HTML 版 …swk/lecture/yaruodsp.pdf · やる夫で学ぶディジタル信号処理...
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HTML: http://www.ic.is.tohoku.ac.jp/~swk/lecture/yaruodsp/main.html
!
, ver. 2016.01.08
http://www.ic.is.tohoku.ac.jp/~swk/lecture/yaruodsp/main.html
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ver. 2016.01.08
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1
1 10
1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2 20
2.1 sin cos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 31
3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.3.1 sinc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 52
4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
5 61
5.1 . . . . . . . . . . . . . . . . . . . . 61
5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
6 68
6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
6.6 4 . . . . . . . . . . . . . . . . . . . . . . . 76
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2
7 (1): 79
7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
8 (2): 84
8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
9 (3): 94
9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
9.3 . . . . . . . . . . . . . . . . . . . . . . . . . . 97
9.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
9.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
10 104
10.1 . . . . . . . . . . . . . . . . 104
10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
10.6 4 . . . . . . . . . . . . . . . . . . . . . . 117
11 120
11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
11.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
12 128
12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
12.2 . . . . . . . . . . . . . . . . . . . . 129
12.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
13 135
13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
13.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
13.4 . . . . . . . . . . . . . . . . . . . . . . . . . . 146
13.5 c+ j s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
13.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
13.7 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
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3
14 z 160
14.1 z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
14.2 z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
14.3 z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
14.4 z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
14.5 es z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
15 172
15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
15.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
15.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
15.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
16 194
16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
16.2 FIR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
16.3 IIR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
16.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
16.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
A 211
A.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
A.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
A.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
A.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
A.5 1/(s )p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220A.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
A.7 1/(1 z1)p z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
B 228
B.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
B.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
, ver. 2016.01.08
-
4
4(2013
) 2
3 II I II
4
II III
()
()
()
2011 8
(2000)
, : MATLAB, , 2000.
Alan V. Oppenheim and Ronals W. Schafer: Discrete-time Signal Processing, Prentice Hall, 1998.
: , , 2004.
H. P. , : , , 1979.
http://www.google.co.jp/search?q=vV[Y
-
5
2011 10twitter
()
()
1. (p. 10)
2.4
2. (p. 20)
3. (p. 31)
(
)
()
4. (p. 52)
5. (p. 61)
4
6. (p. 68)
4
, ver. 2016.01.08
http://topsy.com/www.ic.is.tohoku.ac.jp/~swk/lecture/yaruodsp/main.htmlhttp://b.hatena.ne.jp/entry/www.ic.is.tohoku.ac.jp/~swk/lecture/yaruodsp/main.html
-
6
7. (1): (p. 79)
4
()
8. (2): (p. 84)
()
9. (3): (p. 94)
10. (p. 104)
4
11. (p. 120) FFT
OK
12. (p. 128) () 9
FIR
IIR (2000)
()
13. (p. 135) z
(
) ( s
) ddtf(t)
sF (s) sF (s) f(0)
14. z (p. 160) z
, ver. 2016.01.08
-
7
15. (p. 172)
( Oppenheim & Schafer (1998)
) (2000)
16. (p. 194)
2016.01.08
(2.27) F3 Fn sinc 0 0 (t = 1,2, 0 )
FS ()
2014.11.03
13 u0(t) () u0(0) = 1 13.1
(
)
t = 0 u0(t)
t = 0
4
>>73
4.3 cos 2n
()
>>73
>>81
() Bs () L
2013.11.02
()
(
)
2013.10.27 ()
2 1
()
2013.10.5
, ver. 2016.01.08
http://uni.2ch.net/test/read.cgi/sci/1389290461/
-
8
7
9:
()
14: t = 0 1
2013.05.11
4: (t) ejt1 (t) (t t1) (4.44)(4.45) T1 t1
2012.08.04
7: 3 14: Yi ( wi )
17: rad rad/s
()HTML
2012.07.15
16: x[n] = cos 2n 0 cos 25 n(2n/5 4n/5 ) 17: k
()
2012.07.13 ()
z
2012.01.22
:
, ver. 2016.01.08
http://matsucon.net/material/mona/
-
9
4.4: (
)
7.6: (?)
14.4:
14.5: s = j ? H2(s) H2(n)
2012.01.16 3: k 0 Fk k 0 Fk 10 (10.8) () (
Takayoshi Kawada (@takkaw) )1.2
2011.08.18 17
PDF ()
:
11: Gs/2,s/2() Gs/2,s/2() 14: a c < 1 a c < 0 15: ej X(ej)
2011.08.04 16
2011.07.25 15 z 14 (aF (s) + bF (s)
aF (s) + bG(s) ) 1514
s
2011.07.16 146 F (ej) F (j)
z
2011.07.09 13
2011.07.08 12
2011.07.03 11)HTML ()
HTML PDF
2011.06.23 10 (3)
2011.06.16 8 (1) 9 (2)
2011.06.03 1 typo
(
)
2011.05.27 7
2011.05.20 4
, ver. 2016.01.08
http://twitter.com/takkaw/
-
10
1
1.1
t
t f(t)
x f(x)
2 f(x, y)
f(x, y, t)
1
t x
f(t)
?
! sin cos !
-
1 11
10 Hz20 Hz30 Hz
10 Hz20 Hz30 Hz
?
?
!
117
440 Hz 880 Hz
?
?
1.2
?
f(t)
T0
T0/2 T0/2
t
f(t)
T0/2T0/2T
0
, ver. 2016.01.08
-
1 12
sin cos
f(t) = a0 +k=1
{ak cos
(2k
T0t
)+ bk sin
(2k
T0t
)}(1.1)
?
sin cos
a0
0
OK cos sin
k = 1 cos sin 2/T0 T0 2 1 1 cos
sin
t
f(t)
T0/2T
0/2
a1
b1
k = 2 2
t
f(t)
T0/2T
0/2
a2
b2
k
k
?
k = 1.5 k = 4.3
1 2 3 15
, ver. 2016.01.08
-
1 13
k = 1
0 = 2/T0 T0
?
[s] [Hz] [rad/s] 2
0 = 2/T0 !
0 (1.1)
f(t) = a0 +
k=1
{ak cos (0kt) + bk sin (0kt)} (1.2)
f(t)
sin cos ()
t
f(t)
T0/2T
0/2
a0
k = 0
k = 1
k = 2
a1
cos 0t
b1
sin 0t
a2
cos 20t
b2
sin 20t
0.01 s 100 Hz ( = 2 100 [rad/s]) f(t) 100 Hz 200 Hz 300 Hz
102 Hz 250 Hz
0.01 s
0.01 s
, ver. 2016.01.08
-
1 14
! !!
!
(1.1) =
(1.1)
f(t) ak bk
1.3
?
!
a3 (1.1) a3
a3
a0 T0/2 T0/2
T0/2T0/2
f(t)dt =
T0/2T0/2
{a0 +
k=1
{ak cos
(2k
T0t
)+ bk sin
(2k
T0t
)}}dt (1.3)
T0/2T0/2
f(t)dt =
T0/2T0/2
a0dt+
k=1
{ T0/2T0/2
ak cos
(2k
T0t
)dt+
T0/2T0/2
bk sin
(2k
T0t
)dt
}
(1.4)
a0 0 ?
?
, ver. 2016.01.08
-
1 15
cos sin T0/2 T0/2 ? 0
m
T0/2T0/2
cos(2m
T0t)dt =
T0/2T0/2
sin(2m
T0t)dt = 0 (1.5)
a0 T0/2T0/2
f(t)dt =
T0/2T0/2
a0dt = T0a0 (1.6)
a0
a0 =1
T0
T0/2T0/2
f(t)dt (1.7)
a0 1
a0 sin cos
0 a0
?
(1.5) mn
T0/2T0/2
cos(2m
T0t) cos(
2n
T0t)dt =
T02m,n (1.8)
T0/2T0/2
sin(2m
T0t) sin(
2n
T0t)dt =
T02m,n (1.9)
T0/2T0/2
cos(2m
T0t) sin(
2n
T0t)dt = 0 (1.10)
m,n m = n 1
0
cos sin 2
0 ;
sin cos 0 sin cos
0
, ver. 2016.01.08
-
1 16
0
0
(1.5) (1.8)
22
[f1, f2, f3] [g1, g2, g3] 2 3
3
i=1 figi
?
f1
f2
f3
g1g2
g3
f(t)
g(t)
a b
3
?
a3 a3 cos(
23T0
t) (1.1) cos
(23T0
t)
T0/2T0/2
f(t) cos
(2 3T0
t
)dt =
T0/2T0/2
{a0 +
k=1
{ak cos
(2k
T0t
)+ bk sin
(2k
T0t
)}}cos
(2 3T0
t
)dt
(1.11)
, ver. 2016.01.08
-
1 17
a3 T0/2T0/2
f(t) cos
(2 3T0
t
)dt =
T0/2T0/2
a3 cos
(2 3T0
t
)cos
(2 3T0
t
)dt =
a3T02
(1.12)
a3
a3 =2
T0
T0/2T0/2
f(t) cos
(2 3T0
t
)dt (1.13)
a0
a0 =1
T0
T0/2T0/2
f(t)dt (1.14)
ak =2
T0
T0/2T0/2
f(t) cos
(2k
T0t
)dt (k = 1, 2, 3, ) (1.15)
bk =2
T0
T0/2T0/2
f(t) sin
(2k
T0t
)dt (k = 1, 2, 3, ) (1.16)
f(t) cos(
2kT0
t) f(t) cos
(2kT0
t)
sin(
2kT0
t) f(t) sin
(2kT0
t)
T0 f(t) () (1.1) f(t)
(1.14)
f(t) 1/2 1/3 1/4
?
a0 a0/2
?
(1.14) 3 a0 2
1 2 ?
2 k = 0 cos 1 1 2
, ver. 2016.01.08
-
1 18
2 a0
2 (1.1) 1/2
1.4
?
! !!
(1.1) f(t) sin cos
(1.14)
, ver. 2016.01.08
-
1 19
f(t)
(1.14) (1.1)
f(t) f(t)
f(t) a0 +k=1
{ak cos
(2k
T0t
)+ bk sin
(2k
T0t
)}(1.17)
=
f(t)
f(t)
, ver. 2016.01.08
-
20
2
2.1 sin cos
10 ms
0 Hz100 Hz200 Hz300 Hz
cos sin
100 Hz 100 Hz cos 100 Hz sin
? cos sin
sin cos ?
1/4
0 sin
1 1/4 cos
100 Hz 2.5 ms
200 Hz300 Hz 2
sin cos 1/4
200 Hz300
Hz400 Hz 1.25 ms0.833 ms0.625 ms
2
1 ?
1
ak cos
(2k
T0t
)+ bk sin
(2k
T0t
)(2.1)
1 sin
-
2 21
k
sin k cos
(2k
T0t
)+ cos k sin
(2k
T0t
)= sin
(k +
2k
T0t
)(2.2)
ak bk sin k cos k
sin2 k+cos2 k = 1a2k + b
2k
ak cos
(2k
T0t
)+ bk sin
(2k
T0t
)(2.3)
=a2k + b
2k
(ak
a2k + b2k
cos
(2k
T0t
)+
bka2k + b
2k
sin
(2k
T0t
))(2.4)
(
bka2k+b
2k
, aka2k+b
2k
)k = tan1 akbk
(2.4)
=a2k + b
2k
(sin k cos
(2k
T0t
)+ cos k sin
(2k
T0t
))(2.5)
=a2k + b
2k sin
(k +
2k
T0t
)(2.6)
x
y
1
1
k
cos sin 1 sin
=+
, ver. 2016.01.08
-
2 22
cos ak sin bk 1
a2k + b
2k k = tan
1 akbk
sin cos 1
Akk
f(t) =
k=0
Ak sin
(2k
T0t+ k
)(2.7)
?
?
sin k
sin cos
2.2
j
ej = cos + j sin (2.8)
?
, ver. 2016.01.08
-
2 23
ex+y = exey (2.9)
d
dxeax = aeax (2.10)
ej
Re
Im
1
j
exp(j) = cos + j sin
ej
rej r ej
rejej = rej(+) (2.11)
Re
Im
r
jrr exp(j)
r exp(j{ + })
r
r
(1.1) cos sin
, ver. 2016.01.08
-
2 24
ej ej
ej = cos + j sin (2.12)
ej = cos j sin (2.13)
cos sin
cos =ej + ej
2(2.14)
sin =ej ej
2j(2.15)
f(t) = a0 +
k=1
{ak
ej0kt + ej0kt
2+ bk
ej0kt ej0kt2j
}(2.16)
= a0 +
k=1
{ak jbk
2ej0kt +
ak + jbk2
ej0kt}
(2.17)
2/T0 0
k 1
1 ej0kt k = 1 2 ej0kt k = 1 ej0kt k = 1
a0 ej0kt k = 0
f(t) =
k=
Fkej0kt (2.18)
Fk
? ak bk
?
ck
?
f F
, ver. 2016.01.08
-
2 25
x(t) Xk ?
2.3
ak bk
: (1.5)
(1.8)
T0/2T0/2
ej0mt{ej0nt
}dt = T0m,n (2.19)
?
T0/2
T0/2ej0mtej0ntdt = T0m,n (2.20)
sin cos
(2.19)
m = n 1 T0
m = n 0 = 2/T0 T0/2
T0/2ej0mtej0ntdt =
T0/2T0/2
ej0(mn)tdt (2.21)
=1
j0(m n)[ej0(mn)t
]T0/2T0/2
(2.22)
=1
j0(m n){ej(mn) ej(mn)
}(2.23)
=1
j0(m n){(1)mn (1)mn} (2.24)
= 0 (2.25)
, ver. 2016.01.08
-
2 26
1 (mn)ej(mn) ej(mn)
1 1 (m n) 1 1
Re
Im
Re
Im
(m n) ej(mn) ej(mn) 180
F3
(2.18) {ej03t
} ej03t
T0/2T0/2
f(t)ej03tdt = T0/2T0/2
{ k=
Fkej0kt
}ej03tdt (2.26)
(p. 18)
T0/2T0/2
f(t)ej03tdt = T0/2T0/2
F3ej03tej03tdt = F3T0 (2.27)
Fk
Fk =1
T0
T0/2T0/2
f(t)ej0ktdt (2.28)
T0
Fk =1
T0
T0/2T0/2
f(t)ej 2kT0 tdt (2.29)
ej0kt ej0kt ej0kt
T0 f(t) () (2.18) f(t)
(2.29)
f(t) 1/21/3 1/4
, ver. 2016.01.08
-
2 27
f(t) Fk
f(t)FS Fk (2.30)
FS Fourier Series ()
2.4
sin cos (2.18) ej0kt
ej0kt sin0kt
0k t ?
!
ej0kt 0k sin
sin t 2
3
sin
t
Re
Im
3
ej
ej0kt t = 0 1 t
exp(j
0t)
t
Re
Im
rad0k k f(t)
k
, ver. 2016.01.08
-
2 28
(2.18) ? ?
?
sin cos k
k k
0k ? 10 !
?
?
!exp(j
0t) exp(-j
0t)
t t
Re
Im
Re
Im
(2.14)
2
cos 2
! cos
Re
Im
ej0kt ej0kt
f(t)
Fkej0kt Fkej0kt Fk Fk Fk Fk
Fk
|Fk| = |Fk|
, ver. 2016.01.08
-
2 29
Fk ?
Fkej0kt t = 0 Fk 0
Fk
2
Fk
Fk Fk = Fk
f(t) k 0 Fk k < 0
?
f(t)
cos sin ak bk
Fk Fk
Fk
(2.18)
Fk
akbk cos sin
a0
0 [rad/s]
a1
b1
a2
b2
0
20 30
a3
b3
, ver. 2016.01.08
-
2 30
-30-2
0-
03
02
0
00
|F0 |
[rad/s]
-30-2
0-
03
02
0
00
F0
F1F
2
[rad/s]
|F1 |
|F2 |
|F-1 |
|F-2 |
F-1
F-2
|Fk| Fk
f(t) 2
|Fk|2
2?
2
, ver. 2016.01.08
-
31
3
3.1
?
?
T0 ?
0 = 2/T0
-30
-20
-0
30
20
0
0
2/T0
= 0
T0/2-T
0/2
tFS
FS ?
(2.30)
1
2T0 ?
?
T0
-T0
t
12
-
3 32
2T0
-2T0
t
-
T0
0 = 2/T0
2T0 ?
2
2/2T0 0/2
2 1/2
2
T0
-T0
t
-30
-20
-0
30
20
0
0
2/2T0
= 0 /2
FS
4T0 ?
0/4 4
-30
-20
-0
30
20
0
0
2/4T0
= 0 /4
2T0
-2T0
t
-
FS
?
, ver. 2016.01.08
-
3 33
OK
3.2
(2.18)
f(t) =
k=
Fkej0kt (3.1)
Fkej0kt k ?
OK
0
Fkej0kt Fk
ej0kt
-30
-20
-0
30
20
0
0
f(t)
, ver. 2016.01.08
-
3 34
-30
-20
-0
30
20
0
0
=
0
Fkej0kt ?
Fkej0kt/0 ? 0
f(t)
f(t) =
k=
0Fke
j0kt
0(3.2)
1/2
f(t) =1
2
k=
02Fke
j0kt
0(3.3)
[k] = 0k k
f(t) =1
2
k=
02Fke
j[k]t
0(3.4)
k
2Fk/0 F ([k]) Fk
f(t) =1
2
k=
0F ([k])ej[k]t (3.5)
, ver. 2016.01.08
-
3 35
F () [k] ? F [k]
F [k] k
k [k]
F ([k])
F ([k]) Fk (2.28) 0k [k]
Fk =1
T0
T0/2T0/2
f(t)ej[k]tdt (3.6)
F ([k]) = 2Fk/0
F ([k]) =2
T00
T0/2T0/2
f(t)ej[k]tdt (3.7)
=
T0/2T0/2
f(t)ej[k]tdt (3.8)
0 = 2/T0 F ([k]) [k]
f(t) = = 0
f(t) =1
2
k=
F ([k])ej[k]t (3.9)
0 ?
T0 [k] d
f(t) =1
2
F ()ejtd (3.10)
(3.8) F ([k]) T0
F () =
f(t)ejtdt (3.11)
(3.10)
, ver. 2016.01.08
-
3 36
t t
t
?
?
?
F ()
Fk
?
(3.3) 1/2
, ver. 2016.01.08
-
3 37
?
1/2
1/2
1/2
1/2
?
?
2
< t < f(t) () (3.11) F () f(t) (
)
F () (3.10) f(t) (f(t) F ())
F () f(t) ()
|F ()| F ()|F ()|2 f(t)
?
?
, ver. 2016.01.08
-
3 38
3.3
F () f(t)
f(t)F F () (3.12)
F ()F1 f(t) (3.13)
f(t) F () (3.14)F [f(t)] = F () (3.15)
F1[F ()] = f(t) (3.16)
3.3.1 sinc
a > 0
ra,a(t) =
{1, a t a0, otherwise
(3.17)
t
-a a
1
F [ra,a(t)] =
ra,a(t)ejtdt (3.18)
=
aa
ejtdt (3.19)
ejt (1/j)ejt
= 0
= 0
F [ra,a(t)] =[
1
jejt
]aa
(3.20)
=1
j
{eja eja} (3.21)
?
(2.15) sin = (ej ej)/2j
, ver. 2016.01.08
-
3 39
F [ra,a(t)] = 2 e
ja eja2j
(3.22)
=2
sin a (3.23)
= 0
?
F [ra,a(t)]|=0 = aa
ej0tdt (3.24)
=
aa
dt (3.25)
= 2a (3.26)
?
2/ sin sin
sin
= 0
sin
= 0 2a
a = 4
-2
-1
0
1
2
3
4
5
6
7
8
pipi/20-pi/2-pi
2 / x * sin(4 * x)
, ver. 2016.01.08
-
3 40
0
1
2
3
4
5
6
7
8
pipi/20-pi/2-pi
abs(2 / x * sin(4 * x))
Ga,a() =
{1, a a0, otherwise
(3.27)
t = 0
F1[Ga,a()] = 12
Ga,a()ejtd (3.28)
=1
2
aa
ejtd (3.29)
=1
t
ejat ejat2j
(3.30)
=1
tsinat (3.31)
t = 0
F1[Ga,a()]|t=0 = 12
aa
ej0d (3.32)
=a
(3.33)
sin
sin sin
sinc a = 1
t = 0 1 sinc t
sinc t =sin t
t(3.34)
sinc =sin
(3.35)
?
sinc t t = ,2,3,
, ver. 2016.01.08
-
3 41
sin
t = 1,2,3, a = t = 0 1
sint
t(3.36)
sinc sinc
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-4 -3 -2 -1 0 1 2 3 4
sin(pi * x) / (pi * x)
sinc
sinc
sinc ?
sinc
sinc
sinc
sinc
?
, ver. 2016.01.08
-
3 42
3.3.2
?
0
OK
(t) =
{, t = 00, t = 0 (3.37)
t = 0
t
0
(t)
(t)dt = 1 (3.38)
1 ?
t (t) 0
t = 0 t = 0
1
1
0 1
2 3
2(t) 3(t)
2 3
x(t)
(t)x(t)dt = x(0) (3.39)
, ver. 2016.01.08
-
3 43
t
0
(t)x(t)
x(0)
t = 0 0 x(t)
t = 0 x(0) x(0) ?
t = 0 t1
(t t1) t = t1
(t t1)x(t)dt = x(t1) (3.40)
t
t1
(t t1)
x(t)
x(t1)
t = t1
OK
?
(t) ?
F [(t)] =
(t)ejtdt (3.41)
, ver. 2016.01.08
-
3 44
t = 0
F [(t)] = ej0 (3.42)= 1 (3.43)
? ?
OK ?
(t) 1
?
t
0
(t)
1
1
t = 0
t1
(t t1)
F [(t t1)] =
(t t1)ejtdt (3.44)
= ejt1 (3.45)
?
ejt1 t1
t
t1
(t t1)
exp(-jt1)Re
Im
1
| exp(-jt1) | = 1
exp(-jt1) = -t
1
0
-1
t1
, ver. 2016.01.08
-
3 45
ejt1 (t t1) ?
?
|ejt1 | 1 (t) ejt1 = t1 t1
(t t1)(t) t1 ?
t1 1
2/
2t1
2/= t1 (3.46)
t1
2/
t
0
t = 0
t = t1
ejt1
ejt1 t1 = 0 1 (t t1) ejt1 t1 = 0 1
ejt1 !
F1[ejt1 ] = 12
ejt1ejtd (3.47)
=1
2
ej(tt1)d (3.48)
=1
2
1
j(t t1)[ej(tt1)
]
(3.49)
? ej(tt1)
, ver. 2016.01.08
-
3 46
sinc
?
! !
(t t1) ejt1 ejt1 (t t1)
() 1 ( 1)
F1[( 1)] = 12
( 1)ejtd (3.50)
=1
2ej1t (3.51)
1 = 0
F1[()] = 12
(3.52)
1 = 1
1/2 exp(j1t)
t
Re
Im
1
( - 1)
1 = 0 = 0
, ver. 2016.01.08
-
3 47
t1/2
()
0
cos1t
cos1t =ej1t + ej1t
2(3.53)
=1
2ej1t +
1
2ej1t (3.54)
?
2
OK
ej1t 2( 1) 1
2ej1t +
1
2ej1t (3.55)
F( 1) + ( + 1) (3.56)
1 1
( + 1)
-1
t
cos 1t
1
( - 1)
3.4
?
ej1t cos1t
?
, ver. 2016.01.08
-
3 48
T0
f(t)
T0
t
f(t)
f(t) =
k=Fke
j0kt (3.57)
F [f(t)] =
k=FkF [ej0kt] (3.58)
=
k=
2Fk( 0k) (3.59)
ej1t 2( 1)
0k k 0
F()
0
= 2 / T0
-30
-20
-0
30
20
0
0
f(t)
0 OK ?
(p. 13) 0
T0
0 = 0k = 0k
F (0k) =
f(t)ej0ktdt (3.60)
(2.28)
Fk =1
T0
T0/2T0/2
f(t)ej0ktdt (3.61)
, ver. 2016.01.08
-
3 49
1/T0 1
1
Fk T0
1
? 1
0
?
1
1
Fk
?
2Fk
T0
t
f(t)
-30
-20
-0
30
20
0
0
: Fk
0
= 2 / T0
-30
-20
-0
30
20
0
0
FS
, ver. 2016.01.08
-
3 50
(3.59) Fk(0k) 2Fk(0k) ? 2?
1/2 (p. 34)
Fk 0 F ()
?
F () ejt ?
0
0
0 Fk
2Fk
T0
= 2 / 0
t
f(t)
-30
-20
-0
30
20
0
0
: Fk
-30
-20
-0
30
20
0
0
t
, ver. 2016.01.08
-
3 51
(p. 34)
, ver. 2016.01.08
-
52
4
4.1
11 1
1
OK
20
?
()
()
Ts s sampling s
() fs Hz 2
() s rad/s 20
Ts = 20 106 [s]fs = 50 103 [Hz]s = 100 103 [rad/s]
, f(2Ts), f(Ts), f(0), f(Ts), f(2Ts), (4.1)
-
4 53
, f [2], f [1], f [0], f [1], f [2], (4.2)
t
f(t)
T0
n
f[n]
1
[s] [s]
12
4.2
?
?
1
Hz
?
? ?
1
, ver. 2016.01.08
-
4 54
?
1 rad 1
rad ?
1 rad
rad
rad/sample
() rad
= Ts (4.3)
Ts ?
1 rad Ts 1
1 rad
1
x[n] = cos1n (4.4)
1 = /4 [rad] ?
n n 1 cos /4 n 8
cos 1
n
cos[n/4]
1
OKcos 4n 8
8 = 2/(/4)
, ver. 2016.01.08
-
4 55
?
4.3
x[n] = cos1n (4.5)
1 = /4 1 = 2 ?
? 2 2/2 =
?
?
1 2 [rad]
cos 1 ? cos 1
n
cos[2n]
0
213
54
cos 2n
1
2
, ver. 2016.01.08
-
4 56
?
1 cos
x[n] = cos1n (4.6)
2 ?
1
1 + 2
cos (1 + 2)n (4.7)
?
cos (1 + 2)n (4.8)
= cos (1n+ 2n) (4.9)
= cos1n (4.10)
=x[n] (4.11)
n cos 2n n
x[n] 2 x[n] ? ?
1 sin
2
!
2 ?
1 2
1 1
cos1n 1 1
, ver. 2016.01.08
-
4 57
n
2 1 1
n
! 1 1
2
cost
0 2
, ver. 2016.01.08
-
4 58
= 0.0
= 0.2
= 0.4
= 0.6
= 0.8
= 1.0
= 1.2
= 1.4
= 1.6
= 1.8
= 2.0
?
= 2 = 0
2 4
, ver. 2016.01.08
-
4 59
= 2.0
= 2.2
= 2.4
= 2.6
= 2.8
= 3.0
= 3.2
= 3.4
= 3.6
= 3.8
= 4.0
2 0 2
=
2
=
0 2
0
, ver. 2016.01.08
-
4 60
, ver. 2016.01.08
-
61
5
5.1
? ?
?
?
?
f [n] n
?
?
f(t) f(t) f [n] f
f(t) f [n]
f(t) =
{f [n], t = n
0, otherwise(5.1)
n
f[n]
1
t
f(t)
1
t f [n] 0
f(t)
F () =
f(t)ejtdt (5.2)
? ?
1
-
5 62
f(t) ejt ? f(t)ejt t 0 0 (p. 50)
f [n]
0
5.2
t f(t)ejt
t = n 1/2 t = n+ 1/2 t = n f [n]ejn
t
f[n]e-jn
t = n 1/2 t = n + 1/2
1
F () =
n=f [n]ejn (5.3)
(Discrete-Time Fourier Transform; DTFT)
(3.11)
f [n] t f [n]
0
, ver. 2016.01.08
-
5 63
t
f(t)
1
f(t) =
n=
f [n](t n) (5.4)
1 n n f [n]
n f [n]
(3.11)
F () =
{ n=
f [n](t n)}ejtdt (5.5)
=
n=
f [n]
(t n)ejtdt (5.6)
=
n=
f [n]ejn (5.7)
(5.3)
[n 1/2, n+ 1/2] f [n]ejn
?
(p. 47)
T0
= 2 / 0
t
f(t)
-30
-20
-0
30
20
0
0
, ver. 2016.01.08
-
5 64
?
?
?
0
T0 = 2/0
1 2/1 2
?
2
2
0 3--3
t
f(t)
1
2 ?
?
2 (p. 55)
2
2
0.52.54.5
5.3
F () f [n] (3.10)
?
F () f [n]
f [n]
f [n] ?
, ver. 2016.01.08
-
5 65
? (p. 47)
1
1
?
f [n] =1
2
F ()ejnd (5.8)
f [n] (5.3) F ()
(5.8) f [n]
1
2
F ()ejnd (5.9)
=1
2
{ m=
f [m]ejm}ejnd (5.10)
(5.8) (5.3)
(5.8) n n m
= 12
m=
f [m]
ej(nm)d (5.11)
=1
2
m=
f [m] 2m,n (5.12)
=f [n] (5.13)
f [n]
?
(2.21) (2.25)
m = n
f [n]
(5.8)
, ver. 2016.01.08
-
5 66
n () f [n] ()
(5.3) F () f [n] (
)
F () (5.8) f [n] (f [n] F ())
F () f [n] ()
F () 2 < <
|F ()| F ()|F ()|2 f [n]
F [f(t)] = F () ?
f [n]DTFT F () (5.14)
F ()DTFT1 f [n] (5.15)
f [n] F () (5.16)DTFT[f(t)] = F () (5.17)
DTFT1[F ()] = f(t) (5.18)
?
F ()
F (ej) (5.19)
z
F ()
F (j) (5.20)
, ver. 2016.01.08
-
5 67
F () F ()
z
, ver. 2016.01.08
-
68
6
6.1
?
(5.3) (5.8)
F () =
n=f [n]ejn (5.3)
f [n] =1
2
F ()ejnd (5.8)
(5.3)
(5.8)
n
!
-
6 69
6.2
?
?
?
2
N
2
?
N
2/N 2/N
2
? 2 2/N N
?
1 N
1 N
N 1
, ver. 2016.01.08
-
6 70
N N
(Discrete Fourier Transform; DFT)
6.3
(5.3) (5.8)
F () =
n=
f [n]ejn (5.3)
f [n] =1
2
F ()ejnd (5.8)
2
n
0 3--3
DTFTf[n] F()
1
f [n] F ()
F () ?
(5.3)F () f [n] 1
n
0 3--3
DTFT
: 8 : 2
2/81
f[n] F()
(p. 64)
(p. 63)
f [n] F () 2/N
F () 2 N
f [n]
, ver. 2016.01.08
-
6 71
F ()
F ()
F () =
k=
ck( 2kN
) (6.1)
ck
F () 2/N = 2k/N
( 2k/N) ck ck
ck( 2kN ) = 2k/N ck k = F ()
F () 2 ck N
c0c1 cN1 cN c0
(5.8)
f [n] =1
2
k=
ck( 2kN
)ejnd (6.2)
=1
2
N1k=0
ckej 2knN (6.3)
= 2k/N
? 0 N 1 ? ?
2 N
N k
ck N 0
N 1
? (5.8) 0 2
?
0 2
0 N 1
ck = F [k] 2/N F [k]
, ver. 2016.01.08
-
6 72
(p. 37)
ck
f [n] =1
N
N1k=0
F [k]ej2N kn (n = 0, 1, , N 1) (6.4)
? ?
6.4
?
(5.3)
F () =
n=f [n]ejn (5.3)
f [n] F () = 2k/N
0
f [n] 1
1
F () = 2k/N F (2k/N)
k F [k]
F [k] =
N1n=0
f [n]ej2N kn (k = 0, 1, , N 1) (6.5)
1
(p. 49) (p. 64)
?
, ver. 2016.01.08
-
6 73
1
N
N1k=0
F (k)ej2N kn (6.6)
=1
N
N1k=0
[N1m=0
f(m)ej2N km
]ej
2N kn (6.7)
=1
N
N1m=0
f(m)
N1k=0
ej2N k(nm) (6.8)
=1
N
N1m=0
f(m)Nm,n (6.9)
= f(n) (6.10)
m = n
m = n
(6.5)
ck = F [k] 2/N
(6.5) (6.4)
N
N
nDFT
11
f[n]F[k]
k
0 1 N-1 N-1012 L 2 L
n () N f [n] (6.5) F [k] f [n] (
)
F [k] k N
F [k] (6.4) f [n] (f [n] F [k] )
k F [k] f [n] 2k/N[rad] ()
|F [k]| F [k]|F [k]|2 f [n]
?
, ver. 2016.01.08
-
6 74
k
2k/N k
!? k 2k/N
F [k] N
? k = N k = 0
k = N 2 [rad] 0
k 0 N 0 2
k N k = N/2
k
-
F[0]
F[1]F[2]
F[N-1]
F[N] = F[0]
0
2
k = N/2
k 0 N 1
f [n]DFT F [k] (6.11)
F [k]DFT1 f [n] (6.12)
f [n] F [k] (6.13)DFT[f(t)] = F [k] (6.14)
DFT1[F [k]] = f(t) (6.15)
6.5
, ver. 2016.01.08
-
6 75
(Fast Fourier Transform; FFT)
?
(6.5) ? ?
ej2N kn
F [k] N N
N 1
k k = 0, 1, , N 1 ?
N
N2
O(N2) N2
O(N log2 N)
!!
N =
256 512 1024 N2 65536262144
1048576
626100
N log2 N 2048460810240
10 100
, ver. 2016.01.08
-
6 76
6.6 4
1 4
Fk =1
T0
T0/2T0/2
f(t)ej0ktdt (2.28)
f(t) =
k=Fke
j0kt (2.18)
F () =
f(t)ejtdt (3.11)
f(t) =1
2
F ()ejtd (3.10)
F () =
n=
f [n]ejn (5.3)
f [n] =1
2
F ()ejnd (5.8)
F [k] =
N1n=0
f [n]ej2N kn (6.5)
f [n] =1
N
N1k=0
F [k]ej2N kn (6.4)
2
2 2 2 = 4
4
(p. 64)
?
, ver. 2016.01.08
-
6 77
?
2
!
1/T0
f(t) = Fk = 3
4
1 j () () 0k 2k/N
, ver. 2016.01.08
-
6 78
, ver. 2016.01.08
-
79
7 (1):
7.1
4 4
x(t) FS Xkx(t) F X()x[n] DTFT X()x[n] DFT X [k]
x(t t1) FS ej0kt1Xk (7.1)x(t t1) F ejt1X() (7.2)x[n n1] DTFT ejn1X() (7.3)x[n n1] DFT ej 2N kn1X [k] (7.4)
t n
t1 n1
t1 n1
(7.2) x(t) F X() x(t) X()
-
7 (1): 80
t
x(t)
=+
++
x(t) t1
?
t1 t1
x(t t1)
t
x(t t1)
=+
++
t1
t1
t1
t1
t1
?
1 2/ [s] 2 [rad] 2 t12/ = t1[rad] ?
t1
2/
t
0
?
, ver. 2016.01.08
-
7 (1): 81
(3.46) t1
ejt1 ejt1X()
(3.46)
(t) 1
X() = 1
7.2
ej0k1tx(t)FS Xkk1 (7.5)
ej1tx(t)F X( 1) (7.6)
ej1nx[n]DTFT X( 1) (7.7)
ej2N k1nx[n]
DFT X [k k1] (7.8)
?
0 1
(7.6)ej1t 1
(p. 27) ?
x(t) ej1tx(t)
, ver. 2016.01.08
-
7 (1): 82
t
Re
Im
t
Re
Im
t
ej1t x(t)
ej1t
x(t)
20 Hz 20 kHz
?
A 1000 kHz B 1040 kHz
A A = 1000 2 103 ejAt 980 1020 kHz
0
f
1000
kHz
1040
kHz
A B
, ver. 2016.01.08
-
7 (1): 83
B A = 1040 2 103 ejBt 1020 1060kHz
ejAt ejBt
(mod-
ulation)
(amplitude) amplitude modulation
AM
AM AM ?
AM
frequency modulation
FM
, ver. 2016.01.08
-
84
8 (2):
8.1
2
1
T0
T0/2T0/2
h()x(t )d FS HkXk (8.1)
h()x(t )d F H()X() (8.2)
m=h[m]x[nm] DTFT H()X() (8.3)
N1m=0
h[m]x[nm] DFT H [k]X [k] (8.4)
h()x(t )d
h(t) x(t)
h()x(t )d (8.5)
h[n] x[n]
m=h[m]x[nm] (8.6)
h() x()
-
8 (2): 85
= t m = nm
h()x(t )d =
h(t )x( )(d ) (8.7)
=
h(t )x( )d (8.8)
m=h[m]x[nm] =
m=
h[nm]x[m] (8.9)
h x h x = x h
h(t) x(t) = h(t )x()d (8.1) (8.4) ?
1
(8.1) (8.4)
x h 1
1
(p. 72) ?
1
41
T0h(t) x(t) FS HkXk (8.1)
h(t) x(t) F H()X() (8.2)h[n] x[n] DTFT H()X() (8.3)h[n] x[n] DFT H [k]X [k] (8.4)
2 2
8.2
, ver. 2016.01.08
-
8 (2): 86
x[n] y[n]
x[n] y[n]
x[n] y[n]
n x[n] y[n]
x[n] n y[n] n
x[n] y[n]
x1[n] y1[n] x2[n]
y2[n] a1a2 a1x1[n] + a2x2[n]
a1y1[n] + a2y2[n]
x[n] y[n]
x[n n1] y[n n1]
?
, ver. 2016.01.08
-
8 (2): 87
[n] =
{1, n = 0
0, n = 0 (8.10)
n
[n]
0
0 1 0
1
n
0
n
0
n
0
n
0
h[n]
n
0
n
0
[n] h[n]
, x[1] = 0, x[0] = 2, x[1] = 4, x[2] = 3, x[3] = 0,
n
0
x[n]
n?
n = 0, 1, 2 0
, ver. 2016.01.08
-
8 (2): 88
n = 0, 1, 2 3
n = 0 x[0] = 2
x[0] h[n]x[0] = 2h[n]
n
0
x[0]
n
0
h[n] x[0]
2
n = 1 x[1] = 4 1
4 1
4 h[n 1]x[1] = 4h[n 1]
n
0
x[1]
n
0
h[n - 1] x[1]
n = 2 2 3 h[n2]x[2] =3h[n 2]
n
0
x[2]
n
0
h[n - 2] x[2]
y[n] =
m=
h[nm]x[m] (8.11)
=
2m=0
h[nm]x[m] (8.12)
= 2h[n] + 4h[n 1] + 3h[n 2] (8.13)
n
0
n
0
m
h[n - m] x[m]x[n]
h[n] x[n] =m h[nm]x[m] +
, ver. 2016.01.08
-
8 (2): 89
h[n]
h[0]1
h[1]2 h[2]
n y[n] y[n] n x[n]
h[0]x[n] n 1 x[n 1] 1 h[1]x[n 1] n 2 x[n 2] 2 h[2]x[n 2]
y[n] =
m=
h[m]x[nm] (8.14)
h[n] x[n] =m h[m]x[nm]
h[n]
h[n]
h[n]
x[n] y[n]
h[n]
?
(t)
h(t) t = 0 1
0 1
h(t) x(t) ?
x(t)
t
x(n)
, ver. 2016.01.08
-
8 (2): 90
n t = n x(n)
x(n) (t n)x(n) x(n) n 1
0 1 h(t n)x(n)
h(t) x(n) n
n
y(t) =
n=
h(t n)x(n) (8.15)
0 n
y(t) =
h(t )x()d (8.16)
8.3
h[n]
x[n] = ejn exp(jn)
h[n]
n1
(p. 27)
y[n] h[n]
y[n] =
m=
h[m]x[nm] (8.17)
=
m=
h[m]ej(nm) (8.18)
=
m=
h[m]ejmejn (8.19)
= ejn
m=h[m]ejm (8.20)
, ver. 2016.01.08
-
8 (2): 91
?
m h[m]ejm ? (5.3)
h[n] H()
y[n]
y[n] = H()ejn (8.21)
h[n] ejn H()ejn
H() n
H()
H() ?
(p. 28) |H()| H()
h[n]
H()
exp(jn) H()exp(jn)
h[n]
n|H()|
H()
1 n
?
h[n] x[n] h[n] x[n] x[n] X()
H() H()X()
h[n] H()
, ver. 2016.01.08
-
8 (2): 92
1
0
(p. 42) (t)
1 1?
[n]
n=
[n]ejn (8.22)
=ej0 (8.23)
=1 (8.24)
1
1
?
h(t) H()
(8.20)
x(t) = ejt ?
y(t) =
h()ej(t)d (8.25)
= ejt
h()ej d (8.26)
= H()ejt (8.27)
ejt H()
z
8.4
, ver. 2016.01.08
-
8 (2): 93
h(t)x(t)FS
l=
HXk (8.28)
h(t)x(t)F 1
2
H(W )X(W )dW (8.29)
h[n]x[n]DTFT 1
2
H(w)X( w)dw (8.30)
h[n]x[n]DFT 1
N
N1=0
H []X [k ] (8.31)
, ver. 2016.01.08
-
94
9 (3):
9.1
3
1
T0
T0/2T0/2
|x(t)|2dt =
k=|Xk|2 (9.1)
|x(t)|2dt = 1
2
|X()|2d (9.2)
n=
|x[n]|2 = 12
|X()|2d (9.3)
N1n=0
|x[n]|2 = 1N
N1k=0
|X [k]|2 (9.4)
1
T0
T0/2T0/2
h(t)x(t)dt =
k=HkX
k (9.5)
h(t)x(t)dt =1
2
H()X()d (9.6)
n=
h[n]x[n] =1
2
H()X()d (9.7)
N1n=0
h[n]x[n] =1
N
N1k=0
H [k]X[k] (9.8)
(9.5) h = x (9.1)
? ?
? (9.1) (9.5)
?
-
9 (3): 95
yes (9.1) x h+ x h x (9.5)
(9.5)
h(t) x(t)
(p. 16) [h1, h2, h3] [x1, x2, x3] 3i=1 hix
i
h(t)x(t)dt h(t) x(t)
h1h2
h3
x1x2
x3
h(t)
x(t)
a b
(9.5)(9.8)
1
2 ()
()
(9.1)(9.4)?
, ver. 2016.01.08
-
9 (3): 96
x
x(t)
x2(t)
x(t)
()
9.2
h(t) x(t) i hix
i
h(t)x(t)dt ?
? ?
?
(
?
?
, ver. 2016.01.08
-
9 (3): 97
?
Wikipedia
Wikipedia
h(t) x(t)
x(t) h(t) + x(t)
?
h(t) x(t) (h(t), x(t)) =h(t)x(t)dt
9.3
?
, ver. 2016.01.08
http://ja.wikipedia.org/wiki/%E3%83%99%E3%82%AF%E3%83%88%E3%83%AB%E7%A9%BA%E9%96%93#.E5.AE.9A.E7.BE.A9http://ja.wikipedia.org/wiki/%E8%A8%88%E9%87%8F%E3%83%99%E3%82%AF%E3%83%88%E3%83%AB%E7%A9%BA%E9%96%93#.E5.86.85.E7.A9.8D
-
9 (3): 98
N N
{v1,v2, vN}a1v1+a2v2+ +aNvN {v1, vN}
a1v
1
v2
x = a1v
1 + a
2v
2
a2v
2
v1
(x,y) {e1, e2, , eN} (ei, ej) = i,j
1
0
x {e1, e2, , eN} ei (x, ei)
e1
e2
a1e
1
x = a1e
1 + a
2e
2
= (x, e1)e
1 + (x, e
2)e
2
a2e
2
x ei
x ei ||x|| cos ||ei|| = 1 ||x|| cos =||x|| ||ei|| cos = (x, ei)
x = (x, e1)e1 + (x, e2)e2 + + (x, eN )eN
x =
Ni
(x, ei)ei (9.9)
x x
, ver. 2016.01.08
-
9 (3): 99
[T0/2, T0/2] (x(t), y(t)) =
T0/2T0/2 x(t)y
(t)dt x(t)
{e1(t), e2(t), , eN (t)} (ei(t), ej(t)) = T0/2T0/2 ei(t)e
j (t)dt =
i,j ?
N N
(ei(t), ej(t)) = i,j
?
(2.19) T0/2T0/2
ej0mt{ej0nt
}dt = T0m,n (2.19)
m,n T0
1
T0
{ek(t)} ={
1T0
ej0kt}
k=(9.10)
(9.9) x(t)
x(t) =
k=
(x(t), ek(t))ek(t) (9.11)
(x(t), ek(t)) Xk
Xk = (x(t), ek(t)) (9.12)
=
T0/2T0/2
x(t)
{1T0
ej0kt}
dt (9.13)
=1T0
T0/2T0/2
x(t)ej0ktdt (9.14)
x(t) =
k=Xkek(t) (9.15)
=1T0
k=
Xkej0kt (9.16)
?
, ver. 2016.01.08
-
9 (3): 100
(9.16) (9.14)
Xk = 1T0 Xk
!!
!?
{ej0kt}
L
e1(t)
e2(t)
L
x(t)
X2e
2(t) = (x(t), e
2(t)) e
2(t)
~
X1e
1(t) = (x(t), e
1(t)) e
1(t)
~
(9.10)
{
1T0ej0kt
}k=
?
(p. 18)
(9.10)
9.4
(9.5)
T0/2T0/2
h(t)x(t)dt (9.17)
, ver. 2016.01.08
-
9 (3): 101
h(t) x(t)
(9.16)
T0/2T0/2
h(t)x(t)dt = T0/2T0/2
( k=
Hkek(t)
)( k=
Xkek(t)
)dt (9.18)
ek(t)
{ek(t)} T0/2
T0/2h(t)x(t)dt =
k=
l=
HkXl
T0/2T0/2
ek(t)el (t)dt (9.19)
=
k=
HkXk(ek(t), ek(t)) +
k =l
HkXl (ek(t), el(t)) (9.20)
(ek(t), el(t)) = k,l k = l
T0/2T0/2
h(t)x(t)dt =
k=HkX
k (9.21)
(9.5)
Hk = 1T0 Hk Xk =1T0Xk (9.5)
Hk Xk
h(t) x(t)
L
e1(t)
e2(t)
L
x(t)
h(t)H
1e1(t)
H2e2(t)
X2e2(t)
X1e1(t)
~
~
~
~
(9.18)
HkXk
, ver. 2016.01.08
-
9 (3): 102
L
e1(t)
e2(t)
L
x(t)
h(t)H
1e1(t)
H2e2(t)
X2e2(t)
X1e1(t)
~
~
~
~
H2X
2*
~ ~
H1X
1*
~ ~
0
0
h = x (9.1)
T0/2T0/2
|x(t)|2dt =
k=|Xk|2 (9.22)
x(t) ||x(t)||2 x(t) 2
2N
||x(t)||
| X1 |
| X2 |
| X3 |
~
~
~
(9.18) (9.19)
9.5
( (9.1)(9.5))
?
, ver. 2016.01.08
-
9 (3): 103
x(t)e
jtdt = (x(t), ejt) ejt {ejt}R
(ejt, ejt) = e
jtejtdt = dt 1
?
(ei, ej) = i,j (ej1t, ej2t) =
(1 2)
N
N N N = 4
X [0]
X [1]
X [2]
X [3]
=
W 0 W 0 W 0 W 0
W 0 W 1 W 2 W 3
W 0 W 2 W 4 W 6
W 0 W 3 W 6 W 9
x[0]
x[1]
x[2]
x[3]
(9.23)
W = ej2N
4 4 N N
, ver. 2016.01.08
-
104
10
10.1
n n + 1
n n+ 1
n n+1
?
?
-
10 105
x(t) c X() = F [x(t)]
X() = 0 for || c (10.1)
s = 2Ts
s > 2c (10.2)
xd[n] = x(t)|t=nTs x(t)
2
?
xd[n]
xd[n+ 1] c
x(t) X()
X() = F [x(t)] (10.3)
X()
t
x(t)
s Ts = 2/s
xd[n]
xd[n] = x(t)|t=nTs (10.4)
t
x(t)
Ts
n
xd[n]
1
sampling
t = nTs
, ver. 2016.01.08
-
10 106
Xd() = DTFT[xd[n]] (10.5)
=
n=
x(nTs)ejn (10.6)
Xd() X()
X() = Xd() 2
X() Xd()
2
t
x(t)
Ts
n
xd[n]
1
X()
DTFT
Xd()sampling
--3 3
? ? ?
2
? X() Xd()
(4.3)
= Ts (4.3)
Xd() Ts
Xd(Ts) =
n=
x(nTs)ej(Ts)n (10.7)
t = nTs ejn ej(Ts)n ejt 2 2/Ts = s
, ver. 2016.01.08
-
10 107
t
x(t)
Ts
n
xd[n]
1
X()
DTFTX
d(T
s)sampling
? ? ?
s
= 2 / Ts
-3s/2 -
s/2
s/2 3
s/2
X()
X() Xd(Ts)
Xd(Ts)
?
(p. 63)
(10.7)
Xd(Ts) =
{ n=
(t nTs)}x(t)ejtdt (10.8)
1
Xd(Ts) = F[{
n=(t nTs)
}x(t)
](10.9)
Ts
n= (t nTs) x(t)
, ver. 2016.01.08
-
10 108
t
x(t)
Ts
t
X()
Xd(T
s)
? ? ?
n(t nT
s)
t
{n(t nT
s) } x(t)
s
= 2 / Ts
-3s/2 -
s/2
s/2 3
s/2
( (8.29))
Xd(Ts) =1
2F[ n=
(t nTs)]X() (10.10)
X() !
X() Xd(Ts)
n (t nTs)
n (t nTs)
n (t nTs)
(comb function) (delta train)
Ts(t) =
n=
(t nTs) (10.11)
Xd(Ts) =1
2F [Ts(t)] X() (10.12)
F [Ts(t)] 2
, ver. 2016.01.08
-
10 109
10.2
Ts(t) Ts
?
(p. 64) s = 2/Ts
Ts(t) Ts
?
? s
?
s s
t
n(t nT
s)
Ts
s
?
(9.2)
0
0
?
?
Ts(t)
s
Ts(t) =
k=
[1
Ts
Ts/2Ts/2
(t)ejsktdt
]ejskt (10.13)
=1
Ts
k=
ejskt (10.14)
? [] 1
, ver. 2016.01.08
-
10 110
(3.51) ej1t F2( 1)
Ts(t) =1
Ts
k=
ejskt (10.15)
F 1Ts
k=
2( ks) (10.16)
=2
Ts
k=
( ks) (10.17)
= ss() (10.18)
Ts s = 2/Ts s
Ts(t)F ss() (10.19)
t
n(t nT
s)
Ts
s
1 s
s
k( - k
s)
1
10.3
(10.12)
Xd(Ts) =1
2F [Ts(t)] X() (10.12)
Xd(Ts) =1
2ss() X() (10.20)
=1
Tss() X() (10.21)
Xd(Ts) X()
, ver. 2016.01.08
-
10 111
1
Tss() X() =
1
Ts
[ k=
( ks)]X() (10.22)
=1
Ts
[ k=
(W ks)]X(W )dW (10.23)
=1
Ts
k=
(W ks)X(W )dW (10.24)
=1
Ts
k=
X( ks) (10.25)
?
?
?
(p. 85) h(t) x(t)
h(t) x(t)
x(t) = (t T1) ?
T1
T1
h(t) (t T1) = h(t T1) (10.26)
t
0
t
(t T1) h(t T
1)
T1
0 T1
h(t)
h(t)(t)
t = T1 (t T1) h(t)
x(t) = T1(t) ?
T1
1 1 h(t) T1
h(t) T1(t) =
n=h(t nT1) (10.27)
t t
h(t)
n(t nT
1) n h(t nT1)
(10.25)
, ver. 2016.01.08
-
10 112
X()
X() s
10.4
2
Xd(Ts) =1
Tss() X() (10.21)
=1
Ts
k=
X( ks) (10.28)
s
X()
Xd(T
s)
-3s/2 -
s/2
s/2 3
s/2
X() s/2
Xd(Ts) ?
s/2 s/2 0
X()
Xd(T
s)
-3s/2 -
s/2
s/2 3
s/2
-s/2
s/2
s/2 s/2 X()
X() = Hr()Xd(Ts) (10.29)
, ver. 2016.01.08
-
10 113
Hr() = TsGs/2,s/2() Gs/2,s/2() (3.27)
Gs/2,s/2() =
{1, || s/20, otherwise
(10.30)
X()
Xd(T
s)
-3s/2 -
s/2
s/2 3
s/2
-s/2
s/2
Hr()
xd[n] s/2 s/2 Ts x(t)
Hr()
s/2
s/2
xd[n]
xd[n + 1]
?
?
Hr() hr(t) = F1[Hr()]
(10.29)
x(t) = hr(t) F1[Xd(Ts)] (10.31)
= hr(t) (
n
xd[n](t nTs))
(10.32)
?
Xd(Ts) xd[n]
xd[n] F1xd[n]
, ver. 2016.01.08
-
10 114
Xd(T
s)
-3s/2 -
s/2
s/2 3
s/2
t
nxd[n] (t nT
s)
n
xd[n]
1
DTFT
Ts
(p. 64)
xd[n]
hr(t) Ts
xd[n] xd[n]
hr(t) nTs xd[n] n x(t)
xd[n] hr(t)
hr(t) Hr()
(3.31)
Gs/2,s/2()
F1[Hr()] = TsF1[Gs/2,s/2()] (10.33)
= Ts1
tsin
st
2(10.34)
=sin tTs
tTs
(10.35)
sinc
sinc (3.36) sinc
t t/Ts
sinc t = 1,2,3, 0 sinc t = Ts,2Ts,3Ts, 0
t
1
0 2 4-2-4
t
0 2Ts
4Ts
-2Ts
-4Ts
1
sin t
tsin (t / T
s)
(t / Ts)
, ver. 2016.01.08
-
10 115
sin t = 0
1 (10.32)
1 1 sinc
n
xd[n]
sinc
sinc t
10.5
?
/
2
X()
Xd(T
s)
-3s/2 -
s/2
s/2 3
s/2
100 Hz 150 Hz
100 Hz 2
75 Hz 75 Hz 80 Hz 70 Hz 90 Hz 60 Hz
, ver. 2016.01.08
-
10 116
f [Hz]
150 Hz
0 75 100
75 Hz
75 Hz
70 Hz 80 Hz
70 Hz
80 Hz 70 Hz 80 Hz
2
2
(Nyquist rate) c
() 2c
(Nyquist frequency) s
s/2
? ?
Oppenheim & Schafer (1998)
100 Hz
100 Hz
, ver. 2016.01.08
-
10 117
() 2
10.6 4
x(t) t = ,Ts, 0, Ts, 2Ts,
?
?
x(t) xd[n] = x(t)|t=nTs xd[n]
xd1(t) =
{x(t), t = ,2Ts,Ts, 0, Ts, , 2Ts 0, otherwise
(10.36)
xd2(t) = x(t)n
(t nTs) (10.37)
, ver. 2016.01.08
-
10 118
xd1(t) xd2(t) x(t)
xd[n] ?
xd1(t) xd2(t) t = nTs 0
t = nTs x(t)
xd1(t)
xd[n] xd2(t)
x(t)
xd2(t)
=
4
?
t
x(t) n (t nTs)
?
(p. 84)
X() * k( k
s)
t
x(t) n (t nTs)
?
1 T0
T0
, ver. 2016.01.08
-
10 119
X() k( k
0)
t
x(t) n (t nT0)
1
, ver. 2016.01.08
-
120
11
11.1
?
N N
N
n
N
?
N
N
N
-
11 121
N
n
N
N
N
N L
n
1
11.2
?
(p. 105)
2
?
?
, ver. 2016.01.08
-
11 122
11.3
N
?
N
?
1
n
n
n
N
?
, ver. 2016.01.08
-
11 123
N
!
11.4
xw[n] = wr[n]x[n] (11.1)
wherewr[n] =
{1, n = 0, 1, , N 10, otherwise
(11.2)
n
wr[n]
x[n]
n = 0, 1, N 1 1 0
n = 0 n = N 1 wr[n] 0 11 0 x[n]
wh[n] =
0.54 0.46 cos
2n
N, n = 0, 1, , N 1
0, otherwise(11.3)
n
wh[n]
x[n]
, ver. 2016.01.08
-
11 124
wr[n] wh[n]
?
wr[n] Wr() ?
Wr() =
n=
wr[n]ejn (11.4)
=
N1n=0
ejn (11.5)
=1 ejN1 ej (11.6)
(p. 38)
sin
Wr() =ejN/2
ej/2 e
jN/2 ejN/2ej/2 ej/2 (11.7)
= ej(N1)
2sin N2sin 2
(11.8)
sinc
sinc
, ver. 2016.01.08
-
11 125
0 -
[rad]
1.0
0.5
0.0
| Wr() |
max
| Wr() |
sinc sin 1/(sin/2)
2
2
wr[n] Wr()
?
Wr()
Wr()
?
Wr()
2
2
, ver. 2016.01.08
-
11 126
?
1
0 -
1.0
0.5
0.0
| Wh() |
max
| Wh() |
, ver. 2016.01.08
-
11 127
N cos
N 1 cos
N
, ver. 2016.01.08
-
128
12
12.1
?
?
?
?
(p. 90) ?
ejn (8.21)
H()
y[n] = H()ejn (8.21)
-
12 129
|H()| H() H()
exp(jn) H()exp(jn)
h[n]
n|H()|
H()
1 n
?
? 0
12.2
H()
H() h[n]
h[n] = DTFT1[H()] (12.1)
h[n]
?
n < 0 h[n] = 0
h[n] 0
n
0
n
0
[n] h[n]h[n]
, ver. 2016.01.08
-
12 130
n < 0 h[n] 0 ?
n
0
n
0
[n] h[n]h[n]
?
h[n] x[n] =
m=0
h[m]x[nm] (12.2)
=n
m=h[nm]x[m] (12.3)
1m h[m] 0 m = 0
2?
n m < 0 m > n x[m] m > n x[m]
(12.2) (12.3)?
?
h[n] 0
( 0 ) FIR (Finite Impulse Response)
, ver. 2016.01.08
-
12 131
n
0
n
0
[n] h[n]h[n]
M
h[n] x[n] =M
m=0
h[m]x[nm] (12.4)
( 0 ) IIR (Infinite Impulse Response)
n
0 n0
[n] h[n]h[n]
L
FIR IIR
FIR
FIR
IIR
12.3
h[n]
?
h[n] =
{n, n 00, otherwise
(12.5)
?
0
h[n] = n
, ver. 2016.01.08
-
12 132
h[n] = nu0[n] (12.6)
u0[n]
u0[n] =
{1, n 00, otherwise
(12.7)
u0[n]
u0[n]
IIR
= 0 n h[n] 0 IIR
x[n] y[n] y[n]
y[n] =
m=
mu0[m]x[nm] (12.8)
=
m=0
mx[nm] (12.9)
IIR
?
m = 0
y[n] =
m=1
mx[nm] + 0x[n] (12.10)
l = m 1
y[n] =
{ l=0
l+1x[n (l + 1)]}
+ x[n] (12.11)
=
{ l=0
lx[(n 1) l]}
+ x[n] (12.12)
{ } (12.9) ?
?
l=0 lx[(n 1) l] = y[n 1] ?
y[n] = y[n 1] + x[n] (12.13)
y[n] 1 1
x[n] y[n 1]
, ver. 2016.01.08
-
12 133
y y[0] = 0 y[1] = y[0]+x[1]
y[2] = y[1] + x[2]
(12.9)
1
y[n] =
Nk=1
ky[n k] +Mk=0
bkx[n k] (12.14)
y[n] N M
a0 = 1ak = k (k 1) N
k=0
aky[n k] =Mk=0
bkx[n k] (12.15)
y[n]
M = N = 2
z1 1
z 1 z 1 z 1 z 1
b1
b2
a1
a2
x[n]
y[n]
b0
z1 ?
z
?
FIR
FIR
?
ak = 0 (k 1)
y[n] =
Mk=0
bkx[n k] (12.16)
bk = h[k] (12.4)
, ver. 2016.01.08
-
12 134
FIR
ak 0 IIR
(12.15)
(= )
FIR IIR
IIR (12.15)
, ver. 2016.01.08
-
135
13
13.1
z
?
? z
z
{N
k=0
ak(d
dt)k
}y(t) =
{Mk=0
bk(d
dt)k
}x(t) (13.1)
?
d
dty(t) = ay(t) + x(t) (13.2)
y(0) = 0 x(t) = (t)
-
13 136
(p. 85)
x(t) = 0
d
dty(t) = ay(t) (13.3)
1
y(t)dy(t) = adt (13.4)
1
y(t)dy(t) =
adt (13.5)
log y(t) = at+ c0 (13.6)
y(t) = eat+c0 = ec0eat (13.7)
= Ceat (13.8)
c0 C = ec0
x(t) ?
y(t) = Ceat C
C(t)
y(t) = C(t)eat (13.9)
d
dty(t) =
d
dtC(t)eat + aC(t)eat (13.10)
d
dtC(t)eat + aC(t)eat aC(t)eat = x(t) (13.11)
C(t)
d
dtC(t)eat = x(t) (13.12)
d
dtC(t) = x(t)eat (13.13)
C(t) =
x(t)eatdt (13.14)
x(t) =
(t)
C(t) =
(t)eatdt (13.15)
, ver. 2016.01.08
-
13 137
c1
C(t) = c1 +
t
()ead (13.16)
=
{c1 + 1, t > 0
c1, t < 0(13.17)
t = 0
t = 0 t = 0 ()ea 0 ea0 = 1
u0(t) =
{1, t > 0
0, t < 0(13.18)
u0(0) = 1
t = 0
u0[n]
C(t) = c1 + u0(t) (13.19)
u0(t) (t) (13.16)
y(t) = C(t)eat
y(t) = (c1 + u0(t))eat (13.20)
y(0) = 0 c1 ? u0(0)
?
y(t) 0
(t) t = 0
t = 0 limt0 y(t) y(t)
y(0)
t = 0 t = 0
?
limt0
y(t) = limt0
(c1 + u0(t))eat (13.21)
= c1ea0 (13.22)
= c1 (13.23)
0 c1 = 0
y(t) = eatu0(t) (13.24)
, ver. 2016.01.08
-
13 138
(13.2) y(t) = 0 x(t) = (t)
(13.24)
(13.24)
13.2
?
t x(t)
X(s) =
0
x(t)estdt (13.25)
s X(s)
(13.25) X(s) x(t)
X(s) = L[x(t)] (13.26)x(t)
L X(s) (13.27)
(13.2)
L[d
dty(t)
]= L [ay(t) + (t)] (13.28)
s
(t) 1
eatu0(t)1
s a af(t) + bg(t) aF (s) + bG(s)
d
dtf(t) sF (s) f(0)
h(t) x(t) = 0
h()x(t )d H(s)X(s)
L[d
dty(t)
]= aL [y(t)] + L [(t)] (13.29)
sY (s) = aY (s) + 1 (13.30)
, ver. 2016.01.08
-
13 139
y(t) Y (s)
Y (s)
(s a)Y (s) = 1 (13.31)Y (s) =
1
s a (13.32)
y(t) = eatu0(t) (13.33)
(13.24)
?
13.3
F [x(t)] =
x(t)ejtdt (13.34)
x(t)
sin cos t
0
sin cos
eat
0
eat
2
x(t) ect
?
x(t) = eat a > 0 t x(t) x(t)ect = e(ac)t
c a c < 0 t x(t)ect 0
, ver. 2016.01.08
-
13 140
x(t) = eate-ct
x(t) e-ct = e(a c)t
t0
t
u0(t) x(t)u0(t)ect
x(t) = eate-ct
x(t) u0(t) e-ct
= e(a c)t u0(t)
t0
u0(t)
x(t) ?
2 x(t)
F [x(t)u0(t)ect] =
x(t)u0(t)ectejtdt (13.35)
=
0
x(t)e(c+j)tdt (13.36)
c+ j s (13.25)
t s
c Re{s} = c
, ver. 2016.01.08
-
13 141
t
Re
Im
c
0
Re{s} = c
c
Re
Im
0
c X(s) = X(c+ j)
X(s) s
s
s c
s
c x(t)ect t
c x(t)
s = c+ j X(s)
Re
Im
0 c
, ver. 2016.01.08
-
13 142
(p. 138) (t) eat
L[(t)] = 0
(t)estdt (13.37)
= es0 (13.38)
= 1 (13.39)
0 ?
0 0 0
0 0
X(s) =
0
x(t)estdt (13.40)
0
eat
L[eat] = 0
eatestdt (13.41)
=
0
e(as)tdt (13.42)
=
[1
a se(as)t
]0
(13.43)
=1
a s(limt e
(as)t 1)
(13.44)
lim ? a s
s = c+ j c
Re{a s} < 0
L[eat] = 1a s (0 1) (13.45)
=1
s a (13.46)
(p. 138)
Re{s} > Re{a} c
, ver. 2016.01.08
-
13 143
?
(13.35)
X(c+ j) = F [x(t)u0(t)ect] (13.47)
x(t)u0(t)ect = F1[X(c+ j)] (13.48)
=1
2
X(c+ j)ejtd (13.49)
ect
x(t)u0(t) =1
2
X(c+ j)ectejtd (13.50)
=1
2
X(c+ j)e(c+j)td (13.51)
s
s = c+ j ds = jd
x(t)u0(t) =1
2j
c+jcj
X(s)estds (13.52)
?
Re{s} = c
u0(t) ? x(t) = ?
u0(t)
s
t < 0 0
x(t)u0(t)
x(t)
eatu0(t)L 1sa eat
L 1sa t < 0
1sa (13.52) t = 0 eatu0(t) ?
t = 0 t < 0
0
, ver. 2016.01.08
-
13 144
?
?
?
x(t) ejt
(p. 102) {ejt}R x(t)
x(t)u0(t) e(c+j)t
X(c+ j) = X(s)
?
{e(c+j)t}R {e(c+j)t}R
{ejt}R
=
+
+
+
+
L
{e(c+j)t}R ?
e(c+j)t = ectejt ? ect
c
c > 0
, ver. 2016.01.08
-
13 145
ect
=
+
+
+
+
L
?
t 0
t
(p. 90) ejt
est = e(c+j)t ?
h(t) x(t) = est
y(t) =
0
h()x(t )d (13.53)
=
0
h()es(t)d (13.54)
= est 0
h()esd (13.55)
= H(s)est (13.56)
est est H(s)
x(t) est X(s) H(s)
h(t)
H(s)
H(s)
, ver. 2016.01.08
-
13 146
13.4
?
s
?
ect ?
ect
?
ect
(13.1) {N
k=0
ak(d
dt)k
}y(t) =
{Mk=0
bk(d
dt)k
}x(t) (13.1)
y(t)x(t)
y(t)x(t) Y ()X()
1
2
{N
k=0
ak(d
dt)k
}
Y ()ejtd =1
2
{Mk=0
bk(d
dt)k
}
X()ejtd (13.57)
1
2
{N
k=0
ak(d
dt)k
}Y ()ejtd =
1
2
{Mk=0
bk(d
dt)k
}X()ejtd (13.58)
Y ()ejt Y () X() t
(d
dt)kY ()ejt = Y ()(
d
dt)kejt (13.59)
= (j)kY ()ejt (13.60)
1
2
[{N
k=0
ak(j)k
}Y ()
]ejtd =
1
2
[{Mk=0
bk(j)k
}X()
]ejtd (13.61)
, ver. 2016.01.08
-
13 147
[ ]
{N
k=0
ak(j)k
}Y () =
{Mk=0
bk(j)k
}X() (13.62)
?
Y () =
Mk=0 bk(j)
kNk=0 ak(j)
kX() (13.63)
y(t) y(t)
y(t) = F1[M
k=0 bk(j)kN
k=0 ak(j)kX()
](13.64)
(p. 102) x(t) x(t)
ejt
L
ej1t
ej2t
L
x(t)
X(2)ej2t
X(1)ej1t
ejt 12X()
ejt 12X()12
e a ae
x(t)ejt
jejt
j
, ver. 2016.01.08
-
13 148
L
ej1t
ej2t
L
d/dt x(t)
j2X(2)ej2t
j1X(1)ej1t
? t j
j
x(t) y(t)
x(t) y(t)
ejt
(13.62)
(13.1)
(13.62)
j
Y ()
Y () (13.63)
y(t)
?
x(t)y(t)X(s)Y (s)x(t)
y(t) j c+ j
{N
k=0
ak(c+ j)k
}Y (s) =
{Mk=0
bk(c+ j)k
}X(s) (13.65)
, ver. 2016.01.08
-
13 149
{N
k=0
aksk
}Y (s) =
{Mk=0
bksk
}X(s) (13.66)
s s est s
{ejt}R {e(c+j)t}R
13.5 c+ j s
?
s = c+ j
s
s
(13.66) s
Y (s) =
Mk=0 bks
kNk=0 aks
kX(s) (13.67)
H(s)
H(s) =
Mk=0 bks
kNk=0 aks
k(13.68)
Y (s) = H(s)X(s) (13.69)
Y (s) X(s) H(s)
y(t) x(t) s
H(s)
, ver. 2016.01.08
-
13 150
H(s) X(s) Y (s)
?
(p. 145)
y(t) = h(t) x(t) (13.70)
h(t) h(t)
x(t) = (t) s Y (s) = H(s) 1 = H(s) (t) h(t)
y(t) + 3y(t) + 2y(t) = x(t) (13.71)
ddt y(t) 2
y(t)y(t) y(0) = y(0) = 0
Y (s)
s
s2Y (s) + 3sY (s) + 2Y (s) = X(s) (13.72)
(s2 + 3s+ 2)Y (s) = X(s) (13.73)
Y (s) =1
s2 + 3s+ 2X(s) (13.74)
1s2+3s+2
(p. 138)
?
H(s) =1
s2 + 3s+ 2(13.75)
=1
(s+ 2)(s+ 1)(13.76)
=1
s+ 1 1
s+ 2(13.77)
, ver. 2016.01.08
-
13 151
(p. 138) eatu0(t)L 1sa
L1[
1
s+ 1 1
s+ 2
]= (et e2t)u0(t) (13.78)
2
?
1sa
1(s+2)(s+1)
e2t et
= 0 s = 2,1 e2t et
= 0 s = 3 e3t
?
s = 1 + 2j e(1+2j)t = etej2t 2 [rad/s]
?
e(1+2j)t e(12j)t
= 0
= 0
s = a1 ea1t = 0
(p. 213) a1/
, ver. 2016.01.08
-
13 152
? ?
?
h(t) = (et e2t)u0(t) H(s) = 1(s+2)(s+1) Re{s} > 1
(13.41)
H(s) Re{s} > 1 s = 2 s = 1 ?
Re
Im
-1-2
H(s) =0 h(t)e
stdt H(s) H(s) = 1(s+2)(s+1)
H(s)
?
eatu0(t)L 1sa
1sa Re{s} > a 1sa a eat a
?
1sa a s = a
, ver. 2016.01.08
-
13 153
s
s = c+ j
c+ j = 0
H(s) =
Mk=0 bks
kNk=0 aks
k(13.68)
N > M = 0
(p. 213)
= 0 i (i = 1, , N) wi (i = 1, , N)
H(s) =
Ni
wis i (13.79)
H(s) Hi(s) = 1/(si) N X(s) Yi(s) wi
Y (s)
H1(s)
H2(s)
HN(s)
L
X(s)
Y1(s)
Y2(s)
YN(s)
Y(s)
w1
w2
wN
L
Yi(s) =1
s iX(s) (13.80)
sYi(s) = iYi(s) +X(s) (13.81)
?
d
dtyi(t) = iyi(t) + x(t) (13.82)
(13.2)
, ver. 2016.01.08
-
13 154
eitu0(t) i
= 0
s
s = c+ j s = j
?
/
?
s = j
s = 3 s = 2+ j j
s
?
?
x(t) x(t)
(t)
x(t)
(p. 89)
13.6
s = c+ j
c = 0
t
Re
Im
0
X(j)
, ver. 2016.01.08
-
13 155
0 t < 0 0
s = j
f(t)
F (j) (p. 66) ?
F (s) s j
F () ?
f(t) F (s) f(t)
F () F ()
FFourier() FLaplace(s)
F (j)
F ()
1 cos(1t)
L[cos(1t)] = L[ej1t + ej1t
2] (13.83)
=1
2L[ej1t] + 1
2L[ej1t] (13.84)
=1
2 1s j1 +
1
2 1s+ j1
(13.85)
s j
F [cos(1t)] = 12 1j j1 +
1
2 1j + j1
(13.86)
=1
2j
{1
1 +1
+ 1
}(13.87)
1 cos 1
?
1-1
0 ?
, ver. 2016.01.08
-
13 156
t < 0 cos
s = j
s = j s
cos s = j1
13.7 0
(p. 138)
1
d
dtf(t)
L sF (s) f(0) (13.88)
sF (s) f(0)
0
s F (s)est sF (s)est
s ?
cos
? 0 ?
t = 0 f(0) = 0 f(t) f(t) t = 0 0 F (s)
t0
f(t)
0
f(t) u0(t)
F(s)t
u0(t)
u0(t)
, ver. 2016.01.08
-
13 157
f(t) ddt{f(t)u0(t)} sF (s)
u0(t) 2OK
sF (s) u0(t) t = 0
f(0)(t)
?
u0(t) (t) (p. 137) u0(t)
(t) 1 (t) f(0)
(t) f(0)
t0
f(t)
0
f(t) u0(t)
F(s)t
0t
d/dt [ f(t) u0(t) ]
s F(s)
u0(t)
d/dt
ddtf(t) t = 0
sF (s)
t0
f(t)
0
d/dt f(t)
t
0
t s F(s) f(0)
u0(t)
d/dt
[ d/dt f(t) ] u0(t)
f(0)(t) L f(0) sF (s) ddtf(t)L sF (s) f(0)
, ver. 2016.01.08
-
13 158
0
d
dty(t) = ay(t) + x(t) (13.2)
y(0) 0
sY (s) y(0) = aY (s) +X(s) (13.89)(s a)Y (s) = X(s) + y(0) (13.90)
Y (s) =1
s aX(s) +y(0)
s a (13.91)
y(t)
y(t) =(eatu0(t)
) x(t) + y(0)eatu0(t) (13.92)?
?
1 eatu0(t) x(t)
2
1 0
2
1 2
?
2 3
0
OK (13.1)
, ver. 2016.01.08
-
13 159
?
y(t) = h(t) x(t) h(t) x(t) y(t) y
t [0,) x(t) ((13.25) X(s) x(t) (
)
s = c + j c c
s s
X(s) (13.52) x(t) t < 0 0
(13.25) (13.52)
(13.1)
s
Y (s)/X(s)
= 0 ()
s = j
?
, ver. 2016.01.08
-
160
14 z
14.1 z
z
ect
z
(p. 61)
x[n] ecn 0
DTFT[x[n]u0[n]e
cn] = n=0
x[n]ecnejn (14.1)
=
n=0
x[n]e(c+j)n (14.2)
ec+j z z
X(z) =
n=0
x[n]zn (14.3)
x(t)
(p. 104)
L[x(t)
n=
(t n)]=
0
x(t)
n=
(t n)estdt (14.4)
=
n=0
x(n)esn (14.5)
? 1 2 n = 0 ?
t = 0
n = , , n
n = 0
-
14 z 161
(14.5) z = es z s
c+ j
s = c+ j ? ?
11
z
Z [x[n]] = X(z) (14.6)x[n]
Z X(z) (14.7)
, x[1] = 0, (14.8)x[0] = 1,
x[1] = 3,
x[2] = 2,x[3] = 1,x[4] = 0,
z ?
X(z) =
n=0
x[n]zn (14.9)
= 1 + 3z1 2z2 z3 (14.10)
?
x[n] = n (14.11)
?
X(z) =n=0
nzn (14.12)
=
n=0
(z1)n (14.13)
=1
1 z1 (14.14)
|z1| < 1
, ver. 2016.01.08
-
14 z 162
z |z| > || ||
| |Re
Im
| z | > | |
3
x[n] = [n] (14.15)
X(z) =
n=0
[n]zn (14.16)
= z0 (14.17)
= 1 (14.18)
?
(= )
1 z
1
z
z = ec+j ?
s z
s = c+ j c
z ?
z = ec+j = ecej c z
c c
Re
Im
Re
Im
j
s = c + j
(: const)
c
c z = ec ej
(: const)
s z
, ver. 2016.01.08
-
14 z 163
?
z = ecej ec
= 0 Re{z} = ec = 2 1 c
Re
Im
Re
Im
s = c + j
(c: const)
z = ec ej
(c: const)
ec
s z
c = 0 s z e0 = 1
Re
Im
Re
Im
s = jz = ej
1
(= e0)
s z
s z s
z
Re
Im
Re
Ims z
z
?
c ect
, ver. 2016.01.08
-
14 z 164
z
Re
Im
Re
Ims z
s z
14.2 z
z ?
?
z s
z
X(z) = DTFT[x[n]u0[n]e
cn] (14.19)
x[n]u0[n]ecn = DTFT1 [X(z)] (14.20)
x[n]u0[n]ecn =
1
2
X(z)ejnd (14.21)
ecn
x[n]u0[n] =1
2
X(z)e(c+j)nd (14.22)
z = ec+j z
dz/d = jec+j
x[n]u0[n] =1
2j
X(z)e(c+j)ne(c+j)dz (14.23)
=1
2j
X(z)e(c+j)(n1)dz (14.24)
=1
2j
X(z)zn1dz (14.25)
, ver. 2016.01.08
-
14 z 165
z = ec+j
ec 1
Re{s} = c
Re
Im
Re
Im
ec = -
=
c
j
-j
s z
z (14.25) z x[n] x[n]u0[n]
z
z ?
z
?
z z (14.22)
e(c+j)n
X(z) = X(ec+j)
(p. 144) ejt
e(c+j)t
ejn z e(c+j)n
14.3 z
(p. 128)
, ver. 2016.01.08
-
14 z 166
Nk=0
aky[n k] =Mk=0
bkx[n k] (12.15)
z
?
d/dt s
1D
Dy[n] = y[n 1] (14.26)
Nk=0
akDky[n] =
Mk=0
bkDkx[n] (14.27)
k D k Dk
d/dt D
x y
1
2
Nk=0
akDk
Y ()ejnd =1
2
Mk=0
bkDk
X()ejnd (14.28)
1
2
Nk=0
akDkY ()ejnd =
1
2
Mk=0
bkDkX()ejnd (14.29)
d/dt ejt j
D ejn
Dejn = ej(n1) (14.30)
= ejejn (14.31)
ejn ej
1
2
[N
k=0
ak(ej)kY ()
]ejnd =
1
2
[Mk=0
bk(ej)kX()
]ejnd (14.32)
, ver. 2016.01.08
-
14 z 167
[ ]
Nk=0
ak(ej)kY () =
Mk=0
bk(ej)kX() (14.33)
zejn e(c+j)n
De(c+j)n = e(c+j)(n1) (14.34)
= e(c+j)e(c+j)n (14.35)
1 e(c+j)
Nk=0
ak(e(c+j))kY (z) =
Mk=0
bk(e(c+j))kX(z) (14.36)
z = es = ec+j
Nk=0
akzkY (z) =
Mk=0
bkzkX(z) (14.37)
z
s 1 e(c+j) = es = z1
z1 (p. 133)
z 1 x[n] x[n 1]
Y (z)
Y (z) =
Mk=0 bkz
kNk=0 akz
kX(z) (14.38)
y[n]
H(z) =
Mk=0 bkz
kNk=0 akz
k (14.39)
Y (z) X(z) H(z)
H(z) z
?
[n] z 1 [n] z
Y (z) = H(z) 1 z H(z)
y[n] = h[n]x[n] z Y (z) = H(z)X(z) z (p. 145)
h[n] x[n] = e(c+j)n = zn H(z)zn
, ver. 2016.01.08
-
14 z 168
(p. 154)
X(s) s = jX(z)
z = ej X(ej)
(p. 66)
z
1 2
14.4 z
z Y (z) z
y[n]
z
z
z
[n] 1
nu0(t)1
1 z1 af [n] + bg[n] aF (z) + bG(z)
f [n 1] z1F (z) h[n] x[n] =
m=0
h[m]x[nm] H(z)X(z)
z
H(z) =12z
1
1 32z1 + 12z2(14.40)
z1
?
1sa eatu0(t)
z nu0[n] 11z1
, ver. 2016.01.08
-
14 z 169
z1
H(z) =12z
1
(1 z1)(1 12z1)(14.41)
H(z) =1
1 z1 1
1 12z1(14.42)
z (p. 168)
Z1[H(z)] = 1nu0[n] (12)nu0[n] (14.43)
= (1 (12)n)u0[n] (14.44)
z1
?
1(1z1)k
(p. 225)
14.5 es z
z z = es = ec+j
c+ j s
e(c+j) = es z ?
yi = iyi + x
s 1si
i(si) = 0 i
i
z ?
z1 H(z) = 0
(p. 223)
H(z) =
Ni=1
wi1 iz1 (14.45)
, ver. 2016.01.08
-
14 z 170
z ni u0[n]
Hi(z) = 11iz1
Yi(z) = iz1Yi(z) +X(z) (14.46)
yi[n] = iyi[n 1] + x[n] (14.47)
i |i| > 1 |i| < 1
= 0 eit i i
z ?
H(z) =A(z)N
i=1(1 iz1)(14.48)
= 0 i (i = 1, , N) !
zn
H(z) =znA(z)N
i=1(z i)(14.49)
i z
= 0
z
, ver. 2016.01.08
-
14 z 171
n = 0, 1, 2, x[n] ( (14.3) X(z) x[n] z (
z )
z = es = zc+j c c z
1 z z1
X(z) (14.25) x[n] z n < 0 0
(14.3) (14.25) z
(12.15) z z
z
z Y (z)/X(z)
z
= 0 ()
z = ej
, ver. 2016.01.08
-
172
15
15.1
z
(p. 128)
? z z = ej
h[n] =
{15 , n = 0, 1, 2, 3, 4
0, otherwise(15.1)
?
z
H(z) =
4k=0
1
5zk (15.2)
=1
5
1 z51 z1 (15.3)
h[n] ?
x[n] h[n] y[n]
y[n] =k=0
h[k]x[n k] (15.4)
=
4k=0
1
5x[n k] (15.5)
-
15 173
FIR (12.15) a1, a2, , aN = 0 z
1 z1
Y (z) =
4k=0
1
5zkX(z) (15.6)
H(z) = Y (z)/X(z)
H(z) =4
k=0
1
5zk (15.7)
z = ej
H(ej) =1
5
1 ej51 ej (15.8)
(11.8) N = 5 1/5
H(ej) =1
5ej2
sin 52sin 2
(15.9)
= 2/5,4/5 0
- 0
|H()|
-4/5 -2/5 4/52/5
x[n] = cos 25 n 0
, ver. 2016.01.08
-
15 174
x[n] y[n]
x[n] = cos 10n :
n
x[n]
y[n]
x[n] = cos 5n :
n
x[n]
y[n]
x[n] = cos 25 n :
n
x[n]
y[n]
x[n] = cos 35 n :
, ver. 2016.01.08
-
15 175
n
x[n]
y[n]
y[n] = 0.9y[n 1] 0.81y[n 2] + x[n] + x[n 2] (15.10)
z
Y (z) = 0.9z1Y (z) 0.81z2Y (z) +X(z) + z2X(z) (15.11)
Y (z) =1 + z2
1 0.9z1 + 0.81z2X(z) (15.12)
z = ej
H(ej) =Y (ej)
X(ej)(15.13)
=1 + ej2
1 0.9ej + 0.81ej2 (15.14)
- 0
|H()|
-/2 /2-/3 /3
, ver. 2016.01.08
-
15 176
x[n] = cos 10n :
n
x[n]
y[n]
x[n] = cos 3n :
n
x[n]
y[n]
x[n] = cos 2n :
n
x[n]
y[n]
x[n] = cos 35 n :
, ver. 2016.01.08
-
15 177
n
x[n]
y[n]
?
H(ej)
15.2
0
H(z) =1
5
1 z51 z1 (15.15)
? = 0
= 0
? ?
z = 0 = 0
= 0 z1 z = 0 z = 0 = 0
z z
z5
H(z) =1
5
z5 1z4(z 1) (15.16)
z1 z
, ver. 2016.01.08
-
15 178
= 0 z = 0 (4) z = 1
= 0 z z = rej (r 0, < )
r5ej5 = 1 ej2n (n :) (15.17)
r = 1 = 2n/5
z = 0,ej2/5,ej4/5 (15.18)
4 4
Re
Im
1
j
? z = 1 ?
1
?
z = ej
= 0,2/5,4/5 0
2?
2
H(z) =1 + z2
1 0.9z1 + 0.81z2 (15.19)
z2
H(z) =z2 + 1
z2 0.9z1 + 0.81 (15.20)
, ver. 2016.01.08
-
15 179
= 0
z = 0.9(1
2 j3
2), (15.21)
= 0
z = j (15.22)
Re
Im
1
j
= /2 0
= /3
?
H(z) =(z j)(z + j)
(z 0.9(12 + j32 ))(z 0.9(12 j
32 ))
(15.23)
|H(z)| = |z j| |z + j||z 0.9(12 + j
32 )| |z 0.9(12 j
32 )|
(15.24)
|H(z)|
|H(z)| |z j| |z + j| |z 0.9(12 + j
32 )| |z 0.9(12 j
32 )|
|z j| z j z
, ver. 2016.01.08
-
15 180
Re
Im
z = ej| z j |
| z + j |
| z 0.9(1 + j) / 2 |3
| z 0.9(1 j) / 2 |3
= /3 |z 0.9(12 + j32 )|
H(z)
H(z) = (z j) + (z + j) (|z 0.9(12+ j
3
2)) (z 0.9(1
2 j3
2)) (15.25)
?
z j |z j|e (zj) |z j| H(z)
H(z) = e (zj)e (z+j)
e (z0.9( 12+j
3
2 ))e (z0.9( 12j
3
2 ))(15.26)
= e (zj)+ (z+j) (z0.9(12+j
3
2 )) (z0.9( 12j
32 )) (15.27)
= (z j) + (z + j) (z 0.9(12+ j
3
2)) (z 0.9(1
2 j3
2)) (15.28)
(z j) j z H(z) z
, ver. 2016.01.08
-
15 181
Re
Im
z = ej
(z j)
(z + j)
(z 0.9(1 + j) / 2)3
(z 0.9(1 j) / 2)3
15.3
y[n] = 1.2y[n 1] + x[n] (15.29)
?
Y (z) = 1.2z1Y (z) +X(z) (15.30)
Y (z) =1
1 1.2z1X(z) (15.31)
H(z) = 111.2z1 z = ej
H(ej) =1
1 1.2ej (15.32)
?
|H(ej)|
- 0
|H()|
, ver. 2016.01.08
-
15 182
5
= 0.5
x[n] = cos 10n :
n
x[n]
y[n]
x[n] = cos 3n :
n
x[n]
y[n]
y y 1.2
z H(z) z = ej
0.5 5
H(z) z = ej
H(z) z = ej H(z) = 111.2z1 z h[n] = 1.2nu0[n] h[n] z
Z[h[n]] =
n=0
1.2nzn (15.33)
111.2z1 |z| > 1.2
, ver. 2016.01.08
-
15 183
|z| = 1
H(z) z = ej h[n]
? ?
?
0
|x[n]|
-
15 184
y[n] =
k h[k]x[n k] x[k] = sgnh[k]
y[0] =k
h[k]x[0 k] (15.35)
=k
h[k] sgnh[k] (15.36)
=k
|h[k]| (15.37)
k |h[k]| BIBO
x
|x[n]| < x
|y[n]| = |k
h[k]x[n k]| (15.38)
k
|h[k]| |x[n k]| (15.39)
xk
|h[k]| (15.40)
-
15 185
Re
Im
1
j
?
h[n] = (w1n1 + w2
n2 + + wnnN )u0[n] (15.42)
1, , N
(p. 169)
(p. 223)
?
|i| < 1
n=0 |ni | 1 N
1
z = es
z = es s z
, ver. 2016.01.08
-
15 186
Re
Im
Re
Ims z
1
j
BIBO
15.4
?
?
(p. 90)
(p. 129)
0 0
?
h[n]
H() ?
H() 0 H()
H() h[n]
?
?
, ver. 2016.01.08
-
15 187
? (p. 28)
f(t)
-
h[n]
h[n] ?
h[n]
n
0
h[n]
0
n < 0
h[n] = [n]
0
?
, ver. 2016.01.08
-
15 188
hz[n] nd hz[nnd]n < 0 nd
n
0
h[n nd]
nd
(7.3) Hz() ejnd H() = ejndHz()
?
H() = {ejndHz()} = nd
H()
H() = nd
nd
IIR
IIR 0 0
, ver. 2016.01.08
-
15 189
15.5
IIR ?
? ?
(p. 81)
x[n] ejcn X() X( c)
x[n]
n
n
x[n] ejcn
0 c
X() X( c)
x[n] c
x[n]ejcn c
? ?
, ver. 2016.01.08
-
15 190
H()
n n
n = 0 n = 0
440 Hz
440 Hz
n
x[n] 440 Hz
440 Hz
gd
gd() = dd
H() (15.43)
pd() = H()
(15.44)
, ver. 2016.01.08
-
15 191
H()
c
H(c) : -pd(c)
: -gd(c)
?
?
c
()
c x[n]
X( c)
X() c
H() 1
c H(c) = c X( c) c
c
c
H() = c
H()
c ?
c
, ver. 2016.01.08
-
15 192
c
c
H() = ej(c+(c)c) (15.45)
1 = c
H(c) = ejc (c + ( c)c) c c OK
x[n]ejcn y[n]
Y () = H()X( c) (15.46)= ej(c+(c)c)X( c) (15.47)= ejcej(c)cX( c) (15.48)
ejc
ejcX() (15.49)
c
ejc c ejcn ejcX() x[n c]
y[n] = ejcejcnx[n c] (15.50)
x[n] c
x[n] c ejcejcn = ej(cn)c c
c = gd(c)
? x[n] ? c
x[n c] ?
x[n c] c
x[n]
x(t) c x(t c) x[n c]
, ver. 2016.01.08
-
15 193
x[n]
ejc
(= ) (= )
|H()|
(= )
H()/
d H()/d
= =
IIR
z
0
z
BIBO
FIR
BIBO
H(z) z = ej H(ej)
, ver. 2016.01.08
-
194
16
16.1
? H()
h[n] y[n] = h[n] x[n]?
Gc,c() =
{1, || c0, otherwise
(16.1)
c
cc
| G-c,c() |
c
c <
DTFT1[Gc,c()] =1
2
Gc,c()ejnd (16.2)
=1
2
cc
ejnd (16.3)
(16.4)
(3.31) a
c t n
DTFT1[Gc,c()] =1
nsincn (16.5)
-
16 195
sinc
n
sin cn / n
?
y[n] =
{1
nsincn
} x[n] (16.6)
=
m=
{1
msincm
}x[nm] (16.7)
sinc
(p. 129)
?
FIR IIR
FIR
, ver. 2016.01.08
-
16 196
FIR
y[n] =
Mm=0
h[m]x[nm], (12.4)
H(z) =
Mn=0
h[n]zn (16.8)
M
x[nm]
? IIR
?
IIR
Nk=0
aky[n k] =Mk=0
bkx[n k] (12.15)
H(z) =
Mk=0 bkz
kNk=0 akz