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I know this question has been asked a lot of times before but I cannot really understand how the message is transmitted in single sideband modulation.

See the following images taken from the book "Eletronic communication systems" by Kennedy :

Full AM modulation

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DSB-SC

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SSB-SC

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Now the message is clearly visible in the first two diagrams, but not in the third one.

  • Doesn't the sideband modulated wave look like a regular sine wave from a generator?
  • Where is the message in sideband?
  • How exactly is the message recovered from the third diagram?

Please explain in TIME domain with diagram.

Please focus on conceptual treatment rather than individual techniques of modulation and demodulation.

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  • \$\begingroup\$ In the last figure, Vc is multiplied by Vm. Multiplying two sines produces terms with the sum of the two and the difference of the two frequencies. So, the USB is the frequency of Vc plus the freq of Vm, and the LSB is f of Vc - f of Vm. To see them as shown, they have to be filtered out of the complicated signal formed by the product. \$\endgroup\$ – C. Towne Springer Jul 3 '14 at 4:06
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    \$\begingroup\$ Right. But where is the message ? I mean i do not see anything or any parameter varying according to the message. \$\endgroup\$ – Plutonium smuggler Jul 3 '14 at 5:04
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    \$\begingroup\$ It is a poorly done example. Vm is the modulation signal. It should be changing amplitude. But since the carried is a much higher frequency than Vm, if you show the carrier in a diagram, you can't show enough time to see Vm changing. Use any math package, or a spreadsheet and plot Vm times Vc with one being 100 times higher frequency than the other. Then change Vm with something like a triangle form at an even lower rate. \$\endgroup\$ – C. Towne Springer Jul 3 '14 at 5:34
  • \$\begingroup\$ You mean to say that in the third diagram, in part 3 and 4, the amplitude of the USB and LSB be changing just like in diagram 1 and 2, the difference being the absence of outer envelopes ? Am i correct ? \$\endgroup\$ – Plutonium smuggler Jul 3 '14 at 5:48
  • \$\begingroup\$ There is a subtle difference you can see with plots. In the case of multiplying two sines - say one is ten times the freq of the other - the phase of the carrier (higher freq) does not change. You are Vm is a simple scaling factor. In the sum and difference case where the two sines differ by the freq of the desired sideband, there is a phase change (like in adding two sines of arbitrary phase). \$\endgroup\$ – C. Towne Springer Jul 3 '14 at 19:52
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In the first diagram (broadcast AM) "the message" is a sinewave of frequency \$\omega_m\$. In the second diagram (full AM suppressed carrier modulation) "the message" is also a sinewave of frequency \$\omega_n\$. OK so far? - "the message" signal or modulating signal is just a plain ordinary sinewave.

Either of these two methods produce two sidebands at either side of the carrier frequency. So if the carrier were 1 MHz and the modulation ("the message") were 2 kHz, you would see frequencies of 998 kHz and 1002 kHz on a spectrum analyser. These are the upper and lower sidebands.

If you filtered out the upper sideband you would be left with only the lower sideband and if this was mixed (in a receiver) with a sinewave of 1MHz, you'd recover the original modulating frequency (aka "the message").

Please explain in TIME domain with diagram. Please focus on conceptual treatment rather than individual techniques of modulation and demodulation.

No I'm not going to do that. If you don't understand my words then please let me know.

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  • \$\begingroup\$ So far so good. But considering what Springer just said above, am I correct in assuming that the shape or amplitude of USB or LSB must be varying just like in full AM and DSB-SC, the only difference being the absence of outer envelope, although it is not shown in diagrams ? \$\endgroup\$ – Plutonium smuggler Jul 3 '14 at 9:09
  • \$\begingroup\$ If you mathematically multiply two sinewaves you get sum and difference i.e. 998 kHz and 1002 kHz. When you look at the full signal it will look like the 1st or 2nd picture (depending on the depth of modulation). If you used a tight band pass filter you would find a sinewave at 998 kHz and a sinewave at 1002 kHz. \$\endgroup\$ – Andy aka Jul 3 '14 at 9:50
  • \$\begingroup\$ Ok. That does mean it is a poor example ( it doesnt show variation of amplitude in sidebands, as you described). Thanks. \$\endgroup\$ – Plutonium smuggler Jul 3 '14 at 11:35
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    \$\begingroup\$ Yes it does - the variation is the pure sinewave at 998 kHz (lower sideband) or 1002 kHz (upper s/b). Remember that you are modulating with a pure sinewave (2 kHz in my answer) and that 998 kHz + 1002 kHz = the first 2 pictures seen in your question. \$\endgroup\$ – Andy aka Jul 3 '14 at 11:39
  • \$\begingroup\$ The OP's images of SSB are actually incorrect. Suppose the carrier is 1MHz, the the message is depicted as about 100kHz. So ostensibly a 100kHz tone modulating the 1MHz carrier. The sidebands should be at about 1.1 and 0.9 MHz. Yet the USB and LSB are shown as about 1.5MHz and 0.45MHz.. No wonder this example is incomprehensible -- it fails to link together the key concepts. \$\endgroup\$ – gwideman Mar 21 '19 at 10:59
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I actually asked myself the same question and went through it. The given explanations are correct but they don't give the any feel how it looks like (in the time domain, as the OP requested).

Indeed, modulating a pure sine wave is a trivial waveform and not visible in the RF signal envelope. However, if the baseband signal is sufficiently complex, AM-SSB actually looks similar to normal DSB AM.

Below a random baseband signal of 200kHz bandwidth and 1 MHz carrier for all cases in time domain and frequency domain:

enter image description here

enter image description here

enter image description here

enter image description here

Finally, this shows a zoom into the comparison between DSB and SSB AM:

enter image description here

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