山东大学2014-2015学年2学期数字信号处理（双语） 课程试卷(A)答案与评分细则 4. (20pts)Solution:(方法不唯一，只要答案正确即得分) X12 ∑nwm (a)Fourier transform of n]=(1/2)"un): n= 0+w+w塔+3w赠+wW+5w+6w X(e)= E[nje-jn 三 +=7wi x0+wg+2-1)+3(-w)+(1)+z5w 1-e-加 …5p本) +6(-1)+f可(-w) (b) Now,sample the frequency spectra of n): Y因=X(e0)儿w=2xtno,0≤k≤9 （-2ps) We have the 10-pt DFT: Y( 1-c-2mk/1可 0≤k≤9 9 = ∑nw给 (…4ps) n= n Since 1N DPT 1-()N N-pt 1-se-j(aE/N) …5p） yin]= (严 1-(线)0 0≤n≤9 （…4p) 5. (I0pts)Solution▣ (a)The "gain"along the emphasized path is-W. (b)In general,there is only one path between each input sample and each output sample. (c)[0]to2]:The gain is 1. [1]toX(2):The gain is w2.=-j z2]to X2):The gain is -=-1. 器 z[3]to X2):The gain is-W&W=-W.=j toX2):The gain is W&=1. 5]to x2):The gain is WW=W. 6l to):The gain is -=-1 z[7]toX2):The gain is -W&WRWA=-WR,as in Part (a). 墨 第2项共2页2014-2015 2 数字信号处理（双语） （A）答案与评分细则 2 2 4．(20 pts) Solution：（方法不唯一，只要答案正确即得分） (a) (b) Since 5．(10 pts) Solution： ( 5 ) pts ( 2 ) pts ( 4 ) pts ( 5 ) pts ( 4 ) pts ( 2 ) pts ( 3 ) pts ( 5 ) pts =-j =j