Convolution of gaussian functions

[ElijahRockers]

Homework Statement

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

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

 

[SammyS]

Homework Statement

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

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

On the left hand page, you're fine down to here:

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:

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.

 

[ElijahRockers]

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 $\sqrt{\pi} f_2 (t) = \sqrt{2\pi}e^{-2t^2}$

 

[SammyS]

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 $\sqrt{\pi} f_2 (t) = \sqrt{2\pi}e^{-2t^2}$

I forgot, transforming back, there is a normalization factor to consider.