1 LCR R Q LCR E I I = R 2 + E ( ωl 1 ωc ) 2 (1) ω ω 0 = 1 LC (2) ω 0 = 2πf 0 f 0 = 1/(2π LC) I 0 = E R (3) 1 2 R I 0 Q (sharpness of resonance

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Download "1 LCR R Q 2 2.1 1 LCR E I I = R 2 + E ( ωl 1 ωc ) 2 (1) ω ω 0 = 1 LC (2) ω 0 = 2πf 0 f 0 = 1/(2π LC) I 0 = E R (3) 1 2 R I 0 Q (sharpness of resonance"

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1 E % 23.7 C hpa

2 1 LCR R Q LCR E I I = R 2 + E ( ωl 1 ωc ) 2 (1) ω ω 0 = 1 LC (2) ω 0 = 2πf 0 f 0 = 1/(2π LC) I 0 = E R (3) 1 2 R I 0 Q (sharpness of resonance) I I 0 1/ 2 f 1 = ω 1 /2π f 2 = ω 2 /2π Q = f 0 f 2 f 1 = ω 0 ω 2 ω 1 (ω 1 ω 0 ω 2 ) (4) (1) I 0 / 2 ω 2 ω 1 = R L (5) 1

3 Q Q = ω 0L R (6) 1: 2: 2.2 Q Q 1 I 2 = R 2 + ( E 2 ωl 1 ) 2 (7) ωc I 2 0 = E2 R 2 (8) I0 2 I 2 I 2 = 1 C ωrc 0 C (9) C 0 C C ω 2 LC 0 = 1 1/(ωRC 0 ) = ωl/r = Q I0 2 Q = I 2 C I 2 (10) C 2

4 I 3 1 Q C 0 C (11) 3 (11) Q 3: 3 Q Q 4 V 2 R 0,20,50[Ω] 3 V 1 V 2 I = 2πf(C c + C v ) V 2 V 1 (12) (C c + C v ) I 3 (11) Q 1 3

5 4: 1: 1 KENWOOD CS YEW Type KENWOOD FG YEW CDS V 2 0, 20, 50[Ω] p.9 5 Q C 0 C 2 5 (11) Q R = 0[Ω] Q = 66 R = 20[Ω] Q = 19 R = 50[Ω] Q = 9.6 2: [Ω] [khz] C 0 [pf] C [pf] Q

6 Ω Ω Ω 5: 1 V 1 =1[V] I 0 / 2 5

7 [Ω] Q (6) Q = ω 0L R ω 0 = 1/ LC = 1 ω 0 CR (13) Q 3 3: Q [Ω] Q Q Q r (13) Q = 1 ω 0 (r + R)CR (14) r = 1 ω 0 CQ R (15) 2 3 Q Q 3 r = 17[Ω] Q, R = 0[Ω] Q = 42,R = 20[Ω] Q = 19,R = 50[Ω] Q = 10 6

8 4: [Ω] [Ω] [ 1]? Q Q Q Q = 1 ωcr = ωl R Q (16) (1/ω 0 C) (R) (ω 0 L) (R) Q [ 2] Q P U Q Q Q Q (6) I 2 Q = ωli2 RI = ω LI2 1 (17) 2 RI2 7

9 U P T ω = 2πf = 2π/T Q = 2π U P T (18) 2 1 Q, 1 2π [1] 7 2 8

10 : (R=0 [Ω]) f =254.3 [KHz], R =0 [Ω] C v C v C V 1 V 2 I [pf] [pf] [V] [V] [ma/v]

11 6: (R=20 [Ω]) f =255.7 [KHz], R =20 [Ω] C v C v C V 1 V 2 I [pf] [pf] [V] [V] [ma/v]

12 7: (R=50 [Ω]) f =254.7 [KHz], R =50 [Ω] C v C v C V 1 V 2 I [pf] [pf] [V] [V] [ma/v]

13 [1],,.,, IV.,

~ ~.86 ~.02 ~.08 ~.01 ~.01 ~.1 6 ~.1 3 ~.01 ~.ω ~.09 ~.1 7 ~.05 ~.03 ~.01 ~.23 ~.1 6 ~.01 ~.1 2 ~.03 ~.04 ~.01 ~.1 0 ~.1 5 ~.ω ~.02 ~.29 ~.01 ~.01 ~.11 ~.03 ~.02 ~.ω 本 ~.02 ~.1 7 ~.1 4 ~.02 ~.21 ~.I

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