# Thread: Continous Time Spectral Analysis (Fourier Trasforms)

1. ## Continous Time Spectral Analysis (Fourier Trasforms)

I've learnt Fourier series in the past but have had a year of an awful lecturer teaching Signal Processing and find that I've no idea how to do even the simplest problems. If people could give me some pointers I'd be very grateful.

I'll begin with the easy ones.

Given $\displaystyle f(t) \leftrightarrow F(\omega)$ apply the Fourier transform properties to determine the transforms of the following:

a) $\displaystyle f(T-t)$

For this I get the following answer, haven't expanded the Fourier series:

$\displaystyle f(T-t) \rightarrow e^{jT\omega}F(-\omega)$

b) $\displaystyle f(2t-T)$

I get:

$\displaystyle f(2t-T) \rightarrow \frac{e^{jT\omega}}{2}F(\frac{\omega}{2})$

c) $\displaystyle \frac{d}{dt}\left( f(t)g(t)\right)$

I get:

$\displaystyle \frac{d}{dt} \left( f(t)g(t) \right) \rightarrow \frac{j \omega \left( f * g \right)(\omega)}{\sqrt{2\pi}}$

Though in that one I'm not sure how to represent the convolution of f and g.

So, if those answers are right then let me know, or if as I suspect they're not could someone walk me through one of them to tell me how I'm meant to do the transform? Thanks in advance.

2. The trouble with Fourier transforms is that people use different definitions. The one I prefer is $\displaystyle F(\omega) = \int_{-\infty}^\infty\!\!\!f(t)e^{-i\omega t}dt$. But some people put a factor of 2π (or its square root) outside the integral or in the exponential. From the answers that you have given, it looks as though your lecturer's definition is $\displaystyle F(\omega) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\!\!\!f(t)e^{j\omega t}dt$ (with j instead of i as the square root of –1, since he is an engineer rather than a mathematician, and without the negative sign in the exponential). Is that correct? If so, then your answer to a) is correct. I think that the answer to b) should have an extra factor of 1/2 in the exponential: $\displaystyle f(2t-T) \to \tfrac12e^{jT\omega/2}F(\omega/2)$. I think that the answer to c) is correct (use a plain asterisk to denote a convolution, not a tensor product symbol: $\displaystyle f*g$).

Finally, I don't understand the comment "haven't expanded the Fourier series". These are Fourier transforms, not Fourier series (totally different topics).

3. Thanks for that - nice that you can tell I'm an engineer because of the j :P To be honest I've no idea what the lecturers definition of a Fourier transform is, she has never given us any legible examples of one, just drew a load of graphs that don't tell us much. That's probably why the confusing comment - I assumed $\displaystyle F(\omega)$ was a Fourier series of $\displaystyle \omega$ and could be expanded as such.

I've been messing with the equations trying to better understand how the transform works and have done the following:

$\displaystyle f(T-t) \rightarrow e^{jT\omega}F(-\omega)$ and $\displaystyle F(\omega) = \int_{-\infty}^\infty\!\!\!f(t)e^{-j\omega t}dt$
So
$\displaystyle e^{jT\omega}F(-\omega) = e^{jT\omega}\int_{-\infty}^\infty\!\!\!f(T-t)e^{-jt\omega}dt$
In that integral, when $\displaystyle t=T \rightarrow f(T-T) \rightarrow f(0)$ because the e values cancel out. With Fourier being about frequencies I'd assume this to indicate a fundamental frequency of a signal. Hopefully that made some sense.

I'll email the lecturer and see if those were correct, there'll probably be plenty more to come.

4. Originally Posted by M1ke
I've been messing with the equations trying to better understand how the transform works and have done the following:

$\displaystyle f(T-t) \rightarrow e^{jT\omega}F(-\omega)$ and $\displaystyle F(\omega) = \int_{-\infty}^\infty\!\!\!f(t)e^{-j\omega t}dt$
So
$\displaystyle e^{jT\omega}F(-\omega) = e^{jT\omega}\int_{-\infty}^\infty\!\!\!f(T-t)e^{-jt\omega}dt$ (That is correct.)
In that integral, when $\displaystyle t=T \rightarrow f(T-T) \rightarrow f(0)$ because the e values cancel out. With Fourier being about frequencies I'd assume this to indicate a fundamental frequency of a signal. Hopefully that made some sense.
I don't think it does make much sense. Unlike Fourier series, where you have a fundamental frequency, Fourier transforms don't have anything analogous.

The best picture to have in mind is that Fourier series describe things like the sound of a musical instrument (where there is a fundamental frequency and then a discrete series of harmonics), but Fourier transforms describe things like the light emitted from a star (where there is a continuous spectrum of frequencies from infrared to ultraviolet, none of which can be considered "fundamental").

In mathematics, theories based on Fourier series are often called "harmonic analysis", and theories that come from Fourier transforms are called "spectral analysis". You might think of the Fourier transform as the mathematical equivalent of a prism, that takes a beam of light and splits it into its constituent frequencies.