山东大学2016-2017学年2学期数字信号处理（双语）课程试卷(B)答案与评分细则 For ROC:<0.25 x30=1×2-1×0+1×0-1×0=2:xl]=1×1-1×2+1×0-1×0=1: M=86m+250.25)°-n-1刂-180.5)”4-n-l刂 …3p脑 x3[2=1×1-1×1+1×2-1×0=2:x33=1×2-1×1+1×1-1×2=0: (......3pts) The signal flow graph of parallel-form structure is: x4=1×01×2+1×1-1×1=2:x35-1×0-1×0+1×2-1×1-1: x3[2]=1×0-1×0+1×0-1×2=2:x33=1×0-1×0+1×0-1×0=0: (...pts) ) +n) The results can also be got by figure solution,and akso are given6pois但是无推是过 …4pi 避蒂扣2分 0.75 (e)the minimum N when circular convolution is equal to linear convolution is: .0125g N=4+4-1=7 (..5pts】 6.(15 pts)Solution: 8.(10 pts)Solution: =25 3x+2 x[0]- X0] Since the system is causal,the ROC of Hc(s):Re(s)>-0.5 Find the inverse Laplace transform of Hc(s),we get, X[1] h0=0.5et0+e'ut) (5 pts) x2o→ Since hin]=The(nT)is required,n]=T(0.5e-sa +e-"un] 0.5T T Then H(=)=1 (5 pts) When T=0.1s x[1] 0.05,0.10.15-(0.1e-s+0.05el)=- He)1---++e, >e-aos (5 pts) X[5] X[6] (其中收敏域ROC情操扣3分) W 7.(15 pts)Solution: w x[71 X☑ =1 ax[n=x[n④x,[m Note:N=8. 怨 x0-1×2-1×1+1×1-1×2=0:x1P1×2-1×2+1×1-1×1=0: 图形画正确得5分，其中增益（系数）数值全部正确得5分。若增益（系数）数值不对， 或者位置不对，都要适当扣分(1-4分)，没有注明N-8扣1分. x2]1×1-1×2+1×2-1×1=0:x33=1×1-1×1+1×2-1×2=0: (......4 pts) The resul临can be got by figure solution,and also given4 points但是无推是过程的-2分 墨 b)x[n=x[n⑧x[n 第2页共2页2016-2017 2 数字信号处理（双语） （B）答案与评分细则 2 2 For ROC: z  0.25 [ ] 8 [ ] 25(0.25) [ 1] 18(0.5) [ 1] n n h n n u n u n = + − − − − −  The signal flow graph of parallel-form structure is: 6．(15 pts) Solution: 2 3 2 1 1 0.5 1 ( ) 2 3 1 2 1 1 0.5 1 c s H s s s s s s s + = = + = + + + + + + + Since the system is causal, the ROC of Hc(s): Re(s)> -0.5, Find the inverse Laplace transform of Hc(s), we get, 0.5 ( ) 0.5 ( ) ( ) t t h t e u t e u t − − = + （5 pts） Since h[n] = Thc(nT) is required, 0.5 [ ] (0.5 ) [ ] nT nT h n T e e u n − − = + Then 0.5 1 1 0.5 ( ) 1 1 T T T T H z e z e z − − − − = + − − （5 pts） When T=0.1s 0.05 -0.1 1 0.05 1 0.1 1 -0.1 0.05 1 0.15 2 0.05 0.1 0.15 (0.1 0.05 ) ( ) 1 1 1 ( ) e e z H z e z e z e e z e z − − − − − − − − − − − + = + = − − − + + ， 0.05 z e −  （5 pts） 7．(15 pts) Solution: (a) x3[0]=1×2–1×1+1×1–1×2=0；x3[1]=1×2–1×2+1×1–1×1=0； x3[2]=1×1–1×2+1×2–1×1=0；x3[3]=1×1–1×1+1×2–1×2=0； (The results can be got by figure solution, and also given 4 points 但是无推导过程需扣 1-2 分) (b) x3[0]=1×2–1×0+1×0–1×0=2；x3[1]=1×1–1×2+1×0–1×0=–1； x3[2]=1×1–1×1+1×2–1×0=2；x3[3]=1×2–1×1+1×1–1×2=0； x3[4]=1×0–1×2+1×1–1×1=–2；x3[5]=1×0–1×0+1×2–1×1=1； x3[2]=1×0–1×0+1×0–1×2=–2；x3[3]=1×0–1×0+1×0–1×0=0； (The results can also be got by figure solution, and also are given 6 points 但是无推导过 程需扣 1-2 分) (c) the minimum N when circular convolution is equal to linear convolution is: N=4+4-1=7 8．(10 pts) Solution: Note: N=8. 图形画正确得 5 分，其中增益（系数）数值全部正确得 5 分。若增益（系数）数值不对， 或者位置不对，都要适当扣分（1-4 分），没有注明 N=8 扣 1 分. x n x n x n 3 1 2   =   4   x n x n x n 3 1 2   =   8   (……4 pts) (……3pts) (……5pts) (……3pts) 4 pts 3 pts （其中收敛域 ROC错误扣 3分）