UPDATE : How to continue Computing the Convolution

Question

$$ x(t):=\begin{cases} 1&\text{if $0<t<T$}\\ \\ 0&\text{if otherwise} \end{cases} \qquad\text{and}\qquad h(t):=\begin{cases} t&\text{if $0<t<2T$}\\ \\ 0&\text{if otherwise} \end{cases} $$ Compute $(x*h)(t)$.

My approach: We first note that :$$\displaystyle x(\tau)=\begin{cases} 1&\text{if $0<\tau<T$}\\ \\ 0&\text{if otherwise} \end{cases}\qquad\text{and}\qquad\displaystyle h(t-\tau)=\begin{cases} t-\tau&\text{if $0<t-\tau<2T$}\\ \\ 0&\text{if otherwise} \end{cases}$$ We observe that : $$ 0<\tau<T $$ $$ t-2T<\tau<t $$ Thus, $(x*h)(t)\in\mathcal{R}[\max(0,t-2T),\min(t,T)]$ (i.e. Riemann integrable). Therefore : \begin{align*} (x*h)(t)&:=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)\;\text{d}\tau\\ &=\int_{\max(0,t-2T)}^{\min(t,T)}t-\tau\;\text{d}\tau\\ &=\left. t\tau-\frac{\tau^{2}}{2}\right|_{\max(0,t-2T)}^{\min(t,T)}\\ &=\left[t\min(t,T)-\frac{\min^{2}(t,T)}{2}\right]-\left[t\max(0,t-2T)-\frac{\max^{2}(0,t-2T)}{2}\right]\\ &=t[\min(t,T)-\max(0,t-2T)]-\frac{1}{2}[\text{min}^{2}(t,T)-\text{max}^{2}(0,t-2T)] \end{align*}

Problem: I want to further solve this to obtain the final solution as a piecewise function (if it has such form)

I present to you the MATLAB simulation of the convolution

First one corresponds to $T=1$ and the second one corresponds to $T=2$. Here is the algorithm used:

t = linspace(0,10,1000);   % assume a time span for "t" 
T=1;
m = min(t,T);              % find min of (t-1 , 5) for each "t"
M = max(0,t-2*T);            % find max of (t-3 , 3) for each "t"
vind = M < m;              % find "valid" range of "t"

y = t.*(m-M)-0.5*(m.^2-M.^2);% EVALUATE the CONVOLUTION
y = y.*vind;               % force non-valid range to zero.

figure,plot(t,y);         % DISPLAY:
title('convolution is :');

Credits to $\text{@Fat32}$ for the code in his answer in the link provided in the comment

Answer 1

Some general hints for this type of problem

1. If the function is defined piecewise than chances are your calculation and solution will also have to be done piecewise in sections.
2. Determine first where the result is zero, i.e. where $x(\tau)$ and $h(t-\tau)$ don't overlap. This determines the start and end of the sections that you need to consider. Hint: if both functions are causal than $t=0$ is always a boundary. The length of a convolution is always equal (or smaller) than the sum of the lengths of the signals.
3. Then find the section boundaries: these are where the mins() and maxes() changes from one argument to the other.
4. In each section replace the max()/min() with the correct value for this section. Solve each section individually and splice the solutions together.

It's also super helpful if you can develop some graphical intuition for this type of problem: draw both functions with a few different overlaps and see what happens when they start overlapping, when they are fully overlap and when they stop overlapping.

You already came a long way. Give that a shot and post back if you are still stuck. Happy to help more.

EDIT More help on the piecewise solution. For simplicity let's call the convolution $y(t)$

1. $y(t)$ must be zero for $t < 0$. The convolution cannot start earlier than any of the signals
2. $y(t)$ must be zero for $t > 3T$. The convolution cannot be longer than the sum of the length of the signal.
3. That means you only need to focus on the interval $0 < t < 3t$
4. $min(t,T)$ is $t$ for $ 0 < t < T$ and $T$ for $T < t < 3T$
5. $max(0,t-2T)$ is $0$ for $ 0 < t < 2T$ and $t-2T$ for $2T < t < 3T$

That means that you need to split the solution into three section: $[0,T]$, $[T,2T]$ and $[2T, 3T]$. Within each section, you can replace the min/max functions by the actual value since it doesn't change over the section. Example: in the first section $[0,T]$, we have $max(0,t-2T) = 0$ and $min(t,T)=t$ so you would be integrating over $[0,t]$.