt.grid(True)
plt.tight_layout()
plt.subplot(2,1,2)
plt.plot(w, hwangle, label='Angle')
plt.xlabel("Frequency")
plt.ylabel("Phase")
plt.grid(True)
plt.tight_layout()
plt.show()
（2） 系统具有一个二阶零点和一对共轭极点，分布如下图所示。根据极点所处在的位置， 在 ω = π / 2 \omega = \pi /2 ω=π/2 会存在峰值。 根据零点所处的位置， 在 ω = − π \omega = - \pi ω=−π 幅值处在低谷。 所以系统的幅频特性整体上符合“带通特性”。
该系统的零点位于单位圆外面， 所以系统不属于最小相位系统， 而是最大相位系统， 相角处于一直下降趋势。
H ( z ) = ( 0.96 + z − 1 ) 2 0.36 z − 2 + 1 H\left( z \right) = { {\left( {0.96 + z^{ - 1} } \right)^2 } \over {0.36z^{ - 2} + 1}} H(z)=0.36z−2+1(0.96+z−1)2
▲ 图1.1.9 离散时间系统的零极点分布
下面是使用Python绘制的系统的幅频特性和相频特性。从图中可以看出，由于极点距离单位圆不是特别近，所以带通特性不明显。单位圆外的零点距离单位圆比较近，整体上，该系统的幅频特性更接近于低通滤波器特性。
绘制的相频特性由于需要把相角限制在 ± π \pm \pi ±π 之内，所以中间具有两次跳变，但从变化趋势上来看， 相位始终是下降的。
▲ 图1.1.10 系统的幅频特性和相频特性
1.1.3 绘制幅频特性和相频特性
◎ 求解： 为了给出 H ( s ) H\left( s \right) H(s) 频率特性比较准确的绘制， 将题目中的零极点位置按照比例定义它们的取值，然后通过Python程序绘制出对应的幅频特性与相频特性。 坐标采用 x-轴对数，y-轴对数方式进行绘制，也就是绘制出系统的波特图。
（a） 这是一个单个极点的 H ( s ) H\left( s \right) H(s) ，假设 H ( s ) = 1 s + 1 H\left( s \right) = {1 \over {s + 1}} H(s)=s+11 ，绘制频率特性如下
▲ 图1.1.11 系统的频率特性
（b） 这是一个具有单个零点和极点的系统， 假设 H ( s ) = s + 0.2 s + 1 H\left( s \right) = { {s + 0.2} \over {s + 1}} H(s)=s+1s+0.2 绘制的频率特性如下：
▲ 图1.1.12 系统的频率特性
（c） 根据零极点分布，假设 H ( s ) = s + 1 s + 0.5 H\left( s \right) = { {s + 1} \over {s + 0.5}} H(s)=s+0.5s+1 系统的频率特性如下：
▲ 图1.1.13 系统的频率特性
（d） 根据系统的零极点分布，假设 H ( s ) = s − 0.5 s + 1 H\left( s \right) = { {s - 0.5} \over {s + 1}} H(s)=s+1s−0.5 系统的频率特性对应的波特图为：
▲ 图1.1.14 系统的频率特性
（e） 根据系统零极点分布， 假设 H ( s ) = s + 1 H\left( s \right) = s + 1 H(s)=s+1 系统的频率特性：
▲ 图1.1.15 系统的频率特性
（f） 根据系统零极点分布，假设 H ( s ) = s − 0.25 s + 1 H\left( s \right) = {
{s - 0.25} \over {s + 1}} H(s)=s+1s−0.25 对应的频率特性为
▲ 图1.1.16 系统的频率特性
1.1.4 判断滤波特性
◎ 求解：
（a） 低通；
（b） 带通；
（c） 高通；
（d） 带通；
（e） 带通；
（f） 带通；
1.2 离散傅里叶变换
1.2.1 求DFT
◎ 求解：
（1）解答：
x ( n ) = δ ( n ) x\left( n \right) = \delta \left( n \right)\;\;\;\;\; x(n)=δ(n)
X ( k ) = ∑ n = 0 N − 1 x [ n ] e − j 2 π N n k = 1 X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j{ {2\pi } \over N}nk} } = 1 X(k)=n=0∑N−1x[n]e−jN2πnk=1
（2）解答：
x ( n ) = δ ( n − n 0 ) , ( 0 < n 0 < N ) x\left( n \right) = \delta \left( {n - n_0 } \right),\,\,\,\,\,\,\left( {0 < n_0 < N} \right)\;\;\;\;\; x(n)=δ(n−n0),(0<n0<N)
X ( k ) = ∑ n = 0 N − 1 x [ n ] e − j 2 π N n k = e − j 2 π N n 0 k X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j{ {2\pi } \over N}nk} } \, = e^{ - j{ {2\pi } \over N}n_0 k} X(k)=n=0∑N−1x[n]e−jN2πnk=e−jN2πn0k
（3）解答：
x ( n ) = a n R N ( n ) x\left( n \right) = a^n R_N \left( n \right)\;\;\;\;\; x(n)=anRN(n)
X ( k ) = ∑ n = 0 N − 1 a n R N [ n ] W n k = ∑ n = 0 N − 1 a W k = 1 − ( a W k ) N 1 − a W k = 1 − a N 1 − a e − j 2 π N k X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {a^n R_N \left[ n \right]W^{nk} } = \sum\limits_{n = 0}^{N - 1} {aW^k } = { {1 - \left( {aW^k } \right)^N } \over {1 - aW^k }}\, = { {1 - a^N } \over {1 - ae^{ - j{ {2\pi } \over N}k} }} X(k)=n=0∑N−1anRN[n]Wnk=n=0∑N−1aWk=1−aWk1−(aWk)N=1−ae−jN2πk1−aN
（4）解答：
x ( n ) = e j ω 0 n R N ( n ) x\left( n \right) = e^{j\omega _0 n} R_N \left( n \right)\;\;\;\;\; x(n)=ejω0nRN(n)
X ( k ) = a = e j ω 0 1 − e j ω 0 N 1 − e j ( ω 0 − 2 π N k ) X\left( k \right)\mathop = \limits^{a = e^{j\omega _0 } } { {1 - e^{j\omega _0 N} } \over {1 - e^{j\left( {\omega _0 - { {2\pi } \over N}k} \right)} }} X(k)=a=ejω01−ej(ω0−N2πk)1−ejω0N
（5）解答：
X ( k ) = ∑ n = 0 N − 1 sin ( ω 0 n ) ⋅ e − j 2 π k n N X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {\sin \left( {\omega _0 n} \right) \cdot e^{ - j{
{2\pi kn} \over N}} } X(k)=n=0∑N−1sin(ω0n)⋅e−jN2πkn
= ∑ n = 0 N − 1 1 2 j ( e j ω 0 n − e − j ω 0 n ) ⋅ e − j 2 π k N ⋅ n = \sum\limits_{n = 0}^{N - 1} { {1 \over {2j}}\left( {e^{j\omega _0 n} - e^{ - j\omega _0 n} } \right) \cdot e^{ - j{ {2\pi k} \over N} \cdot n} } =n=0∑N−12j1(ejω0n−e−jω0n)⋅e−jN2πk⋅n
= 1 2 j [ 1 − e j ω 0 N 1 − e j ( ω 0 − 2 π k N ) − 1 − e − j ω 0 N 1 − e − j ( ω 0 + 2 π k N ) ] = {1 \over {2j}}\left[ { { {1 - e^{j\omega _0 N} } \over {1 - e^{j\left( {\omega _0 - { {2\pi k} \over N}} \right)} }} - { {1 - e^{ - j\omega _0 N} } \over {1 - e^{ - j\left( {\omega _0 + { {2\pi k} \over N}} \right)} }}} \right] =2j1[1−ej(ω0−N2πk)1−ejω0N−1−e−j(ω0+N2πk)1−e−jω0N]
= 1 2 j ⋅ ( 1 − e j ω 0 N ) ⋅ ( 1 − e − j ( ω 0 + 2 π k N ) ) − ( 1 − e − j ω 0 N ) ⋅ ( 1 − e j ( ω 0 − 2 π k N ) ) ( 1 − e j ( ω 0 − 2 π k N ) ) ⋅ ( 1 − e − j ( ω 0 + 2 π k N ) ) = {1 \over {2j}} \cdot { {\left( {1 - e^{j\omega _0 N} } \right) \cdot \left( {1 - e^{ - j\left( {\omega _0 + { {2\pi k} \over N}} \right)} } \right) - \left( {1 - e^{ - j\omega _0 N} } \right) \cdot \left( {1 - e^{j\left( {\omega _0 {\rm{ - }}{ { {\rm{2}}\pi k} \over N}} \right)} } \right)} \over {\left( {1 - e^{j\left( {\omega _0 - { {2\pi k} \over N}} \right)} } \right) \cdot \left( {1 - e^{ - j\left( {\omega _0 + { {2\pi k} \over N}} \right)} } \right)}} =2j1⋅(1−ej(ω0−N2πk))⋅(1−e−j(ω0+N2πk))(1−ejω0N)⋅(1−e−j(ω0+N2πk))−(1−e−jω0N)⋅(1−ej(ω0−N2πk))
= sin ω 0 e − j 2 π k N − sin ( ω 0 N ) + sin ( ω 0 N − ω 0 ) e − j 2 π k N 1 − 2 cos ω 0 e − j 2 π k N + e − j 4 π k N = { {\sin \omega _0 e^{ - j{ {2\pi k} \over N}} - \sin \left( {\omega _0 N} \right) + \sin \left( {\omega _0 N - \omega _0 } \right)e^{ - j{ {2\pi k} \over N}} } \over {1 - 2\cos \omega _0 e^{ - j{ {2\pi k} \over N}} + e^{ - j{ {4\pi k} \over N}} }} =1−2cosω0e−jN2πk+e−jN4πksinω0e−jN2πk−sin(ω0N)+sin(ω0N−ω0)e−jN2πk
（6）解答：
X ( k ) = ∑ n = 0 N − 1 a n e − j 2 π N k n X\left( k \right) = \sum\limits_{n = 0}^{N - 1} {a^n e^{ - j{
{2\pi } \over N}kn} } X(k)=n=0∑N−1ane−jN2πkn = 1 − ( a e − j 2 π k N ) N 1 − a ⋅ e − j 2 π k N = 1 − a N 1 − a ⋅ e − j 2 π k N = {
{1 - \left( {ae^{ - j{
{2\pi k} \over N}} } \right)^N } \over {1 - a \cdot e^{ - j{
{2\pi k} \over N}} }} = {
{1 - a^N } \over {1 - a \cdot e^{ - j{
{2\pi k} \over N}} }} =1−a⋅e−jN2πk1−(ae−jN2πk)N=1−a⋅e−jN2πk1−aN
（7）解答：
∑ n = 0 N − 1 n W n = W + 2 W 2 + ⋅ ⋅ ⋅ + ( N − 1 ) W N − 1 \sum\limits_{n = 0}^{N - 1} {nW^n } = W + 2W^2 + \cdot \cdot \cdot + \left( {N - 1} \right)W^{N - 1} n=0∑N−1nWn=W+2W2+⋅⋅⋅+(N−1)WN−1 = W + W 2 + ⋅ ⋅ ⋅ + W N − 1 + = W + W^2 + \cdot \cdot \cdot + W^{N - 1} + =W+W2+⋅⋅⋅+WN−1+ W 2 + ⋅ ⋅ ⋅ + W N − 1 + W^2 + \cdot \cdot \cdot + W^{N - 1} + W2+⋅⋅⋅+WN−1+ W 3 + ⋅ ⋅ ⋅ + W N − 1 + W^3 + \cdot \cdot \cdot + W^{N - 1} + W3+⋅⋅⋅+WN−1+ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ W N − 1 W^{N - 1} WN−1
= W − W N 1 − W + W 2 − W N 1 − W + ⋅ ⋅ ⋅ + W N − 1 − W N 1 − W = { {W - W^N } \over {1 - W}} + { {W^2 - W^N } \over {1 - W}} + \cdot \cdot \cdot + { {W^{N - 1} - W^N } \over {1 - W}} =1−WW−WN+1−WW2−WN+⋅⋅⋅+1−WWN−1−WN = ∑ n = 0 N − 1 W n − N ⋅ W N 1 − W = W − W N 1 − W − ( N − 1 ) W N 1 − W = { {\sum\limits_{n = 0}^{N - 1} {W^n } - N \cdot W^N } \over {1 - W}} = { { { {W - W^N } \over {1 - W}} - \left( {N - 1} \right)W^N } \over {1 - W}} =1−Wn=0∑N−1Wn−N⋅WN=1−W1−WW−WN−(N−1)WN = W − 1 1 − W − ( N − 1 ) 1 − W = − N 1 − W = { { { {W - 1} \over {1 - W}} - \left( {N - 1} \right)} \over {1 - W}} = { { - N} \over {1 - W}} =1−W1−WW−1−(N−1)=1−W−N
∑ n = 0 N − 1 n 2 W n = W + 4 W 2 + 9 W 3 + ⋅ ⋅ ⋅ + ( 2 N − 3 ) W N − 1 \sum\limits_{n = 0}^{N - 1} {n^2 W^n } = W + 4W^2 + 9W^3 + \cdot \cdot \cdot + \left( {2N - 3} \right)W^{N - 1} n=0∑N−1n2Wn=W+4W2+9W3+⋅⋅⋅+(2N−3)WN−1 = W + W 2 + W 3 + ⋅ ⋅ ⋅ + W N − 1 + = W + W^2 + W^3 + \,\,\,\,\,\,\,\,\,\, \cdot \cdot \cdot \,\,\,\,\,\,\,\,\,\, + W^{N - 1} + =W+W2+W3+⋅⋅⋅+WN−1+ 3 W 2 + 3 W 3 + ⋅ ⋅ ⋅ + 3 W N − 1 + 3W^2 + 3W^3 + \;\;\;\; \cdot \cdot \cdot \,\,\,\,\,\,\,\,\,\, + 3W^{N - 1} + 3W2+3W3+⋅⋅⋅+3WN−1+ 5 W 3 + ⋅ ⋅ ⋅ + 5 W N − 1 + 5W^3 + \,\,\,\,\,\,\; \cdot \cdot \cdot \,\,\,\,\,\,\,\,\, + 5W^{N - 1} + 5W3+⋅⋅⋅+5WN−1+ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ⋅⋅⋅⋅⋅⋅⋅⋅+ ( 2 N − 3 ) W N − 1 \left( {2N - 3} \right)W^{N - 1} (2N−3)WN−1
$$$$ = W − W N 1 − W + 3 ( W 2 − W N ) 1 − W + 5 ( W 3 − W N ) 1 − W + ⋅ ⋅ ⋅ + ( 2 N − 3 ) ( W N − 1 − W N ) 1 − W = {
{W - W^N } \over {1 - W}} + {
{3\left( {W^2 - W^N } \right)} \over {1 - W}} + {
{5\left( {W^3 - W^N } \right)} \over {1 - W}} + \cdot \cdot \cdot + {
{\left( {2N - 3} \right)\left( {W^{N - 1} - W^N } \right)} \over {1 - W}} =1−WW−WN+1−W3(W2−WN)+1−W5(W3−WN)+⋅⋅⋅+1−W(2N−3)(WN−1−WN) = 1 1 − W [ ∑ k = 1 N − 1 ( 2 k − 1 ) W k − ∑ k = 1 N − 1 ( 2 k − 1 ) W N ] = {1 \over {1 - W}}\left[ {\sum\limits_{k = 1}^{N - 1} {\left( {2k - 1} \right)W^k } - \sum\limits_{k = 1}^{N - 1} {\left( {2k - 1} \right)} W^N } \right] =1−W1[k=1∑N−1(2k−1)Wk−k=1∑N−1(2k−1)WN] = 1 1 − W [ 2 ∑ n = 1 N − 1 n W n − ∑ n = 1 N − 1 W n − ∑ n = 1 N − 1 ( 2 k − 1 ) ] = {1 \over {1 - W}}\left[ {2\sum\limits_{n = 1}^{N - 1} {nW^n } - \sum\limits_{n = 1}^{N - 1} {W^n } - \sum\limits_{n = 1}^{N - 1} {\left( {2k - 1} \right)} } \right] =1−W1[2n=1∑N−1nWn−n=1∑N−1Wn−n=1∑N−1(2k−1)] = 1 1 − W [ 2 ⋅ − N 1 − W − W − W N 1 − W − ( N − 1 ) 2 ] = {1 \over {1 - W}}\left[ {2 \cdot {
{ - N} \over {1 - W}} - {
{W - W^N } \over {1 - W}} - \left( {N - 1} \right)^2 } \right] =1−W1[2⋅1−W−N−1−WW−WN−(N−1)2] = − 2 N − ( W − 1 ) − ( N − 1 ) ( 1 − W ) ( 1 − W ) 2 = N ( N − 1 ) W − N 2 ( 1 − W ) 2 = {
{ - 2N - \left( {W - 1} \right) - \left( {N - 1} \right)\left( {1 - W} \right)} \over {\left( {1 - W} \right)^2 }} = {
{N\left( {N - 1} \right)W - N^2 } \over {\left( {1 - W} \right)^2 }} =(1−W)2−2N−(W−1)−(N−1)(1−W)=(1−W)2N(N−1)W−N2
1.2.2 圆卷积
（1）必做题部分
◎ 解答：
（1）
（2）
（3）
由于 L=8 大于等于两个序列长度之和减一，所以对应的元卷积结果与线卷积结果相同：
y ⊗ 8 = { 1 , 3 , 6 , 10 , 14 , 12 , 9 , 5 } y_{ \otimes 8} = \left\{ {1,3,6,10,14,12,9,5} \right\} y⊗8={ 1,3,6,10,14,12,9,5}
（2）选做题部分
1.3 离散傅里叶变换性质
1.3.1 求序列的DFT
（1）解答：
X 1 [ k ] = ∑ n = 0 N − 1 x [ N − 1 − n ] W n k = ∑ m = N − 1 0 x [ m ] W ( N − 1 − m ) k X_1 \left[ k \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ {N - 1 - n} \right]W^{nk} } \, = \sum\limits_{m = N - 1}^0 {x\left[ m \right]W^{\left( {N - 1 - m} \right)k} } X1[k]=n=0∑N−1x[N−1−n]Wnk=m=N−1∑0x[m]W(N−1−m)k = ∑ m = 0 N − 1 x [ m ] W − m k ⋅ W − k = X [ − k ] N e j 2 π k N = \sum\limits_{m = 0}^{N - 1} {x\left[ m \right]W^{ - mk} \cdot W^{ - k} } \, = X\left[ { - k} \right]_N e^{j{
{2\pi k} \over N}} =m=0∑N−1x[m]W−mk⋅W−k=X[−k]NejN2πk
（2）解答：
X 2 [ k ] = ∑ n = 0 N − 1 ( − 1 ) n x [ n ] W n k = ∑ n = 0 N − 1 x [ n ] e j n π e j 2 π N k X_2 \left[ k \right] = \sum\limits_{n = 0}^{N - 1} {\left( { - 1} \right)^n x\left[ n \right]W^{nk} } = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{jn\pi } e^{j{
{2\pi } \over N}k} } X2[k]=n=0∑N−1(−1)nx[n]Wnk=n=0∑N−1x[n]ejnπejN2πk = ∑ n = 0 N − 1 x [ n ] e j 2 π N n ( k + N 2 ) = X [ k ± N 2 ] N = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j{
{2\pi } \over N}n\left( {k + {N \over 2}} \right)} } = X\left[ {k \pm {N \over 2}} \right]_N =n=0∑N−1x[n]ejN2πn(k+2N)=X[k±2N]N
（3）解答：
X 3 [ k ] = ∑ n = 0 2 N − 1 x 3 [ n ] W 2 N n k = ∑ n = 0 N − 1 x [ n ] ⋅ W 2 N n k + ∑ n = N 2 N − 1 x [ n − N ] ⋅ W 2 N n k X_3 \left[ k \right] = \sum\limits_{n = 0}^{2N - 1} {x_3 \left[ n \right]W_{2N}^{nk} } = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_{2N}^{nk} } + \sum\limits_{n = N}^{2N - 1} {x\left[ {n - N} \right] \cdot W_{2N}^{nk} } X3[k]=n=0∑2N−1x3[n]W2Nnk=n=0∑N−1x[n]⋅W2Nnk+n=N∑2N−1x[n−N]⋅W2Nnk = ∑ n = 0 N − 1 x [ n ] ⋅ W N n k 2 + ∑ m = 0 N − 1 x [ m ] ⋅ W N ( m + N ) k 2 = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{n{k \over 2}} } + \sum\limits_{m = 0}^{N - 1} {x\left[ m \right] \cdot W_N^{\left( {m + N} \right){k \over 2}} } =n=0∑N−1x[n]⋅WNn2k+m=0∑N−1x[m]⋅WN(m+N)2k = ∑ n = 0 N − 1 x [ n ] ⋅ W N n k 2 + W N k N 2 ∑ m = 0 N − 1 x [ m ] ⋅ W N m k 2 = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{
{
{nk} \over 2}} } + W_N^{
{
{kN} \over 2}} \sum\limits_{m = 0}^{N - 1} {x\left[ m \right] \cdot W_N^{
{
{mk} \over 2}} } =n=0∑N−1x[n]⋅WN2nk+WN2kNm=0∑N−1x[m]⋅WN2mk = [ 1 + ( − 1 ) k ] ∑ n = 0 N − 1 x [ n ] ⋅ W N n k 2 = [ 1 + ( − 1 ) k ] X [ k 2 ] N = \left[ {1 + \left( { - 1} \right)^k } \right]\sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{
{
{nk} \over 2}} } = \left[ {1 + \left( { - 1} \right)^k } \right]X\left[ {
{k \over 2}} \right]_N =[1+(−1)k]n=0∑N−1x[n]⋅WN2nk=[1+(−1)k]X[2k]N
（4）解答：
X 4 [ k ] N 2 = ∑ n = 0 N 2 − 1 { x [ n ] + x [ n + N 2 ] } ⋅ W N 2 n k X_4 \left[ k \right]_{
{N \over 2}} = \sum\limits_{n = 0}^{
{N \over 2} - 1} {\left\{ {x\left[ n \right] + x\left[ {n + {N \over 2}} \right]} \right\} \cdot W_{
{N \over 2}}^{nk} } X4[k]2N=n=0∑2N−1{
x[n]+x[n+2N]}⋅W2Nnk = ∑ n = 0 N 2 j − 1 x [ n ] ⋅ W N 2 n k + ∑ m = N 2 N − 1 x [ m ] ⋅ W N 2 m k = \sum\limits_{n = 0}^{
{N \over 2}j - 1} {x\left[ n \right] \cdot W_N^{2nk} } + \sum\limits_{m = {N \over 2}}^{N - 1} {x\left[ m \right] \cdot W_N^{2mk} } =n=0∑2Nj−1x[n]⋅WN2nk+m=2N∑N−1x[m]⋅WN2mk = ∑ n = 0 N − 1 x [ n ] ⋅ W N 2 n k = X [ 2 k ] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{2nk} } = X\left[ {2k} \right] =n=0∑N−1x[n]⋅WN2nk=X[2k]
（5）解答：
X 5 [ k ] = ∑ n = 0 N − 1 x [ n ] ⋅ W 2 N n k = ∑ n = 0 N − 1 x [ n ] ⋅ W N n k 2 = X [ k 2 ] X_5 \left[ k \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_{2N}^{nk} } = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{
{
{nk} \over 2}} } = X\left[ {
{k \over 2}} \right] X5[k]=n=0∑N−1x[n]⋅W2Nnk=n=0∑N−1x[n]⋅WN2nk=X[2k]
（6）解答：
X 6 [ k ] = ∑ n = 0 2 N − 1 x 6 [ n ] ⋅ W 2 N n k = ∑ n = 0 N − 1 x [ n ] ⋅ W N n k = X [ k ] X_6 \left[ k \right] = \sum\limits_{n = 0}^{2N - 1} {x_6 \left[ n \right] \cdot W_{2N}^{nk} } = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{nk} } = X\left[ k \right] X6[k]=n=0∑2N−1x6[n]⋅W2Nnk=n=0∑N−1x[n]⋅WNnk=X[k]
（7）解答：
X 7 [ k ] = ∑ n = 0 N − 1 x 7 [ n ] ⋅ W N 2 n k = ∑ n = 0 N − 1 x [ n ] ⋅ 1 + ( − 1 ) n 2 W N n k X_7 \left[ k \right] = \sum\limits_{n = 0}^{N - 1} {x_7 \left[ n \right] \cdot W_{
{N \over 2}}^{nk} } = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot {
{1 + \left( { - 1} \right)^n } \over 2}W_N^{nk} } X7[k]=n=0∑N−1x7[n]⋅W2Nnk=n=0∑N−1x[n]⋅21+(−1)nWNnk = 1 2 { ∑ n = 0 N − 1 x [ n ] ⋅ W N n k + ∑ n = 0 N − 1 x [ n ] ( − 1 ) n W N n k } = {1 \over 2}\left\{ {\sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{nk} } + \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]\left( { - 1} \right)^n W_N^{nk} } } \right\} =21{
n=0∑N−1x[n]⋅WNnk+n=0∑N−1x[n](−1)nWNnk} = 1 2 { X [ k ] + X [ k + N 2 ] N } = {1 \over 2}\left\{ {X\left[ k \right] + X\left[ {k + {N \over 2}} \right]_N } \right\} =21{
X[k]+X[k+2N]N}
1.3.2 序列补零扩充长度
◎ 求解：
这个题目有两个求解方式。
（1）
直接利用DFT公式， 得到补零扩充之后序列于原来序列之间的变量 z 域展缩之间的关系。
Y [ k 1 ] = ∑ n = 0 r N − 1 y [ n ] ⋅ W r N n k 1 = ∑ n = 0 N − 1 x [ n ] ⋅ W r N n k 1 Y\left[ {k_1 } \right] = \sum\limits_{n = 0}^{rN - 1} {y\left[ n \right] \cdot W_{rN}^{nk_1 } } = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_{rN}^{nk_1 } } Y[k1]=n=0∑rN−1y[n]⋅WrNnk1=n=0∑N−1x[n]⋅WrNnk1 = ∑ n = 0 N − 1 x [ n ] ⋅ W N n k 1 r = X [ k 1 r ] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_N^{ { {nk_1 } \over r}} } = X\left[ { { {k_1 } \over r}} \right] =n=0∑N−1x[n]⋅WNrnk1=X[rk1]
（2）
利用DFT反变换，或者序列在单位圆的插值结果， 然后再进行均匀抽样获得补零后结果。
Y [ k 1 ] = ∑ n = 0 N − 1 x [ n ] ⋅ W r N n k 1 = ∑ n = 0 N − 1 ( 1 N ∑ k = 0 N − 1 X [ k ] ⋅ W N − n k ) ⋅ W r N n k 1 Y\left[ {k_1 } \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right] \cdot W_{rN}^{nk_1 } } \, = \sum\limits_{n = 0}^{N - 1} {\left( { {1 \over N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right] \cdot W_N^{ - nk} } } \right)} \cdot W_{rN}^{nk_1 } Y[k1]=n=0∑N−1x[n]⋅WrNnk1=n=0∑N−1(N1k=0∑N−1X[k]⋅WN−nk)⋅WrNnk1 = 1 N ∑ k = 0 N − 1 X [ k ] ∑ n = 0 N − 1 W r N − n k r W r N n k 1 = 1 N ∑ n = 0 N − 1 X [ k ] 1 − W r N k 1 N 1 − W r N k 1 − k r = {1 \over N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]\sum\limits_{n = 0}^{N - 1} {W_{rN}^{ - nkr} W_{rN}^{nk_1 } } } = {1 \over N}\sum\limits_{n = 0}^{N - 1} {X\left[ k \right]{ {1 - W_{rN}^{k_1 N} } \over {1 - W_{rN}^{k_1 - kr} }}} =N1k=0∑N−1X[k]n=0∑N−1WrN−nkrWrNnk1=N1n=0∑N−1X[k]1−WrNk1−kr1−WrNk1N
1.3.3 DFT对偶特性
◎ 证明：
根据IDFT公式： x [ n ] = 1 N ∑ k = 0 N X [ k ] ⋅ e j 2 π k n N x\left[ n \right] = {1 \over N}\sum\limits_{k = 0}^N {X\left[ k \right] \cdot e^{
{
{j2\pi kn} \over N}} } x[n]=N1k=0∑NX[k]⋅eNj2πkn
因此： N ⋅ x [ n ] = ∑ k = 0 N X [ k ] ⋅ e − j 2 π k N ( − n ) N \cdot x\left[ n \right] = \sum\limits_{k = 0}^N {X\left[ k \right] \cdot e^{
{
{ - j2\pi k} \over N}\left( { - n} \right)} } N⋅x[n]=k=0∑NX[k]⋅eN−j2πk(−n)
所以： D F T { X [ n ] } = N ⋅ x ( ( − k ) ) N ⋅ R [ n ] DFT\left\{ {X\left[ n \right]} \right\} = N \cdot x\left( {\left( { - k} \right)} \right)_N \cdot R\left[ n \right] DFT{
X[n]}=N⋅x((−k))N⋅R[n]
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