What is the meaning of $Ta_k$ of fourier series or transform? I am taking a course on signal and systems.

In 286 page of my textbook, it says that as T becomes arbitrarily large the original periodic square wave approaches a rectangular pulse. Also it says that all that remains in the time domain is an aperiodic signal corresponding to one period of the square wave. (textbook: siganls and systems sencond edition, author: oppenheim)

I have a difficulty understanding this.. I can't connect this idea with fourier transform.

I suggest a link http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-003-signals-and-systems-fall-2011/lecture-videos-and-slides/MIT6_003F11_lec16.pdf

  • $\begingroup$ Slide 4-6: T is the length (in time) of the period over which you are calculating the Fourier transform. $a_k$ is the $k$th Fourier coefficient (the "amount" of frequency $k$.) In the graph at the bottom they are varying $k$ on the $x$-axis and showing the value of $a_k$ on the $y$-axis. They are multiplying the values of $a_k$ by $T$ because as $T$ increases all the Fourier coefficients get smaller. (They are $1/T$ times an integral.) And then in slide 7 they take $lim_{T\rightarrow \infty} T a_k$, which would be 0 if we didn't multiply back by $T$. $\endgroup$ May 8 '13 at 11:56
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    $\begingroup$ I don't know what is on page 286 of (what book are you using: author, title, year?) But in the linked slides a square wave is not approaching a rectangular pulse. Rather there is a rectangular pulse of width $S$ that is being repeated at interval $T$. As $T$ lengthens to $\infty$ there is just a single pulse (still of width $S$). $\endgroup$ May 8 '13 at 12:03
  • $\begingroup$ Sorry,, I edited my original post. $\endgroup$
    – jakeoung
    May 8 '13 at 12:05
  • $\begingroup$ $Ta_k \to X(jw)$. Is this fact an accidental result? or is there a meaning? $\endgroup$
    – jakeoung
    May 8 '13 at 12:22
  • $\begingroup$ I view it as the definition of $X(j\omega)$. As $T\rightarrow\infty$ the number of frequencies goes from countably infinite to uncountably infinite (i.e., continuous), and rather than a function repeating at intervals $T$ you have a single function that extends (non repeating) to $\infty$. $\endgroup$ May 8 '13 at 12:29

The idea is that a Fourier series is only defined for periodic signals. In the discussion in the linked slides, the author is considering a rectangular pulse train with period $T$. That is, a pulse of width $2S$ repeats periodically with a spacing of $T$ between them. The pulses are therefore centered at:

$$ [\ldots, -2T, T, 0, T, 2T, \ldots ] $$

Now, consider what happens as $T \to \infty$: in the limit, the only pulse that remains is the one centered at zero; the others are infinitely far away. When the author makes the claim that:

$$ \lim_{T\to \infty} T a_k = E(\omega) $$

He or she is trying to show that the Fourier transform, which is defined for suitably well-formed aperiodic signals, can be thought of as the Fourier series of that signal (which typically wouldn't be defined since the signal is not periodic) in the limiting case of an infinite period. Stated a little differently, you can in some way think of an aperiodic signal as a periodic signal with infinite period.

The multiplication by $T$ in the limit is to account for the differences in definition between the Fourier series and Fourier transform: the series representation typically has a factor of $\frac{1}{T}$, while the transform does not. I don't know that there is a lot of insight to be gained via this analysis, but it shows that the series and transform representations are intimately related.


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