1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convolution of gaussian functions

  1. Mar 24, 2015 #1

    ElijahRockers

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Recall that we have defined the Gaussian ##f_s## by ##f_s (t) = \sqrt{s}e^{-st^2}## and shown that ##\hat{f_s}(\lambda) = \frac{1}{\sqrt{2}}e^{\frac{-\lambda^2}{4s}}##.

    Show that ##f_3 \ast f_6 (t) = \sqrt{\pi}f_{1/2}(t) = \sqrt{\pi/2}e^{-t^{2}/2}##

    3. The attempt at a solution

    Not sure what's wrong with my approach, but I'm getting ##i## in both of my attempts answers, and besides that my answers are no where near close to the correct answer. Each of the pages represents a single attempt. I first tried multiplying the Fourier transforms of both functions then taking the inverse, and when that didn't work, I tried using the definition of convolution.

    Image is attached but resized is hard to read... full size is here
     

    Attached Files:

  2. jcsd
  3. Mar 25, 2015 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    On the left hand page, you're fine down to here:
    Capture_ERockers_1.PNG
    The line you have after is in error.
    It would have been better pull out only ##\displaystyle\ e^{-2t^2} \,,\,## leaving an integrand of ##\displaystyle\ e^{\displaystyle-(9\tau^2-12\tau\,t+4t^2)} \ ## .

    The exponent is the negative of a perfect square. Use a substitution to solve the integral.

    On the right hand page, you have the following:
    Capture_ERockers_2.PNG

    Therefore, simply multiply ##\displaystyle\ \hat{f_3}(\lambda) \hat{f_6}(\lambda) \ ## . Then simply transform back using:
    ##\hat{f_s}(\lambda) = \frac{1}{\sqrt{2}}e^{\frac{-\lambda^2}{4s}}##​
    in reverse.
     
  4. Mar 25, 2015 #3

    ElijahRockers

    User Avatar
    Gold Member

    Hmmm.... transforming back using the inverse Gaussian formula should be rather straightforward, but I don't see where the pi comes from in the final answer.

    EDIT: Also the teacher has made a correction to the problem, the answer should be ##f_3 \ast f_6 = \sqrt{\pi} f_2 (t) = \sqrt{\pi/2}e^{-2t^2}## which still doesn't make sense, but I'm inclined to believe he meant ##\sqrt{\pi} f_2 (t) = \sqrt{2\pi}e^{-2t^2}##
     
  5. Mar 25, 2015 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I forgot, transforming back, there is a normalization factor to consider.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Convolution of gaussian functions
  1. Gaussian Function (Replies: 1)

  2. Gaussian function (Replies: 5)

Loading...