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Conceptually when we want to represent a peroidic series, for example a pulse train, we find the Fourier coefficients and get a representation in the time domain.

However what is conceptually wrong with using say an infinite sum of time shifted rect functions to represent it?


Sorry I had a formatting problem, so i am putting this in the bottom post...

My method is as such:

Assuming we have a periodic pulse $x(t)$ such that $x(t) = 1$ for $0 < t< T$ and $0$ for $T < t < T_p$; thus $x(t)$ has a period of $T_p$.

Finding the fourier coefficients Ck via:

$$Ck = \frac{1}{T_p} \int_{-\infty}^{\infty} x(t) * e^\tfrac{-j2\pi kt}{T_p}$$

over 1 period, and thus we can represent x(t) as:

$$x(t) = \sum_k C_k e^\tfrac{j2\pi kt}{T_p}$$

and doing the Fourier transform we will get this form:

$$X(f) = \sum_k C_k \delta(f-kf_p)$$

which is discrete.

However if we consider $x(t)$ to be of this form:

$$x(t) = \sum_n \text{rect} \left(\frac{t-nT_p}{T} \right) $$

Applying fourier transform of $x(t)$ to get (in the form):

$$X(f) = \sum \text{sinc} * e^{-j*W}$$ -- (2)

where sinc() is due to the FT of rect and the e^(-j*W) comes out due to the time shifting property of FT.

Comparing X(f) in (1) and (2), we see that 1 is discrete and the other continuous.

However they come from the same x(t), so isn't this a contradiction?

Sorry for the long post.

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    $\begingroup$ There is nothing wrong with using time-shifted rect functions, and these functions are indeed orthogonal. In fact, discrete-time signal processing in effect replaces a low-pass signal with a sequence of numbers that are the amplitudes of these time-shifted rect functions, and even more so if you think of sample-and-hold circuits which effectively replace a continuous-time waveform with a series of time-shifted rects with different amplitudes, and are simple (but imperfect) form of D/A conversion at the other end. $\endgroup$ – Dilip Sarwate Mar 6 '12 at 12:48
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    $\begingroup$ @Johntan The Z-transform is essentially just that- sums of time shifted rect functions. $\endgroup$ – Jim Clay Mar 6 '12 at 17:55
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    $\begingroup$ You can produce a square wave by summing an infinite number of sine waves, and you can produce a sine wave by summing an infinite number of square waves. $\endgroup$ – endolith Mar 30 '12 at 16:30
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    $\begingroup$ Related: math.stackexchange.com/q/58391 $\endgroup$ – endolith Mar 30 '12 at 17:02
  • $\begingroup$ You can use the orthogonal Haar wavelet transform to decompose a signal into square waves at different scales. $\endgroup$ – Spacey Mar 31 '12 at 3:26
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However what is conceptually wrong with using say an infinite sum of time shifted rect functions to represent it?

There is nothing conceptually wrong with it. Fourier transforms decompose a signal into a sum of complex sinusoids, but you can also decompose a signal into many other things, which may be more useful in certain applications. The Haar wavelet transform, for instance, breaks up a signal a sum of rectangular pulses:

enter image description here source

We use sinusoids in lots of applications because it makes the most sense in those applications. For instance, why do we almost always decompose audio signals into sinusoids? Because our cochleas do the same thing:

enter image description here

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  • $\begingroup$ Very nice diagram of the inner ear - about those images on the right - are they showing where certain wavelengths resonate? $\endgroup$ – Spacey Mar 31 '12 at 3:25
  • $\begingroup$ @Mohammad: I'm not sure what the right side is supposed to show. Just that low frequencies are detected by hair cells deep inside the cochlea? $\endgroup$ – endolith Apr 5 '12 at 2:38
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The main reason for that is that series of cosines and sines form an orthogonal basis. Then, you can use it to represent it in other "space" (frequency "space", for example).

Other things, just to understand other things related to Fourier series and Tranform:

A sine or cosine is just 2 delta functions in frequency representation (Fourier Transform). A rect function, have a sync function representation (that stricly fills all sprectra).

Then, using Fourier representation in frequency, you can easily interpret what are the frequency components of your signal and filter it accordingly.

Another thing to better understand the use of Fourier coefficients, is to understand the relation between the Fourier transform and those coefficients (Explanation1, Explanation2).

We use Fourier series for periodic functions, and Fourier Transform for everything.

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  • $\begingroup$ Thank you for your explanation. So adding on, if we consider the frequency spectrum using the 2 similar representations, we would end up with a discrete frequency spectrum (if we consider the delta functions) and on another hand a continuous spectrum (if we consider the sinc function). Is it not contradictory? $\endgroup$ – John tan Mar 6 '12 at 12:30
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    $\begingroup$ Until you understand quite well Fourier coefficients and so on, I would not mixture continuous (Fourier Trans.) and discrete (DFT). Could you write the steps you are making? $\endgroup$ – Luis Andrés García Mar 6 '12 at 13:06
  • $\begingroup$ We can use Fourier series to locally approximate aperiodic functions by creating a periodic extension. $\endgroup$ – Emre Mar 30 '12 at 22:36
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Sorry I had a formatting problem, so I am putting this in the bottom post.

My method is as such:

Assuming we have a periodic pulse $x(t)$ such that $x(t) = 1$ for $0 < t< T$ and $0$ for $T < t < Tp$; thus $x(t)$ has a period of $Tp$.

Finding the fourier coefficients $Ck$ via:

$$Ck = 1/Tp \int^T ( x(t) * e^{(-j*2*pi*k*t/Tp)})$$

and thus we can represent $x(t)$ as:

$$x(t) = \sum (Ck * e^{(j*2*pi*k*t/Tp)})$$ over all int k

and doing the fourier transform we will get this form:

$$X(f) = \sum (Ck * \delta(f-kfp))$$ over all int k -- (1)

which is discrete.

However if we consider x(t) to be of this form:

$$x(t) = \sum(rect[(t-nTp)/T])$$

Applying fourier transform of x(t) to get (in the form):

$$X(f) = \sum[ sinc * e^(-j*W) ]$$ -- (2)

where sinc() is due to the FT of rect and the $e^{(-j*W)}$ comes out due to the time shifting property of FT.

Comparing $X(f)$ in (1) and (2), we see that 1 is discrete and the other continuous.

However they come from the same $x(t)$, so isn't this a contradiction?

Sorry for the long post.

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  • $\begingroup$ Your $X(f)$ is wrong; you are missing factors of $T_p$ in the argument of the exponentials as well as in what you write simply as $\text{sinc}$. If you do it right, you will get a summation of infinite number of exponential functions which give you impulses of different magnitudes. There is no Fourier transform of periodic functions unless you admit generalized functions or distributions as mathematicians call them, or impulses or "delta functions" as engineers like to say. You use $\mathcal F(\text{sum}) = \text{sum}\ \mathcal F$ which needs justification if the sum is infinite. $\endgroup$ – Dilip Sarwate Mar 6 '12 at 14:23

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