Ssb - question

Hi everybody

Im trying to make SSB modulation with Hilbert transform. And Im using IQ
modulator
here is schema:
http://postimg.org/image/gms5srret/

And I have a question. In Frequency domain there should be one peak, and
on
on my plot there is one - that is good, and second that is
smaller/suppressed.

Is something wrong ?? or is it normal ??

Is Hilbert method for creating SSB signal is good ?? or can it be made
easier ???

thank you very much.
Przemek L.

Sorry I didnt attached second plot.

http://postimg.org/image/h0y3640w5/

2015-09-08 14:48 GMT+02:00 Przemek L. [email protected]:

Your first problem is that you’re using a 32k sample rate with a 20k
sin/cos. This will actually come out as a 12k sin/cos, due to aliasing.
Try the same thing again with a 48k sample rate and see what happens.

Either bandpass filtering or a Hilbert transform can be used to create
SSB.

Not sure what your flowgraph does, but it looks fun.

Jeff.

Unless you’re building the IQ modulator with real signals for
educational purposes, go with the flowgraph Ron mentioned.
Multiplication by a complex sine is much easier. You can also use a
Rotator or Frequency Xlating Filter to do similar things. The bandpass
filter can be built in to the Xlating Filter.

On the additional peaks, the second highest one is about 50 dB lower
than the desired one. From a real hardware point of view, it might as
well not be there (same as 1mW vs 100W). If you use a bandpass filter,
you can make it arbitrarily small, but filtering should be done based on
requirements, not just to make a plot look nice.

  • Jeff

There is an easier way. A DSB modulator and bandpass filter. You can
download a tested flow graph here:

http://www.w6rz.net/ssb.grc

and the test audio file:

The flow graph is set up for LSB. For USB, just swap the Band Pass
Filter block with the one that’s disabled.

The 10 kHz signal source and multiply blocks are optional. I use this to
move the TX signal away from the DC offset spike of my receiver.

Original flow graph is here:

Ron