山东大学：《生物医学信号处理 Biomedical Signal Processing》精品课程教学资源（试卷与答案）B卷-2017_答案

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 By the even symmetry of the sequence as in the figure we know it has the generalized linear 1.(10 pts,2 pts for each)Solution: 1)A: phase.The symmetry is aroundso we know the phase of) 2)B 3)A isarg()=-/2+B (constant). 4)B 37 5)A 2.(15pts)Solution: (--可 (没有推每序列对称性扣2分，不注明广义编性相位扣2分，无常数扣1分) Thus, (.5pi) …3pi gn[xe小-e[xe】=-品-o2*-h (5pi) - s (没有推导过粗扣2分) 5.(15 pts)Solution: Roc:子对 4 1+2z1+2 -7+8= (其中收敏域R0C铺损扣3分) H(9)-1-0752-+0.25✉=8+ -0.75-+0.1252-2 The LTI system is not stable,since the ROC does not include the unit circle. …5pb =8+ -25 18 3.(10 pts)Solution: +1-0.252+1-0.5z 2p昨 H,=1+3 poleinside the ctrcle Since the ROC is not given,the impulse response is not unique.There three possible ROCs which are determined by two poles(0.25,0.5).For three possible ROCs,we get the impulse responses through inverse Z-transform: zero:=-3:outside the unit circle For ROC:>0.5 器 reflected sero:== 2p店 Mn=86mj-250.25)°+180.5)”4可 1+互 Hm(e）=3 ForR0C:0.25<H<0.5 1+一 Mn=86同-250.25)y°)-180.5)°-n- …3p 4.(10 pts)Solution: 第1页共2页

2016-2017 2 数字信号处理（双语） （B）答案与评分细则 1 2 1．(10 pts, 2 pts for each) Solution: 1) A； 2) B 3) A 4) B 5) A 2．(15pts) Solution: The LTI system is not stable, since the ROC does not include the unit circle. 3．(10 pts) Solution: 4．(10 pts) Solution: By the even symmetry of the sequence as in the figure we know it has the generalized linear phase. The symmetry is around n=1/2, so we know the phase of ( ) j X e  is arg / 2 ( ) j X e    = −   +β(constant). （没有推导序列对称性扣 2 分，不注明广义线性相位扣 2 分，无常数扣 1 分） Thus , （没有推导过程扣 2 分） 5．(15 pts) Solution： Since the ROC is not given, the impulse response is not unique. There three possible ROCs which are determined by two poles (0.25, 0.5). For three possible ROCs, we get the impulse responses through inverse Z-transform: For ROC: z  0.5 [ ] 8 [ ] 25(0.25) [ ] 18(0.5) [ ] n n h n n u n u n = − +  For ROC: 0.25 0.5   z [ ] 8 [ ] 25(0.25) [ ] 18(0.5) [ 1] n n h n n u n u n = − − − −  ( ) 1 1 1 1 1 2 1 1 1 1 12 : 1 1 1 1 3 2 1 1 1 1 3 2 3 2 z z X z ROC z z z z z − − − −     −   = + =      + −    + −          1 1 1 3 2     = − − − −         n n x n u n u n   1 1 1 1 , 3 3 1 1 3 −     −     + n Z u n z z   1 1 1 1 1 , 2 2 1 1 2 −   − − −       − n Z u n z z 4 pts 3 pts 3 pts ( ) 1 1 1 1 3 1 1 2 − − + = + z H z z 1 : 2 pole z inside the unit circle = − zero z outside the unit circle : 3: = − 1 : 3 reflected zero z = − ( ) 1 1 1 3 1 1 3 − − + = + ap z H z z ( ) 1 min 1 1 1 3 3 1 1 2 − − + = + z H z z 2 pts 4 pts 4 pts ( ) ( ( ) ) ( ) arg / 2+ 1 2 jw d d j grd X e X e d d        = − = − − =       ( ……5pts ) ( ……5 pts ) 5 pts ( ) 1 2 1 1 2 1 2 1 2 7 8 8 1 0.75 0.125 1 0.75 0.125 z z z H z z z z z − − − − − − − + + − + = = + − + − + 2 pts 1 1 25 18 8 1 0.25 1 0.5 z z − − − = + + − − （其中收敛域 ROC错误扣 3分） 3 pts 3 pts

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:-0.5 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）