Integral Unit Impulse Function

Question

I am having trouble integrating the following unit impulse function.

$$\int_{0}^{4} \delta(t - \tau) \, d\tau$$

I got the answer $$u(t-4) - u(t)$$

but my professor says its $$u(t) - u(t-4)$$

Can some one clarify this for me?

Thanks!

Answer 1

Let us define the function $$ f(\tau)=u(\tau)-u(\tau-4) = \begin{cases} 1 & 0<\tau<4\\ 0 & \tau<0\text{ or }\tau>4. \end{cases} $$ (One way of seeing the second equality is that $u(\tau)$ is zero for $\tau<0$ and one for $t>0$, and $u(\tau-4)$ is zero for $\tau<4$ and one for $\tau>4$. Thus subtracting them, $u(\tau)-u(\tau-4)$ becomes what I wrote above.)

This means that your integral can be written $$ \int_{-\infty}^{+\infty}\bigl(u(\tau)-u(\tau-4)\bigr)\delta(t-\tau)\,d\tau=(f*\delta)(t)=f(t)=u(t)-u(t-4). $$ Here we have seen the integral as a convolution, and used the fact that $\delta$ acts as a unit when it comes to convolutions, i.e. $f*\delta=f$.

Edit

If you prefer, the derivative of $u$ is $\delta$, and hence (this is what it has to be) $$ \int_0^4\delta(t-\tau)\,d\tau=\bigl[-u(t-\tau)\bigr]_{\tau=0}^{4}=-u(t-4)+u(t)=u(t)-u(t-4). $$

Answer 2

In This Answer, I provided a primer on The Dirac Delta. The Dirac Delta is not a function. It is a Generalized Function or Distribution.

The "symbol," $\int_{-\infty}^{\infty} f(\tau)\,\delta(t-\tau)\,d\tau$, is not an integral, although it does share certain properties with the integral. It is, rather, a linear functional that maps a test function $f$ into the number $f(t)$.

Now, the meaning of the "symbol," $\int_a^b f(\tau)\,\delta(t-\tau)\,d\tau$, is the linear functional that maps the test function $f(u_a-u_b)$ as

$$\begin{align} \int_a^b f(\tau)\,\delta(t-\tau)\,d\tau&=\int_{-\infty}^\infty f(\tau)\left(u(\tau-a)-u(\tau-b)\right)\,\delta(t-\tau)\,d\tau\\\\ &=f(t)\left(u(t-a)-u(t-b)\right) \end{align}$$

where $u$ is the unit step function defined by

$$u(t)= \begin{cases} 1&,t>0\\\\ 1/2&,t=0\\\\ 0&,t<0 \end{cases} $$

Therefore, for $f(t)=1$, $a=0$, and $b=4$ we have

$$\bbox[5px,border:2px solid #C0A000]{\int_0^4 \delta(t-\tau)\,d\tau=u(t)-u(t-4)}$$

as expected!

Answer 3

Hint:

Your error is that you have used the ''primitive'' of $\delta(t-\tau)$ with respect to $t$, that is $u(t-\tau)$ but the integral is respect to $\tau$ and the ''primitive'' is $-u(t-\tau)$.