幅值最高为1,最低为0,周期为4的矩形信号的傅里叶系数是什么
时间: 2024-02-09 19:07:17 浏览: 28
周期为4的矩形信号,其数学表达式为:
f(t) =
{
1, 0≤t<2
0, 2≤t<4
}
因为幅值最高为1,最低为0,所以我们可以令A=1/2,即A为矩形信号幅值的一半。
根据公式,可以计算出该信号的傅里叶系数:
A0 = (1/4) * ∫(0,4) f(t) dt
= (1/4) * ( ∫(0,2) 1 dt + ∫(2,4) 0 dt )
= 1/2
An = (1/2) * ∫(0,4) f(t)*cos(nπt/2) dt
= (1/2) * ( ∫(0,2) cos(nπt/2) dt + ∫(2,4) 0 dt )
= (1/2) * ( 2*sin(nπ) / (nπ) )
= 0,当n为偶数时
An = (1/2) * ∫(0,4) f(t)*cos(nπt/2) dt
= (1/2) * ( ∫(0,2) cos(nπt/2) dt + ∫(2,4) 0 dt )
= (1/2) * ( 0 + 2*cos(nπ) / (nπ) )
= (-1)^((n-1)/2) * (1/nπ),当n为奇数时
Bn = (1/2) * ∫(0,4) f(t)*sin(nπt/2) dt
= (1/2) * ( ∫(0,2) sin(nπt/2) dt + ∫(2,4) 0 dt )
= (1/2) * ( 2*(-1)^((n-1)/2) / (nπ) )
= (-1)^((n-1)/2) * (1/nπ),当n为奇数时
综上所述,周期为4的矩形信号的傅里叶系数为:
A0 = 1/2,An = 0(n为偶数),An = (-1)^((n-1)/2) * (1/nπ)(n为奇数),Bn = (-1)^((n-1)/2) * (1/nπ)(n为奇数)