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 n<no,and the system is causal. .…3pi 事Linear: M+1=8%+1-87%1 61z+1=81 (..6pts) T(ari[n]+bza[nl)ginl(ari[n]+bzaln] =aginjziin]+bg(n]zaln] For Blackman window: aT(=1ml)+BT(z2(n]) …3pb (......3 pts) So this is linear. M+1=12%w+1=127%e*1=121 .Not time-invariant:T(zin-nol)gin]in -nol yln -nol =gln-nojzIn -no] which is not TI. …3pi Memoryless:yIn]=T(in])depends only on the nt value of so it is memoryless. 5.(10 pts)Solution: 3p店 a)By the odd symmetry of the sequence as in the figure we know it has the linear phase. 3.(10 pts)Solution: The point of symmetry is n=3.5,so we know the phase of H)is the system function()=- (:-2) (4 pts) ag[H(ee]=-3.5o力 (:+0.3=-02) ROC: >0.3, (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 )