3
$\begingroup$

Request

I am very new to this so please bear with me. I cannot duplicate the answer in the book. I believe I may be making a methodical error. Please correct it for me.

Given:

Find the convolution of $f(t)=1$ and $g(t)=1$.

$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$

My Solution:

$$h(t)=(1*1)(t)=\int_0^t 1\cdot (1-\tau)d\tau=\tau-\frac{\tau^2}{2}|_0^t=t^-\frac{t^2}{2} = \frac{t}{2}$$

Answer in Text:

$$h(t)=t$$

$\endgroup$
4
  • 2
    $\begingroup$ Calculate with $g(t)=1$, as requested, instead of $g(t)=?$. $\endgroup$
    – Thomas
    Feb 15, 2015 at 9:32
  • 1
    $\begingroup$ As @Thomas stated you should replace $g(t)$ with $1$, not with $1-\tau$ $\endgroup$ Feb 15, 2015 at 9:59
  • $\begingroup$ How do I make a large vertical bracket thingy like this one? $|_0^t$ $\endgroup$ Feb 18, 2015 at 2:52
  • $\begingroup$ Which book are you using? $\endgroup$
    – Matt L.
    Feb 21, 2015 at 16:14

2 Answers 2

2
$\begingroup$

The problem is that you're using $1-\tau$ as part of the integrand. However, if $g(t)=1$ then $g(t-\tau)$ also equals $1$, and the integral is simply

$$\int_0^tf(\tau)g(t-\tau)d\tau=\int_0^t1\cdot 1\,d\tau=\int_0^td\tau=t$$

$\endgroup$
2
  • $\begingroup$ @ Matt L. I took differential equations in college about 20 years ago. I didn't complete college because after taking and passing the EIT exam in my junior year I was quickly hired as an engineer and have been practicing since then. But now I am unemployed and no one will hire me without a degree so I am studying many of my core courses so I am ready when I return to complete my degree. The text book (from Math 370A at CSULB) I am reviewing is entitled Advanced Engineering Mathematics 6th Edition by Erwin Kreyszig. $\endgroup$ Feb 21, 2015 at 23:00
  • 1
    $\begingroup$ @JulesManson: OK, thanks, I was just curious if I knew the book. Good luck with your degree! $\endgroup$
    – Matt L.
    Feb 22, 2015 at 9:29
0
$\begingroup$

Convolution is nothing, but multiplication of two functions in different time domain. That means whatever we've done in above two solutions should be wrong, because if we are giving our answer on the basis of the above two solutions that means we have taken that 'our function is existing in only >0 time period, but that's not true a constant function always exists between '$-\infty$ to $+\infty$'. So how can we put the limit between '$0$ to $t$'.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .