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In the frequency domain a sinusoid is a "spike", an impulse function at the sinusoid's frequency. Theoretically at least, a single sinusoid has 0 bandwidth. In practice they actually have a small bu …

I would choose low sample rate, many cycles for a few reasons. Assuming the sine wave doesn't have harmonics, there is no information about the sine wave at the higher frequencies. If you do use a D …

I would do a normalized autocorrelation to determine periodicity. If it is periodic with period $P$ you should see peaks at every $P$ samples in the result. A normalized result of "1" implies perfec …

In general I'd say that you understand the differences pretty well. You nailed the first difference- continuous time vs. discrete time (samples). You were also correct about there being sets of indi …

I would structure the thesis like this: Introduction/Problem Statement (i.e. whatever problem BSS solves) How BSS solves that problem (high level) Theory behind BSS 1) Definition of Independence …

The steps that I would follow to handcraft a bandpass filter are these: Figure out how wide (in Hz) you want the filter to be. Create a low-pass filter with half the one-sided bandwidth as your desi …

I understand that when we take longer time windows and take the fourier transform we would suffer in the time resolution, But Why if we take a shorter time window we would suffer in frequency resol …

With a real cosine a phase shift is equivalent to a time delay. If you delay the signal by $\phi_0$ then you will get the result you want. The number of samples that you will need to delay by will l …

Yes, I believe that your understanding of energy compaction is correct. Does that mean, for any x, if A is DCT matrix, energy of y will be more concentrated when compared to the x. No, it does …

I prefer to think of it as expanding the spectrum due to increasing the sample rate. When you expand the frequency spectrum you reveal the aliases that were already there in the other Nyquist zones. …

The time lag is equal to the index difference times the sample period. The sample period is: $$ T_s = \frac{1}{f_s} = \frac{1}{16000} = 62.5\mu s $$ Thus, the time lag from the first peak to the seco …

Estimating the necessary filter order There are methods for estimating the needed filter order for low/band/high-pass filters, but the fact that you are talking about "colored noise" suggests that yo …