期末考试试卷答案2011

华南农业大学期末考试试卷（A 卷）

2011学年第1学期 考试科目： 数字信号处理

考试类型：（闭卷）考试 考试时间： 120 分钟 学号 姓名 年级专业

一、判断题（本大题共 10 小题，每小题 2 分，共 20 分）

1．任何序列都可以用单位脉冲序列的移位加权和来表示。 （√ ） 2．序列3()2cos(

7)4

x n n π=+是周期序列。 （√ ）

3．序列y (n )= 2x (n )-3，不是移不变系统。 （× ） 4．一个域的离散就必然导致另一个域的连续。 （× ）

5．因果序列的收敛域不包含+∞。 （× ） 6．FIR 滤波器设计可利用模拟滤波器设计的结果。 （× ）

7．最小相位延时系统的零点和极点都在单位圆之内。 （√ ） 8．若滤波器通带内群延时响应特性是一个常数，则为线性相位系统。 （√ ） 9．窗函数的选择原则是在保证阻带衰减的情况下选择主瓣窄的窗函数。 （√ ） 10．全通系统的特点是零极点以单位圆镜像对称。 （√ ）

二、填空题（本大题共 5小题，每小题 4 分，共 20 分） 1．序列()()x n u n =的Z 变换为1

11

1z

z z

-=

--，)3(-n x 的Z 变换是2

31

1

1z z

z z

---=

--

2．线性移不变系统是稳定系统的充要条件是

∞

<=∑

∞

-∞

=P n h n )(。

3．实现FIR 线性相位滤波器的条件是 h(n)= (1)h N

n ±--

4．序列x(n)和h(n)，长度分别为N 和M （N>M ），二者线性卷积的长度为 N+M-1

循环卷积与线性卷积的关系是 。 5．实现一个数字滤波器需要的三种基本运算单元是 加法器，乘法器，延迟器 。

三、简答题（本大题共 6小题，每小题 5分，共 30 分）

1．简述 Z 变换、傅里叶变换、离散傅里叶变换三者之间的关系。

答：傅氏变换是拉氏变换在虚轴S=j Ω的特例,因而映射到Z 平面上为单位圆，即序列在单位圆上的Z 变换为序列的傅氏变换；x(n)的N 点离散傅里叶变换是x(n)的z 变换在单位圆上的N 点等间隔抽样；也是是傅里叶变换在区间[0，2π]上的N 点等间隔抽样。

2. 简述按频率抽选法和按时间抽选法两种FFT 算法的异同？（包括输入输出顺序、基本碟

形、计算量、节点间距离、r W N 因子确定等异同）

答：

3. 设某FIR 数字滤波器的系统函数为： 请画出此滤波器的线性相位结构。

()()

1

2

3

4

1

13535

H z z z

z

z

----=

++++()()N c l N y n y n 点圆周卷积是线性卷积以为周期的周期延拓序列的主值序列。

解：对系统函数求z 反变换，得

得，

该滤波器单位脉冲响应h(n)偶对称，N 为奇数。得线性相位结构如图：

4．简述全通系统的特点以及应用。

答：全通系统的特点是：零点与极点以单位圆为镜像对称。其应用主要有三方面： 1）任一因果稳定系统H (z )都可以表示成全通系统 Hap (z )和最小相位系统H min(z )的级联 2）级联一个全通系统可以使非稳定滤波器变成一个稳定滤波器

3）作为相位均衡器，校正系统的非线性相位，而不改变系统的幅度特性

5. 简述IIR 以及FIR 数字滤波器的区别。（提示：滤波器特点、设计方法及应用的区别）

答：1）从性能上来说，IIR 滤波器传输函数的极点可位于单位圆内的任何地方，因此可用较低的阶数获得高的选择性，所用的存贮单元少，所以经济而效率高。但是这个高效率是以相位的非线性为代价的。选择性越好，则相位非线性越严重。相反，FIR 滤波器却可以得到严格的线性相位，然而由于FIR 滤波器传输函数的极点固定在原点，所以只能用较高的阶数达到高的选择性；对于同样的滤波器设计指标，FIR 滤波器所要求的阶数可以比IIR 滤波器高5~10倍，结果，成本较高，信号延时也较大；如果按相同的选择性和相同的线性要求来说，则IIR 滤波器就必须加全通网络进行相位较正，同样要大增加滤波器的阶数和复杂性。

2）从机构上看，IIR 滤波器必须采用递归结构，极点位置必须在单位圆内，否则系统将不稳定。相

()()()()()()

133112345

5

5

5

h n n n n n n δδδδδ=

+

-+-+

-+

-

反，FIR滤波器主要采用非递归结构，不论在理论上还是在实际的有限精度运算中都不存在稳定性问题，运算误差也较小。此外，FIR滤波器可以采用快速付里叶变换算法，在相同阶数的条件下，运算速度可以快得多。

3）从设计工具上看，IIR滤波器可以借助于模拟滤波器的成果，因此一般都有有效的封闭形式的设计公式可供准确计算，计算工作量比较小，对计算工具的要求不高。FIR滤波器设计则一般没有封闭形式的设计公式。窗口法虽然仅仅对窗口函数可以给出计算公式，但计算通带阻带衰减等仍无显式表达式。一般，FIR滤波器的设计只有计算程序可循，因此对计算工具要求较高。

另外，IIR滤波器虽然设计简单，但主要是用于设计具有片段常数特性的滤波器，如低通、高通、带通及带阻等，往往脱离不了模拟滤波器的格局。而FIR滤波器则要灵活得多，尤其它能易于适应某些特殊的应用，如构成微分器或积分器，或用于Butterworth、Chebyshev等逼近不可能达到预定指标的情况，例如，由于某些原因要求三角形振幅响应或一些更复杂的幅频响应，因而有更大的适应性和更广阔的天地。

6.简述FIR窗函数设计法中，窗函数的选择依据。并分析矩形窗、三角形窗（Bartlett窗）、

汉宁窗、海明窗以及布拉克曼窗的特点。

答：窗函数的选择依据是：1）希望窗谱的主瓣尽量地窄，以获得较陡的过渡带；2）尽量减少窗谱最大旁瓣的相对幅度，也就是使频域的能量能主要集中在主瓣内。这样使肩峰和波纹减少，就可增大阻带的衰减。

相同的阶数下，矩形窗的窗谱主瓣宽度最窄，为N

/

4π,旁瓣幅度最大；三角形窗的窗谱主瓣宽度是矩形窗的两倍，为N

/

8π，旁瓣幅度较矩形窗小；汉宁窗的窗谱主瓣宽度是矩形窗的两倍，为N

/

8π，旁瓣幅度较三角形窗小；海明窗是改进的汉宁窗，其窗谱主瓣宽度也是矩形窗的两倍，为N

/

8π，旁瓣幅度较汉宁窗小；布拉克曼窗的窗谱主瓣宽度是矩形窗的三倍，为N

/

12π，旁瓣幅度最小。因此。用矩形窗设计FIR滤波器时，过渡带最窄，而阻带衰减最小，布莱克曼窗过渡带最宽，但阻带衰减加大。

四、分析计算题（本大题共3小题，每小题10 分，共30 分）

1．设一因果系统的传递函数为

1

12

10.5

()

10.70.12

z

H z

z z

-

--

-

=

-+

，

（1）系统是否稳定？为什么？（2）写出差分方程。

（3）画出系统的极零点分布图。

解：（1）由H(z)的表达式可知，系统有两个极点为：z1=0.3，z2=0.4，所以极点都在单位圆内，因此系统稳定。 （2）1

1

2

10.5()10.70.12z H z z

z

----=

-+=

1

1

2110.310.4z

z

---

--

由于该系统是一个因果，所以，()(2(0.3)(0.4))()n n h n u n =?-

（3）由1

1

2

10.5()10.70.12z H z z

z

----=

-+，可知b0=1；b1=--0.5；a0=1；a1=0.7；a2= -0.12

则差分方程可表示为：

()()0.7(1)0.12(2)0.5(1)y n x n x n x n y n =+---+-

（4）零极点分布如图所示：

2. 下图表示一个5点序列()x n

（1）试画出线性卷积()()x n x n * （2）试画出5点的圆周卷积()x n ⑤()x n ；

解：

()()x n x n *

3．要求用双线性变换法从三阶巴特沃思模拟滤波器导出一低通数字滤波器，已知3dB 截止频率为5Hz ，系统抽样频率为1kHz ，设T=2s 。

表 巴特沃思滤波器分母多项式1

2

2

12211N

N N N N s a s

a s

a s a s ----+++???+++的系数

解：由题意可得3dB 截止频率为5Hz ，则ππ1025=?=Ωc 。去归一化，得

3

2

2

3

3

3

3

2

23

2

2020010001000100050511

)

10(

)10(

2102

11

)(

)(s

s s s

s

s s s s s H s H c

an a +?+?+=

+

+

+

=+++=

Ω=ππππ

π

π

ππ

π

π

将 ()a H s 变换成Butterworth 数字滤波器：

=

+?+?+=

=----+-?

=

+-?

=

1

1

1

1

1123

2

2

3

3

1122020010001000)

()(z z T s z z T s a s

s s s H

z

H π

πππ

3

1

12

1

11

12

3

3

)

11(

)11(

20)11(

20010001000------+-++-++-+z

z z

z z

z ππππ

⑤

()x n ()x n L 点圆周卷积是线性卷积以L 为周期的周期延拓序列的主值区间

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