# Is convolution of sine-squared function, sinusoidal function?

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

• I deleted my answer for integral of sine-squared function that is x/2-sin(2x)/4 +C is not sinusoidal. Regards – George Theodosiou Nov 27 '15 at 13:36
• By above comment I mean have deleted my first answer. Regards – George Theodosiou Nov 28 '15 at 9:27

## 2 Answers

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.

• Mr Ian, many thanks for your immediate answer. A specific question. Consider discrete convolution and first function consists of many periods and other, of one period equal to first's. Regards – George Theodosiou Nov 24 '15 at 15:57
• @GeorgeTheodosiou Can you write a formula out? – Ian Nov 24 '15 at 19:35
• Mr Ian, I am very sorry for the delay. I was meant proof that convolution of sine-squared function is sinusoidal. This aspect is essential in digital filtering I am interested in. Regards. – George Theodosiou Nov 27 '15 at 12:23

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

• Though this only works with $L^1$, hence not for $f$ being a sinusoid. – Ian Nov 27 '15 at 18:18