Sampling Discrete Time Signals

[@its__OVER]

Sir, at 6:30 if we place an ideal low pass filter, then we will get continous signal wrt to our initial discrete signal x[n]. Is that correct?

[@iain_explains]

Good question. I should have made that part clearer. If you convert the discrete-time sampled signal (bottom left) into a continuous-time waveform of impulses, and then put it into a continuous-time LPF, then you'd get the waveform you mention. But I was just talking about discrete-time filtering. A discrete-time filter with a sinc shape impulse response would recover x[n] from the discrete-time sampled signal (bottom left).

[@yuktikumari6042]

wow thank you

[@ethancooper4154]

I’m having a hard time wrapping my head around this one. If you take the first DT signal and just remove 2/3 of the samples, they can be recovered? Example: what if there are two DT signals - [1,2,3,4,5,6] and [1,4,3,9,5,4] - and we want to sample them both at N= 2. The resulting signals would both be [1,3,5], right? How could you differentiate the two in the frequency domain and why, if the later sampled signals are the same, would the have different frequency domain representations?

[@iain_explains]

Excellent question. OK, to start, let's note that the first signal you suggested is a very smooth signal (low pass, small w_M), whereas the second signal you gave has rapid variations from one sample to the next (ie. lots of high frequency components, large w_M). Intuitively, the second signal will not be able to be recovered, because its w_M will be bigger than pi (see 5.40 time mark in the video, but with N=2 in your case). OK so now to your question about them both having the same "sampled" FT because their samples are the same. Yes, that's true. So what it means is that ANY waveform that has the same values at sample times 0,2,4,6,... (in your case with N=2), will have a FT for which the excess part (above pi) will add (alias) to the in-band part (below pi) to give the exact same value as the smooth low pass signal FT. Pretty interesting huh? Great question. For more insights on aliasing, see my video: kzbin.info/www/bejne/eGTRi4h8g9CHfbs

[@ethancooper4154]

@@iain_explains Right, because for my second signal with the higher frequency components sampling at N = 2 is too high a period, and results in aliasing. The aliased signal cannot be used to recover the original signal. Thank you!

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