Теоретические основы электротехники. Часть 2. Колчанова В.А - 30 стр.

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30
2 4
( ) ( 200 2 10 ) 0
B p p p p
= + + ⋅ =
.
1
0
p
=
,
(
)
2,3
1
100 100p j
c
= − ±
,
' 2 4
B p p p p p= + + ⋅ + + ,
3
'
1
( )
( )
( )
к
n
p t
к
к
к
D p
u t e
B p
=
=
= ⋅
∑
,
2
3
4 6
2 4
4 6
2
2 4
2 2 2 2
4 6
3
2 4
3 3 3 3
2 10 0 2 10
( )
(0 200 0 2 10 ) 0(2 0 200)
2 10 2 10
( 200 2 10 ) (2 200)
2 10 2 10
( 200 2 10 ) (2 200)
ot
p t
p t
u t e
p
e
p p p p
p
e
p p p p
⋅ ⋅ + ⋅
= +
+ ⋅ + ⋅ + ⋅ +
⋅ ⋅ + ⋅
+ +
+ ⋅ + ⋅ + ⋅ +
⋅ ⋅ + ⋅
+ =
+ ⋅ + ⋅ + ⋅ +
135 ( 100 100) 135 ( 100 100)
(100 135 ) (100 135 )
100
100 70,5 70,5
100 2 70,5
2
j j t j j t
j t j t
t
e e e e
e e
e
− ° − + ° − −
− ° − − °
−
= + + + =
+
= + ⋅ ⋅ =
100
100 141 cos(100 135 ),
t
e t
−
= + − °
В
.
Пример
Дано:
изображение
:
4 6
2 4
2 10 2 10 ( )
( ) ( ) , ( )
( )
( 200 2 10 )
p D p
F p U p
В
c
B p
p p p
⋅ + ⋅
= = =
+ + ⋅
Определить:
оригинал
.
Решение:
2 4
( ) ( 200 2 10 ) 0
B p p p p
= + + ⋅ =
,
1
0
p
=
,
(
)
2,3
1
100 100p j
c
= − ±
,
' 3 2 4
( ) ( 200 2 10 )
B p p p p
′
= + + ⋅ =
2 4
3 400 2 10 ,
p p+ + ⋅
тогда
3
'
1
( )
( )
( )
к
n
p t
к
к
к
D p
u t e
B p
=
=
= ⋅ =
∑
2
4 6 4 6
0
2
2 4 2 4
2 2
2 10 0 2 10 2 10 2 10
2 Re
0 400 0 2 10 3 400 2 10
p tt
p
e e
p p
⋅
⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅
= + =
+ ⋅ + ⋅ + + ⋅

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