数字信号处理(英文版)复习重点题型和答案
b. (omit)
pp76 Q1.13 Given the discrete time sinusoid x[n] = 2cos(0.2π n + 0.1π ) and the sampling frequency Fs = 2 kHz a. Determine the corresponding continuous time sinusoid in the frequency range 0 < F0 < F / 2 b. Determine two other continuous time sinusoids with the same sample values
⎞ ⎟ X [k ] ⎠
, where X [k ] are computed in the previous question.
b. x[n] = u[n − 2] Ans:
y[n] = h[n] ∗ x[n] = ∑ h[n − k ]x[k ] = ∑ 0.5n−k u[ n − k ]u[ k − 2]
k k
Since u[n] = 0 for n < 0 , u[n − k ] ≠ 0 for k ≤ n , and u[k − 2] ≠ 0 for k ≥ 2 Therefore, when n < 2 , y[n] = 0 . When n ≥ 2
Ans. a. x[n] = 2cos(0.2π n + 0.1π ) Digital frequency: ω0 = 0.2π = 2π F0Ts Therefore, the continuous time frequency:
F0 =
ω0 0.2π 2 = 0.2 kHz = 2π Ts 2π
π
180
=
π
12
π ⎞ π ⎞ ⎛ ⎛ − j ⎜100π t + ⎟ ⎞ 100π t + ⎟ 3 ⎛ j⎜ 12 ⎠ 12 ⎠ ⎝ ⎝ x(t ) = ⎜ e +e ⎟ ⎟ 2⎜ ⎝ ⎠ π π −j ⎞ 3 ⎛ j12 j100π t 12 − j100π t = ⎜e e +e e ⎟ 2⎝ ⎠
a. x = […,1,2,3,…] Ans: y[n] = 0.8 y[n − 1] + 0.2 x[n]
Y ( z ) = 0.8 z −1Y ( z ) + 0.2 X ( z ) Y ( z) 0.2 H ( z) = = X ( z ) 1 − 0.8 z −1 Let z = e jω , we have
xa [n] = 3cos(−1.9π n − 0.2π ) = 3cos(1.9π n + 0.2π )
pp.75 Q1.8 Write the following sinusoids in terms of complex exponentials
a. x(t ) = 3cos(100π t + 15o ) b. x(t ) = 2cos(10π t − 0.1π ) Ans. a. 15o ⇒ 15 ×
ω1 2.2π 2 = 2.2 kHz = 2π Ts 2π ω2 1.8π 2 = 1.8 kHz = 2π Ts 2π
pp.76 Q1.16 In each of the following systems let x(t ) or x[n] be the input and y (t ) or y[n] be the output. Determine whether each system is (1) linear, (2) time invariant, (3) causal, (4) BIBO stable.
Digital Signal Processing Review Questions Apr 15, 2012
pp. 73, Q1.1 Given the sequences shown, write them in terms of unit impulses
x[n]
1
•
1
•
0.5
•
• •
•
n
•
-0.5
+ 3e
− jk
, k = 0,1, 2
k = 0, k = 1, k = 2,
X [0] = 1 + 2 + 3 = 6 X [1] =பைடு நூலகம்1 + 2e
−j 2π 3 4π 3
+ 3e
j
2π 3 −j 8π 3
X [2] = 1 + 2e = 1 + 2e
j
−j
+ 3e
j
2π 3
+ 3e
4π 3
pp. 81 Q1.39 Consider the linear time invariant system with difference equation y[n] = 0.8 y[n − 1] + 0.2 x[n] . Determine the DFS of the output signal for each of the inputs
Pp77 Q1.17 A linear time invariant system has impulse
response h[n] = 0.5n u[n] . Determine the output sequence y[n] for each of the following input signals:
y[n] = ∑ 0.5n−k = ∑ 0.5k =
k =2 k =0
n
n−2
1 − 0.5n−1 , for n ≥ 2 0.5
In summary:
n<2 ⎧ 0, ⎪ y[n] = ⎨1 − 0.5n−1 , n≥2 ⎪ ⎩ 0.5
The output y[n] is the output of the filter h[n] with input x[n] , shown in the following figure
Filter coefficients h[ n] = 0.5n u[n] x[n] = u[n − 2]
z −1 0.5
z −1 0.53
z −1 0.54 +
••• •••
z −1 0.5k
••• 0.5k +1 • • •
1
y[n]
Pp77 Q1.23. Using partial fraction expansion, determine the inverse z-transform of the following functions:
(t 3 + t 2 + t + 1)δ (t )dt
e.
∫
∞
−∞
cos 2 (2π t + 0.1π )δ (t + 1)dt
∫ Ans. d. ∫
Ans a: Ans. e.
∞
−∞ ∞
e − tδ (t − 1)dt = e −1 (t 3 + t 2 + t + 1)δ (t )dt = 1
g. y[n] = x[n] + x[n − 1] + x[n − 2] i. y[n] = nx[n] + x[n − 1] + x 2 [n − 2] Ans. g. linear, time invariant, causal, BIBO stable. i. non-linear, time variant, causal, BIBO unstable Hints: to show y[n] is time variant, substitute x[n] with x[n − L] into the right-hand side of the equation. Can we get y[n − L] ?
Ans. b. x[n] = 2cos(ω x n + 0.1π ) , where ω x = 0.2π + 2π k and k is an integer Select k = ±1 ,
ω1 = 0.2π + 2π = 2.2π ω2 = 0.2π − 2π = −1.8π ⇒ 1.8π
The corresponding continuous time frequencies are: F1 = F2 =
a. x = […,1,2,3,…] Ans: Period: N = 3
X [k ] = DFS { x[n]} = ∑ x[n]e− jk (2π / N ) n
n =0
N −1
= 1 + 2e = 1 + 2e
− jk (2π /3) − jk 2π 3
+ 3e
− jk (2π /3)2 4π 3
H (ω ) =
0.2 1 − 0.8e − jω ⎛ 2π Y [k ] = DFS { y[n]} = H ⎜ k ⎝ 3
k = 0, Y [0] = X [0] ⎛ 2π ⎞ k = 1, Y [1] = H ⎜ ⎟ X [1] ⎝ 3 ⎠ 0.2 = X [1] 2π −j 1 − 0.8e 3 ⎛ 4π ⎞ k = 2, Y [2] = H ⎜ ⎟ X [2] ⎝ 3 ⎠ 0.2 = X [2] 4π −j 3 1 − 0.8e
•
-0.8
Ans:
x[n] = −0.5δ [n + 1] + δ [n] + 0.5δ [n − 1] + δ [n − 2] − 0.8δ [n − 3]
pp.74 Q1.2 Compute the following integrals
a. d.
《数字信号处理》试题库答案
《数字信号处理》试题库答案⼀.填空题1、⼀线性时不变系统，输⼊为 x（n）时，输出为y（n）；则输⼊为2x（n）时，输出为2y(n) ；输⼊为x（n-3）时，输出为y(n-3) 。
2、从奈奎斯特采样定理得出，要使实信号采样后能够不失真还原，采样频率fs与信号最⾼频率f max关系为： fs>=2f max。
3、已知⼀个长度为N的序列x(n)，它的离散时间傅⽴叶变换为X（e jw），它的N点离散傅⽴叶变换X（K）是关于X（e jw）的 N 点等间隔采样。
有关，还与窗的采线性移不变系统的性质有交换率、结合率和分配律。
15.⽤DFT近似分析模拟信号的频谱时，可能出现的问题有混叠失真、泄漏、栅栏效应和频率分辨率。
16.⽆限长单位冲激响应滤波器的基本结构有直接Ⅰ型，直接Ⅱ型，串联型和并联型四种。
17.如果通⽤计算机的速度为平均每次复数乘需要5µs，每次复数加需要1µs，则在此计算机上计算210点的基2 FFT需要10 级蝶形运算，总的运算时间是______µs。
⼆．选择填空题1、δ(n)的z变换是 A 。
A. 1B.δ(w)C. 2πδ(w)D. 2π2、从奈奎斯特采样定理得出，要使实信号采样后能够不失真还原，采样频率f s与信号最⾼频率f max关系为： A 。
A. f s≥ 2f maxB. f s≤2 f maxC. f s≥ f maxD. f s≤f max3、⽤双线性变法进⾏IIR数字滤波器的设计，从s平⾯向z平⾯转换的关系为s= C 。
A.1111zzz-1 1 1 1 z z z ---= +s C. 1 1 21 1 z z T z ---= + D. 1 1 21 1+=-4、序列x1（n）的长度为4，序列x2（n）的长度为3，则它们线性卷积的长度是，5点圆周卷积的长度是。
A． 1/2+δ(n)/2 B. 1+δ(n) C. 2δ(n) D. u(n)- δ(n)12. 下列关系正确的为（ B ）。
《数字信号处理》复习题及答案
《数字信号处理》复习题及答案《数字信号处理》复习题⼀、单项选择题（在每⼩题的四个备选答案中，选出⼀个正确答案，并将正确答案的序号填在题⼲的括号内。
每⼩题2分）1.在对连续信号均匀采样时，若采样⾓频率为Ωs，信号最⾼截⽌频率为Ωc，则折叠频率为( D)。
A. ΩsB. ΩcC. Ωc/2D. Ωs/22. 若⼀线性移不变系统当输⼊为x(n)=δ(n)时输出为y(n)=R3(n)，则当输⼊为u(n)-u(n-2)时输出为( C)。
A. R3(n)B. R2(n)C. R3(n)+R3(n-1)D. R2(n)+R2(n-1)3. ⼀个线性移不变系统稳定的充分必要条件是其系统函数的收敛域包含( A)。
A. 单位圆B. 原点C. 实轴D. 虚轴4. 已知x(n)=δ(n)，N点的DFT［x(n)］=X(k)，则X(5)=( B)。
A. NB. 1C. 0D. - N5. 如图所⽰的运算流图符号是( D)基2 FFT算法的蝶形运算流图符号。
A. 按频率抽取B. 按时间抽取C. 两者都是D. 两者都不是6. 直接计算N点DFT所需的复数乘法次数与( B)成正⽐。
A. NB. N2C. N3D. Nlog2N7. 下列各种滤波器的结构中哪种不是I I R滤波器的基本结构( D)。
A. 直接型B. 级联型C. 并联型D. 频率抽样型8. 以下对双线性变换的描述中正确的是( B)。
A. 双线性变换是⼀种线性变换B. 双线性变换可以⽤来进⾏数字频率与模拟频率间的变换C. 双线性变换是⼀种分段线性变换D. 以上说法都不对9. 已知序列Z变换的收敛域为｜z｜>1，则该序列为( B)。
A. 有限长序列B. 右边序列C. 左边序列D. 双边序列10. 序列x(n)=R5(n)，其8点DFT记为X(k)，k=0,1,…,7，则X(0)为( D)。
A. 2B. 3C. 4D. 511. 下列关于FFT的说法中错误的是( A)。
数字信号处理总复习和习题
∴T[a1x1(n) + a2 x2 (n)] = [a1x1(n) + a2 x2 (n)] = a x (n) + a2 x2 (n) + 2a1a2 x1(n)x2 (n)
2 2 1 1 2 2
2
a1T[x1(n)] + a2T[x2 (n)] = a x (n) + a x (n)
….
y1(n) = ay1(n −1) +δ (n) = an
所以：
y1(n) = a u(n)
n
又 x2 (n) = δ (n −1) 令
则 y2 (0) = ay2 (−1) +δ (−1) = 0 ： y2 (1) = ay2 (0) +δ (0) =1 y2 (2) = ay2 (1) +δ (1) = a
y(n) − ay(n −1) = x(n)
解 （ ） (−1) = 0的 况 ： b y 情
令
Q y(n) − ay(n −1) = x(n)
x1(n) = δ (n)
∴y1(0) = ay1(−1) +δ (2) = ay1(1) +δ (2) = a2
2 1 1 2 2 2
可见： T[a x (n) + a
1 1
2 2
x (n)] ≠ a1T[x1(n)] + a2T[x2 (n)]
故不是线性系统。
(d)
y(n) = 3x(n) + 5 Qy1(n) = 3x1(n) + 5 = T[x1(n)], y2 (n） 3x2 (n) + 5 = T[x2 (n)] = 即，系统操作为乘 加 。 3 5
《数字信号处理》复习思考题、习题(二)答案.doc
一、思考题1、C2、C3、D4、A5、D6、B7、D8、B9、C 10、A 11、C 12、C 13、A 14、A 15、B 16、C 17、A 18、C二、概念填空题1、（1）付氏级数（2） hd （n）（理想的单位脉冲响应）（3） R N（n）（N点矩形窗或N点矩形序列）（4） h （n）（单位脉冲响应）（5）吉布斯（6）波动（不平稳）（7）衰减（最小衰减）2、（8）（9）三角窗、汉宁窗、哈明窗、布莱克曼窗（10）过渡带（11）衰减3、（12）时（13） h （n）（数字滤波器单位脉冲响应）（14） h a（t）（模拟滤波器冲激响应）（15）频谱混叠（16 ）折叠频率（兀/T）4、（17）偶对称（奇对称）（18）奇对称（偶对称）（19）〃二堕二1! （20）线性相位特性25、（21）时（22）窗函数（23）有限长（24）逼近6、（25）某种优化逼近方法（26）逼近（27）频率响应（28）最优三、判断说明题1、判断：正确简述：按照频率采样滤波器结构的推导，上述说法是正确的，这正是频率采样结构的一个优点。
但对于不同的频响形状，N个并联一阶节的支路增益H (k)不同。
2、判断：一致简述：由于对模拟滤波器而言，因果稳定系统传递函数H a(s)的极点均在S平面的左半平面，只要转换关系满足使S平面的左半平面转换到Z平面的单位圆内，就保证了转换后数字滤波器系统函数H (z) 的极点全部在Z平面的单位圆内，从而保证了系统的因果稳定性。
3、判断：不对简述：正确的表述应为：IIR滤波器只能采用递归型结构实现；FIR 滤波器一般采用非递归型结构实现，但也可使结构中含有递归支路。
就是说滤波器结构与特性没有必然的联系。
4、判断：一致简述：由于对模拟域而言，其频率轴就是S平面的虚轴j。
轴，而对数字域来说，其频率轴是z平面的单位圆，因此两者是一致的。
四、计算应用题1、解：1)容易将H (z)写成级联型的标准形式如下：)二(2 + 3广)(3-2广 + 广)H(Z一(4 —广)(1 + 0.9广—0.81厂2)0.5+ 3-2广+疽—— ________ z ______ * ___________________________________1 + 0.9/—0.81厂2显见，该系统的级联结构由一个直接II型一阶节和一个直接II型二阶节级联而成，因此容易画出该系统的级联型结构图如图A-1所示。
数字信号处理复习题带答案
1.若一模拟信号为带限信号，且对其抽样满足奈奎斯特条件，则只要将抽样信号通过_____A____即可完全不失真恢复原信号。
A、理想低通滤波器B、理想高通滤波器C、理想带通滤波器D、理想带阻滤波器2.下列哪一个单位抽样响应所表示的系统不是因果系统___D__A、.h(n)=δ(n)+δ(n-10)B、h(n)=u(n)C、h(n)=u(n)-u(n-1)D、 h(n)=u(n)-u(n+1)3.若序列的长度为M，要能够由频域抽样信号X(k)恢复原序列，而不发生时域混叠现象，则频域抽样点数N需满足的条件是_____A_____。
≥M ≤M≤2M ≥2M4.以下对双线性变换的描述中不正确的是__D_________。
A.双线性变换是一种非线性变换B.双线性变换可以用来进行数字频率与模拟频率间的变换C.双线性变换把s平面的左半平面单值映射到z平面的单位圆内D.以上说法都不对5、信号3(n)Acos(n)78xππ=-是否为周期信号，若是周期信号，周期为多少？A、周期N=37πB、无法判断C、非周期信号D、周期N=146、用窗函数设计FIR滤波器时，下列说法正确的是___a____。
A、加大窗函数的长度不能改变主瓣与旁瓣的相对比例。
B、加大窗函数的长度可以增加主瓣与旁瓣的比例。
C、加大窗函数的长度可以减少主瓣与旁瓣的比例。
D、以上说法都不对。
7.令||()nx n a=，01,a n<<-∞≤≤∞，()[()]X Z Z x n=，则()X Z的收敛域为__________。
A 、1||a z a -<<B 、1||a z a -<<C 、||a z <D 、1||z a -< 。
点FFT 所需乘法（复数乘法）次数为____D___。
A 、2N log NB 、NC 、2ND 、2log 2NN 9、δ(n)的z 变换是AA. 1B.δ(w)C. 2πδ(w)D. 2π 10、下列系统(其中y(n)是输出序列，x(n)是输入序列)中__ C___属于线性系统。
数字信号处理课后答案+第3章(高西全丁美玉第三版)PPT课件
所以
DFT［X(n)］=Nx(N－k) k=0, 1, …, N－1 5． 如果X(k)=DFT［x(n)］， 证明DFT的初值定理
x(0)
1
N 1
X (k)
证: 由IDFT定义式
N k0
x(n)
1 N
N 1
X (k )WNkn
k 0
n 0, 1, , N 1
可知
x(0)
1
N 1
X (k)
教材第3章习题与上机题解答
1． 计算以下序列的N点DFT， 在变换区间0≤n≤N－1内，
(1) x(n)=1
(2) x(n)=δ(n) (3) x(n)=δ(n－n0) (4) x(n)=Rm(n)
0<n0<N 0<m<N
j2π mn
(5) x(n) e N , 0 m N
(6) x(n) cos 2π mn, 0 m N N
sin
(0
2π N
k
)
/
2
k 0, 1, , N 1
或
1 e j0N
X
7
(k
)
1
e
j(0
2 N
k)
（8） 解法一 直接计算：
k 0, 1, , N 1
x8 (n)
sin(0n)
RN
(n)
1 [e j0n 2j
e j0n ]RN
(n)
X8(n)
N 1
x8 (n)WNkn
n0
1
N 1
[e j0n
1 WNk
j π (m1)k
e N
sin
π N
mk
sin
π N
数字信号处理复习题含答案
数字信号处理复习题含答案数字信号处理复习题含答案数字信号处理是一门研究如何对数字信号进行处理和分析的学科。
在现代科技的发展中，数字信号处理已经广泛应用于音频、视频、通信等领域。
为了帮助大家复习数字信号处理的知识，本文将提供一些复习题，并附上答案。
希望这些题目能够帮助大家巩固对数字信号处理的理解。
1. 什么是离散时间信号？答案：离散时间信号是在离散时间点上取值的信号。
离散时间信号可以用数学序列表示，例如x(n)，其中n为整数。
2. 什么是离散时间系统？答案：离散时间系统是对离散时间信号进行处理和变换的系统。
离散时间系统可以用差分方程表示。
3. 什么是离散傅里叶变换（DFT）？答案：离散傅里叶变换是将离散时间域信号转换到离散频率域的一种变换。
DFT可以用来分析信号的频谱特性。
4. 什么是快速傅里叶变换（FFT）？答案：快速傅里叶变换是一种高效计算离散傅里叶变换的算法。
FFT算法可以降低计算复杂度，提高计算速度。
5. 什么是数字滤波器？答案：数字滤波器是对数字信号进行滤波的系统。
数字滤波器可以通过差分方程或差分方程的系数来描述。
6. 什么是有限冲激响应（FIR）滤波器？答案：有限冲激响应滤波器是一种滤波器，其冲激响应具有有限长度。
FIR滤波器可以通过线性组合的方式实现。
7. 什么是无限冲激响应（IIR）滤波器？答案：无限冲激响应滤波器是一种滤波器，其冲激响应具有无限长度。
IIR滤波器可以通过递归的方式实现。
8. 什么是数字信号的抽样和保持？答案：抽样是指将连续时间信号在一定时间间隔内取样得到离散时间信号。
保持是指在抽样的同时，将采样值保持不变。
9. 什么是量化？答案：量化是将连续时间信号的幅值转换为离散的幅值级别的过程。
量化过程中，需要确定量化级别和量化误差。
10. 什么是编码？答案：编码是将量化后的离散信号用一组二进制码表示的过程。
编码可以通过不同的编码方式实现，例如脉冲编码调制（PCM）。
以上是一些关于数字信号处理的复习题及其答案。
数字信号处理答案10
Chapter 10 Solutions10.1 (a) The impulse response is given by h[n] = –0.8h[n –1] + 0.1h[n –2] + δ[n]. The first ten samples are listed in the table.(b) The impulse response contains an infinite number of non-zero terms.10.3 (a)(i) Without pre-warping, the transfer function for the analog filter is15708s 15708)2500(2s )2500(2s )s (H 1p 1p +=π+π=ω+ω=The bilinear transformation 1z 1z f 2s S +-=gives11z00921.01)z1(4954.0157081z 1z 1600015708)z (H ---+=++-=(ii)The analog frequency 2.5 kHz is converted to a digital frequency Ωp1 =8000/)2500(2π =1.9635 rads. This frequency is pre-warped to the analog frequency2tanf 21p S 1p Ω=ω= 23946 rad/sec, which makes the transfer function for the analog filter23946s 23946s )s (H 1p 1p +=ω+ω=After the bilinear transformation, the digital transfer function is obtained:11z1989.01)z1(6.0239461z 1z 1600023946)z (H --++=++-=(b) The magnitude responses for both filters are shown below. The –3 dB frequency for the pre-warped filter is equal to the specified 2.5 kHz.10.4 (a) The cut-off frequency for the analog filter is 1500/(2π) = 238.73 Hz. The digital frequency that corresponds to this analog frequency is Ωp1 = 8000/)73.238(2π = 0.1875 rads. The pre-warped analog cut-off frequency is 2tan f 21p S 1p Ω=ω= 1504.4rad/sec, to give the transfer function 4.1504s 4.1504)s (H +=.Filter with pre-warping |H(f)| f(b) Use the bilinear transformation to get the digital transfer function11z8281.01)z 1(0859.04.15041z 1z 160004.1504)z (H ---+=++-=(c) The difference equation is y[n] = 0.8281y[n –1] + 0.0859x[n] + 0.0859x[n –1]. (d)The frequency response is Ω-Ω--+=Ωj j e8281.01)e 1(0859.0)(H . The magnitude responsemay be found by taking the magnitudes of this expression for several values of Ω. It is plotted below against frequency in Hz.(e) The magnitude for the analog transfer function is224.15044.1504)(H +ω=ωThe shape of the digital filter may be found by substituting 2tan f 2S Ω=ω:2222S 4.15042tan 160004.15044.15042tan f 24.1504)(H +⎪⎭⎫ ⎝⎛Ω=+⎪⎭⎫ ⎝⎛Ω=ΩThis function may be plotted for various values of Ω. The magnitude response is plotted below against digital frequency in rads. When digital frequencies are converted to frequencies in Hz, the plot becomes identical to the one in part (d).10.5 The –3 dB frequency for this filter is 1 kHz. This is the edge of the pass band. The stop band edge can chosen at any convenient place. For example, the gain at 3 kHz is about –46 dB. The stop band ripple is given by 20log δs = –46, or δs = 0.005. The digital frequencies at the edges of the pass band and stop band areΩp1 = π=π=π25.0800010002f f 2S1p radiansΩs1 =π=π=π75.0800030002f f 2S1s radiansωp1 = 4.66272tanf 21p S =Ω rad/secωs1 = 4.386272tanf 21s S =Ω rad/secThe order for the filter is10.6 The stop band attenuation gives 20log δs = –28, or δs = 0.0398. Ωp1 = π=π=π542.02400065002f f 2S1p radians34.66274.38627log 21)005.0(1log log 211log n 21p 1s 2s =⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ≥|H(Ωs1 =π=π=π667.02400080002f f 2S1s radiansωp1 = 4.547912tanf 21p S =Ω rad/secωs1 = 1.832392tanf 21s S =Ω rad/secThe order for the filter can be calculated usingso an order of 8 should be chosen. For this order, the magnitude response for the analog filter is given by14.54791111)(H 16n21p +⎪⎭⎫ ⎝⎛ω=+⎪⎪⎭⎫ ⎝⎛ωω=ωTherefore, the transfer function for the digital filter is14.547912tan 48000114.547912tan f 21)(H 1616S+⎪⎪⎪⎪⎭⎫⎝⎛Ω=+⎪⎪⎪⎪⎭⎫⎝⎛Ω=ΩThe plot of |H(Ω)| versus Ω is the same as the plot of |H(f)| versus f, except that, along the horizontal axis, digital frequencies between 0 and π radians are converted tofrequencies between 0 and 12000 Hz (half the sampling rate). The magnitude response is shown below. The eighth order filter matches the specifications.7.74.547911.83239log 21)0398.0(1log log 211log n 21p 1s 2s =⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ≥10.7 The stop band attenuation gives 20log δs = –35, or δs = 0.01778.Ωp1 = π=π=π625.0800025002f f 2S1p radiansΩs1 =π=π=π95.0800038002f f 2S1s radiansωp1 = 239462tanf 21p S =Ω rad/secωs1 = 2032992tanf 21s S =Ω rad/secThe order for the filter can be calculated usingso an order of 2 should be chosen.10.8 The 600 Hz transition width puts the stop band edge at 1.9 kHz. The digital pass band edge frequencies areΩp1 = π=π=π26.01000013002f f 2S1p radians|H(f)|f9.123946203299log 21)01778.0(1log log 211log n 21p 1s 2s =⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ≥Ωs1 =π=π=π38.01000019002f f 2S1s radiansωp1 = 8.86542tanf 21p S =Ω rad/secωs1 = 0.135922tanf 21s S =Ω rad/secSince the order for the filter is 5,This expression gives96.111log 2s =⎪⎪⎭⎫⎝⎛-δor,3.91101196.12s==-δwhich gives δs = 0.1047, which in turn means a stop band gain of 20log(0.1047) = –19.6 dB. Thus, the stop band attenuation is 19.6 dB.10.9 With 16 kHz sampling, an analog frequency of 3.5 kHz corresponds to a digital frequency of π=π=π=Ω4375.01600035002f f 2S'p rads. Pre-warping gives 'p ω =7.262612tanf 2'p S =Ω rad/sec. The analog transfer function for a first order high passfilter is obtained by transforming a first order low pass transfer function such as⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛-δ=⎪⎪⎭⎫⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ==8.86540.13592log 211log log 211log 5n 2s 1p 1s 2s5.2360s 5.2360s )s (H pp L +=ω+ω=(Any low pass filter with a known cut-off frequency will do.) The high pass transfer function can be found as follows:()()⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫⎝⎛ωω=s 7.262615.2360H s H )s (H L 'pp L H ()()7.26261s s 5.2360s7.262615.23605.2360+=+⎪⎭⎫⎝⎛=The digital counterpart is given by the bilinear transformation:()11H z0985.01z 15492.07.262611z 1z 320001z 1z 32000)z (H ----=++-+-=The difference equation for the high pass filter is y[n] = 0.0985y[n –1] + 0.5492x[n] – 0.5492x[n –1].10.10 A stop band attenuation of 40 dB gives 20log δs = –40 dB, which means δs = 0.01. The low pass prototype for the high pass filter will have a cut-off at f S /2 – 9 = 22 – 9 = 13 kHz. The pass band frequencies are:Ωp1 = π=π=π591.044000130002f f 2S1p radiansωp1 = 1175892tanf 21p S =Ω rad/sec(a)With an order n = 3,⎪⎭⎫⎝⎛ω⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ==117589log 21)01.0(1log log 211log 3n 1s 21p 1s 2s⎪⎭⎫ ⎝⎛ω117589log 1s = 0.667ωs1 = 546219This result can be used to solve for f s1:546219 = 2tan f 21s S Ω411.1207.6tan211s ==Ω-Ωs1 = 2.822 = S1s f f 2πso 748.192f 822.2f S1s =π= kHz. The transition width is 19748 – 13000 = 6748 Hz.(b) With an order n = 6,⎪⎭⎫⎝⎛ω⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ==117589log 21)01.0(1log log 211log 6n 1s 21p 1s 2s ⎪⎭⎫ ⎝⎛ω117589log 1s = 0.333ωs1 = 253143This result can be used to solve for f s1:253143 = 2tanf 21s S Ω667.01s 10117589=ω207.62tan1s =Ω333.01s 10117589=ω877.22tan1s =Ω236.1877.2tan211s ==Ω-Ωs1 = 2.472 = S1s f f 2πso 311.172f 472.2f S1s =π=kHz. The transition width is 17311 – 13000 = 4311 Hz.10.11 The transfer function for a second order low pass analog Butterworth filter is:21p 1p 221p s 2s )s (H ω+ω+ω=The cut-off frequency givesΩp1 = π=π=π625.0800025002f f 2S1p radiansωp1 = 239462tanf 21p S =Ω rad/secfor an analog transfer function573410916s 33865s 573410916)s (H 2++=The bilinear transformation1z 1z 16000s +-=converts this transfer function to a digital transfer function:5734109161z 1z 16000338651z 1z 16000573410916)z (H 2+⎪⎭⎫ ⎝⎛+-+⎪⎭⎫ ⎝⎛+-=2121z20971.0z46295.01z41817.0z83634.041817.0----++++=10.12 (a) An FIR design for this filter requires a Hanning window with N =3.32f S /T.W.= 3.32(8000)/(1500–1000) = 53.1 or 53 terms. This filter requires 53 filter coefficients. (b) The stop band attenuation gives 20log δs = –44, or δs = 0.00631.Ωp1 = π=π=π25.0800010002f f 2S1p radiansΩs1 =π=π=π375.0800015002f f 2S1s radiansωp1 = 66272tanf 21p S =Ω rad/secωs1 = 106912tanf 21s S =Ω rad/secThe order for the filter can be calculated usingso an order of 11 should be chosen. For this order, 2(11) + 1 = 23 filter coefficients are required, many fewer than the FIR design.10.13 The gain at the edge of the pass band for this Chebyshev filter is –2 dB, which means 20log(1–δp ) = –2, or δp = 0.2057. The pass band edge is located at 2 kHz. The stop band edge may be located at any convenient point, perhaps at 3 kHz, where the gain is about –46 dB. Since 20log δs = –46, or δs = 0.005.()()7649.012057.01111122p=--=-δ-=ε()0.2001005.011122s=-=-δ=δ6.10662710691log 21)00631.0(1log log 211log n 21p 1s 2s =⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ≥The digital frequencies at the edges of the pass and stop bands are:Ωp1 = π=π=π4.01000020002f f 2S1p radiansΩs1 =π=π=π6.01000030002f f 2S1s radiansωp1 = 9.145302tanf 21p S =Ω rad/secωs1 = 6.275272tanf 21s S =Ω rad/secThe order for the filter should be99.49.145306.27527cosh 7649.00.200coshcoshcosh n 111p 1s 11=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛ωω⎪⎭⎫ ⎝⎛εδ≥---- or 510.14 (a) The Chebyshev filter has a gain at the edge of the pass band of –0.5 dB, which means 20log(1–δp ) = –0.5, or δp = 0.0559. The pass band edge is located at 10 kHz. The stop band edge is located at 12 kHz, where the gain is about –20 dB. Since 20log δs = –20, or δs = 0.1.()()3492.010559.01111122p=--=-δ-=ε()95.911.011122s=-=-δ=δThe digital frequencies at the edges of the pass and stop bands are:Ωp1 = π=π=π625.032000100002f f 2S1p radiansΩs1 =π=π=π75.032000120002f f 2S1s radiansωp1 = 8.957822tanf 21p S =Ω rad/secωs1 = 7.1545092tanf 21s S =Ω rad/secThe order for the filter should be83.38.957827.154509cosh 3492.095.9coshcoshcosh n 111p 1s 11=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫⎝⎛ωω⎪⎭⎫ ⎝⎛εδ≥---- or 4(b) The magnitude response for a 4th order analog Chebyshev Type I filter is given by()⎪⎭⎫⎝⎛ω+=⎪⎪⎭⎫⎝⎛ωωε+=ω8.95782C 3492.011C 11)(H 2421p 2n2The filter shape for the digital filter is given by⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛Ω+=⎪⎪⎪⎪⎭⎫⎝⎛ω⎪⎭⎫ ⎝⎛Ωε+=Ω8.957822tan 64000C 1219.0112tan f 2C 11)(H 241p S 2n2⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛Ω+=2tan 6682.0C 1219.01124where⎩⎨⎧=--))x (cosh 4cosh())x (cos 4cos()x (C 114 1 |x |1|x |>≤This function can be computed for various values of Ω.The results are shown below, where digital frequencies have been converted to frequencies in Hz using Sf f 2π=Ω. The pass band edge occurs at 10 kHz with a gain of –0.5 dB, as expected. The stop gain gain of –20 dB is achieved slightly before 12 kHz because the order was rounded up to 4.10.15 For both filters, the digital frequencies at the edges of the pass and stop bands are:Ωp1 = π=π=π64.01500048002f f 2S1p radiansΩs1 =π=π=π72.01500054002f f 2S1s radians|H(f)|fωp1 = 472722tanf 21p S =Ω rad/secωs1 = 637532tanf 21s S =Ω rad/sec(a) Note that the pass band ripple for the filter is chosen so that a Butterworth design is possible. The order for the Butterworth filter isso an order of 8 should be chosen. (b) For the Chebyshev version:()()9975.01292.01111122p=--=-δ-=ε()46.12108.011122s=-=-δ=δ96.34727263753cosh 9975.046.12coshcoshcosh n 111p 1s 11=⎪⎭⎫ ⎝⎛⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛ωω⎪⎭⎫ ⎝⎛εδ≥----so an order of 4 should be chosen. Note that the order required for the Chebyshev filter is lower than that required for the Butterworth filter.10.16 The Chebyshev filter has a pass band ripple of –0.5 dB, which means 20log(1–δp ) = –0.5, or δp = 0.0559. Since the center frequency is 5 kHz and the width of the pass band is 1.6 kHz, the pass band edges are located at 4.2 kHz and 5.8 kHz. Because of the 400 Hz transition width, the stop band edges are located at 3.8 kHz, and 6.2 kHz, where the gain is –35 dB. Since 20log δs = –35, or δs = 0.01778.()()3492.010559.01111122p=--=-δ-=ε4.84727263753log 21)08.0(1log log 211log n 21p 1s 2s =⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛ωω⎪⎪⎭⎫ ⎝⎛-δ≥()23.56101778.011122s=-=-δ=δThe low pass prototype for this band pass filter has its pass band edge at (5.8 – 5) = 0.8 kHz. The transition width is 400 Hz, so the stop band is located at 1.2 kHz. Thus, the digital frequencies at the edges of the pass and stop bands are:Ωp1 = π=π=π107.0150008002f f 2S1p radiansΩs1 =π=π=π16.01500012002f f 2S1s radiansωp1 = 3.50902tanf 21p S =Ω rad/secωs1 = 7.77022tanf 21s S =Ω rad/secThe order for the filter should be9.53.50907.7702cosh 3492.023.56coshcoshcosh n 111p 1s 11=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛ωω⎪⎭⎫ ⎝⎛εδ≥---- or 610.17 The high pass filter may be obtained from an arbitrary first order low pass filter such as5.2360s 5.2360s )s (H pp L +=ω+ω=This low pass filter with cut-off ωp = 2360.5 rad/sec may be converted to a high passfilter with cut-off 'p ω using the conversion⎪⎪⎭⎫⎝⎛ωω=s H )s (H 'pp L HThe high pass cut-off of 3 kHz gives:'p Ω = π=π=π75.0800030002f f 2Sl radians'p ω= 4.386272tanf 21p S =Ω rad/secThe conversion formula becomes⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫⎝⎛ωω=s 7.91179977H s H )s (H L 'pp L HThe transfer function for the analog filter is4.38627s s5.2360s 7.911799775.2360)s (H H +=+⎪⎭⎫⎝⎛=The bilinear transformation 1z 1z 160001z 1z f 2S +-=+- produces the digital transfer function: 11z 4142.01)z 1(2929.04.386271z 1z 160001z 1z 16000)z (H --+-=+⎪⎭⎫ ⎝⎛+-⎪⎭⎫ ⎝⎛+-=10.18 The lower cut-off frequency is Ωl = =π8000100020.25π rads , or ωl =4.6627225.0tan f 2S =π rad/sec. The upper cut-off frequency will be Ωu ==π8000150020.375π rads, or ωu = 9.106902375.0tanf 2S =π rad/sec after pre-warping. Theband pass filter may be obtained from an arbitrary first order low pass filter such as5.2360s 5.2360s )s (H pp L +=ω+ω=The transfer function of the band pass filter may be obtained by transforming that of the low pass filter:()()⎪⎪⎭⎫⎝⎛-+=⎪⎪⎭⎫ ⎝⎛ω-ωωω+ω=4.66279.10690s )9.10690)(4.6627(s 5.2360H s s H )s (H 2L l u u l 2pL BP⎪⎪⎭⎫⎝⎛+=s 5.4063)9.10690)(4.6627(s 5.2360H 2L 7.70852870s 5.4063s s5.40635.2360s5.4063)9.10690)(4.6627(s 5.23605.236022++=++=The bilinear transformation gives the digital transfer function:7.708528701z 1z 160005.40631z 1z 160001z 1z 160005.4063)z (H 2BP +⎪⎭⎫ ⎝⎛+-+⎪⎭⎫ ⎝⎛+-⎪⎭⎫ ⎝⎛+-=()212z6682.0z9449.01z 11659.0---+--=10.19 The lower pre-warped cut-off frequency is Ωl = =π20005520.055π rads , or ωl =4.3462055.0tan f 2S =π rad/sec. The upper pre-warped cut-off frequency is Ωu ==π20006520.065π rads , or ωu = 8.4092065.0tanf 2S =π rad/sec. A band stop filter maybe obtained from an arbitrary low pass filter, such as,5.2360s 5.2360s )s (H pp L +=ω+ω=Transforming the low pass transfer function into a band stop transfer function,()()()()⎪⎪⎭⎫⎝⎛+-=⎪⎪⎭⎫ ⎝⎛ωω+ω-ωω=8.4094.346s 4.3468.409s 5.2360H s s H )s (H 2L u l 2l up L BS ()()⎪⎪⎭⎫ ⎝⎛+=8.4094.346s s4.635.2360H 2L ()()7.141954s 4.63s 7.141954s 5.23608.4094.346s s4.635.23605.2360222+++=++=The transfer function for the digital filter is found using the bilinear transformation:7.1419541z 1z 40004.631z 1z 40007.1419541z 1z 4000)z (H 22BS +⎪⎭⎫ ⎝⎛+-+⎪⎭⎫ ⎝⎛+-+⎪⎭⎫ ⎝⎛+-=2121z9691.0z9344.11z9845.0z 9344.19845.0----+-+-=The frequency response of the filter isΩ-Ω-Ω-Ω-+-+-=Ω2j j 2j j BS e9691.0e9344.11e9845.0e 9344.19845.0)(HIt may be used to find the magnitude response for the filter, plotted below against frequency in Hz.10.20 The low pass prototype for the filter has a cut-off of 30 Hz, equal to the bandwidth of the filter. The transfer function for a second order low pass analog Butterworth filter is:21p 1p 221p s 2s )s (H ω+ω+ω=A low pass filter with a cut-off frequency of 30 Hz would giveΩp1 = π=π=π3.0200302f f 2S1p radians|Hωp1 = 8.2032tanf 21p S =Ω rad/secThis gives a low pass analog transfer function41209s 2.288s 41209)s (H 2++=This low pass filter with cut-off ωp = ωp1 = 203.8 rad/sec may be converted to a high pass filter with cut-off 'p ω using the conversion⎪⎪⎭⎫⎝⎛ωω=s H )s (H 'pp L HThe high pass cut-off, f S /2 – 30 = 100 – 30 = 70 Hz gives:'p Ω = π=π=π7.0200702f f 2Sl radians'p ω= 0.7852tanf 21p S =Ω rad/secThe conversion formula becomes⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫⎝⎛ωω=s 159983H s H )s (H L 'pp L HThe analog transfer function for the band pass filter is41209s 1599832.288s 159********)s (H 2+⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=5.621091s 9.1118s s22++=The bilinear transformation1z 1z 4001z 1z f 2s S+-=+-=converts this transfer function to a digital transfer function:5.6210911z 1z 4009.11181z 1z 4001z 1z 400)z (H 22+⎪⎭⎫ ⎝⎛+-+⎪⎭⎫ ⎝⎛+-⎪⎭⎫ ⎝⎛+-=2121z2715.0z7506.01)zz 21(1302.0----+++-=10.21 (a)For a 1 kHz tone at a sampling rate of 4 kHz, the digital frequency isπ=π=π=Ω5.0400010002f f 2S. The z transform for a sine wave, and therefore also thetransfer function for a sine wave generator, is 2122z1z1z z 1cos z 2z sin z )z (H --+=+=+Ω-Ω=.(b)The frequency response for the filter is Ω-Ω-+=Ω2j j e1e)(H . The magnitude response,or filter shape, is plotted below.10.22 (a) Using t = nT, the sampled version of the impulse response is h[n] = nT 2e -With β = e –2T , the third row of Table 6.1 gives the transfer functionH(z) =1T2T2ze11ez z ----=-The filter shape can be obtained from the frequency response H(Ω) =Ω---j T2ee11(i) Ω---=Ωj 2ee11)(H (ii) Ω---=Ωj 1ee 11)(H(b)Analog frequencies in Hz are converted to analog frequencies in rad/sec throughω = 2πf, and to digital frequencies in rads through fT2f f 2Sπ=π=Ω. The figure belowshows the results. Note that the Nyquist frequency for T = 1 is 0.5 Hz, and the Nyquist frequency for T = 0.5 is 1 Hz. This explains the repetitive shapes in the figure. (c) Within Nyquist limits, the faster the sampling rate, the better the approximation to the analog filter, but even T = 0.5 is not sufficient. To follow the analog filter shape for the range shown in the figure, a sampling rate of more than 8 samples per second is needed, that is T < 1/8.10.23 The first order Butterworth transfer function H(s) for a 4 kHz cut-off frequency isπ+π=π+π=ω+ω=8000s 8000)4000(2s )4000(2s )s (H 1p 1pThe filter shape for this analog filter is180001)(H 2+⎪⎭⎫ ⎝⎛πω=ωThis shape is to be duplicated by a digital filter designed with the impulse invariancemethod. The transfer function of the digital filter is111p ze18000ze1)z (H T8000T1p --π-ω--π=-ω=The frequency response of the digital filter isΩ-π-Ω-ω--π=-ω=Ωj T8000j T1p ee18000ee1)(H 1pWith T = 1/f S = 1/32000, the digital filter shape becomesΩ-π-Ω-ω--π=-ω=Ωj 25.0j T1p ee18000ee1)(H 1pUsing the tricks described in the solution to question 10.22, the analog and digital filter shapes can be compared against frequency in Hz, as shown below.10.24 The digital impulse response h[n] obtained by sampling is:h[n] =()T n 300sin e3001nT200-u[n]From Table 6.1, the z transform of ]n [u )n sin(n Ωβ is22cos z 2z sin z β+Ωβ-Ωβ. WithT200e-=β and T 300=Ω, the transfer function for the digital filter becomesH(z) =()()T400T2002T200eT 300cos z e2z T 300sin z e3001---+-With a sampling interval of 2 msec, this transfer function becomesH(z) =()()2118.04.024.0z4493.0z1065.11z00126.0e6.0cos z e2z 6.0sin z e3001------+-=+-The filter shape |H(Ω)| may be found from the frequency responseΩ-Ω-Ω-+-=Ω2j j j e4493.0e1065.11e00126.0)(HFrequency (Hz)The filter shape is shown below. The same trick as was used in the last two solutions is used here to plot the analog and digital filter shapes on the same graph. The 2 msec sampling interval corresponds to a 500 Hz sampling rate, so the Nyquist limit for the digital filter is evident at 250 Hz. Note that the digital filter shape could also be obtained by taking the DTFT of some large number of samples of h[n].10.25 (a) Quantized to 4 bits, the transfer function becomes321321z0625.0z6875.0z0625.11z375.0z0625.1z0625.1375.0)z(H------++++++=(b) Quantized to 5 bits, the transfer function becomes321321z09375.0z6875.0z0625.11z34375.0z0625.1z0625.134375.0)z(H------++++++=The filter shape for the original filter and two quantized copies of it are shown below.。
数字信号处理习题集大题与答案
1设序列x(n)={4，3，2，1} ， 另一序列h(n) ={1，1，1，1}，n=0,1,2,3 （1）试求线性卷积 y(n)=x(n)*h(n) （2）试求6点圆周卷积。
（3）试求8点圆周卷积。
解：1．y(n)=x(n)*h(n)={4,7,9,10,6,3,1}2．6点圆周卷积={5,7,9,10,6,3}3．8点圆周卷积={4,7,9,10,6,3,1,0}2二．数字序列 x(n)如图所示. 画出下列每个序列时域序列： (1) x(n-2); (2)x(3-n); (3)x[((n-1))6],(0≤n ≤5); (4)x[((-n-1))6],(0≤n ≤5);n12340.543210-1-2-3x(3-n)x[((n-1))6]n54321043210.5n12340.5543210x[((-n-1))6]3．已知一稳定的LTI 系统的H(z)为)21)(5.01()1(2)(111------=z z z z H试确定该系统H(z)的收敛域和脉冲响应h[n]。
解:0.52ReIm系统有两个极点，其收敛域可能有三种形式，|z|<0.5, 0.5<|z|<2, |z|>2 因为稳定，收敛域应包含单位圆，则系统收敛域为：0.5<|z|<211111213/25.013/4)21)(5.01()1(2)(--------=---=z z z z z z H )1(232)()5.0(34)(--+=n u n u n h n n4．设x(n)是一个10点的有限序列x （n ）={ 2,3,1,4,-3,-1,1,1,0,6}，不计算DFT ，试确定下列表达式的值。
(1) X(0), (2) X(5), (3)∑=9)(k k X,（4）∑=-95/2)(k k j k X eπ解：（1） （2）（3）（4）5． x(n)和h(n)是如下给定的有限序列 x(n)={5, 2, 4, -1, 2}， h(n)={-3, 2, -1 }(1) 计算x(n)和h(n)的线性卷积y(n)= x(n)* h(n)； (2) 计算x(n)和h(n)的6 点循环卷积y 1(n)= x(n)⑥h (n)； (3) 计算x(n)和h(n)的8 点循环卷积y 2(n)= x(n)⑧h (n)； 比较以上结果，有何结论？14][]0[19===∑=n N n x X W 12][][]5[119180510-=-===⎩⎨⎧-=∑∑====奇偶奇数偶数n n n n n n x n x X n n W20]0[*10][][101]0[99===∑∑==x k X k X x k k 0]8[*10][][101]))210[((][]))[((2)10/2(92)10/2(9010)/2(===-⇔--=-=-∑∑x k X ek X ex k X e m n x k j k k j k m N k j N πππ解：（1）5 2 4 -1 2-3 2 15 2 4 -1 210 4 8 -2 4-15 -6 -12 3 -6-15 4 -3 13 -4 3 2y(n)= x(n)* h(n)={-15,4,-3,13,-4,3,2}(2)5 2 4 -1 2-3 2 15 2 4 -1 210 4 8 -2 4-15 -6 -12 3 -6-15 4 -3 13 -4 3 22-13 4 -3 13 -4 3 2y1(n)= x(n)⑥h(n)= {-13,4,-3,13,-4,3}(3)因为8>(5+3-1),所以y3(n)= x(n)⑧h(n)＝{-15,4,-3,13,-4,3,2，0}y3(n)与y(n)非零部分相同。
《数字信号处理》复习题及答案
《数字信号处理》复习题一、单项选择题（在每小题的四个备选答案中，选出一个正确答案，并将正确答案的序号填在题干的括号内。
每小题2分）1.在对连续信号均匀采样时，若采样角频率为Ωs，信号最高截止频率为Ωc，则折叠频率为( D)。
A. ΩsB. ΩcC. Ωc/2D. Ωs/22. 若一线性移不变系统当输入为x(n)=δ(n)时输出为y(n)=R3(n)，则当输入为u(n)-u(n-2)时输出为( C)。
A. R3(n)B. R2(n)C. R3(n)+R3(n-1)D. R2(n)+R2(n-1)3. 一个线性移不变系统稳定的充分必要条件是其系统函数的收敛域包含( A)。
A. 单位圆B. 原点C. 实轴D. 虚轴4. 已知x(n)=δ(n)，N点的DFT［x(n)］=X(k)，则X(5)=( B)。
A. NB. 1C. 0D. - N5. 如图所示的运算流图符号是( D)基2 FFT算法的蝶形运算流图符号。
A. 按频率抽取B. 按时间抽取C. 两者都是D. 两者都不是6. 直接计算N点DFT所需的复数乘法次数与( B)成正比。
A. NB. N2C. N3D. Nlog2N7. 下列各种滤波器的结构中哪种不是I I R滤波器的基本结构( D)。
A. 直接型B. 级联型C. 并联型D. 频率抽样型8. 以下对双线性变换的描述中正确的是( B)。
A. 双线性变换是一种线性变换B. 双线性变换可以用来进行数字频率与模拟频率间的变换C. 双线性变换是一种分段线性变换D. 以上说法都不对9. 已知序列Z变换的收敛域为｜z｜>1，则该序列为( B)。
A. 有限长序列B. 右边序列C. 左边序列D. 双边序列10. 序列x(n)=R5(n)，其8点DFT记为X(k)，k=0,1,…,7，则X(0)为( D)。
A. 2B. 3C. 4D. 511. 下列关于FFT的说法中错误的是( A)。
A. FFT是一种新的变换B. FFT是DFT的快速算法C. FFT基本上可以分成时间抽取法和频率抽取法两类D. 基2 FFT要求序列的点数为2L(其中L为整数)12. 下列结构中不属于FIR滤波器基本结构的是( C)。