Fourier transform of the Gaussian pulse $f (t) = e^{- z^{2} t^{2}}$ is

\frac{\sqrt{π}}{a} e^{- (2 a ω)^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{a^{2} ω^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{\sqrt{π}}{a} e^{- {(\frac{ω}{2 a})}^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

e^{- (a^{2} ω^{2})}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{\sqrt{π}}{a} e^{- {(\frac{ω}{2 a})}^{2}}$
$F . T [e^{- z^{2} t^{2}}] = \int_{- \infty}^{\infty} e^{- z^{2} t^{2}} . e^{- j ω t} d t$

$= \int_{- \infty}^{\infty} e^{- (z^{2} t^{2} + j ω t)} d t$

Substituting, $a^{2} t^{2} + j ω t = {(a t + \frac{j ω}{2 a})}^{2} + \frac{ω^{2}}{4 a^{2}}$

We get,

$F . T [e^{- a^{2} t^{2}}] = \int_{- \infty}^{\infty} e^{- {(s t + \frac{j ω}{2 a})}^{2}} e^{\frac{ω^{2}}{4 s^{2}}} d t$

$= e^{\frac{- ω^{2}}{4 x^{2}}} \int_{- \infty}^{\infty} e^{- {(a t + \frac{j ω}{2 a})}^{2}} d t$

Putting $u = a t + \frac{j ω}{2 a}, d u = a . d t$

So, $d t = \frac{d u}{a}$

$F . T [e^{- a^{2} t^{2}}] = e^{\frac{ω^{2}}{4 π^{2}}} 2 \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{a} [\int_{0}^{\infty} e^{- n^{2}} d u = \frac{\sqrt{π}}{2}]$

$= e^{\frac{- ω^{2}}{4 s^{2}}} . \frac{2}{a} . \frac{\sqrt{π}}{2}$

$= \frac{\sqrt{π}}{a} e^{\frac{ω^{2}}{4 s^{2}}} = \frac{\sqrt{π}}{a} e^{- {(\frac{ω}{2 a})}^{2}}$

Suggest Corrections

Similar questions

x (t) = e^{- 4 t^{2}}

A e^{- (\frac{π f}{2})^{2}}

F (j ω) = \frac{1}{ω^{2} + 1} e^{2 ω^{2} / (ω^{2} + 1)}

f (t - 2) . e^{f (t)}

The Fourier transform of the complex sinusoidal pulse shown below is z (t) = {\begin{matrix} e^{j 10 t}; & | t | < π 0; & | t | > π \end{matrix}

X (ω) = e^{- ω} u (ω

Fourier transform of e^{- 2 | t - 1 |} is

Join BYJU'S Learning Program