Hello,
can someone explain me this: http://shrani.si/f/2K/I6/1AcwYTok/fourier.jpg
I don't understand it.
Hello,
can someone explain me this: http://shrani.si/f/2K/I6/1AcwYTok/fourier.jpg
I don't understand it.
Unit step function $\displaystyle 1(t) = 1$ if $\displaystyle t\geq 0$ and $\displaystyle 0$, otherwise.
First line is the Fourier transform (FT) of step function.
Second line is the FT $\displaystyle F$ of a generic function $\displaystyle f$ which is shifted by $\displaystyle T$.
Third line is irrelevant for current derivation.
First, the given function can be illustrated as difference of two unit step functions. Note that $\displaystyle 1(t-T) = 1$ if $\displaystyle t\geq T$ and $\displaystyle 0$, otherwise. Therefore, $\displaystyle 1(t) -1(t-T)= 1$ if $\displaystyle t\in [0,T)$ and $\displaystyle 0$, otherwise. Multiply it by $\displaystyle X_0$, you get the desired function.
Using the FT results given in the top, you can write the FT of $\displaystyle 1(t)$ and its shifted version $\displaystyle 1(t-T) $ resulting,
$\displaystyle X(\omega)$
$\displaystyle = X_0[\frac{1}{j\omega}+\pi\delta(\omega)] - X_0[\frac{1}{j\omega}+\pi\delta(\omega)]e^{-j\omega T}$
$\displaystyle = \frac{X_0}{j\omega}[1-e^{-j\omega T}] - X_0\pi\delta(\omega)[1-e^{-j\omega T}]$
Note that the delta function is defined as $\displaystyle \delta(\omega)=0$ for $\displaystyle \omega\neq 0$ and $\displaystyle \int_{-\infty}^{\infty}\delta(\omega) d\omega=1$, i.e. only when $\displaystyle \omega=0$, $\displaystyle \delta(\omega)>0$. Thus, it is clear that $\displaystyle \delta(\omega)[1-e^{-j\omega T}]$ is 0 when $\displaystyle \omega\neq 0$.
However, at $\displaystyle \omega=0$, $\displaystyle 1-e^{-j\omega T}=1-e^{0}=0$. Therefore, the product of $\displaystyle \delta(\omega)$ and $\displaystyle 1-e^{-j\omega T}$ becomes 0 for even $\displaystyle \omega=0$.
Conclusion is that $\displaystyle \delta(\omega)[1-e^{-j\omega T}]$ is always 0.
Thank you, I have another question.
So here:
$\displaystyle = X_0[\frac{1}{j\omega}+\pi\delta(\omega)] - X_0[\frac{1}{j\omega}+\pi\delta(\omega)]e^{-j\omega T}$
$\displaystyle = \frac{X_0}{j\omega}[1-e^{-j\omega T}] - X_0\pi\delta(\omega)[1-e^{-j\omega T}]$
$\displaystyle -\frac{e^{-j\omega T}}{j\omega} = \pi\delta(\omega) ?$
I didn't find this in the Fourier transform table. This transforms are hard to compute.
This has nothing to do with Fourier transform and you are not looking at the while expression. It is just combining terms appear in different places.
$\displaystyle = X_0[\frac{1}{j\omega}+\pi\delta(\omega)] - X_0[\frac{1}{j\omega}+\pi\delta(\omega)]e^{-j\omega T}$
Here, take first and third fraction terms together and, second and forth pi-delta terms together to get the following expression.
$\displaystyle = \frac{X_0}{j\omega}[1-e^{-j\omega T}] - X_0\pi\delta(\omega)[1-e^{-j\omega T}]$
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Be patient, TEX will load, eventually...