山东大学2015-2016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) (6)(I0pts)H()=H()H,() 1)A;2)D;3)B:4)A;5)C, H(e)= 1-0.08=-）0-1242-）_-12421-0.08-）0-0.08=-）(-1-0.08) 2.(45 pts)Solution: （1-0.2-）1-0.4-） (1-0.200-0.4） (1-0.08=-） (2pts) (I)(10pts) -12.421-0.08=-)1-0.08-)） 36.5 48 1-12.52+s-2 （1-0.08=-）1-1242:-）y包 H.()= (4 pts) H(e)=12.5+ *1-0.2-1-0.42-1-0.6:+0.08= (1-0.2-）1-0.4-） (-0.2）0-0.4-）X(E H（日 (-0.08) the poles of H(z):0.2,0.4 (1-0.08=- (4pts) the zeros of H(2):12.42,0.08 (2)(8pts)ROC:>0.4 3.(5 pts)Solution: The system is stable,since the ROC includes the unit circle. (3)(7 pts)The difference equation:y[n]-0.6y[n-1]+0.08y[n-2]=x[n]-12.5x[n-1]+x[n-2]. -2- 5 k=0 otherwise (4)(5pts)Draw the 2nd-order direct form II signal flow graph for the system function H(z). H(e)= 1-12.51+s 0.6 -12.5 1-0.6:+0.08- 4.(15 pts)Solution: -0.081 In the filter design by the impulse invariance method,a pole at s=-s in the s-plane transforms to a pole atin the z-plane and the coefficients in the partial fraction (5)(5 pts)Draw the signal flow graph of 2nd-order parallel-form structure for H(z). expansions of Hc(s)and H(z)are equal,except for the scaling multiplier Ta. H(e）=12.5+ -5-1-11.5 1-0.621+0.08- s+3 1-e 12.5 1 Therefore, s+0.61T3+0.41Ts+0.3s+0.2 (8 pts) 器 x[n] -11.5 2-1 ▣ The above answer is not unique due to periodicity,a more general answer is 0.6 H()= 1 s+(0.3+ 2k)s+0.2+ 2πl、 (7 pts) 暴 -0.08 第1页共2页2015-2016 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) A; 2) D; 3) B; 4) A ; 5) C; 2．(45 pts) Solution： (1)（10 pts） ( ) ( )( ) ( )( ) ( ) ( ) 1 1 1 2 1 1 1 2 1 1 36.5 48 1 12.5 1 0.08 1 12.42 12.5 1 0.2 1 0.4 1 0.6 0.08 1 0.2 1 0.4 z z z z Y z H z z z z z z z X z − − − − − − − − − − − + − − = + − = = = − − − + − − the poles of H(z): 0.2, 0.4 the zeros of H(z): 12.42, 0.08 (2)（8pts）ROC: z  0.4 The system is stable, since the ROC includes the unit circle. (3)（7 pts）The difference equation: y[n]-0.6y[n-1]+0.08y[n-2] =x[n] -12.5x[n-1] +x[n-2]. (4)（5pts） Draw the 2nd-order direct form II signal flow graph for the system function H(z). ( ) 1 2 1 2 1 12.5 1 0.6 0.08 z z H z z z − − − − − + = − + (5)(5 pts) Draw the signal flow graph of 2nd–order parallel-form structure for H(z). ( ) 1 1 2 -5 11.5 12.5 1 0.6 0.08 z H z z z − − − − = + − + (6)（10 pts） ( ) min ( ) ( ) H z H z H z = ap ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1 1 0.08 1 12.42 12.42 1 0.08 1 0.08 0.08 1 0.2 1 0.4 1 0.2 1 0.4 1 0.08 z z z z z H z z z z z z − − − − − − − − − − − − − − − − = = − − − − − (2pts) ( ) ( )( ) ( )( ) 1 1 min 1 1 12.42 1 0.08 1 0.08 1 0.2 1 0.4 z z H z z z − − − − − − − = − − (4 pts) ( ) ( ) ( ) 1 1 0.08 1 0.08 ap z H z z − − − = − (4pts) 3．(5 pts) Solution： 4．(15 pts) Solution: In the filter design by the impulse invariance method, a pole at k s s =− in the s-plane transforms to a pole at k d s T z e − = in the z-plane and the coefficients in the partial fraction expansions of Hc(s) and H(z) are equal, except for the scaling multiplier Td. ( ) 1 N k c k k A H s = s s = +  ←→ ( ) 1 1 1 k d N d k s T k T A H z e z − − = = −  Therefore, ( ) 4 / 2 / 0.6 / 0.4 / d d c d d T T H s s T s T = − + + 2 1 s s 0.3 0.2 = − + + . (8 pts) The above answer is not unique due to periodicity, a more general answer is ( ) 2 1 2 2 (0.3 ) (0.2 ) c d d H s k l s j s j T T   = − + + + + . (7 pts) 12.5−11.5 0.6 -5 −0.08   4 5 0 5, 0 0, kn n k X k W = otherwise  = = =    1 −0.08 −12.5 1 0.6