Спектральный и временной анализ импульсных и периодических сигналов. Кузнецов Ю.В - 92 стр.

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92
а)
[] [ ] []
.2,0;
2
1;0
0
≥=⋅=±=
±
nnCe
A
CAC
j
ϕ
|С[n]|
n
1
А
0
2 0
arg{С[n]}
n
1
-
ϕ
20
А/2 А/2
-2 -1
ϕ
-1
б)
[] [ ] []
.2,0;
2
1;0
2
0
≥=⋅=±=
−±
nnCe
A
CAC
j
π
ϕ
|
С
[n]
n
1
А
0
20
arg{С[n]}
n
1
20
А/2 А/2
-2 -1 -1
-
ϕ
+
π
/2
ϕ
–
π
/2
в)
[] [ ] []
[]
.3,0
;
2
2;
2
1;0
2
21
0
≥=
⋅=±=±=
nnC
e
A
C
A
CAC
j
π
m
|С[n]|
n
1
А
0
2 0
arg{С[n]}
А
1
/2
А
2
/2
-2 -1
π
/2
3 -3
n
1
2
0 -2 -1 3
-
π
/2
г)
[] [ ] []
[]
[]
.4,0;
4
3
;02;
4
1;0
2
0
≥==±
=±=±=
nnC
A
C
C
A
CAC
n
1
А
0
2 0 3
А/4
4
А/4
|С[n]
А/4 А/4
-2 -1-3-4
Задача 3.3
[] []
()
()
nj
Tn
Tn
T
A
nCC
π
τπ
τ
π
τ
−
−⋅⋅
⋅
== e1
sin
,00
0
00
.
n
0
|С[n]
1
n
π
01
arg{С
n
}
-
π
|С[1]
|С[3]
3
5 7
-5
Задача 3.4
2
,
10
A
A
A
A ==
π
.
Задача 3.5
[] []
nj
e
nA
nC
A
C
π
π
−
⋅
⋅==
2
sinc
2
,
2
0
2
.
Задача 4.1 
                            A ± jϕ                                                                                  π
а) C [0] = A0 ; C [± 1] =     ⋅ e ; C[n ] = 0 , n ≥ 2 .                  б) C[0] = A0 ; C[± 1] =        A ± j  ϕ − 2 
                                                                                                          ⋅e              ; C[n] = 0 , n ≥ 2 .
                            2                                                                           2
           |С[n]|                        arg{С[n]}                                      |С[n]                            arg{С[n]}

     А/2     А0 А/2                                                               А/2     А0 А/2             -ϕ +π /2
                                                   ϕ
                            n       -1                           n                                       n                      1            n
 -2 -1 0        1    2            -ϕ     0     1       2                      -2 -1 0          1   2                 -1 0           2
                                                                                                                              ϕ – π /2
                                                             π
                                A1           A mj                                                        A
     C[0] = A0 ; C[± 1] =          ; C[± 2] = 2 ⋅ e 2 ;                       C[0] = A0 ; C [± 1] =        ; C[± 2] = 0 ;
в)                              2             2                          г)                              4
     C[n ] = 0 , n ≥ 3 .                                                      C[± 3] =
                                                                                        A2
                                                                                           ; C [n] = 0 , n ≥ 4 .
                                                                                        4
            |С[n]|                       arg{С[n]}
                                  π/2                                                               |С[n]
        А1/2 А0 А2/2                                                                                 А0
                            n                      2         n                      А/4     А/4       А/4     А/4
 -3 -2 -1 0 1 2 3                 -2 -1 0 1          3                                                                                   n
                                                   -π/2                        -4 -3 -2 -1 0                 1       2    3     4



           Задача 3.3
                                                             A ⋅τ 0 sin (n π τ 0 T )
                                  C[0] = 0, C[n] =                 ⋅                 ⋅ (1 − e − jπ n ) .
                                                              T        nπ τ 0 T
                                       |С[n]                                                           arg{Сn}
                                           |С[1]
                                                                                            π

                                               |С[3]                                                         5 7                    n
                                                                     n                    -5       01
                                   01 3                                                                 -π



           Задача 3.4
                                                                     A            A
                                                           A0 =          , A1 =     .
                                                                     π            2



           Задача 3.5
                                                       A          A          π n  − jπ n
                                             C[0] =      , C[n ] = ⋅ sinc 2      ⋅e      .
                                                       2          2          2 



           Задача 4.1
92

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