1
\$\begingroup\$

I have came across a question where it asks if the input to SSB (LSB) modulator is mh(t) (which is the hilbert transform of m(t)), will the of the modulator be LSB?

What I did :

I did the math and came up with an output of mh(t)cos(wct) - m(t)sin(wct) . Is this the SSB with a carrier of sin instead of cos?

This was my conclusion after trying substituting with diffrent m(t) - is it true?

\$\endgroup\$
1
\$\begingroup\$

No.

But it's also a bad question.

There are fundamentally two differences between upper and lower sidebands:

  • in upper sideband the energy is above the carrier, in lower sideband it is below

  • in upper sideband, modulation components of increasing frequency fall at increasing output frequency, but in lower sideband, modulation signals of increasing frequency fall at decreasing frequency.

So there are two parts to the question:

  1. Can any change to the input cause an SSB modulator to output energy in the wrong sideband? Answer: not if the modulator is functioning properly.

  2. Does the Hilbert transform invert the spectrum of a signal: Answer: no, it does not by itself do that.

Although this does not do so, if you had something which did invert the spectrum of the modulating signal, then while you wouldn't be producing the wrong sideband, you would be producing an "upside down" sideband, so if you changed the sideband mode of the receive and then re-tuned it such that the other sideband fell into the filter, then actually if you got things just right you could recover the original, pre-inverted signal.

In practical terms, SSB gear is often tuned in terms of the center of the sideband energy filter, not the frequency of the supressed carrier. So set your upper/lower switch wrong, then audio of shifted+inverted spectrum is exactly what you get - a sort of garbled sound quite familiar to Ham radio operators. But remember, what the Hilbert transform does by itself is a bit more subtle than invert the frequency spectrum.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ so what i understand is that if we designed it to output LSB of our m(t) then we decided to input the mh(t) instead of m(t) we will get the an inverted version of LSB. true ? and in the receiver we can generate simply sin instead of cos to recover it .. your explainition helped me a lot just still mising those points @Chris \$\endgroup\$ – Mhd Ghd Jul 15 at 18:30
  • \$\begingroup\$ No, because as explained above, a) nothing can make a proper SSB modulator output energy in the wrong place and b) the Hilbert transform does not by itself invert the spectrum. To "generate sin instead of cos" is meaningless, it's not like the transmitter and receiver are phase locked or have any actual reference of phase. \$\endgroup\$ – Chris Stratton Jul 15 at 18:43
2
\$\begingroup\$

You need to show this is the LOWER sideband.

Compare this, with the full math description of an AM signal.

Examine the AM signal, extract the carrier, the upper sideband, and the lower sideband.

Think of the two sidebands as Phasors, rotating, with Inphase and with Quadrature energy components.

Also the phasors may or may not be phase-aligned.

DRAW A POLAR PLOT of each phasor, and add arrows onto the phasor tip to define the direction of rotation; also become clean when the angles are identical and when angles do not agree.

Then decide if your trigonometric result is what you desire.

===========================

Decades ago I taught a Transmitter Design newbie to implement GSM (GMSK) cellphone modulation. Turns out the GMSK (Gaussian minimum shift keying) is a precision single_sideband system, to avoid wasting energy in a carrier. Energy is much better used in reducing Bit Error Rate by improving the data_eye.

The SSB generation method used 0/90 oscillator and 0/90 modulation (4 signals) combined in 2 multipliers and then linearly summed.

Once we cleaned up the onchip capacitive parasitics, the Design produced 50dB carrier and 40+ dB unwanted_sideband suppression. Life was good.

==================================

the only difference between AM and PM is the phasors underlying the two sidebands.

  • AM sidebands are identical in energy (so the trigonometry tells us), but the phasors in the two sidebands rotate in OPPOSITE directions, meeting only at top (90 degrees on a polar plot) and bottom (270 degrees). This rotation in opposite direction results in NO PHASE DEVIATION but plenty of amplitude variation.

  • PM sidebands are identical in energy, but the 2 sidebands are 180 degrees out of phase and rotate in the SAME direction. Result (of being 180 degrees out of phase) is that any inphase energy gets cancelled (thus no amplitude variation), but the orthogonal energy gets to add and become a useful phase variation.

So -------------------- how does ONE cause AM_to_PM conversion? Notice ANY low_pass_filtering (which ANY circuit will provide) results in some residual PM being generated, because the two sidebands no longer have IDENTICAL power. And the slight delay in any circuit, and the slightly different delay between USB and LSB, causes a PhaseShift between our two phasors, so the Quadrature energies no longer exactly cancel.

Thus for highest quality modulation, you need wideband circuits that construct the components of the trigonometric behaviors.

===================================

Let us again view the generation of signals, at least single_tone signals, as an exercise in Phasors, where the Inphase and the Quadrature energies are crucial, and the phase tracking (or precise phase_anti_tracking --- exactly opposite ) is crucial.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ i tried to modulators one was the input is mh(t) and carrier cos while the second input m(t) and carrier sin i got the same output ... but i am a little but not comfortable to have sin as carrier in SSB @anakogsystemsrf \$\endgroup\$ – Mhd Ghd Jul 15 at 17:22

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.