C 解：k,=C,+C, C,=329pF CC2C3 C-C,+C,+CC+CC,+CC, =53+C5 当：Cs=9pF(min) 1 ≈4.5MHz 2πVLC2 当：C,=35pF(max) 1 ≈3.8MHz 2πVLC:1 2 1 C C C k f + 解： = C1 = 329 pF 5 1 2 1 3 2 3 1 2 3 4 5 53 C C C C C C C C C C C C C = + + + Σ = + + 当： C 9 (min) 5 = pF MHz LC fo 4.5 2 1 max = ≈ π Σ 当： 35 (max) C5 = pF MHz LC fo 3.8 2 1 min = ≈ π Σ