山东大学2013-2014学年2学期数字信号处理（双语）课程试卷()答案与评分细则 1.(10 pts,2 pts for each) (e)(5 pts)ROC: The ystm is stable.sie the ROC includes the un circle. 1)C;2)A;3)A4)B,5)D: (f)(7pts) -19- -63 44 2.(40 pts)Solution:H(=)=- 64--.30-126-+24:.y日 (e-2--3) 6-5+2 X() H(间=24+1-56-+W6=24+-5*-万 (a)(5 pts)poles:=1,Zp2=1/3;zeros:4,2=1/5;it isn't a minimum-phase system. The impulse response hin]is (b)(5 pts)The difference equation:6y[n]5y[n-1]+y[n-2]=30x[n]-126x[n-1]+24x[n-2]. n=246-63/2y°+443)}”m. ory[(56w[m-l]+1/6)y[m-2]=5x[间-21xn-1刂+4x[n-21 (c)(5pts)The 2nd-order direct form II signal flow graph is (g)(8 pts)Determine expressions for a minimum-phase system H(and an all-pass H(e) 5-211+4:2 x[m] yin] system()such that H)H(() 1-(56)-+(y6) 5/6 -21 H(e) 6(4-4-1--5）.-20(e-4)1-) -2-3)(1-/21-3例 f21 -1/6 .--a0-月-%-H日H.日 (-20-3)0-4 -19-=1 -63 44 5ps）H(阳间-24+1-6-w6e24+-万*-万 He月）-= -2001-/4)1-5 1-2j1--j The signal flow graph of parallel-form structure (2nd order)is 24 (--4) x[n] -19 H,-两 n Coefficients distributed in each terms may be different. 56,1 3.(15 pts)Proof:If Td=1,the system function of the continuous-time Butterworth filter is -16 Ist order parallel-form structure is 0.70326 0.7032 Hc(s)= where A= 24 器 -63 m 12 The impulse response of the continuous-time Butterworth filter is 暴 ke(0=∑4e0 (4pts) 1 第1页共2页2013-2014 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) 1) C ; 2) A; 3) A; 4) B ; 5) D; 2．(40 pts) Solution： ( ) ( )( ) ( )( ) ( ) ( ) 1 1 1 2 1 1 1 2 6 4 1 5 30 126 24 2 3 6 5 z z z z Y z H z z z z z X z − − − − − − − − − − − + = = = − − − + (a)（5 pts）poles: zp1=1/2 , zp2=1/3 ; zeros: z1=4，z2=1/5 ; it isn’t a minimum-phase system. (b)（5 pts）The difference equation: 6y[n]- 5y[n-1]+ y[n-2] =30x[n] -126x[n-1] +24x[n-2], or y[n]- (5/6)y[n-1] +(1/6) y[n-2] =5x[n] -21x[n-1] +4x[n-2]. (c)（5pts） The 2nd-order direct form II signal flow graph is ( ) ( ) ( ) 1 2 1 2 5 21 4 1 5 6 1 6 z z H z z z − − − − − + = − + (d) （5 pts） ( ) ( ) ( ) 1 1 2 1 1 19 63 44 24 24 1 5 6 1 6 1 2 1 3 z H z z z z z − − − − − − − − = + = + + − + − − The signal flow graph of parallel-form structure（2 nd order） is 1st order parallel-form structure is (e)（5 pts） ROC: 1 2 z  .The system is stable, since the ROC includes the unit circle. (f)（7 pts） ( ) ( ) ( ) 1 1 2 1 1 19 63 44 24 24 1 5 6 1 6 1 2 1 3 z H z z z z z − − − − − − − − = + = + + − + − − The impulse response h[n] is [ ] 24 [ ] 63(1 2) [ ] 44(1 3) [ ] n n h n n u n u n = − +  . (g) (8 pts) Determine expressions for a minimum-phase system H z min ( ) and an all-pass system ( ) H z ap such that ( ) min ( ) ( ) H z H z H z = ap ( ) ( )( ) ( )( ) ( )( ) ( )( ) 1 1 1 1 1 1 1 1 6 4 1 5 20 1 4 1 5 2 3 1 2 1 3 z z z z H z z z z z − − − − − − − − − − − − − = = − − − − ( )( ) ( )( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 min 20 1 4 1 5 1 4 1 2 1 3 1 4 ap z z z H z H z z z z − − − − − − − − − − = = − − − ( ) ( )( ) ( )( ) 1 1 min 1 1 20 1 4 1 5 1 2 1 3 z z H z z z − − − − − − − = − − ( ) ( ) ( ) 1 1 1 4 1 4 ap z H z z − − − = − Coefficients distributed in each terms may be different. 3．(15 pts) Proof: If Td=1, the system function of the continuous-time Butterworth filter is ( ) ( ) 6 6 6 1 1 0.7032 k k k k k c A H s s s s s = = = = − −   , where ( ) 6 6 1 0.7032 k k m m m k A s s =  =  − The impulse response of the continuous-time Butterworth filter is ( ) 6 1 ( ) k S t k k c h t A e u t = =  (4 pts) 5 −16 −21 4 56 -63 44 1/3 1/2 24 z -1 z -1 x[n] y[n]