Please check my solution for this Fourier transform.

[YehiaMedhat]

$$\mathcal{F}\{ e^{-4t^2 -8t} \}$$$e^{-t^2} = \frac{2}{1+\omega^2}$ and $e^{-4t^2 -8t} = e^{-4(t^2+2t+1)} e^4 = e^4 e^{-4(t+1)^2}$
$\therefore\ \text{Applying the time shift property: } e^4\sqrt{\pi}\ e^{i\omega} e^{\frac{\omega^2}{4}}$
$\therefore\ \text{Applying the scaling property: }\mathcal{F}\{ e^4 e^{-(2(t+1))^2} \} = \frac{e^4\sqrt{\pi}}{2}\ e^{i\frac{\omega}{2}} e^{\frac{(\frac{\omega}{2})^2}{4}}$
Final answer:
$$\frac{e^4\sqrt{\pi}}{2}\ e^{i\frac{\omega}{2}} e^{\frac{\omega^2}{16}}$$
Is it right or not, because the solution in the sheets provided by my TA doesn't apply the scaling property to the $e^{i\omega}$, so, the final answer in the sheet is: $\frac{e^4\sqrt{\pi}}{2}\ e^{i\omega} e^{\frac{\omega^2}{16}}$

 

[fresh_42]

I assume that you missed the factor $2$ in the exponent of the shift formula.

 

[mario99]

$$\mathcal{F}\{ e^{-4t^2 -8t} \}$$$e^{-t^2} = \frac{2}{1+\omega^2}$ and $e^{-4t^2 -8t} = e^{-4(t^2+2t+1)} e^4 = e^4 e^{-4(t+1)^2}$
$\therefore\ \text{Applying the time shift property: } e^4\sqrt{\pi}\ e^{i\omega} e^{\frac{\omega^2}{4}}$
$\therefore\ \text{Applying the scaling property: }\mathcal{F}\{ e^4 e^{-(2(t+1))^2} \} = \frac{e^4\sqrt{\pi}}{2}\ e^{i\frac{\omega}{2}} e^{\frac{(\frac{\omega}{2})^2}{4}}$
Final answer:
$$\frac{e^4\sqrt{\pi}}{2}\ e^{i\frac{\omega}{2}} e^{\frac{\omega^2}{16}}$$
Is it right or not, because the solution in the sheets provided by my TA doesn't apply the scaling property to the $e^{i\omega}$, so, the final answer in the sheet is: $\frac{e^4\sqrt{\pi}}{2}\ e^{i\omega} e^{\frac{\omega^2}{16}}$

Original function:
$g(t) = e^{-4t^2 - 8t} = e^{4}e^{-4(t + 1)^2}$

Let $f(t) = e^{-t^2}$

Shifting
$\mathcal{F}\{e^{4}e^{-(t + 1)^2}\} =e^{4}\sqrt{\pi} e^{-i\omega} F(\omega)$, where $F(\omega) = e^{-\frac{\omega^2}{4}}$

The scaling property, you only apply it to the transformed function $F(\omega)$.

Scaling
$\mathcal{F}\{e^{4}e^{-(2[t + 1])^2}\} = \sqrt{\pi}e^{4} e^{-i\omega} \frac{F(\frac{\omega}{2})}{|2|} = \frac{\sqrt{\pi}}{2}e^{4} e^{-i\omega} e^{-\frac{\omega^2}{16}}$

 

[YehiaMedhat]

Original function:
$g(t) = e^{-4t^2 - 8t} = e^{4}e^{-4(t + 1)^2}$

Let $f(t) = e^{-t^2}$

Shifting
$\mathcal{F}\{e^{4}e^{-(t + 1)^2}\} =e^{4}\sqrt{\pi} e^{-i\omega} F(\omega)$, where $F(\omega) = e^{-\frac{\omega^2}{4}}$

The scaling property, you only apply it to the transformed function $F(\omega)$.

Scaling
$\mathcal{F}\{e^{4}e^{-(2[t + 1])^2}\} = \sqrt{\pi}e^{4} e^{-i\omega} \frac{F(\frac{\omega}{2})}{|2|} = \frac{\sqrt{\pi}}{2}e^{4} e^{-i\omega} e^{-\frac{\omega^2}{16}}$

Yes exactly this point why the e that was the reason of shifting wasn't affected by scaling??

 

[mario99]

Yes exactly this point why the e that was the reason of shifting wasn't affected by scaling??

$\displaystyle e^{-i\omega}$ is not affected because:

the Time Scaling Property says

$\displaystyle \mathcal{F}\{f(ct)\} = \frac{1}{|c|}F\left(\frac{\omega}{c}\right)$

It did not say

$\displaystyle \mathcal{F}\{f(ct)\} = e^{-\frac{i\omega}{c}}\frac{1}{|c|}F\left(\frac{\omega}{c}\right)$

To understand the Time Scaling Property of Fourier Transform, you will have to use the definition.

Let $\displaystyle f(t) = e^{-t^2}$

We know that $\displaystyle \mathcal{F}\{f(t)\} = F(\omega) = \sqrt{\pi}e^{-\frac{\omega^2}{4}}$

Now let us scale it by $\displaystyle c$ and use the definition.

$\displaystyle \mathcal{F}\{f(ct)\} = \int_{-\infty}^{\infty} f(ct)e^{i\omega t} \ dt = \int_{-\infty}^{\infty}e^{-(ct)^2}e^{i\omega t} \ dt$


$\displaystyle = \int_{-\infty}^{\infty}e^{-(ct - \frac{i\omega}{2c})^2 - \frac{\omega^2}{4c^2}} \ dt = e^{-\frac{\omega^2}{4c^2}}\int_{-\infty}^{\infty}e^{-(ct - \frac{i\omega}{2c})^2} \ dt$


$\displaystyle = \frac{e^{-\frac{\omega^2}{4c^2}}}{c}\int_{-\infty}^{\infty}e^{-u^2} \ du = \frac{\sqrt{\pi}e^{-\frac{\omega^2}{4c^2}}}{c}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{\sqrt{\pi}} \ du$


$\displaystyle = \frac{\sqrt{\pi}e^{-\frac{\omega^2}{4c^2}}}{c}(1) = \frac{\sqrt{\pi}e^{-\frac{\omega^2}{4c^2}}}{c}$


which is just $\displaystyle \frac{1}{|c|}F\left(\frac{\omega}{c}\right)$

 

[YehiaMedhat]

$\displaystyle e^{-i\omega}$ is not affected because:

the Time Scaling Property says

$\displaystyle \mathcal{F}\{f(ct)\} = \frac{1}{|c|}F\left(\frac{\omega}{c}\right)$

It did not say

$\displaystyle \mathcal{F}\{f(ct)\} = e^{-\frac{i\omega}{c}}\frac{1}{|c|}F\left(\frac{\omega}{c}\right)$

To understand the Time Scaling Property of Fourier Transform, you will have to use the definition.

Let $\displaystyle f(t) = e^{-t^2}$

We know that $\displaystyle \mathcal{F}\{f(t)\} = F(\omega) = \sqrt{\pi}e^{-\frac{\omega^2}{4}}$

Now let us scale it by $\displaystyle c$ and use the definition.

$\displaystyle \mathcal{F}\{f(ct)\} = \int_{-\infty}^{\infty} f(ct)e^{i\omega t} \ dt = \int_{-\infty}^{\infty}e^{-(ct)^2}e^{i\omega t} \ dt$


$\displaystyle = \int_{-\infty}^{\infty}e^{-(ct - \frac{i\omega}{2c})^2 - \frac{\omega^2}{4c^2}} \ dt = e^{-\frac{\omega^2}{4c^2}}\int_{-\infty}^{\infty}e^{-(ct - \frac{i\omega}{2c})^2} \ dt$


$\displaystyle = \frac{e^{-\frac{\omega^2}{4c^2}}}{c}\int_{-\infty}^{\infty}e^{-u^2} \ du = \frac{\sqrt{\pi}e^{-\frac{\omega^2}{4c^2}}}{c}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{\sqrt{\pi}} \ du$


$\displaystyle = \frac{\sqrt{\pi}e^{-\frac{\omega^2}{4c^2}}}{c}(1) = \frac{\sqrt{\pi}e^{-\frac{\omega^2}{4c^2}}}{c}$


which is just $\displaystyle \frac{1}{|c|}F\left(\frac{\omega}{c}\right)$

Thanks
I really appreciate it