Is convolution of sine-squared function, sinusoidal function?

Question

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

Answer 1

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.

Answer 2

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 $\theta:={\rm arg}(A,B)$.