Circuit to retain only the positive frequency components in a signal?

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  • #1
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Hi,
Is there any method(circuit) to retain only the positive frequency components of a signal?

thanks in advance.
 

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  • #2
Averagesupernova
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What is a positive frequency component?
 
  • #3
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when a signal is represented in frequency domain, it has both positive and negative frequency components. example:- cos(2*pi*f*t) in freq domain has 0.5*d(t) at +f and -f.
I am looking to eliminate the component at -f.
(d(t)=delta function)
 
  • #4
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This reference is in the frequency domain. You will only see all of the negative frequency if you mix the signal with a "carrier" frequency that is at the highest frequency of the signal. A bandpass filter that is to the "right" of the carrier frequency will let the positive side of the signal through while impeding the negative side of the signal.
 
  • #5
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This reference is in the frequency domain. You will only see all of the negative frequency if you mix the signal with a "carrier" frequency that is at the highest frequency of the signal. A bandpass filter that is to the "right" of the carrier frequency will let the positive side of the signal through while impeding the negative side of the signal.
I understand what you are trying to say. But, what i was looking for is a means of eliminating the original negative frequency component in the modulating signal without having to shift it by the frequency of the carrier. In other words, i don't want any frequency components towards the left half of the y axis in the frequency spectrum.
 
  • #6
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The only time you will see the negative frequencies (in time domain) is when you mix the signal with a carrier. I am not sure of this, but do filters affect only the positive frequencies or do they have the same mirrored profile in the negative side of the frequency domain. My hunch is that filters do have an even symmetry wrt the y-axis in the frequency domain. And if that is the case, then you would have to mix the signal to bring out negative part of it.

Why the need to eliminate the negative part of a signal?
 
  • #7
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Yes, filters do have a mirrored profile about y axis in frequency domain, hence filtering out the negative frequency component is not possible. Anyway, why i had this question was that i noticed that any periodic signal has symmetric mirrored profile about the y axis in f domain. This, i think is a redundancy and i am trying to find a means of eliminating one half of the freq. spectrum(negative part).
On a related note, do the negative frequencies constitute half of the total signal power?
 
  • #8
Averagesupernova
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Look into methods of generating a single sideband signal. There is a way to do it without brute force filtering.
 
  • #9
chroot
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sanjaysan, the negative frequency components are redundant, in a sense. Consider your time domain signal, [itex]cos(2\pi \omega t)[/itex]. The angular frequency, [itex]\omega[/itex], could be either positive or negative, and the resulting wave would look the same in the time domain. That ambiguity leads to the two-sided, symmetric spectrum.

You can move to a one-sided spectrum if you wish, with no loss of generality, but that's just a mathematical trick. You don't need to design any real, physical device to discard the negative frequencies; they're all in your head from the beginning!

- Warren
 
  • #10
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Warren, Are you saying there is no physical significance attached to negative frequencies?
 
  • #11
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Which energy is defined as negative frequency and which is defined as positive depends on the timing (phase) of the FFT function itself, and is not some independent attribute of a signal. When thinking theoretically, you're in the habit of defining some t=0, which will define what is negative and what is positive. Negative and positive frequencies have physical significance only in a process that already includes a FFT function with a defined timing (a defined t=0 which has some physical meaning, like when the FFT circuit is reset). In an actual circuit you would need to "reset" the FFT process at some point in order to define a t=0, if you want frequency polarity to have physical meaning. And once you do all that, you've got a frequency-domain signal that makes it easy to remove the negative frequencies, anyway. Then do a IFFT. There's no shortcut to doing a FT, since FT, by definition, means "getting the frequency components of a time domain signal" (although admittedly how one gets the frequency components is up for efficiency improvements). Certainly there would be a minor (probably insignificant) simplification of a standard FFT circuit/algorithm when all you wanted was the positive frequencies.
 
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  • #12
sophiecentaur
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Warren, Are you saying there is no physical significance attached to negative frequencies?
"Physical significance"?
The spectrum is just a mathematical way of describing things. Sometimes, the maths suggests the presence of signals but you can't necessarily measure them. This is only like saying that you want a 'Physical Significance' for one of the roots of a (mathematical) equation of motion which doesn't fit the physical problem. You just have to pick the more 'suitable' result.
Try this link to see a thread which touches on just this topic.
 
  • #13
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"Physical significance"?
The spectrum is just a mathematical way of describing things. Sometimes, the maths suggests the presence of signals but you can't necessarily measure them. This is only like saying that you want a 'Physical Significance' for one of the roots of a (mathematical) equation of motion which doesn't fit the physical problem. You just have to pick the more 'suitable' result.
Try this link to see a thread which touches on just this topic.
Sophie, I get that we have been 'picking the more suitable result' for analysing the signal. But, I recently read a paper titled "Negative frequency communication" which says that half of the frequency(negetive) is being wasted and it does have a physical meaning. You can read more about it here-
http://arxiv.org/abs/1012.1403
so is it just a mathematical eventuality or is there more meaning to it??
 
  • #14
sophiecentaur
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I can only read the abstract but that paper seems to be drawing conclusions based on ancient forms of modulation (dsbam, for instance). For decades, modulation systems have been used which do not have a 'redundant' half to their spectrum. SSB has been used since before WW2, afaik and that was 'as crude as you could imagine'.
People have used 'independent sideband' transmissions and suppressed the carrier.

I may be doing the author a disservice here but it certainly looks like that to me. I would need access to the whole paper but I can't seem to find it.
 
  • #15
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I can only read the abstract but that paper seems to be drawing conclusions based on ancient forms of modulation (dsbam, for instance). For decades, modulation systems have been used which do not have a 'redundant' half to their spectrum. SSB has been used since before WW2, afaik and that was 'as crude as you could imagine'.
People have used 'independent sideband' transmissions and suppressed the carrier.

I may be doing the author a disservice here but it certainly looks like that to me. I would need access to the whole paper but I can't seem to find it.
I've read the whole paper. Its not about SSB modulation. The author specifically talks about using the negetive frequency components for communication. Heres the link to the paper-
http://arxiv.org/pdf/1012.1403v5.pdf
Any feedback would be appreciated.
 
  • #16
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Well there is no such thing as negative frequency. In real life. That doesn't make sense.

I only think math behind the fourier transform, and generally math tools are made in that way so that we do operate with negative frequencies.

Cosine is represented with 2 deltas right? (amplitude spectra)

Eliminate one of them, and you no longer have a cosine...

Specifically lets consider this.

[itex] F (cos(\omega _0 t)) = \pi\left[\delta (\omega -\omega _0)+\delta (\omega +\omega _0)\right] [/itex]

Eliminate the negative frequency and you get:

[itex]\pi\left[ \delta (\omega -\omega _0)\right][/itex]

Inverse Fourier transform of this is no longer a cosine. Its complex exponential:

[itex] F^{-1}(\pi\left[ \delta (\omega -\omega _0)\right])=\frac{1}{2}\cdot e^{j\omega _0 t}[/itex]

So its not even a real function anymore.

You can go with using a complex exponential as your carrier.
Design a system, which transfers in parallel a Sine and a Cosine.
 
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  • #17
sophiecentaur
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Looking briefly at the paper I see he describes his chosen form of modulation in terms of multiplying a carrier wave by a modulating signal. Depending upon what the form of the modulating signal (depending upon any constant term) this will produce easily-demodulated AM, Suppressed Carrier AM or something in between. It's clear that this will have redundant frequency components (his 'negative' components). It seems that he is proposing a complex system of modulation to eliminate one of those sidebands. This, again has been done (A Marconi Patent, I believe - or it may be Siemens) with high powered HF transmitters as an efficient way of producing SSB (just one of the sidebands). The modulation consists of a combination of AM and Phase Modulation, in which the sidebands on one side cancel.

If you consider the actual Power involved in his proposed system, if it were anything other than some version of this, then (up to) 100% excess power would have to come from somewhere in the transmitter if information were to be available at the other end of the link by way of this extra available channel.
Using two polarisations for two transmitting channels is, again, nothing new. Every TV transmitter network uses polarisation diversity in its service planning.

This seems to me like a good bit of theoretical bookwork that has been done in a bit of a vacuum. He cites no RF Engineering references to put the work in a real world context and I wonder whether he is actually aware that he may well be re-inventing a wheel.
 
  • #19
sophiecentaur
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That would work fine at low power but I think the (200kW??) transmitter I have seen used just one transmitting valve and achieved the result by AM and PM in a single unit.

Hang on a minute. How would you combine your two signals losslessly from two high power amplifiers? I think that could be a problem.
 
  • #20
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I only know the theory behind it, I never designed one :D
 
  • #21
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Well there is no such thing as negative frequency. In real life. That doesn't make sense.

I only think math behind the fourier transform, and generally math tools are made in that way so that we do operate with negative frequencies.

Cosine is represented with 2 deltas right? (amplitude spectra)

Eliminate one of them, and you no longer have a cosine...

Specifically lets consider this.

[itex] F (cos(\omega _0 t)) = \pi\left[\delta (\omega -\omega _0)+\delta (\omega +\omega _0)\right] [/itex]

Eliminate the negative frequency and you get:

[itex]\pi\left[ \delta (\omega -\omega _0)\right][/itex]

Inverse Fourier transform of this is no longer a cosine. Its complex exponential:

[itex] F^{-1}(\pi\left[ \delta (\omega -\omega _0)\right])=\frac{1}{2}\cdot e^{j\omega _0 t}[/itex]

So its not even a real function anymore.

You can go with using a complex exponential as your carrier.
Design a system, which transfers in parallel a Sine and a Cosine.
Bassalisk, How do i go about generating the complex exponential e^(jwt)? You said a system which transfers sine and cosine waves in parallel. Do you mean sin(wt)+cos(wt)? shouldnt it be cos(wt)+j sin (wt)?
 
  • #22
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Bassalisk, How do i go about generating the complex exponential e^(jwt)? You said a system which transfers sine and cosine waves in parallel. Do you mean sin(wt)+cos(wt)? shouldnt it be cos(wt)+j sin (wt)?
There wouldn't be imaginary current or potential in an actual circuit. The circuit depicted carries two separate products with sine and cosine waves respectively, to represent multiplying by a complex exponential.

You should know that real signals have Hermitian symmetry in the frequency domain. The component at a negative frequency is the complex conjugate of the component at the corresponding positive frequency.

This has to do with the symmetry of cosine and the "odd" symmetry of sine. Consider what happens if we make the frequency negative.

cos(ωt) = cos(-ωt)

but

sin(ωt) = -sin(-ωt)


Anyway, removing the negative frequencies seemingly would imply two outputs instead of one, based on Hermitian symmetry requirement for the frequency domain of real signals.
 
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  • #23
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Bassalisk, How do i go about generating the complex exponential e^(jwt)? You said a system which transfers sine and cosine waves in parallel. Do you mean sin(wt)+cos(wt)? shouldnt it be cos(wt)+j sin (wt)?
Well, imaginary numbers are a pair of real numbers. So that system that I posted in the post above, explains it nicely.

When you go deeper into Signals and Systems, these things are pretty easy to understand :D
 
  • #24
sophiecentaur
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I think we are confusing real, time varying signals with a convenient mathematical representation of them. Neither the exponential notation or the 'cis' notation are any more than models, and 'half' of that model is not relevant to the real world. afaik, one should really prefix the final result of any 'complex' jiggery pokery with the words "The Real Part Of . . . " if you want to get a proper answer.

I think you can do the analysis (albeit in a more lumpy way) without using i at all.
 
  • #25
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I think we are confusing real, time varying signals with a convenient mathematical representation of them. Neither the exponential notation or the 'cis' notation are any more than models, and 'half' of that model is not relevant to the real world. afaik, one should really prefix the final result of any 'complex' jiggery pokery with the words "The Real Part Of . . . " if you want to get a proper answer.
Well put.
 

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