2
\$\begingroup\$

So if I have AM, then to transmit information, I modulate the amplitude of energy peaks in a wave that has a frequency of say 10 kHz.

So normally, the individual peaks in the wave occurring at regular interals would be say 1 Volt high. But by changing the height of the peaks to half volt quarter volt and every thing in between, and having someone tune in to the right frequency, they can hear my voice, because I am now changing their speaker energy output by changing the peaks in my carrier wave.

So SSB says, you can just change the troughs depths instead, or just the peaks height or whichever, it's all the same.

But then suppressed carrier SSB. how can you suppress the carrier?. What are you modulating or changing the peaks of then? And how does the receiver separate your signal from other random noise, without a carrier to check for modulations on?

\$\endgroup\$
3
\$\begingroup\$

SSB is not how you describe but first regular AM as broadcast by the radio stations. This you appear to correctly describe except for the demodulation which you omit.

In AM the amplitude is modulated but not by so much that the carrier amplitude becomes zero at any point. This is regular AM and for instance a \$1V_{peak}\$ carrier might reach (due to the effects of modulation) peaks of 1.8V and troughs of 0.2V: -

enter image description here

A simple diode detector will demodulate this so this method is suitable for public broadcast radio stations because the radio receiver is simplified by using the diode detector.

Suppressed carrier AM is like regular broadcast AM but where the modulating signal pushes the carrier beyond the limits of broadcast AM. The result of this is that the carrier phase becomes reversed every half cycle of the modulating signal. A simple diode detector no longer produces any decent recovery of the audio signal. Here is a picture of what happens when AM is over-modulated (heading towards a suppressed carrier scenario) and you try and use a simple diode detector: -

enter image description here

Why is it called suppressed carrier - the carrier inverts every half cycle of the modualting wave and if you used a spectrum analyser there is no trace of the carrier despite the resultant modulated carrier being centred at the carrier frequency: -

enter image description here

Recovery of the modulating signal (i.e. the original audio signal) is much trickier and that is why this technique is not used in broadcast transmitters in order to keep receiver cost and/or complexity low.

SSB usually is a further modification to suppressed carrier where one of the sidebands is either totally or partially suppressed in order to create a smaller bandwidth in what is being transmitted - more power can be focussed into the single remaining sideband and with a more complex receiver it can be adequately received. To understand what SSB looks like you have to look at the frequency spectrum of DSBSC (Double sideband suppressed carrier) and imagine it is lopped off on one side: -

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Excellent. Allowing the trough of the modulating signal to drive the amplitude of the carrier to zero ( below zero, negative amplitude?!? ) causes the resultant signal to be composed instantaneously of a positive amplitude of the carrier but a 180 degree phase shifted carrier portion \$\endgroup\$ – Andyz Smith Aug 23 '13 at 23:09
  • \$\begingroup\$ So, through destructive interference the resulting RF power envelope contains no power at the carrier frequency? \$\endgroup\$ – Andyz Smith Aug 23 '13 at 23:13
  • \$\begingroup\$ @AndyzSmith Yes indeed and if you fed it to a perfect tuned filter there would become no trace of the carrier. In fact A multiplied by B is precisely what DSBSC is. Folk use microprocessors to do it these days - digitize the audio, digitize the carrier, multiply together and ya get DSBSC whereas if you dc biased your audio so that it doesn't go negative you'll get regular broadcast AM. \$\endgroup\$ – Andy aka Aug 23 '13 at 23:16
  • \$\begingroup\$ Isn't there infinite harmonic energy generated at the exact zero point? And how can the frequency of the signal be the original carrier frequency; the new carrier signal now reaches two peaks in x time. Shouldn't that double the frequency? \$\endgroup\$ – Andyz Smith Aug 23 '13 at 23:16
  • \$\begingroup\$ But it DOES contain power instantaneously. But because it is so fast it cancels as soon as the next phase shifting zero comes? But because that power is harmonically related to the amplitude of the modulating signal? Because the entire RF envelope is harmonically related to the central carrier power? So there is actually power all around the carrier? \$\endgroup\$ – Andyz Smith Aug 23 '13 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.