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高频电子技术复习题(? - 百度文库

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  • 2026/1/10 15:39:00

4֪ıʽΪu01(t) = ( 1 + 0.5cost )cosct,

u02(t) = sint sinct

c = 5uo1(t) u02(t)IJͼƵͼ

5ѵѹʽֱΪ

u1(t) = 2cosv100С10wt + 0.1cosv90С10wt + 0.1cosv110С10wt (V) u2(t) = 0.1cosv90С10wt + 0.1cosv110С10wt (V)

3

3

3

3

3

11ҳ 9 ҳ ƵӼϰ

˵Ϊѵͼֱڵλ裨1ϵıƵʣ

ƽʼƵ׿ȡ P97

6һزΪuct= 4 cosv2С2510wt źΪƵҲƵF = 400Hz

6

Ƶƫfm=10kHz.ֱдƵನıʽ (P140)

7֪źut= 5 cos 2С10t

3

źűʾʽΪuot= 10 cos2С10t10 Sin2С10tV

6

3

11ҳ 10 ҳ ƵӼϰ

ָõźǵƵźŻǵźţ

ָزƵʡԼƵƫΪ٣ P140

8ܵƵźuvtwUm sin (2С10tĵơ֪DzķΪ10V

3

Dz˲ʱƵΪ f(t) = 10 +10cos(2С10)t 1DzǵƵǵನдѧʽ 2ʱƵƫ͵ָ P140

643

11ҳ 11 ҳ ƵӼϰ

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4֪ıʽΪu01(t) = ( 1 + 0.5cost )cosct, u02(t) = sint sinct c = 5uo1(t) u02(t)IJͼƵͼ 5ѵѹʽֱΪ u1(t) = 2cosv100С10wt + 0.1cosv90С10wt + 0.1cosv110С10wt (V) u2(t) = 0.1cosv90С10wt + 0.1cosv110С10wt (V) 33333 11ҳ 9 ҳ ƵӼϰ <

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