Page 253 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

245
6.

ก 4' 6.2 3 " '%

C = 0.7654, L =1.8478, C = 1.8478, L = 0.7654;
1 n n 2 n 3 n 4

*(!
"

C
ก !ก (6.29) C = n
π
2 F R
H L
Ω
(ก
C = C , C = C , R = 50 , F = 8 MHz;
1 n 1 n L H
C 0.7654
C = 1 n = = 304.55 pF
1
π
×
×
6
(2 F R L ) ( 2 3.14 8 10 × 50 )
×
H
C 1.8478
C = n 3 = = 735.23 pF
3 6
π
×
×
×
2 F R 2 3.14 8 10 ×
H L 50
R L
ก !ก (6.28) L = L n
π
2 F
H
Ω
(ก
L = L 2 , L = L n 2 , R = 50 , F = 8 MHz;
H
n
L
×
R L 2 50 1.8478
L = = = 1.83 µH
L n
2
π
× ×
×
2 F H 2 3.14 8 10 6
×
R L 50 0.7654
L = L n 4 = = 0.76 µH
4
π
2 F × × × 6
H 2 3.14 8 10

C = == = 304.55 pF, L = == = 1.83 µH, C = == = 735.23 pF, L = == = 0.76 µH, F = = = = 8 MHz;
1 2 3 4 H
6.3
ก
0! 23

"

ก S # 4 24 (Butterworth High-pass Filters ก 4' !'
2
)
" !7' # ก " !7' S # 4' 6.20 ก ( L C ( L !'" ' ก-

2 " !7' 4 6
" !7' 3
S L ก 23 ก C !'" ' ก-

2 " !7' # 4 6
" !7' # 3
S C R 3 " !7' # ก" ! R
L L

E S

2
# 4' 6.20 ก S # 4 24 ก 31
2
:
ก

ก
ก