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

Method using sine and cosine as carrier.

Frequency spectra.

Full link:

http://ocw.mit.edu/resources/res-6-0...7S11_lec13.pdf

 Recognitions: Gold Member Science Advisor 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.
 I only know the theory behind it, I never designed one :D

 Quote by Bassalisk 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. $F (cos(\omega _0 t)) = \pi\left[\delta (\omega -\omega _0)+\delta (\omega +\omega _0)\right]$ Eliminate the negative frequency and you get: $\pi\left[ \delta (\omega -\omega _0)\right]$ Inverse Fourier transform of this is no longer a cosine. Its complex exponential: $F^{-1}(\pi\left[ \delta (\omega -\omega _0)\right])=\frac{1}{2}\cdot e^{j\omega _0 t}$ 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)?

 Quote by sanjaysan 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.

 Quote by sanjaysan 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

 Recognitions: Gold Member Science Advisor 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.

 Quote by sophiecentaur 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.

 Recognitions: Gold Member Science Advisor That's where that paper on "negative Frequencies in Modulation" seems to be skating on thin ice.

 Quote by sophiecentaur That's where that paper on "negative Frequencies in Modulation" seems to be skating on thin ice.
Yes its somewhat like talking about negative time. But nevertheless, this thread was really interesting!

 Quote by Bassalisk 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
So, is x(t)cos(wt)+y(t)sin(wt) the required complex exponential modulated signal?

 Quote by sanjaysan So, is x(t)cos(wt)+y(t)sin(wt) the required complex exponential modulated signal?
x(t)cos(wt)+j*x(t)sin(wt) ****

y(t) would have a real part x(t)cos(wt)
y(t) would have an imaginary part x(t)sin(wt)

They come in pairs.

http://ocw.mit.edu/resources/res-6-0...me-modulation/

This video is very much addressing that subject.

 Recognitions: Gold Member Science Advisor If I have a room of 12 square metres area and I want a carpet for it, I would choose one with dimensions 3m by 4m. This would be a REAL carpet that had Real dimensions, made of Real Wool. The Maths would tell me that a carpet with dimension -3m by -4m would also do the job. Only a loony would go out to look for one of those in a shop. Why look for anything more significant when some Maths suggests that a Negative Frequency component could exist for a signal? At the beginning of the thread there was a question about the existence of Power in this 'Mirror' signal. Clearly not. A square wave oscillator takes power from its (DC) power supply (real, measurable Joules from a battery). This Power is exactly the same as the power in the square wave - as you can see by heating up an element or by integration. There is no other power in the system. The 'negative frequency' component is just an artifact of the Maths - just like the negatively dimensioned carpet. The same applies to a sinewave. So what about the two sidebands which are generated in AM? They are at (absolute) positive frequencies and carry energy - along with the carrier and, if you add up the three Powers, you get the value of Power which the power supply delivers (less a measurable / calculable factor due to the efficiency of the modulator. The power in each sideband can be used to carry other information and the power of the carrier can be reduced to zero - giving you two SSB transmissions, each of which has the same SNR as the original dsbam signal, half the signal spectrum occupancy and may save the power of a carrier, depending on how you actually produce the ssb signal. When you draw diagrams of signal spectra and show them shifting around and being filtered, there is no explicit mention of the Power involved. So the diagram proves nothing about the existence or otherwise of 'components'. I, personally, was a bit disappointed with that MIT Movie. It would be very easy to get some wrong messages from it, I think. OK, as far as it went but strictly an undergraduate treatment of the topic, I should say, aimed at getting predictable results from a comms system.

 Quote by chroot sanjaysan, the negative frequency components are redundant, in a sense. Consider your time domain signal, $cos(2\pi \omega t)$. The angular frequency, $\omega$, 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
Then how do you explain the recovery of baseband signal from single sided passband signal. Suppose we have only upper sideband of a signal then in recovery of message signal the mirrored band of the signal contributes to form the spectrum of the message signal. What do you think happens physically here........

 Recognitions: Gold Member Science Advisor Look at the block diagram of an ssb receiver. A local oscillator at the original carrier frequency will mix with the sideband and produce a baseband signal - and other mixing products, of course but they will be at non-baseband frequencies and their power level is not relevant. If you're concerned about the SNR of the demodulated signal then it would be 3dB lower than when both sidebands are demodulated because the noise bandwidth would be half but the demodulated signal would be half the level - giving 3dB net loss. BUT there was 3dB less power transmitted for a start so there would is overall disadvantage (as long as the transmitter can be made efficient. It is important to get the Maths and the Physical World reconciled properly. Like I said earlier. All the answers to the calculations should start off with "The real part of".

 Tags electronics, hilbert transform, modulation
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