5.6习题15.6习题5-1设Fi(a)是线性相位FIR系统函数H(=)的一个因子。试确定满足条件的最低阶的H(=)。(1) F (=) = 1+ 2=- + 3=2(2) F(=) =3 + 5=-l 4z 2 22-3解:设F(=)=a+bz-l+cz-2,则零点为其倒数的多项式Fz()为F,(=)=(a+ bz+cz2 )=- =c+bz-l +az-2(1) H(=)= (1+ 22-* +32)(3+22-* +2~)= 3 +82 +14=2 + 82-* + 32-(2)H(=)=(3+52-1 -42-2 -2z-)(-2 -4z- +5=-2 +3z3)= 6222" +3-~ + 542 +3- 22--5 6-5-2某三阶设FIR滤波器的系统函数Hi(a)为H,(=) =(6 - z-1 -12 2- )(2 + 52-l)(1)试确定幅度响应和H(=)相同的所有FIR滤波器的系统函数。(2)上述FIR滤波器哪个是最大相位系统,哪个是最小相位系统?解:设H(=)=α+bz-是一个实系数的1阶FIR系统,则H(=)=z"H(=")=b+az"是和H(=)幅度响应相同的FIR系统H,(=) =(2 32" )(3 +42-')(2 +52-) =12 + 28≥- - 292-2 602-3(1) H,(=) =(-3 + 2-' )(3 +4z-')(2+ 52-) = -18 57=" 142~ +402-3H,(2) =(2 3-")(4 + 3z-)(2 + 5z") = 16 + 28=-l = 48-2 452-3H,(=) = (2 32")(3 +42)(5 + 22-") = 30 + 72 622 24=-3H, (=) = (3 + 2=- )(4 + 3z~* )(2 + 5=) = 24 62 = + 7= + 30 =-3
5.6 习题 1 5.6 习题 5-1 设 F1(z)是线性相位 FIR系统函数 H(z)的一个因子。试确定满足条件的最低阶的 H(z)。 (1) 1 2 1 ( ) 1 2 3 - - F z = + z + z (2) 1 2 3 1 ( ) 3 5 4 2 - - - F z = + z - z - z 解:设 1 2 1 ( ) - - F z = a + bz + cz ,则零点为其倒数的多项式 ( ) 2 F z 为 2 2 1 2 2 ( ) ( ) - - - F z = a + bz + cz z = c + bz + az (1) ( ) (1 2 3 )(3 2 ) -1 -2 -1 -2 H z = + z + z + z + z 1 2 3 1 3 8 14 8 3 - - - - = + z + z + z + z (2) ( ) (3 5 4 2 )( 2 4 5 3 ) -1 -2 -3 -1 -2 -3 H z = + z - z - z - - z + z + z 1 2 3 4 5 6 6 22 3 54 3 22 6 - - - - - - = - - z + z + z + z - z - z 5-2 某三阶设 FIR 滤波器的系统函数 H1(z)为 ( ) (6 12 )(2 5 ) 1 2 1 1 - - - H z = - z - z + z (1)试确定幅度响应和 H1(z)相同的所有 FIR 滤波器的系统函数。 (2)上述 FIR 滤波器哪个是最大相位系统,哪个是最小相位系统? 解: 设 1 ( ) - H z = a + bz 是一个实系数的 1 阶 FIR 系统,则 1 1 1 0 H (z) z H ( ) z b az - - - = = + 是和 H z( ) 幅度响应相同的 FIR 系统 1 1 1 1 2 3 1 ( ) (2 3 )(3 4 )(2 5 ) 12 28 29 60 - - - - - - H z = - z + z + z = + z - z - z (1) ( ) ( 3 2 )(3 4 )(2 5 ) 1 1 1 2 - - - H z = - + z + z + z 1 2 3 18 57 14 40 - - - = - - z - z + z ( ) (2 3 )(4 3 )(2 5 ) 1 1 1 3 - - - H z = - z + z + z 1 2 3 16 28 48 45 - - - = + z - z - z ( ) (2 3 )(3 4 )(5 2 ) 1 1 1 4 - - - H z = - z + z + z 1 2 3 30 7 62 24 - - - = + z - z - z ( ) ( 3 2 )(4 3 )(2 5 ) 1 1 1 5 - - - H z = - + z + z + z 1 2 3 24 62 7 30 - - - = - - z + z + z
H,()=(2-3--")(4 +32)(5+2--)= 40 142 572-218--3H,(=)=(-3 +2--)(3+4z")(5 + 2z) =45 48- +282 +16-H,(=)=(-3+2z")(4+3z)(5+2z) =-60-29z- +282- +122-(2) H,(=)=(2-3=)(3+4=)(2+5=-")是最大相位系统H(=)=(-3+2=")4+3=-)5+22-")是最小相位系统5-3已知8阶I型线性相位FIR滤波器的部分零点为:z1=2,22=j0.5,3-j(1)试确定该滤波器的其他零点。(2)设h[0]=1,求出该滤波器的系统函数H(a)。解:(1)z4=1/ z/=1/2;25=1/ 22=-j2,26=22=-j0.5, 27-25-j2≥8=1/ 23= -j(2)H(=)= (1-==)=1+ 8-2.5(+ 27)+6.25(22+ 26)-13.125(-3+ -5)+10.5 5-4已知9阶ⅡI型线性相位FIR滤波器的部分零点为:=1=2,22=j0.5,23-j(1)试确定该滤波器的其它零点。(2)设h[0]=1,求出该滤波器的系统函数H()。解(1)z4=1/ =/=1/2;z5=1/ 22= -j2,z6==2'= -j0.5, 27-z5=j2;≥8=1/ 3= j,29= -1;(2)H(=)= (1-=)=1+ 29-1.5(2'+ 8)+3.75(-2+ 7)- 6.875 (3+ 2) -2.625(2*+ 25)5-5已知8阶IⅡII型线性相位FIR滤波器的部分零点为:2=-0.2,22=j0.8(1)试确定该滤波器的其他零点。(2)设h[0]=1,求出该滤波器的系统函数H(=)。解:(1)23=1/ 2,--5;
( ) (2 3 )(4 3 )(5 2 ) 1 1 1 6 - - - H z = - z + z + z 1 2 3 40 14 57 18 - - - = - z - z - z ( ) ( 3 2 )(3 4 )(5 2 ) 1 1 1 7 - - - H z = - + z + z + z 1 2 3 45 48 28 16 - - - = - - z + z + z ( ) ( 3 2 )(4 3 )(5 2 ) 1 1 1 8 - - - H z = - + z + z + z 1 2 3 60 29 28 12 - - - = - - z + z + z (2) ( ) (2 3 )(3 4 )(2 5 ) 1 1 1 1 - - - H z = - z + z + z 是最大相位系统 ( ) ( 3 2 )(4 3 )(5 2 ) 1 1 1 8 - - - H z = - + z + z + z 是最小相位系统 5-3 已知 8 阶 I 型线性相位 FIR 滤波器的部分零点为:z1=2,z2=j0.5,z3=j (1)试确定该滤波器的其他零点。 (2)设 h[0]=1, 求出该滤波器的系统函数 H(z)。 解: (1)z4=1/ z1=1/2; z5=1/ z2= -j2,z6=z2 * = -j0.5,z7=z5 * = j2; z8=1/ z3= -j (2) ( ) (1 ) 1 8 1 k k H z z z - = = Õ - =1+ z -8 -2.5(z -1 + z -7 )+6.25(z -2 + z -6 )-13.125(z -3 + z -5 )+10.5 z -4 5-4 已知 9 阶 II 型线性相位 FIR 滤波器的部分零点为:z1=2,z2=j0.5,z3=-j (1)试确定该滤波器的其它零点。 (2)设 h[0]=1, 求出该滤波器的系统函数 H(z)。 解: (1)z4=1/ z1=1/2; z5=1/ z2= -j2,z6=z2 * = -j0.5,z7=z5 * = j2; z8=1/ z3= j; z9= -1; (2) ( ) (1 ) 1 9 1 k k H z z z - = = Õ - =1+ z -9 -1.5(z -1 + z -8 )+3.75(z -2 + z -7 )- 6.875 (z -3 + z -6 ) -2.625(z -4 + z -5 ) 5-5 已知 8 阶 III 型线性相位 FIR 滤波器的部分零点为:z1= -0.2,z2=j0.8 (1)试确定该滤波器的其他零点。 (2)设 h[0]=1, 求出该滤波器的系统函数 H(z)。 解: (1)z3=1/ z1=-5;
5.6习题3z4=1/22=-j1.25,zs=z2 =-j0.8,z6=z4 = j1.25;27= -1;≥8= 1;H(=)=(2)Z.ke=128+5.2(-17)+2.2025 (2-)- 6.253 (-3-25)5-6已知9阶IV型线性相位FIR滤波器的部分零点为:2=-1,22=0.8,23=0.5j,(1)试确定该滤波器的其他零点。(2)设h[0]-1,求出该滤波器的系统函数H(=)。解:(1)z4=1/ 22=1.25;z5=1/ 23= -j2,26=23 =-j0.5, 27=z5 = j2;28=21= -1;z9= 1;(2)H(=)=I (1-==)=1- 21.05 (21- 28)+ 2.2 (2- 7)- 2.4125 (3- 26) -6.6625(24- 25)5-7试证明式(5-7)和式(5-8)。证:M/2-1H(e"°)=217h[k]e-soh[k]e-ji + h[M / 2]e-i(M/2)0 +h[k]e-ikNk=01-0k=M /2+1令l=M-k,则有M/2-1M/2-h[M-]]e-i(M-1)2h[k]e-i(M-k)ah[k]e-nnWK=0k=M/2+l所以H(ej°)=e-(M/2)n[)[e-ie ;(M2)0 +e-(M/2-0) ]+[M /2]=e-j(M2)n2 h[k]cos[(0.5M-k)2]+h[M /2]=MI2=e-(M/2)n2h[0.5M -k]cos(k2)+h[M /2]>k=记:L=M/2,则有A2h[L-k]cos(k2)+h[L]A(2)=)?21=IH(ej0)=e-(M/2) A(2)
5.6 习题 3 z4=1/ z2= -j1.25,z5=z2 * = -j0.8,z6=z4 * = j1.25; z7= -1; z8= 1; (2) ( ) (1 ) 1 8 1 k k H z z z - = = Õ - =1- z -8 +5.2(z -1 - z -7 )+ 2.2025 (z -2 - z -6 )- 6.253 (z -3 - z -5 ) 5-6 已知 9 阶 IV 型线性相位 FIR 滤波器的部分零点为:z1= -1,z2=0.8,z3= 0.5j, (1)试确定该滤波器的其他零点。 (2)设 h[0]=1, 求出该滤波器的系统函数 H(z)。 解: (1)z4=1/ z2=1.25; z5=1/ z3= -j2,z6=z3 * = -j0.5,z7=z5 * = j2; z8=z1= -1; z9= 1; (2) ( ) (1 ) 1 9 1 k k H z z z - = = Õ - =1- z -9 -1.05 (z -1 - z -8 )+ 2.2 (z -2 - z -7 )- 2.4125 (z -3 - z -6 ) -6.6625(z -4 - z -5 ) 5-7 试证明式(5-7)和式(5-8)。 证: W W W W kW M k M k M M k k M k H h k h k h M h k j /2 1 j j( /2 ) /2 1 0 j 0 j (e ) [ ]e [ ]e [ / 2]e [ ]e - = + - - - = - = = å = å + + å 令l = M - k ,则有 W W j( ) W /2 1 0 j( ) /2 1 0 j /2 1 [ ]e [ ]e [ ]e M k M k M l M l k M k M h k h M l h k - - - = - - - = - = + å = å - = å 所以 [ ] þ ý ü î í ì = + + - - - - = - (e ) e å [ ] e e e [ / 2] j j( 2 ) j( / 2 ) /2 1 0 j j( / 2 ) H h k h M k M/ M k M k W M W W W W [ ] þ ý ü î í ì = å - + - = - e 2 [ ]cos (0.5 ) [ / 2] /2 1 0 j( /2 ) h k M k h M M k M W W þ ý ü î í ì = å - + = - e 2 [0.5 ]cos( ) [ / 2] /2 1 j( / 2 ) h M k k h M M k M W W 记:L=M/2,则有 ( ) 2 [ ]cos( ) [ ] 1 A h L k k h L L k = å - + = W W (e ) e ( ) j j( / 2) W W W H A - M =
5-8试证明式(5-10)和式(5-11)。证:M--)/2学H(ei°)=2h[k]e-jkah[k ]e-inh[k]e-jia7k=01=0k=(M+1)2令l=M-k,则有(M-1V2(M-1)/2产Z h[k Je-i(M- )n2 h[M-1]e-i(M-1)2h[k]e-ik?0k=(M+1)/2所以[(M-)/2h[[e-ej(M2) +e-(M/2-k)=e-(M/2)QIH(ejn)k=0=e-j(M2)o(M-)22h[k]cos[(0.5M -k)2]2(M-1)/2=ej(M/2)02h[K]cos[(M -1)/2 - k+0.5)2]NE(M-1)/2=e-(M/2)Q2 h[(M -1)/2- k]cos[(k + 0.5) 2)]7keo记:L=(M-1)/2,则有A(2)=≥2h[L-k]cos[(k+0.5)2]=0H(ejg)=e-(M/2)2 A(2)5-9试证明式(5-13)和式(5-14)。证:对III型线性相位系统,h[M/2]=0,所以M /21MMh[k]e-knh[k]e-ia +h[k]e-jikaH(ejn)=)k-01-0k=M/2+1令l=M-k,则有2M/2-M/2-Ih[k]e-jh[M -/]e-i(M-1)2h[kJe-j(M-k)a>>10A=0A=M/2+I所以112-H(ejQ)=e-(M/2)Qh[ke-iej2)a-e~(M/2-t)0212=e-(M/2)n2jh[k ]sin [(0.5M-k)2]2N=e-j(M/2)42jh[0.5M-k|sin(kQ)2记:L=M/2,则有
5-8 试证明式(5-10)和式(5-11)。 证: W W W kW M k M k M k k M k H h k h k h k j ( 1 )/ 2 j ( 1 ) /2 0 j 0 j (e ) [ ]e [ ]e [ ]e - = + - - = - = = å = å + å 令l = M - k ,则有 W W j( )W ( 1 )/ 2 0 j( ) ( 1 ) /2 0 j ( 1) / 2 [ ]e [ ]e [ ]e M k M k M l M l k M k M h k h M l h k - - - = - - - = - = + å = å - = å 所以 [ ] þ ý ü î í ì = + - - - - = - å W W j W j( 2) W j( /2 )W ( 1)/2 0 j j( / 2) (e ) e [ ] e e e k M/ M k M k M H h k [ W ] W e 2 [ ]cos (0.5 ) ( 1) /2 0 j( /2) h k M k M k M = å - - = - [( )W ] W e 2 [ ]cos ( 1)/ 2 0.5 ( 1) /2 0 j( / 2) = å - - + - = - h k M k M k M e 2 [( 1)/ 2 ]cos[( 0.5) )] ( 1) /2 0 j( / 2) W W = å - - + - = - h M k k M k M 记:L=(M-1)/2,则有 ( ) 2 [ ]cos[( 0.5) )] 0 W = å - + W = A h L k k L k (e ) e ( ) j j( / 2) W W W H A - M = 5-9 试证明式(5-13)和式(5-14)。 证: 对 III 型线性相位系统,h[M/2]=0,所以 W W W kW M k M k M k k M k H h k h k h k j / 2 1 j / 2 1 0 j 0 j (e ) [ ]e [ ]e [ ]e - = + - - = - = = å = å + å 令l = M - k ,则有 W W j( ) W /2 1 0 j( ) /2 1 0 j /2 1 [ ]e [ ]e [ ]e M k M k M l M l k M k M h k h M l h k - - - = - - - = - = + å = å - = - å 所以 [ ] W W j W j( 2) W j( / 2 ) W / 2 1 0 j j( / 2) (e ) e [ ] e e e k M/ M k M k M H h k - - - - = - = å - [ W ] W e 2j [ ]sin (0.5 ) /2 1 0 j( / 2) h k M k M k M = å - - = - e 2j [0.5 ]sin( ) /2 1 j( /2) W W h M k k M k M = å - = - 记:L=M/2,则有
5.6习题5A(2)=2 2h[L-k]sin(k2)1=lH(ej0)=e-(M/2)em/2 A(2)5-10试证明式(5-16)和式(5-17)。证:( M)/2H(eja)=^Mh[k]e-ioZ h[k]e-inh[k ]e-ikak=0k=0k=(M+1V2令1=M-k,则有(M-)/2(M-1)/2AZ h[k]e-i(M-1)ah[M - 1]e-i(M-1)2h[k]e-ikaM1=010k=(M+1)/2所以-j(M/2)Q-j(M /2-k)-jQj(M/2)Oh[keH(e>1-0(M-1)/2=e(M/2)Q2jh[k ]sin [(0.5M - k)2]Mk=0(M1)/2=e-j(M/2)o2jh[k]sin [(M -1) /2 -k +0.5)2]A=0(M-1)/2=e-j(M2)02jh[(M -1) /2-k]sin[(k + 0.5)2)]2k=0记:L=(M-1)/2,则有2h[L-k]sin[(k +0.5)2]A(2) =1=0H(eig)=e-(M/2)aem/2 A(2)5-11Hi(=)、H2()、Hs(=)和H4(=)分别表示I型、IⅡI型、IⅡII型和IV型线性相位FIR滤波器,试判断下列级联是否为线性相位系统?如是线性相位系统试确定其类型。(1)Gi()= Hi() H(日)(2) G2(2)= H(=) H(=)(3) G;(2)= H() H(=)(4)G4(=)= Hi() H4(=)(5) Gs(=)=H2() H2(=) (6) G6(=)= H2() Hs(=)(7)G(=)= H2(=) H4(=) (8) Gs(=)= Hs() Hs(=) (9) Gg(=)= Hs(=) H4(=)(10)G10(=)= H4(=) H4(=)解:设 Hi()、H2(=)、Hs(=)和 H4(2)的阶数分别为 K(偶数)、L(奇数)、M(偶数)与 N(偶数)(1)-(K+G(=")= 2-*H(=") -*H(=")= H(+) H()= G(c)由于2K为偶数,所以是I型线性相位系统(2) -(K+D)G2(=-")= *H;(=") ="H2(=-")= Hi() H2(2)= G2()由于K+L为奇数,所以是II型线性相位系统
5.6 习题 5 ( ) 2 [ ]sin( ) 1 A W h L k kW L k = å - = (e ) e e ( ) j j( / 2) j / 2 W W W H A - M p = 5-10 试证明式(5-16) 和式(5-17)。 证: W W W kW M k M k M k k M k H h k h k h k j ( 1 )/ 2 j ( 1 ) /2 0 j 0 j (e ) [ ]e [ ]e [ ]e - = + - - = - = = å = å + å 令l = M - k ,则有 W W j( ) W ( 1) / 2 0 j( ) ( 1 ) /2 0 j ( 1) / 2 [ ]e [ ]e [ ]e M k M k M l M l k M k M h k h M l h k - - - = - - - = - = + å = å - = - å 所以 [ ] þ ý ü î í ì = - - - - - = - å W W j W j( 2) W j( /2 ) W ( 1)/ 2 0 j j( / 2) (e ) e [ ] e e e k M/ M k M k M H h k [ W ] W e 2j [ ]sin (0.5 ) ( 1) /2 0 j( /2) h k M k M k M = å - - = - [( )W ] W e 2 j [ ]sin ( 1)/ 2 0.5 ( 1) /2 0 j( /2) = å - - + - = - h k M k M k M e 2j [( 1) / 2 ]sin [( 0.5) )] ( 1) / 2 0 j( /2 ) W W = å - - + - = - h M k k M k M 记:L=(M-1)/2,则有 ( ) 2 [ ]sin [( 0.5) )] 0 W = å - + W = A h L k k L k (e ) e e ( ) j j( / 2) j / 2 W W W H A - M p = 5-11 H1(z)、H2(z)、H3(z)和 H4(z)分别表示 I 型、II 型、III 型和 IV 型线性相位 FIR 滤波器, 试判断下列级联是否为线性相位系统?如是线性相位系统试确定其类型。 (1)G1(z)= H1(z) H1(z) (2) G2(z)= H1(z) H2(z) (3) G3(z)= H1(z) H3(z) (4)G4(z)= H1(z) H4(z) (5) G5(z)= H2(z) H2(z) (6) G6(z)= H2(z) H3(z) (7)G7(z)= H2(z) H4(z) (8) G8(z)= H3(z) H3(z) (9) G9(z)= H3(z) H4(z) (10)G10(z)= H4(z) H4(z) 解: 设 H1(z)、H2(z)、H3(z)和 H4(z)的阶数分别为 K(偶数)、L(奇数)、M(偶数)与 N(偶数) (1)z -(K+K)G1(z -1 )= z -KH1(z -1 ) z -KH1(z -1 )= H1(z) H1(z)= G1(z) 由于 2K 为偶数,所以是 I 型线性相位系统 (2) z -(K+L)G2(z -1 )= z -KH1(z -1 ) z -LH2(z -1 )= H1(z) H2(z)= G2(z) 由于 K+L 为奇数,所以是 II 型线性相位系统