true@elbereth.rutgers.edu (Dr. Destruction) (12/16/90)
I am working on a project which requires (among other things) the inversion of frequencies in a baseband of approximately 1.6kHz. Normally, I am aware, the easiest method of performing this would be to modulate the baseband signal (surpressed carrier of course) up to 1.6kHz and then sharply filter it at 1.6kHz, the result being the inverted spectrum from 0 to 1.6kHz. Currently, though, I am scrubbing the modulation and performing the inversion by sampling the baseband signal at 1.6kHz using natural sampling (ie: gating) with a gate width of approximately 10% of the period (at 1.6kHz). After this is done, the original baseband signal in inverted, scaled to account for the spectral energy loss due to sampling, and added to the sampled waveform. The result of this is as follows: the sampled signal is the typical spectral replication of the baseband (assume it to be bandlimited at 1.6k) at intervals of 1.6k, scaled by a rather wide sinc function (due to the convolution with the pulse train -> natural sampling). The baseband signal, however, is still present. So, in order to achieve the inversion, the remaining baseband signal is subtracted from the sampled signal, leaving only spectral replications at 1.6k and higher, which, after filtering, leaves the inversion and nothing else (theoretically that is :-) Ok, now: the problem is that it is *very* difficult to attenuate the inverted baseband signal to *precisely* the correct amplitude to cancel the residual baseband remaining in the sampled waveform. Failure to subtract the baseband signal (or subtracting too much for that matter) will of course leave traces of the baseband signal in the spectrum, which is highly undesirable for me. My question is that it seems that the result of what I am doing could be achieved as follows: 1) Gate (sample) the signal at 90% of the period 2) Invert it 3) Filter sharply at 1.6k This seems to be (thinking in the time-domain here) accomplishing the same thing as the 10% sampling and subtraction of the baseband. The plus is that I wouldn't need to subtract anything now! Do you agree with this?? I figured before I went and tried it I'd get a 2nd (or Nth) opinion. What problems can you see with this that I am not taking into account? Is there any reason a system such as this would be difficult to physically realize? Any feedback (no pun) will be appreciated! ----------------------------------------------------------------------------- Frederick True true@elbereth.rutgers.edu ftrue@attmail.att.com -----------------------------------------------------------------------------
jgk@osc.COM (Joe Keane) (12/19/90)
In article <Dec.15.11.52.08.1990.12495@elbereth.rutgers.edu> true@elbereth.rutgers.edu (Dr. Destruction) writes: >[explains usual method of frequency inversion] > >My question is that it seems that the result of what I am doing could be >achieved as follows: > > 1) Gate (sample) the signal at 90% of the period > > 2) Invert it > > 3) Filter sharply at 1.6k > >This seems to be (thinking in the time-domain here) accomplishing the same >thing as the 10% sampling and subtraction of the baseband. The plus is that >I wouldn't need to subtract anything now! Do you agree with this?? No, it's not subtracting the right amount of the baseband signal. >I figured before I went and tried it I'd get a 2nd (or Nth) opinion. What >problems can you see with this that I am not taking into account? Yes, the gated singal is too weak. Suppose you gate a signal with a 10% duty cycle, and then filter it. Then to get the signal back to its original amplitude, you have to multiply it by the reciprocal of the duty cycle, in this case 10. Of course there is still a wide sinc envelope, but at least the DC response is correct. >Is there any reason a system such as this would be difficult to physically >realize? No, but it won't do what you want. >Any feedback (no pun) will be appreciated! How about this. To do what you want, you should use a stream of alternating gating pulses. As long as the pulses have equal width and opposite amplitude, there is no baseband feedthrough. You can send my consulting fee now :-).