purnapragna
Advanced Member level 4
We know the diferentiation property of the Fourier Transform that
\[f(t)\rightarrowF(\omega)\]
then
\[$\dfrac{df(t)}{dt}\rightarrow j\omegaF(\omega)$\]
Now let us try to find FT of the signal \[$x(t)=u(t)$\] using differentioation property.
we know that \[$\dfrac{du(t)}{dt}=\delta(t)$\]. So FT of \[$u(t)$\] using differentiation property turns out to be \[$\frac{1}{j\omega}$\] which we know that to be false. Note that it also turns out that FT of \[$\frac{sgn(t)}{2}$\] is also \[$\frac{1}{j\omega}$\]. So it turns out that we have to use FT properties carefully with the signals which have DC component.
I want some comments from the members regarding this statement.
Thank you!
Purna!
\[f(t)\rightarrowF(\omega)\]
then
\[$\dfrac{df(t)}{dt}\rightarrow j\omegaF(\omega)$\]
Now let us try to find FT of the signal \[$x(t)=u(t)$\] using differentioation property.
we know that \[$\dfrac{du(t)}{dt}=\delta(t)$\]. So FT of \[$u(t)$\] using differentiation property turns out to be \[$\frac{1}{j\omega}$\] which we know that to be false. Note that it also turns out that FT of \[$\frac{sgn(t)}{2}$\] is also \[$\frac{1}{j\omega}$\]. So it turns out that we have to use FT properties carefully with the signals which have DC component.
I want some comments from the members regarding this statement.
Thank you!
Purna!
