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enter image description here

For the first part of the above problem, I copied an example from my book and I got the answer to be

$$t(t-1)+t(t-2)=t^2-3t$$

considering that the integral is the sum of the area of the rectangles in the graph. Does this make sense? I don't quite understand how this answer would work or if it's right and I'm not sure how to do the second part to see if I would get the same answer to confirm.

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We are asked to graphically find the convolution of the two pulses:

enter image description here

We see that for $f(t), t \in [0,5]$ and for $g(t), t \in [0, 1]$, thus we need to evaluate the convolution integral in the range $[0 + 0, 5 + 1] = [0, 6]$, that is $0 \le t \le 6$.

We need to flip one of the pulses about the vertical axis and since they are pulses, we can choose either one, but it usually easier to choose the one with smaller duration, so lets choose to flip $g(t)$. We also now write these as $f(\tau), g(\tau)$. A plot of these is:

enter image description here

If we shift $g(- \tau)$ to the left, we get $g(t-\tau)$, which we can see in the plot above, we have no overlap between the two curves, thus the integral is zero in this region (which we will call it region $1$). Note that we label $g(-\tau)$ over the range $(t-1, t)$ for all calculations (notice what you get for $t=1$).

Next, we start shifting the signal $g(t-\tau)$ to the right $(t \gt 0)$. Consider first the interval $0 \le t \le 1$, which is:

enter image description here

In this interval, the signals overlap as shown, hence the product is not zero in this interval and the convolution is given as (which we call region $2$):

$$\displaystyle f(t) \ast g(t) = \int_0^t 1 \times 2 ~d\tau = 2t, 0 \le t \le 1$$

By shifting $g(t-\tau)$ further to the right, we get the same kind of overlap for $1 \le t \le 5$ as:

enter image description here

In this interval, which we will call region $3$, the integral yields:

$$ f(t) \ast g(t) = \displaystyle \int_{t-1}^t 1 \times 2 ~d\tau = 2, 1 \le t \le 5$$

By shifting $g(t-\tau)$ further to the right , for $5 \le t \le 6$, we get:

enter image description here

Again, we have overlap (which we will call region $4$), so the integral is given by:

$$ f(t) \ast g(t) = \displaystyle \int_{t-1}^5 1 \times 2 ~d\tau = 2(6-t), 5 \le t \le 6$$

Lastly, we shift $g(t-\tau)$ further to the right for $t \ge 6$ and we have no overlap (we will call this region $5$), so the integral is zero, as:

enter image description here

Putting all of these regions together, we have:

$$f(t) \ast g(t) = \begin{cases}\begin{array}{cc} 0 & t\leq 0 \\ 2 t & 0\leq t\leq 1 \\ 2 & 1\leq t\leq 5 \\ 2 (6-t) & 5\leq t\leq 6 \\ 0 & t \ge 6 \end{array} \end{cases} $$

A plot of the convolution $f(t) \ast g(t)$ is:

enter image description here

For the second part, we can write the functions as:

$$f(t) = u(t) - u(t-5), g(t) = 2(u(t) - u(t-1))$$

Now, we need to find:

$$f(t) \ast g(t) = \displaystyle \int_{- \infty}^{\infty} f(\tau) \ast g(t-\tau)~ d\tau = \int_0^6 f(\tau) \ast g(t-\tau)~ d\tau$$

We can also solve this as:

$$f(t) \ast g(t) = \int_{- \infty}^{\infty} g(\tau) \ast f(t-\tau)~ d\tau = \int_0^6 g(\tau) \ast f(t-\tau)~ d\tau $$

I will let you work out those details as you already know the answer.

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