$$\begin{align} y(2.34) &= h \star x\,\big\vert_{t=2.34}\\ &= \int_{-\infty}^\infty h(\tau)x(2.34-\tau)\,\mathrm d\tau &\scriptstyle{\text{from the definition of the convolution integral}} \\ &= \int_0^\infty e^{-\tau}x(2.34-\tau)\,\mathrm d\tau &\scriptstyle{\text{because } h(\tau)=0\text{ for }\tau < 0}\\ &= \int_0^\infty e^{-\tau}u(1.34-\tau)\,\mathrm d\tau &\scriptstyle{\text{because }x(2.34-\tau) = u((2.34-\tau)-1)}\\ &= \int_0^{1.34} e^{-\tau}\,\mathrm d\tau &\scriptstyle{\text{because }u(1.34-\tau) = 0 \text{ for }\tau < 1.34}\\ &= 1-e^{-1.34} \end{align}$$

给定具有脉冲响应的连续时间 LTI 系统$h(t) = e^{-t}\text{u}(t)$输出$y(t)$对于任何输入$x(t)$通过卷积积分找到：

$$y(t) = x(t) \star h(t) = \int_{-\infty}^{\infty} {x(\tau)h(t-\tau)} d\tau$$

具体什么时候$x(t) = \text{u}(t-1)$，我们可以采用最一般的方法，首先取积分，然后插入 t=2.34，否则正如@DilipSarwate 建议的那样，您可以先将 t=2.34 代入被积函数，然后再取积分。我更喜欢这里的第一个：

$$\begin{align} y(t) &= \int_{-\infty}^{\infty} {\text{u}(\tau-1)e^{-(t-\tau)} \text{u}(t-\tau)d\tau}\\ y(t) &= \int_{\tau=1}^{\tau=t} {e^{-(t-\tau)}d\tau}\\ y(t) &= 0 &\scriptstyle{\text{for t less than 1}}\\ y(t) &= e^{-(t-\tau)} \big\vert_{\tau=1}^{\tau=t} = 1 - e^{-(t-1)} &\scriptstyle{\text{ for t greater than 1, hence:}}\\ y(t) &= [1 - e^{-(t-1)}] \text{u}(t-1) &\scriptstyle{\text{, for all t}}\\ \end{align}$$

我希望很清楚积分限制是如何被$\text{u}(t)$阶跃函数。

并插入$t=2.34$产量：

$$y(2.34) = [1 - e^{-(2.34-1)}] \text{u}(2.34-1) = [1 -e^{-1.34}]\text{u}(1.34) = 0.7382$$