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If sampling at a rate of 48 kHz, with no anti-aliasing filter on the input, and the input signal is contaminated by a 100 kHz sine, at what frequency would the sinewave signal appear in the output spectrum?

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    $\begingroup$ Sorry for asking, but this sounds like a homework question; if that is the case, can you show what have you been able to do so far, and be more specific about what you're having trouble with? In any case, you can find the answer here: en.wikipedia.org/wiki/Aliasing#Sampling_sinusoidal_functions $\endgroup$ – MBaz Apr 20 '15 at 23:45
  • $\begingroup$ It's a final exam question and we've only begun studying DSP recently. One approach has been that the first harmonic of the sampling frequency is 48kHz *2 = 96kHz then using the difference 100kHz -96kHz = 4kHz . $\endgroup$ – Donald Hawken Apr 21 '15 at 0:10
  • $\begingroup$ Wait, is the interference 100 MHz, or 100 kHz? In your first comment, it appears it is 100 kHz. $\endgroup$ – MBaz Apr 21 '15 at 0:13
  • $\begingroup$ Yes that's right. Correction :I would have thought that the 100 kHz contamination would yield a noisey waveform with a fundamental frequency of Fs and unpredictable harmonic content based on when the samples of the 100 MHz sine were taken. $\endgroup$ – Donald Hawken Apr 21 '15 at 0:22
  • $\begingroup$ so it appears you have answered your own question, Donald. $\endgroup$ – robert bristow-johnson Apr 21 '15 at 3:52
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In your case, since 100kHz is more than twice 48 kHz, it will be folded twice in the sampled version, then there are two pairs of positions: +/-4 KHz and +/- 44Khz.

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