2008数字信号处理期末考试A卷答案

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电子科技大学二零零 八 至二零零 九 学年第 一 学期期 末 考试

数字信号处理课程考试题 半期考卷 （ 120 分钟） 考试形式： 开卷 考试日期 2008年12月 30 日

课程成绩构成：平时 20 分， 期中 20 分， 实验 20 分， 期末 40 分

1. Given a causal system with the transfer function being 1

12

3()252z H z z z

----=-+. (1) Try to calculate it’s differen ce equation.

(2) Try to give the zero-pole plot of the system, then determine whether the system is stable. (3) Calculate it’s impulse response.

(4) Give a canonic direct realization of the system and plot it. Solution: （共20分）

(1) Difference equation is: 2y[n]-5y[n-1]+2y[n-2]=-3x[n-1]（3分） (2) zero-pole plot:（5分）

The ROC is |z|>2, so the system is not stable.（2分）

(3) 111

1221

1

33120.52()25225221210.5

12z z z z H z z z z z z z z z -------

--===-=--+-+----

11[]0.5(0.5)[]2(2)[]{(0.5)(2)}[]n n n n h n n n n μμμ--∴=-=- （5分）

(4)

（5分）

2.Consider a length-9 sequence x [n ] = {2, -3, -1, 0, -4, 3, 1, 2, 4}, -2 ≤ n ≤ 6. The z -transform X (z ) of x [n ] is sampled at N points ωk = 2 k /N, 0 ≤ k ≤ N -1, on the unit circle yielding the frequency samples:

2/[]()|,0,1,

,1j k N z e X k X z k N π===-

Determine the periodic sequence []x n whose discrete discrete Fourier series coefficients are given by

[]X k when N = 4 and N = 11, respectively ( without evaluating []X k ). Solution:（共15分）

According to the definition of z-transform,

6

2

()[]n

n X z x n z

-=-=

∑ (0.1)（5分）

Then, []X k can be written as

26

2

[][],0,1,,1kn j

N

n X k x n e

k N π-=-=

=-∑ (0.2)（5分）

According to the definition of DFS

21

[][],,0,1,,1kn N j

N

n X k x n e

n k N π--===-∑ (0.3)（5分）

when N = 4, [][4]x n x n r =+, and its principle period is {0, 2, 2, 0}, 0 ≤ n ≤ 3

when N = 11, [][11]x n x n r =+, and its principle period is {-1, 0, -4, 3, 1, 2, 4, 0, 0, 2, -3}, 0 ≤ n ≤ 10

3．Knowing the transfer function of a causal system as 1

11()10.9z H z z --+=-, we get a new transfer

function 41()()H z H z =. Please:

(1) Sketch the magnitude response of 1()j H e ω

.

(2) Determine the ω values of peaks and dips in magnitude of 1()j H e ω

.

(3) If the transfer function H(z) is cascaded with a system 11

()(1)(10.9)G z z z --=+- to get a new system

transfer function H 2(z), try to compute the phase function and the group delay of 2()H z .

Solution:（共20分）

(1) 4141()10.9z H z z --+=-→4141()10.9j j j e H e e ωω

ω--+=-→4141()10.9j j j e H e e ωωω

--+=-

magnitude response:

（共9分）

(2) ω values of peaks in magnitude of 1()j H e ω:-π/2,0,π/2, π, （2分）

ω values of dips in magnitude of 1()j H e ω:,－3π/4, －π/4，π/4, 3π/4（2分） （3）12122()()

()(1)12H z H z G z z z z ---==+=++