Nyquist Frequency Phase Shift

Question

The figure below shows in dashed lines sinusoidal signals of the same frequency at three different phase shifts. The signals are then sampled such that the sinusoidal frequency is exactly a half of the sampling frequency, i.e. the frequency of all the sinusoids is the Nyquist frequency. The samples taken from this signal are represented by the circles.

From the figure, it seems that the amplitude of the digital sinusoidal signals is dependent upon the sampling rate and instant of sampling. In fact if the sampling times coincide with the zero-crossings of the sinusoid, then no signal will be detected at all.

I had initially thought that sampling a bandlimited analog signal at the appropriate sampling frequency would enable perfect reconstruction, but this counter-example has left me stumped. It seems that this sinusoid will generally not be reconstructed properly if digitized and then reconstructed at this rate. Have I gone wrong in my understanding, and if so, can someone please point me in the right direction?

Cheers!

Answer 1

The sample rate needs to be GREATER than (NOT just equal to) twice the highest non-zero frequency content of the signal being sampled. Just a little bit greater might work, but the closer the sample rate is to twice the signal frequency, the longer in time you may need to sample to raise the signal above the noise and complex conjugate image in a DFT/FFT result.

Answer 2

The sampling theorem states that $f_\mathrm{S} \geq 2f_\mathrm{max}$, where $f_\mathrm{S}$ and $f_\mathrm{max}$ are the sampling and maximum signal freuqency, respectively. But there's an additional condition: The equal sign only holds if the signal spectrum does not contain a dirac impulse at $f_\mathrm{S}/2$ which is clearly the case in your example. Therefore $f_\mathrm{S} > 2f_\mathrm{max}$ has to be fulfilled.

A reference for this statement can be found here, for example:

If a signal contains a component at exactly B hertz, then samples spaced at exactly 1/(2B) seconds do not completely determine the signal, Shannon's statement notwithstanding. This sufficient condition can be weakened, as discussed at Sampling of non-baseband signals below.

More recent statements of the theorem are sometimes careful to exclude the equality condition; that is, the condition is if x(t) contains no frequencies higher than or equal to B; this condition is equivalent to Shannon's except when the function includes a steady sinusoidal component at exactly frequency B.

The above is an excerpt of an earlier version of the according Wikipedia article.

Answer 3

By definition, band-limited signals in the sense of the sampling theorem have finite energy. Sine waves are periodic and thus have infinite energy. So any dirac pulse in the Fourier transform is not permissible.

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To be more precise, the sampling theorems only applies to signals that can be represented as

$$x(t)=\int_{-f_s/2}^{f_s/2} X(f)\,e^{2\pi i\,ft}\,df$$

with $X\in L^2$. In the class of $L^2$ functions, values at specific points do not matter, so the values of $x(t)$ do not depend on the specific values of $X(\pm f_s/2)$.

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Any realistically occuring signal $x(t)$ is of finite length and thus has a continuous Fourier transform. For any approximation of "band-limited", the Fourier transform $X(f)$ needs to have negligible values at $\pm f_s/2$ and beyond.

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Addendum: Any finite signal that looks for some time segment like a sine wave $\sin(2\pi ft)$ and sports a sufficiently smooth fade-in and -out has smooth peaks in its spectrum around the frequencies $\pm f$. Outside the peaks, the amplitude goes towards zero, but never does reach it to stay at zero. Thus one has to define a threshold of where the amplitude is for practical purposes identical to zero, absolute or relative to the peak value of the amplitude. But that leads to the frequency contents of the signal to be contained in between $\pm (f+h)$, so that even the strict rule requires a sampling frequency of at least $f_s=2(f+h)>2f$.

Answer 4

again, even with the Addendum, i think Lutz's answer misses the point. the point is (quoting Wikipedia):

To illustrate the necessity of $f_s \ > \ 2B$, consider the family of sinusoids (depicted in Fig. 8) ) generated by different values of $\theta$ in this formula:

$$x(t) = \frac{\cos(2 \pi B t + \theta )}{\cos(\theta )}\ = \ \cos(2 \pi B t) - \sin(2 \pi B t)\tan(\theta ), \quad -\pi/2 < \theta < \pi/2.$$

With $f_s = 2B$ or equivalently $T = 1/(2B)$, the samples are given by:

$$x(nT) = \cos(\pi n) - \underbrace{\sin(\pi n)}_{0}\tan(\theta ) = (-1)^n$$

regardless of the value of $\theta$. That sort of ambiguity is the reason for the strict inequality of the sampling theorem's condition.

Fig. 8

Answer 5

Your observation is correct, and it has been noted before. One example, for instance, is mentioned on pages 160-161 of "Principles of Communication Systems" by H. Taub and D. Schilling, McGraw-Hill Book Company, 1971. After introducing the sampling theorem, the authors state:

"An interesting special case is the sampling of a sinusoidal signal having the frequency Fm. Here, all the signal power is concentrated precisely at the cutoff frequency of the low-pass filter, and there is consequently some ambiguity about whether the signal frequency is inside or outside the filter passband. To remove this ambiguity, we require that Fs (be greater than) 2Fm rather than that Fs (be greater than or equal to) 2Fm. To see that this condition is necessary, assume that Fs = 2Fm but that an initial sample is taken at the moment the sinusoid passes through zero. Then all successive samples will also be zero. This situation is avoided by requiring Fs (be greater than) 2Fm."

Answer 6

I have an analog sine wave generator that outputs 17kHz. And I am sure that it is realistically occurring. If I wanted to use this signal for some sound effect for a youthful audience, for, say, a finite twenty periods, and the first sample exactly coincides with a zero crossing, then I'm pretty sure that it is not going to reach its full amplitude in the reconstruction. So much for faithful reconstruction, I might say, even at a signal frequency well below half the sampling rate. Seems like my old open reel would do a better job than a 44k1 sound adapter.

Answer 7

And that, rbj, seems to be answering OP's question. It is the 'ringing' of the reconstruction filter that fills the 'holes' made by the 'beat' of the sampling rate (to put it mildly simplistic). The samples still coincide with the frequency of the signal to be reproduced, so the filter keeps getting energy at that frequency. And, in an ironic contrast with previous assertions regarding 'infinite signals', it would actually be impossible to correctly quantize a signal that just 'jumps from the origin' for a few samples, as such a signal has infinite bandwidth. So OP's example cannot be the whole signal. Formulas for bandwith used in amplitude modulation can explain.

Answer 8

I think the OP is on to something here. The complete parameterization of a band-limited signal of bandwidth B will be 3-dimensional surface (magnitude as a function of complex freq). Sampling at 2B yields only a 2-dimensional surface (magnitude as a function of 'real' freq). Will there not be overlap in the mapping between the two, and thus loss of information? $S(f) \{f \in C \} \rightarrow S(|f|) \{f \in C \}$?

It appears that to completely capture the signal the I and Q components need to be sampled.