山东大学2011-2012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 1.(10pts,2 pts for each)Solution: 4.(15 pts)Solution: 1)D 2)D (a) g6n=201ogo0.05=-26dBlg6=200g60.1=-20dB 3)B 4)A 1g 6=minimum(6,,6,)=-26dB 5)c The peak approximation errors of the Hanning window,Hamming window and Blackman window are littler than Ig=minimum(6.6.)=-26dB,so Hanning window,Hamming 2.(15pts)Solution: window and Blackman window can be used to meet this specification (..6pts) (b) .Stable:Let n]l M then ITn]lgnM.So,it is stable if gnll is bounded. …3pb △w=g,-w。=0.lz .Causal:vin]=ginlziln]and yaln]=ginjraln],so if sin]=n]for all n no,then For Hanning window and Hamming window: n]=yaln]for all n0.3, (3p) T1111111 012345678 The poles and zeros ofthe system function in thez-plane: (2pi) je） Thus, 器 8n[xe小=-[He∬=-品-3oj=35 (5pt) 0 b)The system is type II FIR linear-phase system. (3p肱) 墨 (3ps) 第1页共2页
2011-2012 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) D 2) D 3) B 4) A 5) C 2．(15 pts) Solution: 3．(10 pts) Solution: the system function ( 2) ( ) ( 0.3)( 0.2) z z H z z z − = + − （4 pts） ROC： z 0.3， （3 pts） The poles and zeros of the system function in the z-plane: Re( )z j z Im( ) -0.3 0 0.2 2 （3 pts） 4．(15 pts) Solution: (a) The peak approximation errors of the Hanning window, Hamming window and Blackman window are littler than , so Hanning window, Hamming window and Blackman window can be used to meet this specification . (b) For Hanning window and Hamming window: For Blackman window: 5．(10 pts) Solution: a) By the odd symmetry of the sequence as in the figure we know it has the linear phase. The point of symmetry is n=3.5, so we know the phase of ( ) j H e is arg 3.5 ( ) j H e = − ). Thus , b) The system is type II FIR linear-phase system. 3 pts 3 pts 3 pts 3 pts 3 pts (……6 pts) 10 lg 20log 0.1 20 s lg 20log 0.05 26 p = = − 10 dB = = − dB 0.1 = − = w w w s p lg minimum( , ) 26 p s = = − dB lg minimum( , ) 26 p s = = − dB 8 8 1 1 1 81 0.1 M w + = + = + = 12 12 1 1 1 121 0.1 M w + = + = + = (……3 pts) (……6 pts) ( ) (arg 3.5 3.5 ( ) ) ( ) jw d d j grd X e H e d d = − = − − = ( ……2 pts ) ( ……5 pts ) ( ……3 pts )
山东大学2011-2012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 8.(10 pts)Solution: 6.(15pts)Solution: x0] 0+oXI0例 (a)The difference equation relating x[n]and y[n]:(7pts) x nm+1.9n-月-1.08fn-2]+0.18fn-3]=7.96-10.13[n-刂+2.87xn-2] +0X2] (b)Rearrange the system function: W He=2870.18-X235-e 6] 2.871.18-2.35-e4 x3]9 1-0-0.320-0.621-*1-0.32-*1-0.6e✉ x[4] 0→一0XT10 The signal flow graph of cascade structure of the system with Ist-order sections is x[5] x(n). 2.87 1.18 2.35 yn) x[6] →-0X3 x(7 (8 pts) 7.(15 pts)Solution: 因-2- …7p otherwise 四[网=x0冈=9-h5sn≤9 [n+l,0≤n≤4 器 x[0]=l,x[=2,x[2]=3,x[B]=4,x[4=5,x[5=4, x[6]=3,[]=2,x[8=l[9]=0 ,8 暴 第2项共2页
2011-2012 2 数字信号处理（双语） （B）答案与评分细则 2 2 6．(15 pts) Solution: (a) The difference equation relating x[n] and y[n]： （7pts） y n y n y n y n x n x n x n [ ] 1.9 [ 1] 1.08 [ 2] 0.18 [ 3] 7.96 [ ] 10.13 [ 1] 2.87 + − − − + − = − − + − [ 2] (b) Rearrange the system function: 1 1 1 1 1 2.87(1.18 )(2.35 ) ( ) (1 )(1 0.3 )(1 0.6 ) z z H z z z z − − − − − − − = − − − 1 1 1 1 1 2.87 1.18 2.35 1 1 0.3 1 0.6 z z z z z − − − − − − − = − − − The signal flow graph of cascade structure of the system with 1st-order sections is （8 pts） 7．(15 pts) Solution: (a) . (b) 8．(10 pts) Solution: x n( ) y n( ) −1 2.87 1.18 1 z − 1 1 z − 0.3 −1 1 z − 0.6 2.35 4 1 5 0 5, 0 0, = = = = kn n k X k W otherwise x n x n x n 3 1 2 = 10 3 3 3 3 3 3 3 3 3 3 0 1, 1 2, 2 3, 3 4, 4 5, 5 4, 6 3, 7 2, 8 1, 9 0 = = = = = = = = = = x x x x x x x x x x 1, 0 4 9 , 5 9 + = − n n n n 7 pts 8 pts