# Convolution Question

1. Apr 25, 2012

### gauss mouse

1. The problem statement, all variables and given/known data
This is a question from a sample exam, rather than a homework problem.

$\text{Let }f \text{ be the periodic function with period } 1 \text{ defined for }-1/2\leq t<1/2 \text{ by }f(t)=t^2\\ \text{ and let }g \text{ be the periodic function with period } 1 \text{ defined for }-1/2\leq t<1/2 \text{ by } g(t)=1. \\ \text{(a) What is the period of }f\ast g? \\ \text{(b) Find the expression of }f \text{ for }1/2\leq t<3/2, \text{ and }-2\leq t\leq -1. \\ \text{(c) Show that }f\ast g\text{ is given for }0\leq x\leq 1 \text{ by } \\ (f\ast g)(x)= \frac{5}{4}-\frac{1}{3}\left(x-\frac{1}{2}\right)^3+\frac{1}{3}\left(x+\frac{1}{2}\right)^3-\left(x+\frac{1}{2}\right)^2+x.$

2. Relevant equations
The convolution of two functions $f_1$ and $f_2$ both with period $L$ is given by
$\displaystyle (f_1\ast f_2)(x)=\frac{1}{L}\int_{-L/2}^{L/2}f_1(x-t)f_2(t)dt= \frac{1}{L}\int_{-L/2}^{L/2}f_1(t)f_2(x-t)dt$

3. The attempt at a solution
(a) Is it so that the convolution will have the same period as the two functions which you convolve?
(b) Clearly the answers are $(t-1)^2\text{ and }(t+3/2)^2$.

(c)There seems to be something wrong with the question. The first thing that looks wrong is that the last expression is,
when you expand it, nothing more than the constant $\displaystyle\frac{13}{12}$.

I calculated the convolution as follows
$\displaystyle \int_{-1/2}^{1/2}f(t)g(x-t)dt=\int_{-1/2}^{1/2}f(t)dt=\int_{-1/2}^{1/2}t^2dt=\frac{1}{12}$

It seems that the lecturer wants us to compute it as follows
$\displaystyle \int_{-1/2}^{1/2}f(x-t)g(t)dt$
I think that he wants us to use answers from part (b) to figure out what values $f$ should take. I don't see why we should do it this way instead of my way or why the answers would be different between his way and my way. I also can't see exactly what to do for his way. I don't see how I would always know where $x-t$ was and hence I don't see how I would always know the value of $f$ at $x-t$. I know that's what part (d) is for but I can't see what to do.