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Finding the signal x(t) when given properties of fourier series coefficients

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose we are given the following information about a continuous-time periodic signal with period 3 and Fourier coefficients [tex]a_{k}[/tex]

    1. [tex]a_{k} = a_{k+2}[/tex]
    2. [tex]a_{k} = a_{-k}[/tex]
    3. [tex]\int_{-0.5}^{0.5}x(t)dt = 1[/tex]
    4. [tex]\int_{0.5}^{1.5}x(t)dt = 2[/tex]

    Determine x(t)

    2. Relevant equations

    if x(t) is a periodic input signal it can be expressed as

    [itex]x(t) = \sum_{k=-\infty}^{+\infty}a_{k}e^{jk\omega_{0} t}[/itex]

    which is called the fourier series of x(t), and [tex]\omega_{0}[/tex] is the fundamental frequency of x(t)

    also [tex]a_{k} = \int_{-\infty}^{+\infty} x(t) e^{-j \omega_{0} k t} dt [/tex]


    3. The attempt at a solution

    I've tried to solve this problem many times, when I was reading about it in the fourier series chapter.. I thought if I moved onto the next chapter which was fourier transform it would help me but it didn't, so here I am again..

    what I first found is that [tex]\omega_{0} = \frac{2\pi}{3}[/tex]

    hence [tex]x(t) = \sum_{k=-\infty}^{+\infty}a_{k}e^{jk\frac{2\pi}{3} t}[/tex]

    now from the second property we can understand that x(t) is an even signal... hence

    if

    x(t) <-> ak
    x(-t) <-> a-k
    x(t) = x(-t) => ak = a-k

    now from this we can simplify the summation

    [tex]x(t) = \sum_{k=0}^{+\infty}a_{k}(e^{jk\frac{2\pi}{3} t}+e^{-jk\frac{2\pi}{3} t})=\sum_{k=0}^{+\infty}a_{k} 2cos(\frac{k 2\pi t}{3})[/tex]

    now from the first property since [tex]a_{k} = a_{k+2}[/tex]

    we see from the frequency shifting property that

    [tex] x(t) = x(t) e^{\frac{-j 4\pi t}{3}} [/tex]

    good, we have 2 other properties

    from the third property since [tex]\int_{-0.5}^{0.5}x(t)dt = 1[/tex] then x(t) must be the dirac delta function in this interval, hence [tex] x(t) = \delta (t)[/tex] for [tex] -0.5<=t<=0.5 [/tex]

    and from the fourth property, again it's a shifted dirac delta with amplitude changed hence

    [tex] x(t) = 2 \delta (t-1)[/tex] for [tex] 0.5 <= t <= 1.5 [/tex]

    well I have these results which I hope to be correct, how can I use them in order to find the final x(t)?

    I've solved many such exercises but they were a lot easier.. there was parseval involved, some other properties, for example it was stated that x(t) was real and even etc.. this is the only exercise that I have difficulty to find a solution and I need to learn how to solve it because our professor likes these kind of exercises

    thanks in advance
     
    Last edited: Jul 21, 2011
  2. jcsd
  3. Jul 21, 2011 #2
    This is more of an algebra problem. As [tex]a_{k} = a_{k+2}[/tex]
    and [tex]a_{k} = a_{-k}[/tex] , two constants determine the series. (3) & (4) are precisely the equations which make this a problem of two equations in two unknowns.
     
  4. Jul 21, 2011 #3
    which are the equations?

    I mean, from the first two, we have equations about [tex]a_{k}[/tex] but these just tell us some properties about the frequency shifting and also about the symmetry of the signal..

    about the 3 and 4 I'm not sure at all, I could only figure out what the x(t) would be in a specific range.. now T is 3 and we know from -0.5 to 0.5 and from 0.5 to 1.5

    if I knew what happened from 1.5 to 2.5 then I would have the signal, since it's periodic with fundamental period 3
     
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