3

I understand what SSB and DSB mean in frequency domain.

For SSB, the signals appear at one side of the carrier frequency.

For DSB, the signals appear at both sides of the carrier frequency.

So now suppose the signal is a sinusoidal Acos(w1t) at baseband and the carrier frequency is a sinusoidal Bcos(w2*t) where w2 > w1. If these two signals are applied to an SSB mixer and a DSB mixer, then how the resulting SSB and DSB signals look like in time domain?

Underdog
  • 377
  • 2
  • 12

4 Answers4

4

If these two signals are applied to an SSB mixer and a DSB mixer, then how the resulting SSB and DSB signals look like in time domain?

If you have an upper sideband signal, you will have a sinusoid at the sum of the carrier frequency and the modulation frequency.

If you have a lower sideband signal, you mill have a sinusoid at the difference of the carrier frequency less the modulation frequency.

If you have a double sideband signal suppressed carrier, you will have the linear sum of the two cases; at practical ratios of carrier to modulation frequency you would not really be able to even notice that there was more than one frequency component there when examining on a timescale short enough to view the actual waveform.

Importantly, the amplitude of the modulated signal would strictly vary with that of the modulation - no input, no output. In contrast an "AM" modulator with carrier would show an output even without an input, with an envelope that would then vary depending on the input.

TL;DR the time domain view of SSB/DSB modulated signals is not interesting, you need a frequency domain instrument with logarithmic power (instead of linear amplitude) to see the kinds of things that really matter.

Chris Stratton
  • 33,491
  • 3
  • 44
  • 90
  • @Chirs Stratton Thanks for your reply. But may I know why you said "the time domain view of SSB/DSB modulated signals is not interesting". I think it is still necessary to check whether the output signal is distorted and whether there is enough voltage swing. Could you please further advise? Thanks. – Underdog Sep 12 '20 at 00:52
  • 2
    You'll see "distortion" in the log frequency domain much more readily than the time domain, where there simply isn't enough range to distinguish even fairly severe cases. – Chris Stratton Sep 12 '20 at 01:28
  • Thanks for your reply – Underdog Sep 12 '20 at 01:31
  • @ChrisStratton that's very true, but excuse my inexperience there, I've not seen one that doesn't simply produce or consume a complex baseband (or a full-band IF), making the mixing process independent of the signal's actual twosidedness? – Marcus Müller Sep 12 '20 at 15:16
  • But you sparked my curiosity: how/why would one employ an SSB receiver to demod FSK? – Marcus Müller Sep 12 '20 at 15:19
  • @MarcusMüller it might be useful to learn about these things before making inaccurate comments. While SSB had been known for a while, it didn't really start to take off until the end of the 1940's and really came into dominance in the 50's and 60's, and continues there today for efficient non-digital non-encrypted voice. In terms of receiving FSK with an SSB rig, you literally get FSK out of the audio port. Zero IF IQ largely makes sense in a DSP context; a sideband signal at a zero IF is not intelligble audio, it must be mixed back up after the zero-centered filter. – Chris Stratton Sep 12 '20 at 15:24
  • SSB modulation is more complicated than it seems because to cutoff the negative spectrum of a signal and then move it in frequency is a nontrivial operation and from a theoretical point of view can't actually be done in finite time. Same goes for demodulation.... – FourierFlux Sep 13 '20 at 18:36
  • @FourierFlux an SSB modulator with good suppression of the undesired sideband is indeed a bit tricky to build, which is part of why there were decades between when the technique became known and when it became popular. It's relatively more practical when you include the typical lower (as well as upper) limit on the frequency range of the modulation input. – Chris Stratton Sep 13 '20 at 18:43
1

@ChrisStratton has described time-domain properly. Since OP has asked what it looks like, here's an example of two sinusoidal waves after DSB modulator.

  • "Carrier wave" V(a) is 20kHz
  • "modulation" V(b) is 1kHz.

Calling them carrier vs. modulation is arbitrary - usually we consider the lower frequency to be modulated onto the higher frequency.
For DSB, note that there is a phase inversion of the carrier wave when the modulating signal passes through a zero-crossing. If a DC offset is added to the 1kHz signal (in this case \$\pm1V\$) then you end up with AM rather than DSB.
DSB sine modulationLTSpice schematic
It may look like there is significant energy at 20kHz: there is not. For this sinusoidal case, there is energy only at 19kHz, and an equal amount at 21kHz. Perhaps this is why time-domain plots of DSB or SSB are not particularly useful to your eye & brain.

  • If you pass this DSB waveform through a bandpass filter, whose passband includes 21000Hz, and whose stopband rejects 19000Hz, you get SSB: a sinewave @ 21000Hz.

  • If you pass this DSB waveform through a bandpass filter whose passband includes 19000Hz, and whose stopband rejects 21000Hz, you get similar SSB: a sinewave @ 19000Hz.

There are a few other ways to generate SSB than the filter method described above: phasing method, and FM (Weaver method).

glen_geek
  • 25,809
  • 1
  • 23
  • 53
  • You put a great effort in your answer, embracing PSD, phasing method and Weaver method, but stopped short of showing that the SSB waveform still carries a modulating signal. Why? This is not evident, considering your observation "if you pass this DSB waveform through a bandpass filter, ... you get SSB: a sinewave @ 21000Hz". – V.V.T Sep 14 '20 at 09:05
  • OP constraint is limited to cos modulation, and cos carrier.,,, For this specific case SSB output is a sinusoidal wave that appears unmodulated. You could measure no difference from the output of a simple crystal oscillator. I think it pointless to show a plot of a plain sinusoidal wave. The DSB signal (containing both sidebands) does appear modulated. – glen_geek Sep 14 '20 at 13:55
  • To communicate a useful and relevant knowledge, you could, for example, discuss this "OP constraint" in your answer before you did it in the comment. Also, are you sure you understood it right? The OP did not posit that A is real and constant and w1 is constant. – V.V.T Sep 14 '20 at 15:25
  • @V.V.T Yes, I've assumed "A" & "B" are constant, and don't vary with time. If they do, SSB becomes rather complex...running simulations is instructive, as you've suggested. – glen_geek Sep 14 '20 at 16:14
0

However it may seem unintersting to some experienced engineers, examining time-domain waveforms is very instructive as compared to reading a verbal description of those. Whether you capture the waveforms from a mixer device or plot the result of a simulation (in my opinion, you better start with simulation), you learn not just about how signals in question look, but also mixer circuitry and measurement techniques (if you're lucky to have access to lab technique) bzw. programming and debugging. There is nothing unwieldy in writing a simulation for any flavor of mixers and modulation kinds, in computer language of your choice. Also, you can examine the simulator output both in time domain and in frequency domain and compare your impression from both. To make your undertaking easier, the EE departments publish helpful handouts. See, for example, V3: DOUBLE SIDE BAND SUPPRESSED CARRIER MODULATION AND DEMODULATION By Laurence G. Hassebrook.

Still, if you are interested in leisurely watching pre-captured oscilloscope pictures, YouTube is a your service.

Any way, seeing is believing!

V.V.T
  • 4,154
  • 7
  • 10
-3

Let us examine AM versus PM.

AM has 2 equal_amplitude sidebands, in exact opposite phase, rotating in opposite direction.

By having exactly opposite phases, the orthogonal energy cancels, ensuring there is no Phase Deviation of the energy.

PM has 2 equal_amplitude sidebands, that rotation in OPPOSITE DIRECTION, that produce NO AMPLITUDE VARIATIONS.

To achieve NO AMPLITUDE Variations, the two sidebands must start 180 degrees apart.

analogsystemsrf
  • 34,253
  • 2
  • 19
  • 48