2-10周期三角波的傅里叶级数展开式为2-112-122-132-142-152-16求正弦信号x(t)=Asin(at+ϕ)的绝对均值μ|x|，均方根值xrms(t)解μ|x|=1T∫−T/2T/2|x(t)|dt=1T∫−T/2T/2A|sin(at+ϕ)|dt¿2AT∫0T/2sinatdt=−Aπcosat|0T/2=2Aπψx2=1T∫0TA2sin2atdt=A2T∫0T1−cos2at2dt=A22x...

时间:2024-11-02 11:22栏目:行业资料