2-1$ as opposed to $0$ to $N-1$?

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Oct 13, 2016 at 15:00	comment	added	robert bristow-johnson		@GrowinMan - when you have a continuous-time function $x(t)$ and you uniformly sample it by multiplying it by $$ s(t) = T \sum\limits_{n=-\infty}^{\infty} \delta(t-nT) $$, that will cause the spectrum or Fourier Transform of $x(t)$, which we call "$X(f)$" to be made into a periodic function of $f$. $$ \mathcal{F} \{ x(t) \cdot s(t) \} = \sum\limits_{k=-\infty}^{\infty} X(f - k f_s) $$ where $f_s = \frac{1}{T}$ that is discretely sampled in $t$ and periodic in $f$. and this can be turned around the other way.
Oct 13, 2016 at 7:36	comment	added	GrowinMan		Months later I'm reading this again and I'm still a little lost on the Math, so I'm going to try asking for clarifying questions. What do you mean by `the discreteness in one domain causes the periodicity in the other`
Mar 31, 2016 at 18:46	comment	added	robert bristow-johnson		what i am saying is that you do not know and you cannot know both the amplitude and the phase of a frequency component with frequency precisely at Nyquist. you can assume one and calculate the other based on that assumption. for all $\|k\| < N/2$, you know everything you need for each real sinusoid from the magnitude and angle of $X[k]$. but not at $\|k\| = N/2$. there, you don't know enough about the sinusoid to describe it completely. you know something about it, but not everything, and the reason is because of aliasing.
Mar 31, 2016 at 18:36	comment	added	Gilles		What I am saying is that for certain values of $k$ it is possible to know both the amplitude and the phase of $X[k]$. Which your statement seem to refute.
Mar 31, 2016 at 18:03	comment	added	robert bristow-johnson		forgot to qualify the rhetorical question i asked in the previous comment. other than $\theta = \pm \frac{\pi}{2}$ or $\pm \frac{3\pi}{2}$ or $\pm \frac{5\pi}{2}$, etc., do the samples come out differently as $\theta$ varies? so we have a term in the continuous-time signal: $$ \frac{X[N/2]}{N \cos(\theta)} \cos(2 \pi (f_\text{s}/2) t + \theta) $$ at precisely the Nyquist frequency. if you know the angle $\theta$, then you know the amplitude $\frac{X[N/2]}{N \cos(\theta)}$, or vise versa. but you don't know both.
Mar 31, 2016 at 17:54	comment	added	robert bristow-johnson		sorry, you're wrong on all counts. if $x[n]$ is real, then we know that $X[N/2]$ and $X[0]$ are real as well. but they can be positive or negative. how does that affect their angles? also try $$ \begin{align} x(t) & = \frac{X[0]}{N} + \frac{X[N/2]}{N \cos(\theta)} \cos(2 \pi (f_\text{s}/2) t + \theta) \\ & \quad + \frac2N \sum\limits_{k=1}^{\frac{N}{2}-1} \|X[k]\| \cos(2 \pi (k/N) f_\text{s} t + \arg\{X[k]\}) \end{align}$$ where $t = n/f_\text{s}$. do the samples come out differently as $\theta$ varies? this is why there is aliasing unless the sample rate exceeds twice the bandwidth.
Mar 31, 2016 at 8:23	comment	added	Gilles		I don't agree with your statement that at $k=-N/2$ "you cannot know both the amplitude and phase" for real-valued sequences with $N$ even. Not only $\angle X[-N/2]=0$, but also $\angle X[0]$, $\angle X[N/2]$, and $\angle X[\pm N]$. And you also know their amplitudes.
Mar 31, 2016 at 6:23	history	edited	robert bristow-johnson	CC BY-SA 3.0	added 105 characters in body
Mar 31, 2016 at 6:19	comment	added	robert bristow-johnson		$X[N/2]$ could be negative. then the angle is not zero. but the sinusoidal component at Nyquist could have virtually any phase, as long as the amplitude is adjusted so that it hits the same alternating points. a pure $\sin()$ term is not possible at Nyquist if $X[N/2] \ne 0$.
Mar 31, 2016 at 6:10	history	edited	robert bristow-johnson	CC BY-SA 3.0	fix mistake
Mar 31, 2016 at 5:29	history	edited	robert bristow-johnson	CC BY-SA 3.0	added 1018 characters in body
Mar 30, 2016 at 22:24	history	edited	robert bristow-johnson	CC BY-SA 3.0	added 287 characters in body
Mar 30, 2016 at 21:50	history	edited	robert bristow-johnson	CC BY-SA 3.0	added 287 characters in body
Mar 30, 2016 at 21:36	history	edited	robert bristow-johnson	CC BY-SA 3.0	added 1148 characters in body
Mar 30, 2016 at 21:18	comment	added	Gilles		For real-valued sequences with $N$ even, you know both $\left\|X[N/2]\right\|$ and $\angle X[N/2]$. And $\angle X[N/2]=0$.
Mar 30, 2016 at 20:50	history	answered	robert bristow-johnson	CC BY-SA 3.0

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