Ladies, Gentlemen
By sinusoidal function, I mean function of the form Asin(x) or Acos(x) for A real number. I make note that am beginner in convolution process.
Regards
Ladies, Gentlemen
By sinusoidal function, I mean function of the form Asin(x) or Acos(x) for A real number. I make note that am beginner in convolution process.
Regards
Not quite. For instance, if the two sinusoids have exactly the same frequency, then the convolution operation will cause the amplitude to grow without bound. For instance the convolution of $\sin(t)$ with itself is $\frac{1}{2} \left ( \sin(t) - t \cos(t) \right )$. This is a resonance effect; it is commonly treated in elementary differential equations, since it is the solution to the equation
$$y''+y=\sin(t),y(0)=0,y'(0)=0.$$
Also, it is not necessary that the two sinusoids be in phase.
Let $f$ be an $L^1$-function on ${\mathbb R}$. Then $g:=f*\cos\>$ is defined by $$g(x):=\int_{-\infty}^\infty f(t)\cos(x-t)\>dt=\int_{-\infty}^\infty f(t)(\cos x\cos t+\sin x\sin t)\>dt\ .$$ Put $$\int_{-\infty}^\infty f(t)\cos t\>dt=:A,\qquad \int_{-\infty}^\infty f(t)\sin t\>dt=:B\ .$$ It follows that $$g(x)=A\cos x+B\sin x=\sqrt{A^2+B^2}\cos(x-\theta)\qquad(-\infty<x<\infty)\ ,$$ with $\theta:={\rm arg}(A,B)$.