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Why use this instead of a positive frequency? I asked on the math forum but it just wasn't helpful.

I understand the idea of a negative frequency is important in general since many real signals like cosine can be constructed using complex exponentials of two opposite rotating frequencies. But why in the FT? Also would it be correct to view this as 'frequency' since in the complex domain it can be negative? I recall someone arguing it is not frequency but index: https://www.quora.com/What-is-the-meaning-of-negative-frequencies-in-the-Fourier-transform

An interesting explanation I had never seen was from wikipedia:

> A reason for the negative sign in the exponent is that it is common in electrical engineering to represent by {\displaystyle f(x)=e^{2\pi i\xi _{0}x}}{\displaystyle f(x)=e^{2\pi i\xi _{0}x}} a signal with zero initial phase and frequency {\displaystyle \xi _{0}.}{\displaystyle \xi _{0}.}[3][remark 5] The negative sign convention causes the product {\displaystyle e^{2\pi i\xi _{0}x}e^{-2\pi i\xi x}}{\displaystyle e^{2\pi i\xi _{0}x}e^{-2\pi i\xi x}} to be 1 (frequency zero) when {\displaystyle \xi =\xi _{0},}{\displaystyle \xi =\xi _{0},} causing the integral to diverge. The result is a Dirac delta function at {\displaystyle \xi =\xi _{0}}{\displaystyle \xi =\xi _{0}}, which is the only frequency component of the sinusoidal signal {\displaystyle e^{2\pi i\xi _{0}x}.}{\displaystyle e^{2\pi i\xi _{0}x}.}

But can any signal input be modeled as a complex exponential form as above? For example, if your input was a rectangle function.

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  • $\begingroup$ Unfortunately no, I looked at that thread like 30 times. It doesn't answer why the FT contains negative exponent, it just discusses why negative frequency is necessary. $\endgroup$ – NimbleTortoise May 13 at 9:29
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    $\begingroup$ it does answer that – it's simply the projection onto a $e^{jft}$ with a negative $f$. $\endgroup$ – Marcus Müller May 13 at 9:39
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    $\begingroup$ and to answer the closing question: yes, that's one of the main results: any bounded time function can be modeled through an integral over complex exponentials (maybe containing diracs, if the signal is periodic). That's the whole point of the Fourier transform, from a function theory point of view: it's just a change of base, nothing special. I know the weights for these complex exponentials by heart, for the rectangle, because that's so essential: it's the sinc function. $\endgroup$ – Marcus Müller May 13 at 9:40
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    $\begingroup$ I specifically answer that question in my previous comment. Yes you can.. I don't know why you're so agressive. $\endgroup$ – Marcus Müller May 13 at 9:58
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    $\begingroup$ It's not condescending. Sorry if that happens to be your impression of me, it wasn't intended. I'll, however, stop trying to help you here, I've got the feeling you feel antagonized by me, and that's nothing I want to encourage. $\endgroup$ – Marcus Müller May 13 at 10:41
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I think you may be confused about the use of negative frequencies and the sign of the integration coefficient.

Let's take a look at the inverse Fourier Transform

$$ x(t) = \int_{-\infty}^{+\infty}F(\omega)\cdot e^{ 2 \pi i \omega} d \omega $$

That basically means that you can construct any time domain signal as the from a set of complex exponential. So you can break down a complicated signal into a number of constituent parts that are hopefully easier to work with.

That's typically called an orthogonal base. Complex exponentials are not the only choice, you could use a orthogonal polynomials, wavelets, etc. Complex exponentials are popular since they are easy to work with, specifically when you are dealing with linear time invariant systems

Note two things

  1. The integration goes from $-\infty$ to $+\infty$, i.e you need both negative and positive frequencies to generate any possible time domain signal
  2. The integral variable is positive, but that does only mean than you first integrate the negative frequencies and then the positive ones.

Of course you need to know what the weights for the complex exponentials, so you need to invert the formula above. You get

$$F(\omega) = \int_{-\infty}^{+\infty}x(t)\cdot e^{ -2 \pi i \omega} d t$$

That's just how the math works out. The integration variable is negative, but that doesn't mean anything. The formula needs to produce results for both positive and negative values of \omega since we need both positive and negative frequencies to represent a time domain signal.

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That is a very poorly written article IMO.

In simple rule of thumb terms:

For real valued tones, there isn't a qualitative distinction between negative and positive frequencies, so there is no point in using negative frequencies. Choosing not to use them by convention is quite different than they don't exist.

For complex valued tones, the negative and positive denote the chirality of the signal, whether you have a left handed or right handed screw, and thus is qualitatively distinctive. Their existence is unquestionable.

The negative sign in the exponent in the definition makes it so a complex tone with a given frequency corresponds to the same value on the frequency scale instead of its negative. With a real valued tone, there is no distinction as you get a conjugate pair of frequency values. In the complex tone, it makes a difference.

The negative sign is by convention. The forward and inverse operations are equivalent in nature.

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  • $\begingroup$ You had the only relevant answer here. I still have yet to get an intuition for it however. You say the negative exponent makes a difference when the input signal is complex rather than real. So if my input is cos(4hz)+jsin(4hz), how does that make a difference when multiplied by e^- as opposed to e^+? $\endgroup$ – NimbleTortoise May 14 at 5:29
  • $\begingroup$ @NimbleTortoise Look at the definitions: The negative sign in the exponent is the same as saying $(k)$ or $(-k)$ in the discrete case and $(\omega)$ or $(-\omega)$ in the continuous. Thus, the effect flippling the negative is simply flipping the values about zero on the frequency scale. For a complex tone, there is one peak, it matters which way the frequency values are flipped if you want to match. For a real tone there is a pair of peaks the same distance from zero and flipping them is the same as conjugating them. There is no way to tell the sign of the frequency parameter. $\endgroup$ – Cedron Dawg May 14 at 12:25
  • $\begingroup$ For an intuitive understanding, check out dsprelated.com/showarticle/768.php. The sign of the exponent defines the direction the "wrapping" gets done in. $\endgroup$ – Cedron Dawg May 14 at 12:25
  • $\begingroup$ $$ \cos( 4 \theta ) + i sin( 4 \theta) = \frac{e^{i4 \theta}+e^{-i4 \theta}}{2} + i \cdot \frac{e^{i4 \theta}-e^{-i4 \theta}}{2i} = e^{i4 \theta} $$ Each real tone (cosine and sine) produces two peaks in the frequency domain. The peak pairs from the two tones align. Once set cancels, the other reinforces. If you remove the negative sign in the exponent in the defintion, the location of the reinforced peak in the complex case won't match the signed value of the frequency parameter. $\endgroup$ – Cedron Dawg May 14 at 12:33
  • $\begingroup$ So you are saying if the input frequency is 4 for a complex signal, in a FT definition with positive exponent the amplitude will show up at frequency -4? That seems like more than just convention. $\endgroup$ – NimbleTortoise May 14 at 22:17
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While reading the above few answers, it’s been stated like negative frequency has no physical significance, but it is completely wrong. Fourier transform is used to pull out - that component of a signal (under test)which is having a particular repetition frequency( or rotation frequency in that sense). If you agree to the fact that complex exponential represents a unit circle ( if the exponent value varies with time or any independent variable ). \begin{equation}  e^{ j \omega t} \end{equation} As the independent variable varies the point moves along a circle and it will have a particular frequency of rotation. A circular rotation can be defined with two things, the radius of rotation( amplitude or Fourier coefficient in that manner) and frequency of rotation ( number of 2 pi rotation completed in a second ), but if you look we need one more parameter to define the sense of rotation, whether the ‘thing’ is rotating clockwise or anti-clockwise.

Negative or Positive frequency defines that! It gives us the sense of rotation.

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In the FT "negative frequencies" means nothing physically, however they are essential in the maths behind the FT. The FT is nothing more than a change of base.

If we take an example with a door function p(t), \begin{equation} p(t) \begin{cases} 1 & \text{ if } -\frac{1}{2}\leq t\leq\frac{1}{2} \\ 0 & \text{ elsewhere } \end{cases} \end{equation}

We apply the fourier transform on this signal, \begin{equation} \mathcal{F}(\nu)=\int_{-\infty}^{+\infty}p(t)e^{-2i\pi\nu}dt \end{equation}

Since the signal is present between $-\frac{1}{2}\leq t \leq\frac{1}{2}$

\begin{equation} \mathcal{F}(\nu)= \frac{sinc(\pi\nu)}{\pi\nu} \end{equation}

The absolute value of the fourier transform is displayed below.

enter image description here

Imagine you want to retrieve your original signal from the FT, $\mathcal{F}(\nu)$, you will need to apply the inverse fourier transform defined as :

\begin{equation} p(t)=\int_{-\infty}^{+\infty}\mathcal{F}(\nu)e^{2i\pi\nu}dt \end{equation}

The signal we are trying to integrate is defined between -$\infty$ and $+\infty$ and that's why you need to keep the negative part, otherewise you won't be able to go back to the initial basis.

In the article you saw, they call "negative frequencies", indexes because it is directly derived from the discrete fourier transform.

To answer your last question, i guess that every signal can be represented as a sum of complex sinuoïds if its area under the curve is integrable. The Fourier Transform can be seen as a development of the fourier series of a signal with an infinite period. (Correct me if i'm wrong)

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