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Help understanding magnitude and phase of sinc functions?

  1. Mar 1, 2014 #1
    help understanding magnitude and phase of sinc functions??

    1. The problem statement, all variables and given/known data
    prob1.png

    2. Relevant equations
    pτ(t) ⇔ τsinc([itex]\frac{τω}{2\pi}[/itex])
    Δτ(t) ⇔ [itex]\frac{τ}{2}[/itex]sinc2([itex]\frac{τω}{4\pi}[/itex])
    sinc(x) = [itex]\frac{sin(\pi x)}{\pi x}[/itex]

    3. The attempt at a solution
    My instructor had posted solutions to the problems:
    prob2.png

    I understand how to get the fourier transforms, I was just having problems finding magnitude and phase graphs. For a) I don't know how they got amplitude near 1.5, but I know that in general for a single sinc function sinc(x), you will have zeroes at x-1. I also know that in general phase graphs of single sinc function will be +180 deg (doesnt really matter what the sign is, I guess) when you have zeroes. I just don't understand how this works when there are multiple sinc or sinc2 functions. And why the phase graphs are all sawtooth etc., when I was expecting just + 180deg.
     
    Last edited: Mar 1, 2014
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  3. Mar 1, 2014 #2

    vela

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    The way it's expressed, it's not easy to see how the amplitude and phase vary with ##\omega##, so you want to rewrite it:
    \begin{align*}
    X(\omega) &= -\mathrm{sinc }\left(\frac{\omega}{2\pi}\right) e^{-j\omega/2} + \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) e^{-j3\omega/2} \\
    &= \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) (e^{-j3\omega/2} - e^{-j\omega/2}) \\
    &= \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) (e^{-j\omega/2} - e^{j\omega/2}) e^{-j\omega} \\
    &= \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) \left[-2j \sin\left(\frac{\omega}{2}\right)\right] e^{-j\omega} \\
    &= 2\mathrm{sinc }\left(\frac{\omega}{2\pi}\right) \sin\left(\frac{\omega}{2}\right) e^{-j(\omega+\frac{\pi}{2})}
    \end{align*} The factors out front are real, and the complex behavior arises from the exponential. Now it should be clearer that how you can get a maximum amplitude near 1.5. Also, you can see how the phase varies with ##\omega##. Do you understand why there's a discontinuity in the phase at ##\omega=0##?

    This is example is relatively easy to analyze, but sometimes you're just going to have to brute force it and resort to figuring out the real and imaginary parts of X separately.
     
  4. Mar 1, 2014 #3
    I am guessing the discontinuity at ω=0 is because of the sin function, sin(0)=0. In the phase at ω=0 is probably because of the exponential, there is imaginary value j. I still don't understand amplitude of 1.5 though (don't know how the exp changes amplitude/phase graph). If I set w=[itex]\pi[/itex] (since thats where it seems amplitude is highest) and do sinc-> sin I have:

    = 2(2sin([itex]\frac{ω}{2}[/itex]))[itex]\times[/itex]sin([itex]\frac{ω}{2}[/itex])[itex]\times[/itex]e-j[itex]\frac{3\pi}{2}[/itex]
    = j4sin2([itex]\frac{ω}{2}[/itex])

    but maybe I'm doing this wrong? I'm also wondering how the e-j(w+[itex]\frac{\pi}{2}[/itex])causes the sawtooth curve in the phase graph.
     
  5. Mar 1, 2014 #4

    vela

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    Why would the fact that sin 0 = 0 imply the existence of a discontinuity?

    This sentence makes no sense.

    You need to review polar notation of complex numbers. You seem to be missing some pretty basic facts about complex numbers, which is probably why you're so confused.

    I'm not sure why you're replacing sinc with sin, and if you're setting ##\omega=\pi##, why does ##\omega## still appear in the final line?

     
  6. Mar 1, 2014 #5
    I guess I was trying to replace sinc with sin because to understand it easier, my book shows how a basic sinc function looks but I guess i was confused because at ω=0 in this problem it is 0, however in the book the amplitude is highest at ω=0 for sinc. so I thought this could be because of multiplying by sin, at ω=0 sin is 0 so the entire eq. would be 0 (maybe I don't understand it at all). I think I understand the exponential, its magnitude would be 1, its phase is everything after the j. But I guess I am confused because I don't know how curve of e^-jω would look, my calc won't let me graph that because it has imaginary parts. so I am unsure how it affects both graphs.

    I think the main trouble I am having: how does that phase in exp (ω+[itex]\frac{\pi}{2}[/itex]) change the phase graph so its look like sawtooth instead of horizontal lines at 180. If I had to guess it is subtracting and adding 90 deg so that is why it looks like that. Also at ω=[itex]\pi[/itex] it seems an amplitude of 1.5 (I still don't understand this). If I plug in [itex]\pi[/itex] as ω for the exponential I get 'j' which I am unable to graph.
     
    Last edited: Mar 1, 2014
  7. Mar 1, 2014 #6

    vela

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    Why don't you write down expressions for the amplitude and phase of ##X(\omega)##?

    Why would the phase be a horizontal line at 180 degrees? That would imply that X(ω) is a negative real number.

    I'm not sure what your confusion is about the amplitude.
     
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