ʱ : 高频电子技术复习题(? - 百度文库ϿʼĶ
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
ƽʼƵȡ 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