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What are negative frequencies with respect to communication systems?

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sridhara

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hi guys...
could any one explain me what is negative frequencies with respect to communication systems....
 

eulerian frequency omega

Hi,
The concept of negative frequency is purely mathematical. This can be easily proven by some simple trigonometric math. However in comms systems the 'negative' frequencies is simply copies of you signal symetric about the sampling frequency. Thus if you sample (for the carrier) at 433MHz and your data sits at 433.1MHz then you see an image at 432.9Mhz as well. If you move this to the 0Hz point negative frequencies needs to be used.

However, as mentioned negative frequency is not a physical phenomenan. Imagine a sime wave of -100Hz - I can't.

Hope this helps
Cheers
Slayer
 

phasor 8 point dft matrix

The vector rotates the other direction. You can detect this with quadrature methods. One example is a direct down conversion.
 

negative frequency in rotating phasor

Euler's identity: exp(I*w*t) = cos(w*t) + j*sin(w*t)

From which:
cos(w*t) = [exp(exp(I*w*t)) + exp(exp(I*(-w)*t))]/2

The real cos is constructed from a positive w and negative w. In the complex plane we have a forward and backward rotating vector with a phase relationship such that the imaginary portion cancels out.
 

negative omega t

In communication systems we come across signals like Asinωt and Acosωt. These signals are projections of a rotating phasor on the real and imaginay axes. The phasor has a magnitude A and angular frequency ω.

Rotation in the anti-clockwise direction is considered positive and clockwise negative.

So the frequency -ω means the phasor is rotating in the clockwise direction.

Cheers....
 

concept of negative frequency

Hello to sridhara and to everyone !

As you already might know, when you're making a Fourier Transform from Time Domain into Frequency Domain of harmonic single tone signal (like sin(t) or cos(t) - which is also called a phasor) it results of two delta functions (δ): one at the negative side of the axes and one in the postive side of it and also they scaled by half (both).
This is a mathematical result of the transformation.
In the real world, when you want to see the frequency content (spectrum) of this signal (on Spectrum Analyzer for example) you will see only the positive part which refers to the physical postive frequency content.

Glad to help.

Itzik
 

Re: negative frequencies

The concept of negative and positive frequency can be as simple as a wheel rotating one way or the other way. A signed value of frequency indicates both the rate and direction of rotation. The rate is expressed in units such as revolutions (aka cycles) per second (hertz) or radians/second (where 1 cycle corresponds to 2π radians).
Contents
[hide]

* 1 Sinusoids
* 2 Complex sinusoids
* 3 Sampling of positive and negative frequencies and aliasing
* 4 Negative frequency as a matched filter for positive frequencies
* 5 Negative frequency in Doppler radar
* 6 External links

[edit]

Sinusoids

A sinusoid is a function of an angular argument, and its amplitude varies cyclically as the angle (aka phase) steadily increases or decreases. When the angle is a function of time, the concept of negative frequency is sometimes used to distinguish a decreasing angle from an increasing one. But sinusoids are not monotonic functions. Consequently, \cos(\omega t + \theta)\, does not preserve the sign of \omega \,, just as f(x)=x^2\, does not preserve the sign of x\,. [We note that \theta \, represents a usually unknown, random phase offset.] In most cases of dealing with a single, real-valued sinusoid, it is sufficient to assume that \omega \, is positive. It represents the frequency, in units of radians/sec.


Sometimes there are two sinusoids with the same frequency, and a known phase difference, for instance:

R(t) = \cos(\omega t + \theta)\,
and
I(t) = \cos(\omega t + \theta -\begin{matrix}\frac{\pi }{2} \end{matrix}) = \sin(\omega t + \theta )\,

When \omega > 0\,, R(t)\, appears to lead I(t)\, by \begin{matrix} \frac{1}{4} \end{matrix}\, cycle (=\begin{matrix}\frac{\pi }{2} \end{matrix}\, radians). But when \omega < 0\,, the roles are reversed. So in that case it is possible to distinguish negative and positive frequencies. The diagram depicts a negative frequency. R(t)\, and I(t)\, are referred to as real and imaginary, respectively. And \theta = 0\,.


A parametric plot of imaginary vs real would trace a circular path (like the rotating wheel). The addition of a time dimension creates a corkscrew pattern. A negative frequency (decreasing phase) causes a clockwise rotation in a right hand coordinate system as time increases:
[edit]

Complex sinusoids

The complex function: \cos(\omega t) + j\cdot \sin(\omega t)\, facilitates many kinds of mathematical operations involving \cos(\omega t)\,, due in large part to Euler's simplification:

e^{j \omega t} = \cos(\omega t) + j\cdot \sin(\omega t)\,

This very useful form is often referred to as a complex sinusoid. And of course it preserves the distinction between positive and negative \omega \,.

* For positive values, it is also called the analytic representation of \cos(\omega t)\,.

The Fourier transform of e^{j \omega t}\, produces a non-zero response only at frequency \omega \,.

* The transform of \cos(\omega t)\, has responses at both \omega \, and -\omega \,, which reflects the fact that \cos(\omega t)\, is insufficient to determine the sign of \omega \,.
o So just as \sqrt{4} = \pm 2 \,, the interpretation of the ambiguity often depends on collateral information.
o An alternative, and surprisingly useful, viewpoint is that both frequencies are present, as implied by the inverse of Euler's formula: \cos(\omega t) = \begin{matrix} \frac{1}{2} \end{matrix}(e^{j \omega t}+e^{-j \omega t})\,.


[edit]

Sampling of positive and negative frequencies and aliasing

When a complex sinusoid is sampled at regular intervals, its frequency becomes indistinguishable from certain other frequencies, including negative ones (referred to as aliasing). The adjacent figure illustrates this effect for several cases. The red indicates 0 Hz (aka DC). Successively higher frequencies are indicated by orange, blue, purple, violet, black, and blue. Note that some frames depict "R" and "I" for the same frequency, and others depict the "I" samples of different frequencies that are aliases of each other.


For instance, the fourth frame (purple and green) compares samples of the imaginary component of the fractional frequency +\begin{matrix}\frac{5}{8}\end{matrix} with those of negative frequency -\begin{matrix}\frac{3}{8}\end{matrix}, to illustrate that they are indistinguishable. Or in other words: e^{j2\pi \left(+\begin{matrix}\frac{5}{8}\end{matrix}\right) n} = e^{j2\pi \left(-\begin{matrix}\frac{3}{8}\end{matrix}\right) n}\, for integer values of n, representing the sample number. The underlying waveforms are just the imaginary components of: e^{j 2 \pi \left(+\begin{matrix}\frac{5}{8}\end{matrix}\right) F_s t}\, and e^{j 2 \pi \left(-\begin{matrix}\frac{3}{8}\end{matrix}\right) F_s t}\,, where F_s \, is the sample rate (samples/sec).


Likewise +\begin{matrix}\frac{7}{8}\end{matrix} is indistinguishable from -\begin{matrix}\frac{1}{8}\end{matrix}. And \begin{matrix}\frac{8}{8}\end{matrix} (last plot) is indistinguishable from \begin{matrix}\frac{0}{8}\end{matrix} (first plot).


[edit]

Negative frequency as a matched filter for positive frequencies

The rows of the DFT matrix begin at zero frequency, and get more negative as we move downward, row by row. This is because each of these rows functions as a matched filter to measure increasingly positive frequencies in the signal under test. For example, the top row of the 8 point DFT matrix measures DC in the signal, while the next row, which is a signal of fractional frequency −1/8, measures the strength at +1/8 fractional frequency in the signal under test.
 
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    tenso

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Re: negative frequencies

Negative frequncies imply only the phase of the signal...i.e in the case of a real signal it just means whether the signal is cosine signal or a sine....
 

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