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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

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

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  • $\begingroup$ 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 $\endgroup$ – George Theodosiou Nov 24 '15 at 15:57
  • $\begingroup$ @GeorgeTheodosiou Can you write a formula out? $\endgroup$ – Ian Nov 24 '15 at 19:35
  • $\begingroup$ 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. $\endgroup$ – George Theodosiou Nov 27 '15 at 12:23
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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 $\theta:={\rm arg}(A,B)$.

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

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