Convolution of two piecewise functions

Question

$$\phi(t)=\begin{cases}1&t\in[0,1)\\0&\text{otherwise}\end{cases}$$ and $$\psi(t)=\begin{cases}1&t\in[0,1/2)\\-1&t\in[1/2,1)\\0&\text{otherwise}\end{cases}$$

I know that $\mathcal{F}^{-1}[\hat{\phi}\cdot\hat{\psi}]=\frac{1}{\sqrt{2\pi}}\phi*\psi$, but I am supposed to find the convolution using the definition $f*g=\int_{-\infty}^\infty f(t-x)g(x)dx$. My biggest problem is how do I go about finding the bounds of integration? How do I know when $t-x\in[0,1/2)$?

Answer 1

By definition, $$\phi\ast\psi(t)=\int_{-\infty}^{\infty}\phi(t-x)\psi(x)\;dx=\int_{-\infty}^{\infty}\phi(x)\psi(t-x)\;dx$$ $\phi(x)=0$ unless $0\leq x<1$, in which case $\phi(x)=1$. Therefore $$ \int_{-\infty}^{\infty}\phi(x)\psi(t-x)\;dx=\int_0^1\psi(t-x)\;dx$$ Now consider a few cases. If $t\geq 2$ then $t-x\geq t-1\geq 1$ for all $x\in[0,1]$, so the integral is zero. Similarly, if $t<0$ then $t-x\leq t<0$ for all $x\in[0,1]$, so again the integral is zero.

So that leaves $0\leq t<2$. If $0\leq t<1$ then $t-x\in [0,1)$ when $0\leq x\leq t$, so $$ \int_0^1\psi(t-x)\;dx=\int_0^t\psi(t-x)\;dx=\int_0^t\psi(x)\;dx$$ and this integral is equal to $t$ if $0\leq t<\frac{1}{2}$, and is equal to $\frac{1}{2}-(t-\frac{1}{2})=1-t$ if $\frac{1}{2}\leq t<1$.

Finally, if $1\leq t<2$, then $t-x\in[0,1)$ when $t-1<x\leq 1$, hence $$\int_0^1\psi(t-x)\;dx=\int_{t-1}^1\psi(x)\;dx$$ which is equal to $\frac{1}{2}-(t-1)-\frac{1}{2}=1-t$ if $1\leq t<\frac{3}{2}$, and is equal to $-[1-(t-1)]=t-2$ if $\frac{3}{2}\leq t<2$.

Answer 2

You may consider this fact: Convolution commutes with translations, i.e. writing $T_xf(z) = f(y-x)$, and hence $supp(T_xf) = x + supp(f)$, you have $$ T_x(f \ast g) = (T_x f) \ast g = f \ast (T_x g) $$. Since the second function consists of two shifted copies of the indicator function of the half-interval you only have to compute the convolution of $\phi$ with the indicator function of $[0,1/2]$, i.e. the dilated version of that function.

This also requires some changes, but you will see that you get a trapezoidal function. So you have to look at the end at the difference of two trapezoidal functions, which should be easy. In fact, they are all piecewise linear over between the half-integers if I see it correctly.