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Circuit to retain only the positive frequency components in a signal? |
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| Jan21-12, 12:53 PM | #18 |
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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 |
| Jan21-12, 12:57 PM | #19 |
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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. |
| Jan21-12, 01:00 PM | #20 |
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I only know the theory behind it, I never designed one :D
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| Jan21-12, 09:25 PM | #21 |
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| Jan21-12, 10:05 PM | #22 |
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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. |
| Jan22-12, 06:22 AM | #23 |
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When you go deeper into Signals and Systems, these things are pretty easy to understand :D |
| Jan22-12, 06:49 AM | #24 |
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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. |
| Jan22-12, 06:51 AM | #25 |
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| Jan22-12, 06:52 AM | #26 |
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That's where that paper on "negative Frequencies in Modulation" seems to be skating on thin ice.
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| Jan22-12, 08:18 AM | #27 |
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| Jan22-12, 11:39 AM | #28 |
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| Jan22-12, 11:44 AM | #29 |
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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. |
| Jan23-12, 04:52 AM | #30 |
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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. |
| Jan26-12, 12:16 PM | #31 |
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| Jan26-12, 12:43 PM | #32 |
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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". |
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