How to find the convolution of e^t and sin(2t)

  • Thread starter Thread starter errordude
  • Start date Start date
  • Tags Tags
    Convolution
errordude
Messages
17
Reaction score
0

Homework Statement


Hi, can anyone help me to determine the convolution for the 2π periodic function f(t)=e^t and sin2t
 
Physics news on Phys.org


Someone can probably help you but I doubt they will until you show some effort at working this for yourself.

According to the rules of this forum (https://www.physicsforums.com/showthread.php?t=5374),
On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.​
 


well i know how to write e^t as a 2π periodic Fourier series but then what?
 


What's the definition of the convolution of two functions? When you posted this problem, you deleted sections 2 and 3 about Relevant equations and your efforts at solving the problem.
 


You don't need to use FOurier transforms for this problem. It is quite simple indeed, just go to the definition of the convolution of two functions, as was suggested here all along:

f\ast g=\int_0^tf(\alpha)g(t-\alpha)d\alpha

Now just substitute your functions instead of g and f. Hint: inside the integral i would let g(t-alfa)=e^(t-alfa), it makes the integration process easier. You know how to integrate, right?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top