# Calculate the convolution of two constants. (5.6-1)

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:

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

$$h(t)=t$$

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

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
$$\int_0^tf(\tau)g(t-\tau)d\tau=\int_0^t1\cdot 1\,d\tau=\int_0^td\tau=t$$
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 $t$'.