2008数字信号处理期末考试A卷答案

学院_______________________ 系别____________ 班次_____________ 学号__________ 姓名________________………….……密…..……….封……..……线………..…以………..…内………....答…………...题…………..无…….….效…..………………..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 1123()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) 1111221133120.52()25225221210.512z 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,62()[]nn X z x n z-=-=∑ (0.1)（5分）Then, []X k can be written as262[][],0,1,,1kn jNn X k x n ek N π-=-==-∑ (0.2)（5分）According to the definition of DFS21[][],,0,1,,1kn N jNn X k x n en 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 ≤ 3when 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 ≤ 103．Knowing the transfer function of a causal system as 111()10.9z H z z --+=-, we get a new transferfunction 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 systemtransfer 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 ωωω--+=-→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 ---==+=++22()122(cos 1)j j j j H e e e e ωωωωω---∴=++=+The phase function ()θωω=-，The group delay ()1g d d θωτω=-=（共7分）ing windowed Fourier series method, design a causal low-pass FIR digital filter with the followingspecifications:Pass band edges f p : 6kHz Stop band edges f s : 8kHz Pass band ripple αp : 1dB Stop band attenuation αs : 40dBSampling frequency FT: 20kHzSolution:（共15分）(1) Perform some specifications transform.22347,,,555210p p s s p s s p c TT f f F F πωωπππππωωωωωω+====∆=-===（每个1分，共4分）(2) The ideal impulse response issin [],c LP nh n n nωπ=-∞<<∞（1分） (3) According the requirement of stop band attenuation, the Hann （or Hamming,Blackman ） window should be chose. The transition bandwidth of hann window is 3.11π/M, i.e.,3.1115.555M M ππ≤⇒≥（2分） So, M = 16 and the Hann window function is2[]0.50.5cos ,21n w n M n M M π⎛⎫=+-≤≤ ⎪+⎝⎭（2分）(4) Windowing: h t [n ] = h LP [n ]w [n ], -16 ≤ n ≤ 16（2分） Casual system: h [n ] = h t [n -16], 0 ≤ n ≤ 33（2分）(5) Test ：Compute H(e jw )=DTFT(h[n]) to verify.（1分）5. The normalized transfer function of an order-2 low-pass Butterworth analog filter is as follows:121)(2++=s s s H an .With bilinear transform, an order-2 low-pass IIR Butterworth digital filter with -3dB cutoff frequency at 1000Hz and sampling frequency at 4000Hz has been designed. The structure of the filter is sketched in figure 1. Determine the value of constant K and M and expression of prototype analog filter H a(s).Solution:（共20分）10002240002C C T f F πωππ==⨯=（3分） Prewarp ()124cC tg tgωπΩ===（2分）Prototype analog filter()()()A an an C s H s H H s ===Ω（5分）Bilinear transform1111121111()()|11())111A z s z G z H s z z z z -----=+--==--++++1222--=- ),)K M ∴==分分6. In practice ，digital filters are often used to process analog signal. The processing procedure is shown as figure 2, where T denotes sampling interval ，and satisfies Nyquist Theorem . And we can regard it as an equivalent analog filter.Figure 2If the digital filter h[n] is a bandpass filter with the cutoff frequencies 121,,1563c c kHz T ππωω===，give the cutoff frequency 1c f and 2c f of the equivalent analog filter.Solution ：11112222,366,33c c c c c c c c TT T TT TTωωππωωππω=Ω∴=Ω=Ω===Ω=Ω==（分）（3分） Because the last step is an ideal lowpass filter with cutoff frequency being Tπ，the cutoff frequencies ofthe equivalent system is 12,63c c TTππΩ=Ω=rad/s.So, 12111150001150001250,250021212266c c c c f Hz f Hz T T ππΩΩ========（2分）（2分）。