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）A卷-2018_答案

山东大学2017-2018学年2学期数字信号处理（双语）课程试卷(A)答案与评分细则 1.(40 pts)Solution: 6)(10 pts)The signal flow graph for an 8-point decimation-in-frequency FFT algorithm with normal ordering of the input time sequence x[n]and bit-reversed ordering of 1)5pt) (a)Yes,the aytem is allpass,tince it is of the appropriate form the output sequence X[k]. x09 (b)No,the yrtem is not allpass,since tbe sero does ot occur at the conjugate reciprocal location of the pole. (c)Yes,the systemis allpass,since it is of the appropriate form W x29 0X2 (d)Yes,the ystem is allpam.This system conists of an allpass ystem in cascade with a pole at sero. The pole at zero is simply a delay,and does not change the magnitude spectrum. W x31Q -1 oX6 2)(5 pts)The sequences in(f)has all the zeros of its z-transform inside the unit circle w 4 (minimum phase system),The sequences in (c)has all the zeros of its z-transform outside the unit circle(maximum phase system). ] 3)(5 pts) WS By the symmetry ofan we know it has linear phase.The symmetry is around n=1/2 so we X3 know the phase of Xa(e)is arg(Xa(e=-w/2.Thus, W we 7] oX17] Ke1=-是ue明=-兰{-= d 32 2.(15 pts)Solution: 7 a)(5 pts)the system function H(=)= 1+2+:2 1-0.50.52 The signal flow graph of the system with 2dorder transposed direct form II structure: 4)(5 pts) Using long division,we get y H()= 1-盛210 0.5 1- 的 Taking the inverse z-transform.hn]= n=0,1,2,,9 t0.5 0. otherwise Since hin]is 0 for n <0,the system is causal. Since the ROC include the unit circle,the system is stable. b)(5 pts)Find the impulse response of the system,hin] 5)(10pts The system function is:H(z)(1+z)+2(z+5)(+)+3z3. 1+20-1+22 The signal flow graph of direct form structure with the number of coefficient multipliers 可-2++ should be smallest is: 怨 Taking the verse:=-2a+-r网+3时 tn] c)(5 pts)Find the output of this system,y[n forthe inputn=2".olutio1: a-=三-三=2++} 第1项共3页

2017-2018 2 数字信号处理（双语） （A）答案与评分细则 1 3 1 1 2 0.5 0.5 1．(40 pts) Solution: 1) (5 pts) 2) (5 pts) The sequences in (f) has all the zeros of its z-transform inside the unit circle (minimum phase system); The sequences in (c) has all the zeros of its z-transform outside the unit circle(maximum phase system). 3) (5 pts) 4) (5 pts) Since the ROC include the unit circle, the system is stable. 5) (10 pts) The system function is: H(z)=(1+z-6 )+2(z-1+z-5 )-(z-2+z-4 )+3z-3 . The signal flow graph of direct form structure with the number of coefficient multipliers should be smallest is: 6) (10 pts) The signal flow graph for an 8-point decimation-in-frequency FFT algorithm with normal ordering of the input time sequence x[n] and bit-reversed ordering of the output sequence X[k]. 2．(15 pts) Solution: a) (5 pts) the system function ( ) 1 2 1 2 1+2 1 0.5 -0.5 z z H z z z − − − − + = − The signal flow graph of the system with 2 nd -order transposed direct form II structure: b) (5 pts)Find the impulse response of the system, h[n]. c) (5 pts) Find the output of this system, y[n], for the input Solution 1:           k y n h n x n h k x n k  =− =  = −    ( ) 2 n k k h k  − =− =  0 2 2 1 2 1 8 -2 [ ] k 3 3 n k k  k  = −     =     −        + +

山东大学2017-2018学年2学期数字信号处理（双语）课程试卷(A)答案与评分细则 調》 Since hin]=The(nT),T=1, M=T(0.5e-tfnT+0.5e-fnT]）=0.5e-hm]+0.5e-m 1+211 0.5 0.5 Solution2:y[可=He儿2" 242"_182 Then e1-ggt1-s,e (5 pts) 1-0505 Since 3.(15 pts)Solution: -24-2saot4-2o5405c the em ncto日产e2习oc号Hk3 0.5-e）.0.5l-e） 1-e-刃 I-el-wn) (4 pts) Sinee H(z)=1 when z=0,we get k=-3/8. So H()=-3(-4) 3-0-4- Anothersoltion fore[可-立h个=h小*u[可一 (6 pts) 8(e-0.5X:-3)81-0.51-32- sC)-m()=1 0.5 H(e)= 3e--4e-10- 80-05---w0-3s-20-05X0-写0--月 0.5 0.5 10.5 1 0.5 严-e+可+-e可卡-e阿可1-e-小：en1-en soH=- u- 1 050.5 -e-0.5 -e人i 0.5 20-05X0-59 --e-可t-e可-e1-e ie阿1-e1 0.5 -7en+-0sen i-2Ttmal H(e)= 0.sl-e）.0.s-e） 1-e- 1-e- 5.(15 pts)Solution: N-1 (a)Using the analysi6 qioofE4,x内=∑njw n=0 Since is also periodic with period 3N, 3N- 附= ∑nw数 n- 器 -05 0 0.5 1 N-1 2N-1 Real Part 孙w索+艺w安 pole-zero plots for Hmi(z)and Hap(z) pole-zero plots for Hmin(z)and Hap(z) n=2N Performing a change of variables in the second and third summations of X 4.(15pts)Solution: 墨 1)(9ps)H,(=+2s+2+1++1-寸 s+1 0.5 405,Resp-1→h)=0.5e-u0）+0.5e-0 X因= ,nw+w龄 n+NW+W 定+2N哈 第2页共3页

2017-2018 2 数字信号处理（双语） （A）答案与评分细则 2 3 0 1 1 -2 4 2 2 1 8 k 3 3 n k k  =       + −                 =        + 4 16 18 -2 15 3 5 2 2 n n   = =     + + Solution 2:   2 1 1 1+2 2 4 ( ) 2 2 2 1 1 1 0.5 -0.5 8 4 1 5 2 z n n n y n H z = + = = = − 3．(15 pts) Solution: The system function ( ) ( 4) ( 0.5)( 3) k z H z z z − = − − , 1 : 3 2 ROC z   Since H(z)=1 when z=0, we get k=-3/8. So ( ) 1 1 1 1 3 ( 4) 3 (1 4 ) 8 ( 0.5)( 3) 8 (1 0.5 )(1 3 ) z z z H z z z z z − − − − − − = − = − − − − − (6 pts) ( ) 1 1 1 1 1 1 1 1 3 ( 4) (1 4 )( 3) 8 (1 0.5 )( 3) ( 4)(1 3 ) z z z z H z z z z z − − − − − − − − − − − = − − − − − 1 1 1 1 1 1 1 1 1 1 1 (1 ) ( )(1 ) 1 4 4 3 2 1 1 1 (1 0.5 )(1 ) (1 )( ) 3 4 3 z z z z z z z z − − − − − − − − − − − = − − − − − So 1 min 1 1 1 (1 ) 1 4 ( ) 2 1 (1 0.5 )(1 ) 3 z H z z z − − − − = − − − , 1 : 2 ROC z  . 1 1 1 1 1 1 1 ( )(1 ) 4 3 ( ) 1 1 (1 )( ) 4 3 ap z z z H z z z − − − − − − − = − − , 1 : 3 4 ROC z   (6 pts) pole-zero plots for Hmin(z) and Hap(z) pole-zero plots for Hmin(z) and Hap(z) 4．(15 pts) Solution: 1)（9 pts） 2 1 ( ) 2 2 a s H s s s + = + + 0.5 0.5 s j s j 1 1 = + + + + − , Re(s)>-1→ ( 1 1 ) ( ) ( ) 0.5 ( ) 0.5 ( ) j t j t h t e u t e u t − − − + = + Since h[n] =T hc(nT) ,T=1, ( ) ( ) ( ) 1 1 [ ] 0.5 [ ] 0.5 [ ] j nT j nT h n T e u nT e u nT − − − + = + ( 1 1 ) ( ) 0.5 [ ] 0.5 [ ] j n j n e u n e u n − − − + = + Then ( 1 1 ) 1 1 ( ) 0.5 0.5 ( ) 1 1 H z j j e z e z − − − + − − = + − − ， 1 z e −  （5 pts） Since     =− =  n k s n h k ( ) ( ) ( ) 1 1 0.5 [ ] 0.5 [ ] − − − + =− = +  n j k j k k e u k e u k ( ) ( ) ( ) 1 1 0 0.5 0.5 − − − + = = +  n j k j k k e e ( ) ( ) ( ) ( ) ( ) ( ) 1 ( 1) 1 ( 1) 1 1 0.5 1 0.5 1 1 1 j n j n j j e e e e − − + − + + − − − + − − = + − − （4 pts） Another solution for s[n]: Since     =− =  n k s n h k =  h n u n     → 1 ( 1 1 ) 1 1 ( ) 1 0.5 0.5 ( ) ( ) ( ) 1 1 1 j j S z H z U z z e z e z − − − − + − −   = = +   −   − − , z 1 1 ( 1 1 1 1 1 1 ) ( ) ( ) ( ) 1 1 ( ) ( ) 1 0.5 0.5 1 0.5 1 0.5 1 1 1 1 1 1 1 j j j j j j z e e e e z e e z − − − − + + − − − − + − −   = + + +   −   − − − − − − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1 1 1 0.5 0.5 0.5 0.5 1 1 1 1 1 1 1 j j j j j j j j e e z e e e e z e e z − − − + − − − − + − − − − − + − + − −   − − = + + +   −   − − − − − − , z 1 →   ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 1 0.5 0.5 0.5 0.5 [ ] [ ] [ ] 1 1 1 1 j j j n j n j j j j e e s n u n e u n e u n e e e e − − − + − − − + − − − + − − − +   − − = + + +     − − − − ( ) ( ) ( ) ( ) ( ) ( ) 1 ( 1) 1 ( 1) 1 1 0.5 1 0.5 1 1 1 j n j n j j e e e e − − + − + + − − − + − − = + − − 5．(15 pts) Solution:

山东大学2017-2018学年2学期数字信号处理（双语）课程试卷(A)答案与评分细则 Since is periodic with period N,and W=W 元内=（1+e+e4） (（1+e-2)+e2()）k/3 3k/3,k=3 = 0, otherwise (b)Using N=2 andn]as in the Fig : 树= 时 R-0 = * n=0 0+0e“ =1+2(-1) 3,k=0 -1,k=1 Observe,from the Fig,that n]is also periodic with period 3N=6: 3N-1 内= nw缺 5 = “ n-0 =(1+e*+e-w)1+2(-1)) =(1+e*+e-)/3例 器 9,k=0 -3,k=3 0,k=1,2,4,5 第3负共3项

2017-2018 2 数字信号处理（双语） （A）答案与评分细则 3 3