Convolution Question Homework: Period & Expression Calculation

[gauss mouse]

Homework Statement

This is a question from a sample exam, rather than a homework problem.

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

Homework Equations

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

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.

 

[mighty2000]

Thank you for your question. I can offer some insights and clarification on this problem.

Firstly, yes, the period of the convolution of two functions is the same as the period of the original functions. This can be seen from the convolution integral formula, where the limits of integration are from -L/2 to L/2, where L is the period of the functions being convolved.

For part (b), you are correct in your answers. The expressions for f in the given intervals are indeed (t-1)^2 and (t+3/2)^2. These can be obtained by shifting the original function f(t)=t^2 by the given intervals.

For part (c), there is a typo in the given expression for (f*g)(x). It should be (x-1/2)^3 instead of (x+1/2)^3. This can be seen from the given formula for the convolution of two functions, where the second function is shifted by x instead of -x. With this correction, the given expression will simplify to the constant 13/12.

I understand your confusion about using the formula \int_{-1/2}^{1/2}f(x-t)g(t)dt to compute the convolution. The reason for this is to make the limits of integration for the convolution integral match the limits of integration for the original functions, which are -1/2 to 1/2. This is a common practice in mathematics, where we use the dummy variable t for the function being integrated, and x for the variable we are integrating with respect to.

I hope this helps to clarify the problem. Let me know if you have any further questions. Best of luck on your exam!