Some SSB theory help. - Page 1
Forum section: electronics / beginners

author: firewalker
author_rank: Super Contributor
timestamp: January 29, 2014, 03:33:25 pm
content: The last week or so I am trying to understand SSB modulation and demodulation. I have created the following simulation circuit.An 1 kHz audio signalM(2 Vpp).A carrier 10 MHz signalC(10 Vpp).The output of a balanced oscillator will be a product ofV(C)*V(M).The output after a crystal ladder 3 kHz BPF is the SSB signal.To demodulate the SSB signal we have to multiply it with the BFO's output (same frequency and phase with the carrier).Is everything correct up until now?Alexander.

author: firewalker
author_rank: Super Contributor
timestamp: January 29, 2014, 08:20:33 pm
content: As far LtSpice. Is my proccess correct? After the BFO?Alexander.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 31, 2014, 04:13:21 am
content: You have a pretty good handle on it,but I would suggest:Try to visualise the process in your head,write down the steps,hand draw a block diagram.The LTspice simulation schematic is messy & has additional junk which is not necessary for an understanding of the process.( But,then again,Ihatesimulations!)"To demodulate the SSB signal we have to multiply it with the BFO's output (same frequency and phase with the carrier)."In most SSB applications,the "BFO's" phase relationship is not important.In fact,it's funny you should mention BFOs as these were normally free running tunable oscillators incorporated in early Communications Receivers to provide an audible note when receiving keyed carrier Morse Code transmissions (commonly,if erroneously,called CW).With the advent of SSB,these were used to provide a local carrier to allow demodulation,which they did quite well,considering their simplicity.More modern designs use a fixed local carrier,& shift the received signal in the IF passband to adjust for best intelligibility & most natural voice.This local signal is still not in phase with the original carrier,though.It is possible to phase lock the local carrier by not reducing the original carrier to zero,but instead transmitting a reduced carrier.This is most often used in RF systems that use ISB--two sidebands with different information.PS:There is something weird with your last waveform--if this is after the demod,it should be the same as your first one--a 1kHz sinewave.

author: firewalker
author_rank: Super Contributor
timestamp: January 31, 2014, 11:23:05 am
content: Thanks. I finally got it clearly in mi head.I borrowed a spectrum analyzer. With my oscilloscope, some simple circuits with the LM1496 I build, some crystal filters and a bunch of signal generators I was able to understand what is happening.It is something that you can visualize only by simulating it.P.S. The weird wave form is due a mistake in my circuit.Alexander.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 31, 2014, 01:19:36 pm
content: "It is something that you can visualize only by simulating it".People have been visualising it for 70 years or so without simulators.Ultimately,you learnt more by making a real one!

author: W7NGA
author_rank: Regular Contributor
timestamp: January 31, 2014, 02:08:35 pm
content: The product of the signal with the BFO or Product Detector contains additional terms that must be filtered. In a receiver, those terms are above the audio range and the audio stage and our ears do the filtering. Using a simulation, I suspect you have to pick off the audio terms separately to yield an identifiable audio waveform.daniel

author: G0HZU
author_rank: Super Contributor
timestamp: January 31, 2014, 02:12:12 pm
content: The other way to look at it is to use the basic trig identities freely available in reference books for things like sinA.sinB etc to see the equations for a multiplier.You can then plot the equations in an excel spreadsheet and graph the results. You can 'filter' the equations by simply removing the unwanted terms from the excel spreadsheet. This is the same as filtering an unwanted mixer/multiplier term with an RF filter. eg filtering off the unwanted sideband or filtering off the unwanted (sum) product detector term.But you should end up with the same waveforms in excel that you would get with a simulation or a real test. However, you will also get some unwanted terms in the real tests because the components used will not be perfect.The reason I suggest this type of analysis (I feel it is better than the simulation or real tests) is because you can also analyse how to generate SSB using mixers in quadrature. This is known as the phasing method where you don't need the crystal filter to remove the unwanted sideband. The phasing method does it for you by rejecting the unwanted image (sideband) term. eg you can work out how image rejecting mixers work by simply looking at a few basic trig identities and then play with them in excel.

author: Rory
author_rank: Frequent Contributor
timestamp: January 31, 2014, 02:26:04 pm
content: Quote from: firewalker on January 31, 2014, 11:23:05 amWhat was the mistake in your circuit?

author: firewalker
author_rank: Super Contributor
timestamp: January 31, 2014, 02:42:19 pm
content: Quote from: Rory on January 31, 2014, 02:26:04 pmI hadn't put the SSB signal through a product detector. I had behavioral voltage source and just multiplied the SSB with the BFO.Alexander.

author: firewalker
author_rank: Super Contributor
timestamp: January 31, 2014, 02:48:05 pm
content: Quote from: G0HZU on January 31, 2014, 02:12:12 pmYes you are correct. I could also useOctave. But my problem wasn;t the math behind SSB. I wanted to clear in my head in order to teach to my student with simple words what SSB is. Now with the help the simple circuits and the lab's spectrum analyzer I believe that they will understand it.Alexander.