# Graph of $[u(\tau)-u(\tau-4)]\cdot[u(t-\tau)-u(t-\tau-4)]$

I'm trying to visualize what would the following signal be like $$[u(\tau)-u(\tau-4)]\cdot[u(t-\tau)-u(t-\tau-4)]$$ Consider that $\tau$ is the independent variable here, $t$ is the shift variable.

Note: I'm doing this to calculate the continuous-time convolution

As this signal (function) exists inside an integration operator, and also assuming that there are no impulses residing at any discontinuities introduced by $u(t)$ step functions, then we can simply disregard the isolated discontinuity points and rely on the following:

[u(\tau)-u(\tau-4)][u(t-\tau)-u(t-4-\tau)] = \begin{cases} \begin{align} &0 &\scriptstyle{\text{for t < 0}}\\ &u(\tau)-u(\tau-t) &\scriptstyle{\text{for 0 < t < 4}}\\ &u(\tau-t+4)-u(\tau-4) &\scriptstyle{\text{for 4 <t < 8}}\\ &0 &\scriptstyle{\text{for t > 8}}\\ \end{align} \end{cases}

Where the intervals are interpreted with the $t$ viewed as a parameter for an integrand argument function of a dummy variable $\tau$.

This convolution is reminiscent of convolving a rectangular pulse of duration 4 units with itself, whose output result should be a triangular pulse of duration 8 units, extending from $t=0$ to $t=8$ with an apex at $t=4$. Assuming that this reminiscence is correct, I would define the convolution result with this more apparent form as (defining the pulse as $p(t) = u(t) - u(t-4)$) : y(t) = p(t)*p(t) = \int_{-\infty}^{\infty} {p(\tau)p(t-\tau)d\tau} = \begin{cases} \begin{align} &0 &\scriptstyle{\text{for t < 0}}\\ &\int_{\tau=0}^{\tau=t}{1d\tau} = t &\scriptstyle{\text{for 0 < t < 4}}\\ &\int_{\tau=t-4}^{\tau=4}{1d\tau} = 8-t &\scriptstyle{\text{for 4 <t < 8}}\\ &0 &\scriptstyle{\text{for t > 8}}\\ \end{align} \end{cases}

Where the limits of the integrals are deduced from the integrand arguments given at the first set of definition above: For example, for the interval $0<t<4$ the limits of integral are choosen based on the integrand $u(\tau) - u(\tau -t)$. Looking at this integrand one can see that for this to be nonzero (for a given fixed t) we need both $\tau > 0$, (from the first $u(\tau)$) and also $\tau < t$ from the second $u(\tau-t)$. So combining these two conditions requires that $0<\tau<t$ must be the limits of integration. The other cases are treated similarly.

On the other hand the operation is most easily understood and visualied by the help of an application of "flip & drag" technique to evaluate the intervals and to deduce the resulting integrand functions, instead of the way I presented as a piecewise definition form.

• Again you saved my life :D plus your edit make it way much clearer. THANKS – Hossam Houssien May 25 '16 at 0:40
• @HossamHoussien you're welcome! ;) – Fat32 May 25 '16 at 0:44
• @HossamHoussien I have also added about how to determine the limits of integration, hoping that it is more clear now... – Fat32 May 25 '16 at 1:02
• yea it more and more obvious now :) – Hossam Houssien May 25 '16 at 1:19