电子线路 非线性部分 第五版 谢嘉奎 课后答案

f2 = 30 kHz

Vm02 = 1VMa2 = 0.4

P02

Vm0 = 5V

P012Vm0212.5W Pav = P0 + Pav1 + Pav2 = 18.08 W

BWAM

Fmax = 33 kHz

BWAM = 2F = 2(1033 4-3

(1) v(t) = 5coscost(V)

(2) v(t) = 5cos(+ t

(3) v(t) = (5 + 3cos) cost (1)(2)

>>12Vm0220.5WPav22P02(112Ma2)1.08W 2fc = 1000 kHz (b)

(3)(c)

4-6

4-8 igDv

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