山东大学2011-2012学年2学期数字信号处理（双语）课程试卷(4)答案与评分细则 1.(10pts,2 pts for each)Solution: 1)Compute the 8-point DFT G[k]and ofthe two sequence gin]and n]; 1)B 2)Compute G[]]; 2)A 3)B 3)Compute IDFTof G[]H[]and y.(m)=IDFTG[]] (5pts) 4)A 5)B 4.(15 pts)Solution: 2.(10 pts)Solution: 3x+2 110.51 H,)=2+35+2+*+s+055+1 a)The system is generalized linear-phase system.Since h[4-n]=-hInl,M=4,By the odd (5 pts) symmetry of the sequence as in the figure we know it has the generalized linear Ht)=0.5e05+e phase. (.2亦) Since hn]=he(nT)is required,n=0.Seasr+e The point of symmetry is n=2,so we know the phase of H)is 0.5 (5 pts) ag[H(em】=-2o+π/2人 Then H()严1-eFti-erF (.3pt) When T=0.1s 1.5-(e-aas +0.5eal)= H()= em(5 ps) 0.5 1 Thus 5.(15 pts)Solution: 8w[x(e小-[l】小=-品-2a*2-2 (..2亦 a)System function:H()=7 2/52/5 -2可40.5可-21+052可 b)The system is type III FIR linear-phase system. (3p The poles and zeros of the system function in the z-plane: im 1 zero at Z= 3.(15 pts)Solution: e 怨 a)片(m)={-6,16,0,-19,2,4n=0,12,34,5} (5 pts) (5pts) b)y.(m)={-4,20,0,-19,n=0,12,3} (5pts) b)The ROC and the impulse response h[n]. 墨 e)yu[n]can be computed by using DFT when length N of DFT is not less than 6.The steps Ifthe system is stable,then the ROC includes the unit circle: <H<2.(3ps) are as following: 第1页共2页2011-2012 2 数字信号处理（双语） （A）答案与评分细则 1 2 1．(10 pts, 2 pts for each) Solution: 1) B 2) A 3) B 4) A 5) B 2．(10 pts) Solution: a) The system is generalized linear-phase system. Since h[4-n]= -h[n], M=4,By the odd symmetry of the sequence as in the figure we know it has the generalized linear phase. The point of symmetry is n=2, so we know the phase of ( ) j H e  is arg 2 / 2 ( ) j H e    = − +     ). Thus , b) The system is type III FIR linear-phase system. 3．(15 pts) Solution: a) y n n L ( ) 6,16,0, 19,2,4, 0,1,2,3,4,5 = − − =   （5 pts） b) y n n c ( ) 4,20,0, 19, 0,1,2,3 = − − =   （5 pts） c) yL[n] can be computed by using DFT when length N of DFT is not less than 6 . The steps are as following: 1) Compute the 8-point DFT G k[ ] and H k[ ] of the two sequence g n[ ] and hn[ ] ; 2) Compute G k[ ] H k[ ] ; 3) Compute IDFT of G k[ ] H k[ ] and ( ) L y n = IDFT{ G k[ ] H k[ ] }. （5 pts） 4．(15 pts) Solution: 2 3 2 1 1 0.5 1 ( ) 2 3 1 2 1 1 0.5 1 a s H s s s s s s s + = = + = + + + + + + + 0.5 ( ) 0.5 t t h t e e − − = + （5 pts） Since h[n] = hc(nT) is required, 0.5 [ ] 0.5 nT nT h n e e − − = + Then 0.5 1 1 0.5 1 ( ) 1 1 H z T T 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 0.5 1 1.5 ( 0.5 ) ( ) 1 1 1 ( ) e e z H z e z e z e e z e z − − − − − − − − − − − + = + = − − − + + ， 0.05 z e −  （5 pts） 5．(15 pts) Solution: a) System function： ( ) ( ) ( ) 1 1 2 5 2 5 1 2 1+0.5 H z z z − − = − − ( )( ) 1 1 1 1 2 1+0.5 z z z − − − = − The poles and zeros of the system function in the z-plane: （5 pts） b) The ROC and the impulse response h[n]. If the system is stable, then the ROC includes the unit circle: 1 2 2   z 。（3 pts） ( ……3 pts ) ( ) (arg 2 / 2 2 ( ) ) ( ) jw d d j grd X e H e d d        = − = − − + =       ( ……2 pts ) ( ……2 pts ) ( ……3pts )