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Define the Fourier Transform of a certain signal in the time domain FT[$x(t)$]=$X(j\omega)$

$X(j\omega)$ = $\int$ $x(t)$ $e$^($j\omega$$t$)$ $dt

I'd like to ask what is the meaning of the value obtained from $X(j\omega)$ with certain frequency $\omega$

for example if we have a voltage signal of 1 Volt and found that $$X(j100) = 100$$ - the unit is weird for me which will be voltage*sec- what does that 100 mean here?

Also: Why there is a factor of $\frac{1}{T}$ difference between the units of Fourier series and Fourier transform ? I've asked at Signals processing/Math stack exchange but no answer

I've read this answer but it says:

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.

which didn't satisfy me.

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    $\begingroup$ Define all symbols in your question. What is $X$? Is that the Fourier transform of some time signal $x(t)$? Why is there a complex argument inside the $X$? Define everything explicitly. $\endgroup$ – DanielSank Apr 18 '16 at 18:29
  • $\begingroup$ I'm sorry but i've assumed the reader is already familiar with Fourier transform, it transforms the signal from the time domain to the frequency domain which is characterized but the summation/integrals of complex exponential so it's complex function after all, i've also used the convention with those symbols. $\endgroup$ – Mohamed Osama Apr 18 '16 at 18:31
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    $\begingroup$ I'm familiar with transforms, but it is helpful to define quantities. We may after all use different conventions and variables. $\endgroup$ – anon01 Apr 18 '16 at 18:35
  • $\begingroup$ Yes @anon0909 is correct. In fact as you become more familiar with Fourier transforms and associated techniques, you will find that there are so many different conventions and notations that defining them explicitly in each document is absolutely necessary. For example, engineering and physics books tend to use opposite convention for the sign in front of the $\omega$. There are also different conventions with respect to the $1/(2\pi)$ prefactors. $\endgroup$ – DanielSank Apr 18 '16 at 18:39
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    $\begingroup$ I still do not see an explicit definition of the symbols in the question. Do not try to describe what you mean in a comment. Update the question and define the symbols explicitly with equations, not words. $\endgroup$ – DanielSank Apr 18 '16 at 18:40
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I'd like to ask what is the meaning of the value obtained from X(jω) with certain frequency ω

Consider for a moment, the synthesis equation where we 'construct' $x(t)$ out of a weighted 'sum' (integral) of the orthonormal basis functions of time: $e^{j\omega t}$

$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}\omega\:X(\omega)\:e^{j\omega t}$$

Here, it is clear that $X(\omega)$ is the amount or weight of $e^{j\omega t}$ that goes into 'constructing' $x(t)$ from these basis functions. We haven't specified what $X(\omega)$ is but we assert that, for a large class of $x(t)$, there is an appropriate $X(\omega)$ such that the above holds.

As you may have already concluded, there is a way to find $X(\omega)$ given an $x(t)$ which is

$$X(\omega) = \int_{-\infty}^{\infty}\mathrm{d}t\:x(t)\:e^{-j\omega t}$$

which (left as a fun exercise for the reader) can be verified by substituting the later expression into the former expression.

Since, in the first expression, we're integrating with respect to $\omega$ which has units of $\mathrm{s}^{-1}$, it must be that $X(\omega)$ has the units of $x(t)$ multiplied by $\mathrm{s}$.


Why there is a factor of $1/T$ difference between the units of Fourier series and Fourier transform ?

If we were to make a discrete approximation of the first expression, it would be something like

$$\tilde{x}(t) = \frac{1}{2\pi}\sum_{n = -\infty}^\infty \frac{2\pi}{T} \:X(n\frac{2\pi}{T})e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:\frac{X(n\frac{2\pi}{T})}{T}e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:a_n e^{jn\frac{2\pi}{T}t}$$

which is periodic with period $T$ (thus the tilde over). Using the usual method of integrating the product of both sides with $e^{-jm\frac{2\pi}{T}t}$ over a period $T$, we arrive at

$$\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jm\frac{2\pi}{T}t} = Ta_m$$

since the integral on the right hand side vanishes for $n \ne m$. And so,

$$ a_n= \frac{X(n\frac{2\pi}{T})}{T} = \frac{1}{T}\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jn\frac{2\pi}{T}t}$$

So, the $\frac{1}{T}$ comes from the fact that the $a_n$ are the 'sampled' Fourier transform divided by the period $T$ as shown above.

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  • $\begingroup$ Can you help me a bit how you converted the integral into summation? i'm confused on how you transformed dw? $\endgroup$ – Mohamed Osama Apr 23 '16 at 19:05
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    $\begingroup$ @MohamedOsama, I just turned the integral into a Riemann sum. More precisely, let $\Delta \omega = \frac{2\pi}{T}$ and then $$\lim_{1/T \rightarrow 0} \sum_{n = -\infty}^{\infty} \Delta \omega X(n\Delta \omega) e^{jn\Delta \omega t} = \int_{n = -\infty}^{\infty} d\omega\:X(\omega)e^{j\omega t}$$ $\endgroup$ – Alfred Centauri Apr 23 '16 at 20:16
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Since the Fourier transform is a way to express the time evolution of your quantity such as the voltage in frequency space, you can interpret it as the value as the density distribution of your quantity in frequency space. Note that it is a continuous distribution, thus there is a density. The reason for the unit is simple: it's Volt/Hertz = Volt * Second. I like to think in terms of power, which is closely related to voltage if you have an Ohmic load. For example you investigate some radiofrequency signal. You could look at time traces or at the Fourier Transform. The values of the latter tells you how much of the power is distributed over which frequencies. So given a frequency, if you integrate that signal around that frequency you get a value proportional to how much power is in these frequencies. As you say in your comment, the value with more meaning is the locally integrated value. Again with regard to units. Ask yourself: how much power (voltage) is between frequency f and f + 1Hz. That's the value of the Fourier Transform at point f.

You can check your definition with a sine voltage with 1 Volt. For an infininte sine, it's a Dirac delta function but for finite time the delta peak is finite and broadened. You can integrate it over different boundaries and investigate the results.

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    $\begingroup$ Beware that the FT of a sine is a Dirac function. $\endgroup$ – L. Levrel Apr 20 '16 at 20:37
  • $\begingroup$ Can you explain the Volt/Hertz? do you mean that it's like the probability distribution function , the value at any point carry no significance but the integral has the significance ? what will happen if i integrated X(jw) dw from certain frequency to other without multiplying by e^jwt ? $\endgroup$ – Mohamed Osama Apr 20 '16 at 20:47
  • $\begingroup$ @Hans: Nice answer, but beware that the power is usually the square of the signal. $\endgroup$ – L. Levrel Apr 21 '16 at 19:40
  • $\begingroup$ That's right. In this context I'd like to share the dBm - W - V conversion table from Mini-Circuits www.minicircuits.com/pages/pdfs/dg03-110.pdf $\endgroup$ – Hans Apr 23 '16 at 16:42
  • $\begingroup$ very good answer , so the number corresponds to the density of the signal which corresponds to the power , but is the value itself important or the integral of the continous function that matter more? $\endgroup$ – Mohamed Osama Apr 23 '16 at 19:07
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  1. Your Fourier transform eq. is incorrect. it should be $e^{-jwt}$ not $e^{jwt}$
  2. Physical meaning: Split your integral into real and imaginary parts, the Fourier transform equation is split into integral with sine and cosines instead of $e^{jwt}$, but with same frequency. $F = \int \Re{x(t)} cos(wt) dt + j \int \Im {x(t)} sin(-wt) dt$.

You could think of each term in the sum as dot (or inner) product of your function with a sin and cosine at the frequency. If your function is orthogonal to the $cos(wt)$ and $sin(-wt)$, only then the Fourier transform at that frequency is zero. Else it is a magnitude of the projection of your function on the sine and cosine at that frequency.

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  • $\begingroup$ his transform is fine: it is merely convention whether or not you use a minus sign (though inverse transform must have opposite sign) $\endgroup$ – anon01 Apr 21 '16 at 4:00
  • $\begingroup$ did n't get you actually with the part of dot product , is like the component in the x direction , so it's the vector dotted with i unit vector? $\endgroup$ – Mohamed Osama Apr 23 '16 at 19:09
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Generally put, Fourier Transform is a decomposition of your signal to its harmonic components. Each point in Fourier space correspond to a certain frequency (be it temporal, or spatial, at the most common cases), and the higher the value of the transform there, that frequency becomes more dominant in the actual signal.

You can look at it like a 'histogram of harmonies' - the closer your signal comes to a pure sine wave, the distribution of the Fourier transform will
'want' to populate a single point in the Fourier space (which represents the frequency of the sine wave). On the other hand, if your signal is complex (meaning, rich in harmonies), the population of the Fourier transform will be more spread out in the Fourier space.

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