According to method described in a paper [1] if we want to find points of maximum slope in a signal $f(t)$, then one has to do following
- Convolve $f(t)$ with $g(t)$ where $g(t)=-cos(\omega t).(1/\sigma\sqrt{2.\pi}).e^{(-0.5.t^2/\sigma^2))}$ a modulated gaussian curve.
- Local maximas in $f(t)*g(t)$ will correspond to points with maximal slope in $f(t)$
Here is the simulation result done to check the veracity. Blue curve is sine wave ($f(t)$). Green curve is $g(t)$. Red curve is $f*g$. As one can see, points of maximal slope in $f(t)$ indeed do occur at maximas in $f*g$.
Question How to mathematically show that local maximas in $f*g$ correspond to points of maximum slopes in $f$?
I did this much analysis - Points with maximum slope will be points at which $df/dt$ will be maximum i.e. $d^2f/dt^2 = 0$
On the other side, $(f*g)(t) = \int_{-\inf}^{+\inf}g(\tau).f(t-\tau)d\tau$
I don't know how to write it in math notation but we have to prove that $t$ at which $d^2f/dt^2 = 0$ is same $t$ at which $d(f*g)(t)/dt = 0$ occurs. How to proceed?
[1] Prasanna, SR Mahadeva, and B. Yegnanarayana. "Detection of vowel onset point events using excitation information." INTERSPEECH. 2005. Page 2, Right Column http://speech.iiit.ac.in/svlpubs/conference/MahadevaYegna2005.pdf
2 Matlab code for plotting
>> n=1:160;
>> g=(1/16*sqrt(2*pi))*exp(-0.5*((n.^2)/16^2));
>> plot(sin(n/5));hold on;plot(conv(sin(n/5),-g.*cos(n/10)),'r');plot(-g.*cos(n/10),'g');