2. Re: Fourier transform - Question

Unit step function $1(t) = 1$ if $t\geq 0$ and $0$, otherwise.
First line is the Fourier transform (FT) of step function.
Second line is the FT $F$ of a generic function $f$ which is shifted by $T$.
Third line is irrelevant for current derivation.

First, the given function can be illustrated as difference of two unit step functions. Note that $1(t-T) = 1$ if $t\geq T$ and $0$, otherwise. Therefore, $1(t) -1(t-T)= 1$ if $t\in [0,T)$ and $0$, otherwise. Multiply it by $X_0$, you get the desired function.

Using the FT results given in the top, you can write the FT of $1(t)$ and its shifted version $1(t-T)$ resulting,
$X(\omega)$
$= X_0[\frac{1}{j\omega}+\pi\delta(\omega)] - X_0[\frac{1}{j\omega}+\pi\delta(\omega)]e^{-j\omega T}$
$= \frac{X_0}{j\omega}[1-e^{-j\omega T}] - X_0\pi\delta(\omega)[1-e^{-j\omega T}]$

Note that the delta function is defined as $\delta(\omega)=0$ for $\omega\neq 0$ and $\int_{-\infty}^{\infty}\delta(\omega) d\omega=1$, i.e. only when $\omega=0$, $\delta(\omega)>0$. Thus, it is clear that $\delta(\omega)[1-e^{-j\omega T}]$ is 0 when $\omega\neq 0$.
However, at $\omega=0$, $1-e^{-j\omega T}=1-e^{0}=0$. Therefore, the product of $\delta(\omega)$ and $1-e^{-j\omega T}$ becomes 0 for even $\omega=0$.
Conclusion is that $\delta(\omega)[1-e^{-j\omega T}]$ is always 0.

3. Re: Fourier transform - Question

Sorry, but does someone understand this? The link does not contain viruses or backdoors.

4. Re: Fourier transform - Question

Originally Posted by Nforce
Sorry, but does someone understand this? The link does not contain viruses or backdoors.
I have replied, but you have to wait till a moderator approves it.

Sent from my HUAWEI GRA-L09 using Tapatalk

5. Re: Fourier transform - Question

Thank you, I have another question.

So here:
$= X_0[\frac{1}{j\omega}+\pi\delta(\omega)] - X_0[\frac{1}{j\omega}+\pi\delta(\omega)]e^{-j\omega T}$
$= \frac{X_0}{j\omega}[1-e^{-j\omega T}] - X_0\pi\delta(\omega)[1-e^{-j\omega T}]$

$-\frac{e^{-j\omega T}}{j\omega} = \pi\delta(\omega) ?$

I didn't find this in the Fourier transform table. This transforms are hard to compute.

6. Re: Fourier transform - Question

Originally Posted by Nforce
Thank you, I have another question.

So here:
$= X_0[\frac{1}{j\omega}+\pi\delta(\omega)] - X_0[\frac{1}{j\omega}+\pi\delta(\omega)]e^{-j\omega T}$
$= \frac{X_0}{j\omega}[1-e^{-j\omega T}] - X_0\pi\delta(\omega)[1-e^{-j\omega T}]$

$-\frac{e^{-j\omega T}}{j\omega} = \pi\delta(\omega) ?$

I didn't find this in the Fourier transform table. This transforms are hard to compute.
I am not quite sure what is your question.
if you are asking whether $-\frac{e^{-j\omega T}}{j\omega} = \pi\delta(\omega) ?$ is true, then it is not.

The reason for latter term to become 0 is as I explained in my first reply.

7. Re: Fourier transform - Question

Sorry, but how did you get from this:

$= X_0[\frac{1}{j\omega}+\pi\delta(\omega)]$

to this:

$= \frac{X_0}{j\omega}[1-e^{-j\omega T}]]$

This is my first encounter with Fourier transforms.

8. Re: Fourier transform - Question

Originally Posted by Nforce
Sorry, but how did you get from this:

$= X_0[\frac{1}{j\omega}+\pi\delta(\omega)]$

to this:

$= \frac{X_0}{j\omega}[1-e^{-j\omega T}]]$

This is my first encounter with Fourier transforms.
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.

$= X_0[\frac{1}{j\omega}+\pi\delta(\omega)] - X_0[\frac{1}{j\omega}+\pi\delta(\omega)]e^{-j\omega T}$

Here, take first and third fraction terms together and, second and forth pi-delta terms together to get the following expression.

$= \frac{X_0}{j\omega}[1-e^{-j\omega T}] - X_0\pi\delta(\omega)[1-e^{-j\omega T}]$

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9. Re: Fourier transform - Question

Sorry, not 'while expression' just 'whole expression'
Auto correct issues.

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