∫Tδ(t-1)e^-jkttdt
时间: 2024-05-30 19:15:34 浏览: 85
I'm sorry, as an AI language model, I cannot solve indefinite integrals analytically. However, if the limits of integration were provided, I could try to solve it using integration techniques.
相关问题
某LTI系统,当输入f(t)=e^2tε(t)时,其零状态响应为yzs(t)=(4e^-2t-e^-t-2e^-3t)ε(t)当f(t)=ε(t),y(0_)=y'(0_)=1时,求系统的全响应。
根据线性时不变系统的性质,系统的全响应等于其零状态响应加上其零输入响应。因此,我们需要先求出系统的零输入响应。
对于输入为单位阶跃函数 ε(t),其拉普拉斯变换为 F(s) = 1/s。因此,系统的传递函数为:
H(s) = Y(s)/F(s) = Y(s)/[1/s] = sY(s)
其中,Y(s)为系统的拉普拉斯变换。
根据系统的零状态响应,我们可以求出其初始状态:
y(0+) = yzs(0+) = (4e^0-e^0-2e^0)ε(0+) = 1
y'(0+) = s*y(0+) = s*1 = s
因此,系统的初始状态为 y(0) = 1,y'(0) = s。
接下来,我们需要求解系统的全响应。由于输入为单位阶跃函数,因此其单位冲激响应为 h(t) = dy(t)/dt,即:
h(t) = d/dt [e^2tε(t)] = 2e^2tε(t) + e^2tδ(t)
其中,δ(t)为单位冲激函数。
根据卷积定理,系统的零输入响应为:
yzi(t) = h(t) * u(t) = ∫[0,t] h(τ) dτ
将 h(t) 和 ε(t) 的表达式带入上式,得到:
yzi(t) = ∫[0,t] (2e^2τε(τ) + e^2τδ(τ)) dτ
对于 t>0,上式的第二项为零,因此:
yzi(t) = ∫[0,t] 2e^2τε(τ) dτ = 2∫[0,t] e^2τ dτ = (e^2t-1)/2
因此,系统的全响应为:
y(t) = yzi(t) + yzs(t) = (e^2t-1)/2 + (4e^-2t-e^-t-2e^-3t)ε(t)
最终答案为:
y(t) = (e^2t-1)/2 + (4e^-2t-e^-t-2e^-3t)ε(t)
求δ’(t)*e^-3tδ(t)
根据单位冲激函数的性质,其导数是一个导数为δ'(t)的脉冲函数,即:
δ'(t) = d/dt δ(t)
因此,可以将δ'(t)*e^(-3t)δ(t)看作是一个脉冲函数在t=0处的值乘以e^(-3t),即:
δ'(t)*e^(-3t)δ(t) = δ'(0)*e^0*δ(0) = δ'(0)
因此,问题转化为求δ'(0)。由于δ(t)在t=0处的值为无穷大,因此需要使用广义函数的定义。根据广义函数的导数定义,有:
∫δ'(t)ϕ(t) dt = -∫δ(t)ϕ'(t) dt
其中,ϕ(t)是一个测试函数,即具有无限可导性和紧支集的实函数。将ϕ(t)取为e^(-3t),则有:
∫δ'(t)e^(-3t) dt = -∫δ(t)(-3e^(-3t)) dt
化简可得:
δ'(0) = -3δ(0)
由于δ(0)的值为无穷大,因此δ'(t)*e^(-3t)δ(t)不存在有限的解。
阅读全文
相关推荐













