Solve Funky Integral: Simplify g(t)

[e(ho0n3]

[SOLVED] A Funky Integral

Let

$$f(t) =\left\{\begin{array}{lll} 0 &\mbox{ if }t < 0 \\ 1 &\mbox{ if }0 \leq t \leq 1 \\ 0 &\mbox{ if }t > 1 \end{array}\right$$

I need a simplified version of $g(t)$ shown below:

$$g(t) = \int_{-\infty}^\infty f(u) \cdot f(t - u) \, du$$

I divide the integral into the following sum:

$$g(t) = \int_{-\infty}^0 f(u) \cdot f(t - u) \, du + \int_{0}^1 f(u) \cdot f(t - u) \, du + \int_{1}^\infty f(u) \cdot f(t - u) \, du$$

$f(u)$ in the first and third term is 0 according to the definition of $f$. Hence

$$g(t) = \int_0^1 f(t - u) \, du$$

Is this right? What do I do know? Should I consider the cases where $t < 0$, $0 \le t \le 1$ and $t > 0$? Or should that be $t - u$?

 

[G01]

What you have so far looks good. Now, consider the cases for when f(t-u) is 0 and 1. Remember that f(t-u) is just f(t) shifted by u units on the t axis.

 

[e(ho0n3]

When f(t-u) is 0, g(t) is 0. When f(t-u) is 1, g(t) is 1. Is that what you mean?

 

[G01]

When f(t-u) is 0, g(t) is 0. When f(t-u) is 1, g(t) is 1. Is that what you mean?

Almost. What I mean is that you should split up f(t-u) into a piecwise function. Find the range of t values for which f(t-u) is 1 and 0. In other words, write f(t-u) in the way f(t) is given to you. After you do this you should be able to break up the integral on your last line into smaller integrals that you can easily evaluate.

 

[e(ho0n3]

When t < u, f(t - u) = 0. When t is in the range [u, 1 + u], f(t - u) is 1. If t > 1 + u, f(t - u) is 0.

If t < 0, given that the limits of the integral are 0 and 1, t - u < 0 and f(t - u) is 0 so g(t) = 0 right?

If t is in [0, 1], then whenever u > t, t - u < 0 and f(t - u) is 0. So in this case, the integral would break like this:

$$\int_0^1 f(t - u) \, du = \int_0^t f(t - u) \, du \: + \int_t^1 f(t - u) \, du$$

On the right-hand side, f(t - u) in the first term is 1 and in the second term 0 so g(t) is 1 in this case.

When t > 1, the integral breaks up exactly like above except that f(t - u) in the first term is 0 and 1 in the second term. g(t) is 1 in this case.

In summary, g(t) = 0 if t < 0; otherwise g(t) = 1. Right?

 

[e(ho0n3]

If t is in [0, 1], then whenever u > t, t - u < 0 and f(t - u) is 0. So in this case, the integral would break like this:

$$\int_0^1 f(t - u) \, du = \int_0^t f(t - u) \, du \: + \int_t^1 f(t - u) \, du$$

On the right-hand side, f(t - u) in the first term is 1 and in the second term 0 so g(t) is 1 in this case.

I messed up. g(t) is not 1. If f(t - u) is 1 in the first term, then that integral becomes t so g(t) = t in this case.

 

[e(ho0n3]

When t > 1, the integral breaks up exactly like above except that f(t - u) in the first term is 0 and 1 in the second term. g(t) is 1 in this case.

I made a mistake here too. I realized this after working out some examples where t > 1. For example, g(10) = 0 and g(1.1) = 0.9. It seems that if 1 < t < 2, then g(t) = 2 - t. If t is greater than or equal to 2, then g(t) is 0.

 

[e(ho0n3]

Correction. g(t) = 0 when t >2 and g(t) = 2 - t whenever t is in the range (1, 2].