Calculating integral consisting of two piecewise functions

Question

There is a system with response:

$h(t)= \begin{cases} 1, & \text{if}\ 0 \leq t \leq 2 \\ 0, & \text{elsewhere} \end{cases}$

I need to find the response of the system to the following input:

$u(t)= \begin{cases} 1-t, & \text{if}\ 0 \leq t \leq 1 \\ 0, & \text{elsewhere} \end{cases}$

I know that in order to get the response $y(t)$ I need to calculate the integral $\int_0^t{u(\tau)h(t-\tau)\,d\tau}$. From this integral I get $y(t)=t-\frac{1}{2}t^2$ for $0 \leq t \leq 1$, however this is not the complete answer and incorrect (has to be $<$ not $\leq$). So my question is how to calulate the other part of the solution? And why are the other parts not $0$ if $u(t)$ is $0$ in the integral?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mbox{Lets}\ \mrm{f}\pars{t} & =\int_{0}^{t}\mrm{u}\pars{\tau}\mrm{h}\pars{t - \tau}\,\dd\tau = \int_{0}^{t} \braces{\vphantom{\large A}\pars{1 - \tau}\bracks{0 \leq \tau \leq 1}} \bracks{0 \leq t - \tau \leq 2}\,\dd\tau \\[5mm] & = \int_{0}^{t}\pars{1 - \tau}\bracks{0 \leq \tau \leq 1} \bracks{t - 2 \leq \tau \leq t}\,\dd\tau\qquad \pars{~\mbox{Obviously}\, \,\mrm{f}\pars{t} = 0\ \mbox{if}\ t \leq 0~} \end{align}

Then,

\begin{align} \left.\vphantom{\large A}\mrm{f}\pars{t}\,\right\vert_{\ t\ >\ 0} & = \int_{0}^{t}\pars{1 - \tau}\bracks{0 \leq \tau \leq 1} \bracks{t - 2 \leq \tau \leq t}\,\dd\tau \\[5mm] & = \bracks{t < 1}\int_{0}^{t}\pars{1 - \tau}\,\dd\tau + \bracks{t > 1}\int_{0}^{1}\pars{1 - \tau} \bracks{t - 2 \leq \tau \leq t}\,\dd\tau \\[1cm] & = \bracks{t < 1}\pars{t - {1 \over 2}\,t^{2}} + \bracks{t > 1}\bracks{t - 2 < 0}\int_{0}^{1}\pars{1 - \tau}\,\dd\tau \\[2mm] & + \bracks{t > 1}\bracks{0 < t - 2 < 1}\int_{t - 2}^{1}\pars{1 - \tau}\,\dd\tau \\[1cm] & = \bracks{t < 1}\pars{t - {1 \over 2}\,t^{2}} + \bracks{1 < t < 2}{1 \over 2} + \bracks{2 < t < 3} \pars{{1 \over 2}\,t^{2} - 3t + {9 \over 2}} \end{align}

---

$$\bbx{% \mrm{f}\pars{t} =\int_{0}^{t}\mrm{u}\pars{\tau}\mrm{h}\pars{t - \tau}\,\dd\tau = \left\{\begin{array}{lcl} \ds{-\,{1 \over 2}\,t^{2} + t} & \mbox{if} & \ds{0 < t \leq 1} \\[2mm] \ds{1 \over 2} & \mbox{if} & \ds{1 < t \leq 2} \\[2mm] \ds{{1 \over 2}\,t^{2} - 3t + {9 \over 2}} & \mbox{if} & \ds{2 < t < 3} \\[2mm] \ds{0} && \mbox{otherwise} \end{array}\right.} $$

Answer 2

For t>1 Let $\tau$ =x (easier on my typing) . If x < t-1 we have h(t-x)=0.

Also u(x)=0 so for x>1 so you need to calculate y(t)= $\int_{t-1}^1 (1-x) dx$ .I will let you do it.