数字信号处理第5章习题答案完整版.pdf

《数字信号处理第5章习题答案完整版.pdf》由会员上传分享，免费在线阅读，更多相关内容在行业资料-天天文库。

1、习题51.（1）δ=0.017，δ=0.0089PS(2)δ=0.026,δ=0.0002PS2.(1)A=0.087dBPA=40dBS(2)A=0.31dBPA=32.8dBS'3A=2AA'=2Appss4ω'=π−ω,ω'=π−ω,ω'>ω'ppssps5．NAk−0.53−82j3+82j6（.1）H(Z)=∑=++1−esKTZ−11−e−0.2Z−1−0.2(1−2j)−1−0.2(−1+2j)−1k=14(1−eZ)4(1−eZ)3−82j3+82jNA−0.5k44(2)H(Z)=∑1−esK

2、TZ−1=1−e−0.2Z−1+−0.2(1−2j)−1+−0.2(−1+2j)−1k=11−eZ1−eZ5S+177（1）H(S)=(S+3)(S+4)0.50.5（2）H(S)=+S+2−3jS+2+3j3S+18（1）H(S)=2S+4S+332238[(1+S)+3(1+S)(1−S)+3(1+S)(1−S)+(1+S)]（2）H（S）=32233[7(1+S)+13(1+S)(1−S)+9(1+S)(1−S)+3(1+S)]−39ωp=0.5π×10rad310fp=10rad/sΩpf=p2π11.

3、采用脉冲响应不变法：−1−1−1−0.2565+0.3944Z1.9146−1.1086Z−1.6581−0.6384ZH（Z）=++−1−2−1−2−1−21−1.3962Z+0.7224Z1−1.1580Z+0.4120Z1−1.0756Z+0.2971Z双线性变换法：−120.106（1+Z）H()z=−2−1−2−1−1−2[−0.2193Z+0.212Z+0.4313][0.905Z−1.788Z+1.413][1.6323−1.788Z+0.5797Z]jω频率响应：H(e)=H(z)jωz=e12

4、.a）用脉冲响应不变法：−0.39j0.39jH(Z)=+−0.17+0.9j−1−0.17−0.9j−11−eZ1−eZjω−0.39j0.39jH(e)=+−0.17+0.9j-jω−0.17−0.9j-jω1−ee1−eeb）用双线形变换法：0.07H(Z)=1−Z−11−Z−12()+0.17+0.04−1−11+Z1+Z−16（1−Z）13H(Z)=−2−2−2()2.5716+1.4824Z（(3.412+0.588Z（)(3.9319+0.0681Z)jω频率响应H(e)=H(Z)jωZ=e(−2

5、−4)0.10561−2z+z14H（Z）=−1−2−3−41.1289−2.5783z+0.3865z−2.5783z+0.957z115H(Z)=220.94(Z-1)20.94(Z-1)[]−1.4142+122Z-1.24Z+1Z-1.24Z+1jωH(e)=H(Z)Z=ejω2−1⎛zˆ−α⎞0.223⎜1+⎟⎜−1⎟19．⎝1−αz⎠H（Z）=2−1−1zˆ−α⎛zˆ−α⎞1−0.2952+0.18⎜⎟−1⎜−1⎟1−αz⎝1−αz⎠−120.223()1+zH(zˆ)(=Hz)zˆ−1+α20．H

6、()z=LPLPz−1=LP1−0.2952z−1+0.18z−2−11+αzˆ21−211.998+1.998z22.H（Z）=−221+0.998z