Problem with convolution, insecure

Question

$$f(t)= t^2\cdot u(t),\quad g(t)=t^4\cdot u(t)$$

I know that I need to use convolution theorem to solve this problem, but I really don't know what to do with step functions. Do I need to include them in the equation or? Any help appreciated.

Answer 1

To answer how to perform convolution with step functions, I would say there are two common ways: directly with integration (which is feasible for this problem but not for many others), or with the Laplace transform. However, doing integration with step functions is tricky and worth explaining, in my opinion, so I'll answer for future readers:

Direct Integration

By noting that step functions are piecewise, and thus so are $f(t)$ and $g(t)$, we can break the integral up into the two "pieces" and evaluate $f$ on each piece:

$$\begin{align} f(t)\ast g(t) &= \int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau \\ &=\int_{-\infty}^0f(\tau)g(t-\tau)d\tau + \int_{0}^{\infty}f(\tau)g(t-\tau)d\tau \\ &=\int_{-\infty}^0(\tau^2\cdot 0)g(t-\tau)d\tau + \int_{0}^{\infty}(\tau^2\cdot 1)g(t-\tau)d\tau \\ &=0 + \int_{0}^{\infty}\tau^2g(t-\tau)d\tau \\ \end{align}$$

Change variables with $v=t-\tau$ to get:

$$\begin{align} f(t)\ast g(t) &= \int_{-\infty}^{t}(t-v)^2g(v)dv \end{align}$$

Here, we can do the same thing and break up the integral, but we have a problem: does the break point ($v=0$) happen between $-\infty$ and $t$, or above $t$? We can split the two cases apart:

$$\begin{align} f(t)\ast g(t) &= \begin{cases} \int_{-\infty}^{t}(t-v)^2g(v)dv, \quad t<0 \\ \int_{-\infty}^0(t-v)^2g(v)dv + \int_{0}^t(t-v)^2g(v)dv, \quad t\ge 0 \end{cases} \\ &= \begin{cases} \int_{-\infty}^{t}(t-v)^2(v^4\cdot 0)dv, \quad t<0 \\ \int_{-\infty}^0(t-v)^2(v^4\cdot 0)dv + \int_{0}^t(t-v)^2(v^4\cdot 1)dv, \quad t\ge 0 \end{cases} \\ &= \begin{cases} 0, \quad t<0 \\ 0 + \int_{0}^t(t^2v^4-2tv^5+v^6)dv, \quad t\ge 0 \end{cases} \\ &= \begin{cases} 0, \quad t<0 \\ \frac{t^7}{5}-\frac{t^7}{3}+\frac{t^7}{7}, \quad t\ge 0 \end{cases} \\ &= \frac{1}{105}t^7u(t) \end{align}$$

Laplace Transform

If we use a basic Laplace Transform table, we see that $\mathcal{L}\{t^nu(t)\}=n!/s^{n+1}$, as noted in the comments. This can be proved inductively using integration by parts on the transform integral. Once we have this, though, convolution in $t$ is just multiplication in $s$, so we get:

$$\begin{align} \mathcal{L}\{f(t)\ast g(t)\} &= \frac{2!}{s^3}\cdot \frac{4!}{s^5} \\ &=\frac{7!}{7!}\cdot \frac{2\cdot 4!}{s^8} \\ &=\frac{2}{5\cdot 6\cdot 7}\cdot\frac{7!}{s^8} \\ &=\frac{1}{105}\cdot \frac{7!}{s^8} \end{align}$$

Taking the inverse Laplace transform of both sides gives us the same answer as above:

$$\begin{align} f(t)\ast g(t) &= \mathcal{L}^{-1}\Big\{\frac{1}{105}\cdot\frac{7!}{s^8}\Big\} \\ &=\frac{1}{105}\cdot\mathcal{L}^{-1}\Big\{\frac{7!}{s^8}\Big\} \\ &=\frac{1}{105}t^7u(t) \end{align}$$