信号分析与处理_杨西侠_课后
二三五章
2-1 画出下列各时间函数的波形图,注意它们的区别
t?u(t) 1)x(t) = sin ,1
x1
(1t 0 234π
t) - 2)x(t) = sin[ , ( t – t ) ]?u(t) 1 20π π π
x21
t (t) 0 t
- 01 )x3(t) = sin , t?u ( t – t ) 30
x3
1
(t
t 0 t)
0
4)x(t) = sin[ , ( t – t ) ]?u ( t – t ) 200
x41
t (t) 0 t
- 01
2-2 已知波形图如图2-76所示,试画出经下列各种运算后的波形图
x(t)1
t -1 0 1 2 3
图 2-76
1
(1)x ( t-2 )
x 1
( t-2
t
) -1 2 3 4 0
1
(2)x ( t+2 )
x
1
( t+2 )
t ---0 1 -
4 3 2 1
(3)x (2t)
x(2t)1
t -1 0 1 2 3
(4)x ( t/2 )
x 1
( t/2
t
) 0 1 2 3 4 --
2 1
(5)x (-t)
x (-t)1
t
-2 -1 0 1 2 -3
2
(6)x (-t-2)
x (-t-2)1
t
-----0 1
5 4 3 2 1
(7)x ( -t/2-2 )
x ( -t/2-2 ) 1
t
-8 -7 -6 -5 -4 -3 -2 -1 0 1
(8)dx/dt
dx/dt1
t -2 -1 0 1 2 3
-δ (t-2) 2-3 应用脉冲函数的抽样特性,求下列表达式的函数值
,,
x(t,t)0δ(t) dt = x(-t) (1)0,,,
,,
x(t,t)0(2)δ(t) dt = x(t) 0,,,
,,tt00,(t,t)0(3) u(t -) dt = u() ,,,22
,,
,(t,t)0(4) u(t – 2t) dt = u(-t) 00,,,
,,t,,,e,t2(5)δ(t+2) dt = e-2 ,,,
3
1,,,,
,,t,sintδ(t-) dt = + (6),,,662
,,jt,,,,,,,,e,t,,t,tdt0(7) ,,,
,,,,jtjt,,,,,,e,tdte,(t,t)dt0=– ,,,,,,
,j,t0e= 1- = 1 – cosΩt+ jsinΩt0 0
2-4 求下列各函数x(t)与x(t) 之卷积,x(t)* x(t) 1212
-at(1) x(t) = u(t), x(t) = e ? u(t) ( a>0 ) 12
t,,1,a,,a,,ated,u,()eu(t,,)d,(1,e)x(t)* x(t) = = = 12,,0,,a
,(2) x(t) =δ(t+1) -δ(t-1) , x(t) = cos(Ωt + ) ? u(t) 124
,,,
[cos(,t,)u,()],[(t,,,1),,(t,,,1)]d,x(t)* x(t) = 12,,,4
,,
= cos[Ω(t+1)+]u(t+1) – cos[Ω(t-1)+]u(t-1) 44(3) x(t) = u(t) – u(t-1) , x(t) = u(t) – u(t-2) 12
,,
[u,(),u,(,2)][u(t,,),u(t,,,1)]d,x(t)* x(t) = 12,,,
当 t <0时,x(t)* x(t) = 0 12
t
d,当 0
2
E2,,,, x(t) = [δ( t +) +δ( t–)–2δ(t)] ,22
由微分特性
,,j,j,,2E2,E,222 (,,2),(2cos,2)ee ( jΩ) X(Ω) = 2,,
E,,,2Sa() X(Ω) = 24
2-13已知矩形脉冲的傅里叶变换,利用时移特性求图2-82所示信号的傅里叶变换,并大致画出幅度谱
,,G(t) 解: = E [ u( t + )–u( t–)] ,22
,,E Sa(),G(,) = ,2
,,GG x(t) = ( t + )–( t–) ,,22
由时移特性和线性性
14
,,,,,,,,j,j22ESaESa () (),,ee– X(Ω) = 22
,,j,j,22,ee,,,,,,E Sa()E Sa(),,sin = ?2j = 2j 2j222
2Eτ
Ω ,,,,22- - 0 ,,,,
2-14已知三角脉冲x(t)的傅里叶变换为 1
E,,,2Sa() X(Ω) = 124
,
试利用有关性质和定理求x(t) = x(t–) cosΩt的傅里叶变换 2102
解:由时移性质和频域卷积定理可解得此
由时移性质
,-j,,2X (,) e F [x (t–)] = 112
由频移特性和频域卷积定理可知:
1F [x(t )cosΩt]= [X(Ω–Ω)+ X(Ω+Ω)] 0002
,
X (Ω) = F [x (t–)cosΩt] 2102
,,,,,,00,,,j,j122ee = [ X (Ω–Ω) + X(Ω+Ω) ] 1002
15
,,,,,,00jj,,,,()(),,,,,,,,E,002222ee = [Sa+ Sa] 444
)的傅里叶逆变换x(t) 2-15求图2-82所示X(Ω
|X(Ω)| |X(Ω)|
A A
Ω Ω -Ω 0 Ω -Ω 0 Ω 0000
φ(Ω) φ(Ω)
π/2 π/2
Ω Ω -Ω 0 Ω -Ω 0 Ω 0000
-π/2 -π/2 a) b)
,j,(,)e 解:a) X(Ω) = | X(Ω)|
,jt0G(,)e = ,20
由定义:
,,1j,tX(,)ed, x(t) = ,,,,2
1,0jtjt,,0Aeed, = ,,,2,0
A,0j(tt),,ed0, = ,2,,,0
Aj,(t,t),00e| = ,,0,,2j(tt)0
A
sin[,(t,t)] = 00,(t,t)0
16
A,0Sa[,(t,t)] = 00,
,,1jt,x(t),X(,)ed, b) ,,,,2
,,,11jj,00,jtjt2,2AeedAeed,, =+ ,,,,002,2,
,,(),,AAj(t)j,t,,002eded2,, =+ ,,,,002,2,
,,(),,(),,AAjtjt2,002e|e| =+ ,,00,,,,2j2j
,AA,j(,t,)0,2e = ,2,j2(,,),jt20
,AAj(,t,)0,2e – ,2,j2(,,),jt20
A,A,Sa[,t,]sin[(,t,)] == 00,22,(,t,),02
2-16确定下列信号的最低抽样频率与抽样间隔 (1) Sa(100t)
2(2) Sa(100t)
2(3) Sa(100t)+ Sa(100t)
解:(1)由对偶性质可知:
Sa(100t)的频谱是个矩形脉冲,其脉宽为[-100,100]
即Ω = 100 =2πfmm
50
? f= m ,
由抽样定理 f ? 2fsm
50100? f? 2× = s ,,
,T? s100
17
(2) 由对偶性质可知
Sa(100t)的频谱是个矩形脉冲,其脉宽为[-100,100]
又由频域卷积定理可知
2(100t)的频谱是脉宽为[–200,–200]的三角形脉冲 Sa
即Ω = 200 =2πfmm
100 ? f= m ,
由抽样定理 f ? 2fsm
200100? f? 2× = s ,,
,T? s200
(3) 由线性性质可知
22Sa(100t)+ Sa(100t) 的频谱是Sa(100t)和Sa(100t)之和
?其Ω =2πf= 200 mm
100即 f= m ,
200则f ? 2f= m s,
,T? s200
2-17已知人的脑电波频率范围为0,45Hz,对其作数字处理时,可以使用的最大抽样周期T是多少,若以
T = 5ms抽样,要使抽样信号通过一理想低通滤波器后,能不是真的回复原信号,问理想低通滤波器的截
至频率f应满足什么条件, c
解:由已知条件,可知f = 45Hz m
由抽样定理f ? 2f= 90Hz sm x(f)
1? T ? 90
10001f T = 0.005 ? f= = = 200 s T5-45 0 45 由抽样定理和低通滤波可知
45 ? f? 200-45 = 155 c
x(f) 即45 ? f? 155 c
f 2-18若F[a(t)] = X(Ω), 如图2-85所示,当抽样
-45 0 45 200
脉冲p(t)为下列信号时,试分别求抽样后的抽样
Ω) X(1 信号的频谱X(Ω), 并画出相应的频谱图 s
(1) p(t) = cos t
Ω
18 -1 0 1
图 2-85
(2) p(t) = cos2 t
,,
,(t,2,n) (3) p(t) = ,n,,,
,,
,(t,,n) (4) p(t) = ,n,,,
解:由抽样特性可知
x= x(t) p(t) s
由频域卷积定理可知
1 X(Ω) = X(,)*P(,) s (Ω) X s,21
(1) P(Ω) = [δ(Ω+1)+δ(Ω-1)]
1/2 1? X(Ω) = X(,)*P(,) s Ω ,2
1-2 -1 0 1 2 = [X(,,1),X(,,1)]218 (1) (Ω) X s(2) P(Ω) = [δ(Ω+2)+δ(Ω-2)] 1
1? X(Ω) = X(,)*P(,) s 1/2 ,2
Ω 1 = [X(,,2),X(,,2)]2-3 -2 -1 0 1 2 3
18 (2)
,,,2,(,,n)(3) P(Ω) = ,1X (Ω) s2, n,,, ,2
,,
,(,,n) = ,Ω n,,,
-3 -2 -1 0 1 2 3 1? X(Ω) = X(,)*P(,) s 18 (3) ,2
,,1X(,,n) = ,,2n,,,
,,,2,(,,2n)(4) P(Ω) = ,,n,,,1 (Ω) X s ,,,
2,(,,2n) = ,n,,,Ω
1-3 -1 1 2 3 -2 0 ? X(Ω) = X(,)*P(,) s ,218 (3)
19
,,1X(,,2n) = ,,n,,,
X (1) = 2, X (2) = 0, X (3) = 2 ppp
3-1 解:序列频谱的定义为
,,,jn,j,x(n)eX(e), =
n,-,
,,,jn,j,,(n)eX(e),(1) = = 1
n,-,
,,,jn,j,-j3,,(n,3)eX(e),e(2) = =
,,n-
,,,jn,j,[0.5,(n,1),,(n),0.5,(n,1)]eX(e),(3) =
n-,,
j,,j,e,ej,-j, 0.5e0.5ecos , = + 1 += 1 + = 1 + 2
,,,njn,j,au(n)eX(e),(4) =
n,-,
,,,,,,njn,j,nae(ae),, = = (?0 < a < 1, ?收敛)
,n,0n0
1 = ,j,1,ae
,,jN,,N,11,e,,jnjn,,j,eR(n)e X(e),,N(5) = = = ,j,1,en,-,n,0
,N,,,NNN,jj,jsin222N-1ee,e2,-j 2 e,,, = ?= ,,jj,jsin222ee,e2
,,,jn,x(n,n)e,03-2 (1) DTFT[x(n-n)] = 0,,n-
,,,jn,,jm,0,,jnj,0m,n,nx(m)eeX(e)e,0=
,,m-
20
,,,,*,jn,jn,**x(n)e[x(n)e],, (2) DTFT[x(n)] = =
n,-,,,n-,,,,jn(,)**-j,[x(n)e]X(e),= =
,,n-
,,,,,,,jn,jm,()-j,x(-n)em,,nx(m)eX(e),, (3) DTFT[x(-n)] = =
n,-,,,,m
,,,jn,[x(n) * y(n)]e,(4) DTFT[x(n)* y(n)] =
n,-,
,,,,,,,,,,jn,,jnx(m)y(n,m)ex(m)y(n,m)e,,,,= =
,,,,,,,,,,n--mmn,,,,j,,jm,,j,jm,x(m)Y(e)eY(e)x(m)e,, = = m,,,m,,,
j,j,X(e)Y(e) =
,,,jn,x(n)y(n)e, (5) DTFT[x(n) y(n)] =
n,,,
,,,1,jjn,,jn,[X(e)ed,]y(n)e, = ,,,2,n,,,
,,,1,,,(,,)jjnX(e)[y(n)e]d,, = ,,,2,,,,n
,11,jj()j,j,,,,X(e)*Y(e)X(e)Y(e)d, = = ,,,,22,
,,d,jn,j,nx(n)ej[X(e)], (6) DTFT[nx(n)] = = d,n,,,
,,,jn,x(2n)e, (7) DTFT[x(2n)] =
n,,,
,,,,jm2m,2nx(m)e ,
m,,,
21
jmjm,,,,m1,,22m取整数[x(m)e,(,1)x(m)e],
m2,,,,,,,,,j11,jm,m22x(m)ex(m)(,e),, = +
22m,,,m,,,,,jj1122X(e)X(,e) = + 22
1j,j,2X(e)*X(e) (8) DTFT[x(n)] = ,2
,,,,,jn,,j2n,x(n)ex(2n)e,,aa (9) DTFT[x(n)] = = an,,,n,,,
,,,j2n,j2,x(n)eX(e), = =
n,,,
1,,10j,jn,jn,X(e)ed,ed,3-3 解:x(n) = = ,,,,,,202,,
jn,,jn,001ee1,jn,,0e|, = , = ,0n,2j,2jn
,n,n,,sinsin0000,Sa(n,) = = = 0,n,n,,0
3-4解: 由DFS的定义
,1Nnkx(n)W,pN X (k) = p,0n
,2 N,1,jn,02x(n)e ? X (0) = ,pp1 n,0n 3
x(n)-2 -1 0 1 2 3 4 5 , = p = 4
n,0图3-44
,3,jn2x(n)e, X (1) = = 2 + (–j ) ppn,0
+ 0 + j = 2
22
3,jn,x(n)e,pX (2) = = 2 + (–1 ) + 0 + (–1 ) = 0 p0n,
,33,jn2x(n)e X (3) = = 2 + j + 0 + (–j ) = 2? X (k)是周期函数,其周期长度N=4 ,pppn,0
, ? X (k) = Z[1+cos(k)]或 X (0) = 4, X (1) = 2, X (2) = 0, X (3) = 2 ppppp2
3-5 解: 由DFS的定义
2,N,1,jnkN X (k) = xe()np1,pn,0
2,21N,,jnk2N X (k) = xe()np2p,n,0
2,2,N,121N,,jnk,jnk2N2N+ =xe()nxe()npp,,n,0nN,
2,k2,kN,1N,1()mN,,jn(),jN2N2 mnN,,+ xe()nxe()mN,,,ppn,0m,0
2,kN,1m,jk,jk,N2 = X () + xe()mep1,p2m,0
kkk,jk,,jk, == ()(1 ),e()() ,eXXXp1pp11222
为奇数0 , k,, = ,kX为偶数2() , kp1,,2
3-6。解:与3-4答案相同,可由定义求出。
x(n) 只不过此时的x(k)非周期的。
2
, X (k) = Z[1+cos(k)]R(k) 1 p42n
或 X (0) = 4, X (1) = 2, X (2) = 0, X (3) p1p1p1p1-2 -1 0 1 2 3 4 5 = 2
图3-45 离散时间信号
3-8 解:(1)由定义得,
2,2,jnk3 Xk(),e,n,0
23
20? X(0),e,3,n,0
2,24,,2,jn,,jj333 X(1),e,,,10ee,,n,0
4,42,,2,jn,,jjn333 Xee(2),,e,,1,0,n,0
2,m (2)? Nm,,4,
2
?只要,N就取整数 m,1N,4
2,3,jnk,N ? Xk(),cosne,2n,0
3,,j0 ? X(0),cosne,,,,,10100,2n,0
,3,jn,2 X(1),cosne,,,,,10102,2n,0
3,,jn, X(2),cosne,,,,,10100,2n,0
3,3,jn,2 X(3),cosne,,,,,10102,2n,0
Xkk()1cos, k=0,1,2,3,,,
2,3,jnkN(3) Xk(),xne(),n,0
3
? X(0),xn(),5,n,0
,3,jn2 X(1),xn()ejjj,,,,,,,1(2)132,n,0
3,jn, X(2),xn()e,,,,,,,,,1(2)(1)(3)5,n,0
3,3,jn2 X(3),xn()ejjj,,,,,,,121(3)2,n,0
24
2,N,1,jnkNDFTxnXk()(),,xne(),,,n,03-9 解:(1)
2,N,1,jnkN? . . . X(0)1,X(1)1,XN(1)1,,DFTnXk,()1(),,,,()ne,,,n,0
? XkR()1,,(), 0,1,2,...,1kkN,,N
2,6,N,1,jk,jnkNN(2) DFTn,(3),,,,(3)ne,e,,,n,0
2,N,1NjkN,2,,jnk11,,aeannN,,(3) aeDFTa,,,,22,,,,,,jkjkn,0NN11,,aeae
22,,NN,,11,,jnkjwkn()0jwnjwn00NN,,(4) DFTe,,eee,,,,nn,,00
jwNjwNjk2,0011,,eee ,,22,,,,jwkjwk()()00NN11,,ee
2,2222,,,,NN,,11jnjnjnkjkn,,(),,NNNNN (5) DFTeeee,,,,,,nn,,00,,
jk2(1),,Nk , 1,,,1,e ,,,,,Nk(1),,2,jk(1),0 , 1 k,,,N1,e
,N,,N11,z,,nnZxnxnzzz()() (1),,,,3-10 解:(1) ,,,,,11,znn,,,,0
2,,jk2,N,1,jnk1,eNDFTxnxneNk()()(),,,, (2) ,,,2,,jkn,0N1,e
,,N1jwjnwjnw,,DTFTxnXexnee()()(),,, (3) ,,,,nn,,,,0
NNN,,jwjwjwjNw2221(),,eeee,,www,jw,,jjj1,e222eee(),
NsinwN,1,jw()22,ewsin2
25
jwXeN(), 当时, w,0
2,jwXe()0, 当时, ,wkN
11(4)由(3)可得,当x(n)由4点通过补零扩为10点时,此时的圆卷积和线卷积的
结果相同。由于线卷积的长度为4+4-1=7 ?可知x(n)由4点通过补零扩为最少7点时,圆卷积和线卷积相等。
,ln,,IDFTXklRkxnW()()(),,3-12 证明:频移定理为 pNN,,
由IDFT的定义可知,
,,IDFTXklRk()(),pN,,
2,N,1jnk1N,,Xkle(),pNk,0
22Nl,,1,,jnmj,ln,,1NN,Xmee()p,,,Nkl,,,,
2jln,,lnN,,xnexnW()()N
3-13 解:频移定理
,ln,,IDFTXklRkxnW()()(),, pNN,,
22,,jmnjmn,211,,mnmnNNmneeWW,,,,cos()()() (1)? NNN22
211,,,,mnmn,,,,DFTxnmnDFTxnWDFTxnW()cos()()(),, ? NN,,,,,,N22,,
由频移特性:
21,,,,,DFTxnmnXkmXkmRk()cos()()()(),,,, ppN,,,,N2,,
22,,jmnjmn,211,,mnmnNNmneeWW,,,,sin()()() (2)? NNNjj22
211,,,,mnmn,,,,DFTxnmnDFTxnWDFTxnW()sin()()(),, ? NN,,,,,,Njj22,,
由频移特性:
21,,,,,DFTxnmnXkmXkmRk()sin()()()(),,,, ppN,,,,Nj2,,
3-14
解:由DFT的定义可知,
26
22,,rNN,,11,,jnkjnkrNrNDFTynynexne()()(),,,,,,nn,,00 2kN,1,,jn()kNr,,xneX()(),rn,0
3-15 证明:频域圆卷积定理, ynxnhn()()(), 若 则 YkXkHk()()(),,
N,11 ()()(),,XlHklRl ,pNNl,0
N,11 ()()(),,HlXklRl,pNNl,0
N,1,nkYkDFTynxnhnW()()()(),,,,,Nn,0
NNN,,,1111,,,,nklnnk,, ()()()(),,xnIDFTHkWxnHkWW,,,,,NNN,,,,Nnnl,,,000,, NN,,111()kln, ()(),HlxnWN,,Nln,,00
N,11 ()()(),,HlXklRlpN,Nl,0
N,11YkXlHklRl,,()()()() 同理可证 ,pNNl,0
3-16 证明:由卷积的定义可知
,
xnnxmnmxn()()()()(),,,,,,(1) ,m,,,
,xnnnxmnnmxnn()()()()(),,,,,,,,,(2) ,000m,,,
1111,3Ts,,,0.02Ts,,,,3-19解:(1)(2)0.510 max1minF50f,221000,,n,,
T0.0261 (3) ?N,,264 ,,,N40min,3,T0.510
7N,,2128 (4)分辨力提高一倍,则,则N,80,取 Ts,0.041min
Nnn,,,,DFTRnNk()(),,3-17解:18..又 (1)(1)()()DFTDFTRnXk,,,,,,,NN,,,,2
Nn,, ? (1)(),,,,DFTNk,,2
27
(0) (0) (0) x(0) xxxX(0) 123x(1) x(1) x(1) x(8) X(1) 123x(2) x(2) x(2) x(4) X(2) 123x(3) x(3) (3) X (3) x(12) x123x(4) x(4) x (4) X (4) x(2) 123x(5) x(5) x (5) X (5) x(10) 123x(6) (6) (6) X(6) x(6) xx123x(7) x(7) (7) X (7) x(14) x123x(8) x(8) x (8) X(8) x(1) 123(9) x(9) x (9) X (9) x(9) x123x(10) x(10) (10) X (10) x(5) x123
x(13) x(11) x(11) x (11) X (11) 123x(12) (12) (12) X (12) x(3) xx123x(13) x(13) x (13) X (13) x(11) 123x(14) x(14) x (14) X (14) x(7) 123x(15) x(15) x (15) X(15) x(15) 123
NN,,11,1N222221(21),,nnkrrkrrk3-18 解: DFTXNWWW()(1)(1)(1),,,,,,,,,,,NNN,,,,,000nrr
NN,,1122rkkrk ,,WWW,,NNNrr,,0022
Nnn,,,, (1)(1)()()DFTDFTRnXk,,,,,N,,,,2
28
5-1 用冲击响应不变法求相应的数字滤波器系统函数H(z)
s,3 1)H(s) = a2s,3s,2
s,12)H(s) = a2s,2s,4
解:由H(s)分解成部分分式之和 a
s,3s,312 1)H(s) = ==– a2(s,2)(s,1)s,1s,2s,3s,2
,T,T,1121,e(1,2e)z?H(z) = –= ,T,1,2T,1,T,T,1,3T,21,ez1,ez1,e(1,e)z,ez
11
s,1222)H(s) = =+ a,,2j,js,2s,433s,2es,2e
11
,T,1221,ecos(3T)z,,?H(z) = += ,jj,T,1,2T,233,2Te,1,2Te,11,2ecos(3T)z,ez1,ez1,ez
5-2 设h(t)表示一个模拟滤波器的单位冲击响应 a
,0.9t
e , t?0
h(t)= a
0 , t,0
(1)用冲击响应不变法,将此模拟滤波器转换成数字滤波器,确
定系统函数H(z)(以T作为参数)
(2)证明,T为任何值时,数字滤波器是稳定的,并说明数字滤
波器近似为低通滤波器,还是高通滤波器
,0.9te 解:(1)? h(t)= u(t) a
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1 ? H(s) = as,0.9
1 ? H(z) = ,0.9T,11,ez
1 (2)? H(z) = ,0.9T,11,ez
,0.9Te 则其极点为z=
? T > 0 ? |z| < 1
,jej,H(z)|j,e) = = H(j,,0.9Tz,ee,e
j,e可以看出当ω?时,| H() |?
? 是低通滤波
5-3 图5-40是由RC组成的模拟滤波器,写出其系统函数H(s),并a选用一种合适的转换方法,将H(s)转换成数字滤波器H(z) a
解:由回路法可知(这是一个高通滤波器)
C
dUtdxtdyt()()()caa(t) R y(t) xRCRCRCy(t)== – aaadtdtdt
Y(s)RCs? == H(s) aX(s)1,RCs
由于脉冲响应不变法只适宜于实现带通滤波器,所以最好用双线
性变换法实现H(z)
,12RC1,z,,1,1T1,z2RC(1,z)H(s)|,1a21,z,1?H(z) === ,1,2s,,2RC1,z(T,2RC),(T,2RC)z,z,1T1,,1,z,1T1,z
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,c5-4 设模拟滤波器的系统函数为H(s)= ,式中Ω是模拟滤波acs,,c器的3dB带宽,利用双线性变换,设计一个具有0.2π的3dB带宽的
单极点低通数字滤波器
解:由预畸可知
210.65tan(,0.2,),== cTT2
0.65
T H(s) = ?a0.65s,T
由双线性变换法可得
0.65
,1T0.65(1,z)H(s)|,1a21,z,1H(z) === ,1s,,21z0.65,2.65,1.35z,1T1,z,,,1T1zT,
5-5 要求通过模拟滤波器设计数字滤波器,给定指标:3dB截至角频
率ω=π/2,通带内ω=0.4π处起伏不超过1dB,阻带内ω=0.8π处cps衰减不小于20dB,用Butterworth滤波特性实现
(1)用冲击响应不变法
(2)用双线性变换法
解:(1)用冲击响应不变法
? 先将数字指标转换为低通原型模拟滤波器指标
,0.4,p,== pTT
,0.8,s,== sTT
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?设计模拟滤波器,求出H(s) aButterworth的频响函数为
12|H(j,)|= a,2n1,()
,c
111,10H(j,)10? === ap,,pp2n2n1,()1,()
,,cc
2011,10H(j,)10=== as,,2n2nss1,()1,()
,,cc
2,101lg()110,101? n = =2.14 ,s2lg(),p
? 取 n = 3
,? 求 c
12,2|H(j,)|10= = a,2ns1,()
,c
,,0.8s
? ω= rad/s = = 0.372π 6c 629910,1
,c,,? = 设T = 1, 则 = 0.372π ccT
? 求H(s)查表可得 a
1,H(s), a2,,,(s,1)(s,s,1)
1,()|,Hs? H(s) = asa2,s,sss,c(,1)(,,1)2,,,ccc
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? 由冲击响应不变法
先将H(s)分解成部分分式 a
AAA123
++ H(s) =as,ss,ss,s123
=
AAA123
则H(z) =++ ,sT,1sTsT,,,1,13121,ez1,ez1,ez
=
(2)用双线性变换法
?由预畸求模拟滤波器原型指标
1.453,2ptan,== pT2T
0.155,2stan,== sTT2
?设计模拟滤波器,求出H(s) a
Butterworth的频响函数为
12|H(j,)| = a,2n1,()
,c
11,10H(j,)10? == ap,p2n1,()
,c
201,10H(j,)10== as,2ns1,()
,c
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2101,lg()1
10101, ? n = =1.51 ,s2lg(),p
取n =2
, ?求 c
12,2|H(j,)|10 == 取T=1 as,2n1,()
,c
,6.155s
, ? =rad/s = = 2.862 6c269910,1
?求H(s) a
查表可得:
1
,H(s) = 2a,,s,1.4142s,1
1,H(s)| H(s) = = asa2,s,ss,c,1.4142,12,,cc
=
?由双线性变换法求
H(s)|,1a21,z H(z) == s,,,1T1,z
5-6 已知图5-41h(n)是偶对称序列N=8,h(n)是h(n)圆周位移后的121
序列。设H(k)=DFT[h(n)], H(k)=DFT[h(n)] 1122
(1) 问|H(k)| = |H(k)|是否成立,θ(k)与θ(k)有什么关系, 1212
(2) h(n),h(n)各构成低通滤波器,试问它们是线性相位的,延时12
34
是多少,
(3) 这两个滤波器的性能是否相同,为什么,若不同谁优谁劣,
解:(1) 由DFT的时移定理
mkWX(k) DFT[x(n-m)R(n)]= 可知 pNN
H(k)和H(k)只有相位差,幅值相等,即有 12
|H(k)| = |H(k)| 12
mkWθ (k)和θ(k)相差12N
,2,j4k,jk,4k8eWe即θ(k)–θ(k)= == 218
(2) ? 无论h(n),h(n)都是偶对称序列 12
? 所以他们构成的低通滤波器具有线性相位
N,18,1
延时 α===3.5 22
(3) 不相同,相位相差kπ
h(n)要优于h(n),因为其相位滞后时间少 12
5-7用矩形容器设计一个近似理想频率响应的FIR线性相位的数字滤
,j,,e , 0, |,|,,波器 c
j,e H() = d
,, |,|,, 0 , c
(1) 求出相应于理想低通的单位脉冲响应h(n) d
(2) 求出矩形窗设计法的h(n)表达式确定τ与N之间的关系
(3) N取奇数或偶数对滤波特性有什么影响,
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,1jjn,,H(e)ed, 解:(1) h(n)= dd,,,2,
sin[,(n,,)]1,c,cjjn,,,eed, = = ,,,(n,,),2c,
(2) h(n)= h(n) R(n), h(n)只能取偶对称序列,由线性相位 dN
N,1 τ= 2
(3) 由于N无论取奇数还是偶数,都可实现低通滤波,而且只
N,1要N的取值使h(n)为关于的偶对称函数,就能保证线性相关,2
另外N的大小,只影响余振的多少和过滤带的窄宽,不会影响阻带良域。
5-8用矩形容器设计一个线性相位高通FIR数字滤波器
j,,,e , ,, |,|,, c
j,e H() = d
0, |,|,, 0 , c
(1) 求出响应于理想高通的单位脉冲响应h(n) d
(2) 求出矩形窗口设计法的h(n)表达式,确定τ与N之间的关系
(3) N的取值有什么限制,为什么,
,1jjn,,H(e)ed, 解:(1) h(n)= dd,,,2,
11,,,cjjn,jjn,,,,,,,eed,eed, = + ,,c,,,22,,
11,,,j(n,)j(n,),,,,ed,ed, = + ,,cc,,22,,
,1,j(n,)j(n,),,,,[e,e]d, = ,c,2,
36
,1cos[,(n,,)]d,= ,c,,
1,sin[,(n,,)]|= ,c(n,),,
,,sin[(n,)]1csin[(n,,),]= – (n,),(n,,),,
,cSa[,(n,,)]Sa[,(n,,)]= – c,
? h(n)仍然是偶函数 d
(2) h(n)= h(n) R(n) dN
? h(n)为偶对称序列,要保持滤波器具有线性相位,则须有
N,1τ= 2
(3) 这是一个高通滤波器,由于h(n)为偶对称,而当N取偶数时,所得到的滤波器不能实现高通特性
? N只能取奇数
5-9考虑一个长度为M=15的线性相位FIR滤波器,设滤波器具有对称单位样值响应,并且它的幅度响应满足条件
1, k = 0, 1, 2, 3
2k,
H() = 15
0, k = 4, 5, 6, 7 确定该滤波器的系数h(n)
,jH(e)|a2, 解:由于H(k) = ,k,N
37
2,N,1jnk1NH(k)e ? h(n) = IDFT[H(k)] = ,Nk,0214,j0k115H(k)e ? h(0) = = 1 ,15k,0
28j,21415,jk11,e115,H(k)e h(1) = =2 ,,15j15k,0151,e
56j,41415,jk11,e115,H(k)e4 h(2) = = ,,15j15k,0151,e
84j,61415,jk11,e115,H(k)e6 h(3) = = ,,15j15k,0151,e
h(4) = 0
h(5) = 0
h(6) = 0
h(7) = 0
由频率特性可知,这是一个低通滤波器
N,115,1
? 要取h(n)关于α===7这一点偶对称时,可实22现低通滤波(奇对称时,无法实现低通滤波)
? 取 h(8) = h(6)
h(9) = h(5)
h(10) = h(4)
h(11) = h(3)
h(12) = h(2)
h(13) = h(1)
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h(14) = h(0)
5-10设FIR滤波器的系统函数为
-1-2-3-4H(z) = 0.1(1+0.9z+2.1z+0.9z+z) 求出滤波器的单位抽样响应,判断是否具有线性相关,并求出其幅度特性和相位特性,画出其直接型结构和线性相位型结构
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