# Convolution integral

• Mar 26th 2011, 10:04 AM
Mppl
Convolution integral
The exercise I tried to solve in the picture I attached is a convultion integral that should be solved for 0<=t<3 where u(t) is the step function (0 for all negative numbers and one for other numbers) and d(t) is the Dirac function. Can anyone see anything wrong with my resolution cause the coeficient i called A should be independent of t and the fact is that I cant see where my mistake is. Anyone can see my error?
thank you
• Mar 26th 2011, 12:15 PM
Opalg
Quote:

Originally Posted by Mppl
The exercise I tried to solve in the picture I attached is a convultion integral that should be solved for 0<=t<3 where u(t) is the step function (0 for all negative numbers and one for other numbers) and d(t) is the Dirac function. Can anyone see anything wrong with my resolution cause the coeficient i called A should be independent of t and the fact is that I cant see where my mistake is. Anyone can see my error?
thank you

Why do you think that A should be independent of t? Your calculation looks mostly correct to me, except for one thing: k is supposed to be an integer, so the condition $k\leqslant t/3$ means that the greatest allowable value of k is $k_0 = \lfloor t/3\rfloor$ (the greatest integer not exceeding t/3). Then the convolution (note the spelling!) $e^{-t}\mu(t) \ast \sum_{k=-\infty}^\infty \delta(t-3k)$ is the function

$\displaystyle \sum_{k=-\infty}^{k_0}e^{-t+3k} = \frac{e^{-t+3k_0}}{1-e^{-3}}.$

At first sight, that looks like a constant times $e^{-t}$. But in fact $k_0$ depends on t and is not a constant. (However, it only changes value at integer values of t/3, and is constant in the intervening intervals.)
• Mar 27th 2011, 11:41 AM
Mppl
in fact the purpose of that exercise is to find the coeficient A for t between 0 and 3 (does it change anything? I don't even know why that information is there since it seems to me it is of no use...)and in the solutions manual the solution says that A is t independent thats why I told you that it should be that way...seems the solutions are wrong then...
No, the solutions manual is correct. I said that the value of $k_0$ is constant in any interval in which the integer part of t/3 does not change. For t between 0 and 3, its integer part is 0. Therefore on that interval the solution is $\dfrac{e^{-t}}{1-e^3}$, which is a constant multiple of $e^{-t}.$