数字信号处理翻译

姓名班级学号成绩

考试内容：2010年第一学期数字信号处理

第一部分解答题

1.Please give the stages of digital processing of analog signals and the basic

components of DSP system.

2. Please describe Sampling Theorem Compute the z-transform of the following

sequences x(n)

x(n) = (-0.5)n u(n)

3. Try to test the linearity and time invariance of the discrete time systems defined

as follows:

x

=n

n

n

y

x

-

(

)1

(-

)

(

)

4. Given a causal IIR discrete-time system described by the difference equation

y[n]-0.4y[n-1]=x[n].

And it is known that the input sequence is x[n]= x[n]=(0.3)nμ[n]. .

(1)Determine the output sequence y[n] using the z-transform.

(2)Determine the expression of the frequency response H(e jω)in the form

|H(e jω)|e jϕ(ω)

第二部分文献翻译

参考文献：the scientist and engineer's guide to digital signal processing

具体内容：第123 页第二段---第127 页

原文：

Part 1. 解答下列习题

1. Please give the stages of digital processing of analog signals and the basic

components of DSP system. 答：

(1)信号的数字化需要进过采样、保持、量化和编码四个过程

(2)数字信号系统的基本组成为：前置预滤波器、A/D 变换器、数字信号处理器、D/A 变换器、模拟滤波器。

2. Please describe Sampling Theorem Compute the z-transform of the following

sequences x (n )，x (n ) = (-0.5)n u(n ) 解：(1)

采样定理：在进行模拟/数字信号的转换过程中，当采样频率大于信号中最高频率的2倍时，采样之后的数字信号完整地保留了原始信号中的信息，一般实际应用中保证采样频率为信号最高频率的5～10倍；采样定理又称奈奎斯特定理。

(2)

由题有：

()()()101

0.51

0.510.5n

n n

n X z x n z z z z

∞

-=-∞

∞

-=-=

⎡⎤=-⎣⎦=

>-+∑

∑

3. Try to test the linearity and time invariance of the discrete time systems defined

as follows:)1()()(--=n x n x n y 解：

验证线性：

()()()

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

12012121211x n ax n bx n y n ax n bx n ax n bx n ay n by n →+=+----=+

验证时不变性：

()()

()()()

()

0 n m 1n n m x n x n m y n x n m x n m y n m +→+∈→+=+-+-=+ 、

综上，此系统为线性时不变系统。

4. Given a causal IIR discrete-time system described by the difference equation y[n]-0.4y[n-1]=x[n].

And it is known that the input sequence is x[n]= x[n]=(0.3)n μ[n]. . (1)Determine the output sequence y[n] using the z-transform.

(2)Determine the expression of the frequency response H(e j ω) in the form |H(e j ω)|e j ϕ(ω)

解：

(1)

先对元方程做z 变换，有：

()()()10.4Y z z Y z X z --=

其中：

()1

1

0.310.3X z z z

-=

>- 我们得到：

()11

11

0.410.310.4Y z z z z --=

>--

对其做逆z 变换，即得所求：

()()()()()

()1

0.30.43140.4 n 0

314

n n

n n y n n n μμ+=*⎛⎫- ⎪⎝⎭=≥-

(2)

由上可知：

()1

1

0.410.4H z z z -=

>-

代换变量后有：