I want to deconvolve $p(x)$ from the following expression:
$$f(x) = (p\cdot p)(x)$$
where $f(x)$ and $p(x)$ are both real functions. Additionally,
- $f(x)$:
- has odd symmetry
- $\lim\limits_{x\rightarrow\infty}f(x)=\infty$
- $\lim\limits_{x\rightarrow\infty}\dfrac{df}{dx}=\lim\limits_{x\rightarrow-\infty}\dfrac{df}{dx}=1$
- $p(x)$:
- has even symmetry
- $\lim\limits_{x\rightarrow\infty}p(x)=1$
- $\lim\limits_{x\rightarrow\infty}\dfrac{dp}{dx}=\lim\limits_{x\rightarrow-\infty}\dfrac{dp}{dx}=0$
Currently, I'm trying to use Fourier transforms, i.e.:
$p(x) = \mathcal{F}^{-1}\left\{\sqrt{\dfrac{-j}{\omega}\left[\mathcal{F}\left\{\dfrac{df}{dx}-1\right\}+\delta(0)\right]\right\}$
But is there an easier way? Am I missing some obvious convolution identities?