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当前位置:首页 > 建筑/环境 > 设计及方案 > 第五章有限长单位脉冲响应滤波器的设计方法数字信号处理习题答案
第五章有限长单位脉冲响应滤波器的设计方法1.解:(a)cndeededeeHnhnjjjnjjdd)(221)(21)()()(])[()1(])[(]|)(cos|)([sin)(2)()](sin)([cos)(2cacncacjjjnSnSenjnnendnjnneccc)(])[()1()()()(nnSnnhnhRcacnRd(b)为了保证线性相位)(])[()1()(2112,12,,21nknSnhkkkNNNRcacn为奇数若h(n)的类型取决于n)1(,N为奇数h(n)为偶对称第一类,h(n)必须偶对称于n=处,否则不满足N为奇数的已知条件若N为偶数。即N=2k,则)(])[()1()(21212nnSnhkkRcacnh(n)必须奇对称于n=处,否则不满足N为偶数的已知条件(c))()]12cos1)(([2)1()(nNnnSnhRcacn2.解:(a)ccdejedejenhnjjnjjd)()(2121)()(]2/)[(sin)1()cos(2)1(]||[2][221)()()()(nnnnneneedjedjeecncnnjnjinjnjjcccc)()(]2/)[(sin)1()(*)()(21nnnnnhnhRcnRd(b)为了保证线性相位21N若N为奇数,设N=2k+1则=k)()(]2/)[(sin2)1()(21nknknnhRcnh(n)满足奇对称,即h(n)=-h(N-1-n)属于第III类FIR滤波器若N为偶数,设N=2k则=k-1/2)2/1(]2/)2/1[(sin2)1()(21knknnhcnh(n)满足偶对称,即h(n)=h(N-1-n)属于第II类FIR滤波器(c))(]12cos1[)(]2/)sin[()1()(1nNnnnnhRcn3.解:ccccdeedeenhnjjnjjd00002121)()()sin()cos(2)()])(sin[()])(sin[(000nnnnnnccc)()()(nnhnhRd(a)N为奇数时,设N=2k+1,kN21)()()sin()cos(2)(0nknknknnhRch(n)满足于偶对称,属于第I类FIR滤波器(b)N为偶数时,设N=2k,=k-1/2)()2/1()2/1sin()2/1cos(2)(nknknknnhRch(n)满足偶对称,属于第II类FIR滤波器(c)N为奇数时,用升余弦窗设计)(]12cos46.054.0[)()sin()cos(2)(0nRNnknknknnhNcN为偶数,用升余弦窗设计)(]12cos46.054.0[)()2/1sin()2/1cos(2)(0nRNnknknknnhNc4.解:与第三题相比知)(由-变为-/2,所以只需将上题)(nhd由偶对称变为奇对称即可)()()()sgn()()sin()cos(2)(0nnhnhnnnnnhRdc(a)N为奇数,=k)()sgn()()sin()cos(2)(0nknknknknnhRc奇对称属于第III类滤波器(b)N为偶数,=k-1/2)()2/1sgn()2/1()2/1sin()2/1cos(2)(0nknknknknnhRc奇对称属于第IV类滤波器(c)用改进升余弦窗设计N为奇数)sgn()(]12cos46.054.0[)()sin()cos(2)(0knnRNnknknknnhNcN为偶数)(]12cos46.054.0[)2/1sgn()2/1()2/1sin()2/1cos(2)(0nRNnknknknknnhNc5.解:(a)一个带阻滤波器相当于一个全通滤波器减去一个带通滤波器全通)()(jjeeH带通)()()(jBjBeHeH则带阻)()()()](1[)()(jBjBjjreHeHeeH(b)因是线性相位滤波器,不妨设deeHnhnjjBr)()](1[21)()()()sin()(21)(nhnnnhdeEEnj6.解:(a))()sgn()()sin()()()()sgn()()sin()(sin21)(2/)(,1|)(|)()2/(nananannnhnhananannhnndenhaeHejeeHRRddnjdjdajajjd(b)N为奇数时,=(N-1)/2=kanannh,0,1)(N为偶数时,=(N-1)/2=k-1/2)()2/1sgn()2/1()2/1sin()(nknknknnhR显然N为偶数时性能好(c))())]1/(1[1()sgn()()sin()(020INnIananannh7.解:(a))()1()()1(21)(nannhandeejnhRanannjad(b)N为奇数时,a=k)()1()(nknnhRknN为偶数时,a=k-1/2)(2/1)1()(nknjnhRknN为奇数时性能好(c))())]1/(21[1()1()(020INnIknnhkn9.解:(a))1(5.0)14(13,,3,2,0)()(5.0)1(,1)0(),/11()21(2)(*15/)115(14)15/11(djdddjddHeHkkNHkHeHHNkNNkk102)(1)(NnknNjdekHNnh)5.05.01(151152815141521514njjnjjeeee1415/22151)(151)(okkjjjjeekHeeH(b)横截型x(n)1z1z1z1z)0(h)1(h)2(h)3(h)14(h频率采样型H(0)jex(n)H(1)1/15y(n)152jeje15jeH(14)1528jeje(c)横截型用的乘法器多,频率采样型用的加法器多10.解:(a))(2nh为)(1nh的圆周移位kkkkHkHekHWkHkHjkkmN)()(|)(||)(|)()()(2121112(b)如图所示,又5.32/)1(,8NaN知)(1nh,)(2nh均关于n=3.5偶对称,所以属于线性相位滤波器时延为3.511.解:其他,00,1|)(|cjdeH图略第一类线性相位FIR滤波器,121),21)((2210),21(2)(,121|,)(||)(|,211]2int[,041]2int[0,1|)(|)11(2)21()(,e|H(K)|H(K)(K)jNkNNkNNNkNkNKNkNkNHKHNkNNNkKHNkkNNKcc偶对称3217),33(3332160,3332)(,3225,12417169,080,1|)(|kkkkKkkkkKH及设计的过渡带宽3322N如果边沿设定(k)为一点,即令(9)=(24)=0.39]})332sin(33)]332(33sin[)332sin(33)]332(33sin[[2sin33233sin{)(3342*21610,09,39.080,1|)(|16kkkkeeHNKKKKHjj则过渡带宽为12.解:(a)N=333217),33(3332160,3332)(,2013,121,12,39.03222,110,0|)(|kkkkkkkkkkH(b)N=323318),34(3433170,3433)(,2113,122,12,39.03323,110,0|)(|kkkkkkkkkkH13.解:(a)N=33,因为N为奇数,所以可能是第I,III型滤波器2821,125,13229,2013,40,0|)(|kkkkkkH第I型3217),33(3332160,3332)(kkkkk第III型3217),33(33322160,33322)(kkkkk(b)N=34,可能是第II,IV型滤波2922,125,13330,2113,40,0|)(|kkkkkkH第II型3318),34(3433170,3433)(kkkkk第IV型3318),34(34332170,34332)(kkkkk14.解:(a)N为偶数,上面正交网络可设计成第IV型滤波器1,,1]21[),)(2(212]21[,,0,2212)(NNkkNNNNkNkNk(b)N为奇数,纯虚数幅度响应样本为:1,,1,0,0)(NkjkkjHr由于这是一个III型线性相位滤波器,在处振幅响应应为零,即0kH为了减少波动,在靠近处(即中点两旁)设过渡点,不妨选值为0.4j1,1,2/)1(,2/)3(,4.022/)3(,2/)5(2,2/)1(,0,0NNNkjNkNnkjNkkHk1,,1]21[),)(2)(21(2]21[,0),2)(21(2)(NNkkNNNNkNkNk15.解:(a)(虚数)幅度样本为:1,,1]21[),(2]21[,,0,2)(NNkkNNjNkkNjkjHrN为奇数时没有突变边沿N为偶数时没有突变边沿(b)N为偶数时1,,1]21[),)(2)(21(2]21[,,0),2)(21(2)(1,,12/),(22/),12(212,,0,2NNkkNNNNkNkNkNNkkNNNkNNNkkHk
本文标题:第五章有限长单位脉冲响应滤波器的设计方法数字信号处理习题答案
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