How Does the Fourier Transform Relate to Derivatives in Signal Processing?

[nhrock3]

cant understand this transformation
i know that each derivative pops iw
and
$$\hat{y}' ->-ixy(x)$$
$$\hat{y}'(\omega) ->-ixy(x)$$
x is a signs of derivative

but i don't know how its been done in here
$$-ixy'(x)=(i\omega \hat{y}(w))'$$
how they decided that is the derivative of this whole expression
muliplying by x means derivative
but here it something else
$$f[xy'(\omega )]=i\frac{\mathrm{d} }{\mathrm{d} \omega}f[y'(x)]=i(i\omega \hat{y}(\omega))'=i(i \hat{y}(\omega)+i\omega \hat{y}'(\omega))$$
i can't see what laws they follow here
?

 

[LCKurtz]

cant understand this transformation
i know that each derivative pops iw
and
$$\hat{y}' ->-ixy(x)$$
$$\hat{y}'(\omega) ->-ixy(x)$$
x is a signs of derivative

but i don't know how its been done in here
$$-ixy'(x)=(i\omega \hat{y}(w))'$$
how they decided that is the derivative of this whole expression
muliplying by x means derivative
but here it something else

I think you need to be more careful about the relation between derivatives and transforms. That last equation doesn't make any sense as it is written because you have a function of x = a function of omega. I also don't think it is transcribed correctly. You must mean one is the transform of the other.

Using the hat notation for the transform you have the following three relationships between transforms and derivatives:

$$y(x) \leftrightarrows \hat y(\omega)$$

$$-ixy(x) \leftrightarrows \hat y'(\omega)$$

$$y'(x) \leftrightarrows i\omega \hat y(\omega)$$

To get your identity you start with the second one above:

$$-ixy(x) \leftrightarrows \hat y'(\omega)$$

Now apply the third one to this last one by multiplying the right side by $i\omega$:

$$(-ixy(x))' \leftrightarrows i\omega \hat y'(\omega)$$

Maybe that will help you.

[Edit - added later]

Here's your identity:

Start with 3:

$$y'(x) \leftrightarrows i\omega \hat y(\omega)$$

Now multiply the left side by -ix and apply the second one at the top:

$$-ixy'(x) \leftrightarrows (i\omega \hat y(\omega))'$$