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Homework Help: Sampling Sinusoid Problem

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    See figure attached.

    2. Relevant equations

    3. The attempt at a solution

    I'm quite sure what he's getting at with this problem...

    The minimum sampling frequency would be 2 times the maximum frequency component of the signal. Since this is a sinusoidal signal, it should only have 1 frequency component, that is ωo which in our case is a mess of a's and k's.

    Can someone give me a nudge in the right direction, or clarify what needs to be done?

    Thanks again!

    Attached Files:

  2. jcsd
  3. Dec 11, 2012 #2


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    Staff: Mentor

    It's hard to tell, but I think the frequency is going to vary with time. That term with the square root of the sum of two squares is not going to be linear with time. I'd try to figure out if the frequency goes up or down with time, and then use the start time or the end time to calculate the 2x sampling frequency...
  4. Dec 11, 2012 #3
    I tried plotting the function within the sinusoid across values of t where I arbitrarily chose k = 1 and a = 2 and it is an exponential decreasing function for t>0.

    But where does that get me with start and end times?

    If I control everything, I could just put the start time where the frequency is almost zero and I would need an extremely small sampling frequency to sample the function, right?
  5. Dec 11, 2012 #4
    The question is really weird! First of all, sampling is generally defined for signals that exist for all time, yet you're told you only care about signals between [itex] t_o [/itex] and [itex] t_f [/itex]. Also, sampling is usually based on Fourier theory, but this signal doesn't have a Fourier transform if you're integrating it over all time.

    You could try to base your answer on the highest instantaneous frequency (i.e. twice that for unaliased sampling), but remember that instantaneous frequency is defined as the rate of change of phase, that is

    [tex] \omega = \frac{\text{d}\theta}{\text{d}t} [/tex]

    where, in your case,

    [tex] \theta = \frac{k}{2}\left(\sqrt{t^2+a^2}-t\right). [/tex]

    All in all, as berkeman indicates, I don't think the question is answerable without further background information.
  6. Dec 11, 2012 #5


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    Staff: Mentor

    So does that mean that the frequency is decreasing during the interval? Which end of the time interval gives the highest frequency? And so then what sampling frequency should you use?
  7. Dec 12, 2012 #6
    What I think you're suggesting is going to underestimate the bandwidth.

    Is this a question from a section on FM or PM modulation? My first thought was to use Carson's Rule to estimate the bandwidth. I'd also look at the effect of finite sampling interval -- that's like multiplying the function by a rectangle in the time domain and will further expand the bandwith if it's significant.
  8. Dec 13, 2012 #7
    You need to sample at [tex]2*\frac{k}{2}\left(\sqrt{t_{o}^2+a^2}-t_{o}\right)[/tex]

    Because since the frequency is decreasing, the maximum frequency will lie at to.
  9. Dec 13, 2012 #8

    rude man

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    In the last sentence of the problem I think they mean 'the rate of change of PHASE is constant ..." , not rate of change of FREQUENCY...'.
  10. Dec 15, 2012 #9
    Yes, I suspect this may the approach the question is designed to test. But, of course, Carnson's Rule only gives about 98% of the bandwidth - there's no indication in the question that this is enough.

    That's actually the instantaneous phase, not the frequency. To get the instantaneous frequency this needs to be differentiated with respect to time.

    Perhaps, although if the rate of change of phase is constant, so will the rate of change of frequency (it will be zero!).

    I'm convinced this question is not answerable. Consider an FM signal with peak frequency deviation of [itex] \Delta f [/itex]. This means the instantaneous frequency can change about the carrier frequency [itex] f_{\text{c}} [/itex] according to [itex] f_{\text{c}} \pm \Delta f [/itex], yet the bandwidth occupied by the signal is not [itex] 2\Delta f [/itex] as you might think, it is, in fact, infinte - even for the case of sinusoidal modulation.

    So unless there is some predefined and agreed approximation, or some short time Fourier, or even non-Fourier sampling theory assumed, I'll venture that the bandwidth of the signal is infinite, and hence no sampling frequency can possibly capture the signal without some finite energy loss.

    Another way to look at this is that the time domain signal is finite in duration, and hence, by the same idea as is used to derive Hesinberg's Uncertainty principle, the frequency domain signal must be infinite in duration i.e. time and frequency are Fourier conjugates.
  11. Dec 16, 2012 #10
    The bandwidth is infinite as you say but practically there is a bandwidth where most energy lies and hence the rules of thumb like Carson's Rule which I think is adequate to answer the question.

    The infinite bandwidth can be shown by representing the FM signal as the product of two exponentials:

    Vfm = cos([wc+km(t)]t)=cos[wct+ka(t)]
    where wc=carrier frequency, m(t)=message, and a(t)=∫m(t)dt (changed message to phase)



    ejka(t) can be expanded in a power series in a so you have a modulating carrier at wc multiplied by a power series in a, and each term of the power series is thus modulated to wc in the frequency domain. a(t) has the same bandwidth as m(t) so the power series terms a, a2, a3, ... have bandwidth B, 2B, 3B, ... and the result is the sum so the bandwidth is infinite.

    Carson's Rule comes from sampling m(t) at the the minimum frequency and representing m(t) using a staircase. Then the FM signal consists of rectangular bursts of sinusoids of duration equal to the sample period of m(t) and frequency equal to the carrier plus the constant amplitude of the sampled km(t) in that interval. In the frequency domain this is a sinc function (from the time domain rectangle) modulated to the frequency during the time period. The main lobe of the sinc is assumed to contain most of the signal energy and the summation of all these sincs leads to the bandwidth estimate.

    The question is further complicated by saying they would only sample the signal during the interval (to,tf). This is the same as multiplying the signal by a rectangle of width (tf-to) in the time domain and in the frequency domain this means convolving a sinc with the spectrum of the signal. This also makes the bandwidth infinite, if it wasn't already, but practically we can estimate the bandwidth of the sinc function as the width of the main lobe. So the sinc adds its main lobe width to the bandwidth of the function found with Carson's Rule. The longer the sample period (tf-to), the skinnier the sinc lobe and the less it adds to the bandwidth of the resulting signal. *** Note this assumes we want to try to reproduce the signal outside the sample interval and if we only try to reconstruct the signal in the interval (tf-to) then there is no multiplication by a rect in the time domain. Reproduction of a signal outside the sampling interval depends on finite bandwidth of the signal and is therefore not possible to achieve perfectly.

    So the question is arithmetically simple to answer since the equations are very simple. But the theory behind it is more sophisticated than it seems. That's why I was wondering what sort of text this question came from. The question's hint is actually wrong (and was in fact the same mistake early pioneers in FM made!) so maybe the question comes from an elementary signals text and the 'answer' they are looking for really is twice the highest instantaneous frequency, which is approximately correct for loud FM signals (Δf >> bandwidth of m).

    Also notice we've been talking about the bandwidth occupied around the central carrier frequency which is k/2. If there won't be any demodulation or bandpass sampling, then the highest frequency component will be k/2+B/2 and sampling must occur at twice that rate. (The other half of B is to the left of k/2)
    Last edited: Dec 16, 2012
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