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Verify spectrum of the signal

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Make a sketch of the spectrum of the signal defined by:

    [itex] x(t) = \sum_{k = -3}^{3}\frac{1}{1+j\pi k}e^{j4\pi kt} [/itex]

    Use polar notation for the phasors on the plot, and sketch the frequency axis in Hz.

    2. Relevant equations

    3. The attempt at a solution

    [itex]a_{k} = \frac{1}{1 + j\pi k}[/itex]

    [itex]a_{-k}^{*} = a_k[/itex]

    [itex]a_1 = \frac{1}{1+j\pi} [/itex]

    [itex]a_2 = \frac{1}{1+j2\pi} [/itex]

    [itex]a_3 = \frac{1}{1+j3\pi} [/itex]

    Convert them to polar, I get the general form:

    [itex] \frac{1}{\sqrt{(k\pi)^2 + 1}}e^{-j arctan(k\pi)} [/itex]

    My frequency stems will be at -6, -4, -2, 0, 2, 4, 6 with the above amplitude for each of the corresponding k values.

    Can you double check my work, the arctan is makeing me question my results.
  2. jcsd
  3. Oct 8, 2013 #2

    rude man

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    This all looks good. The frequencies are +/- 0, 2, 4 and 6 Hz as you post.
    Did you remember a0/SUB] = 1?

    If all you're supposed to do is plot the magnitude spectrum, you'd be done.

    The exp(-jtan-1kπ) term is the phase expression. I.e. phase = tan-1(kπ) and you could plot that vs. the 7 frequency spots as well.
  4. Oct 8, 2013 #3
    Thank you for checking my work. I was concerned about the arctan, i had not come across that before.

    the other way i was thinking if we look at k=+/-1

    [itex]\frac{1}{1+j\pi}e^{j\pi t} + \frac{1}{1-j\pi}e^{-j\pi t}[/itex]

    Getting a common denominator

    [itex]\frac{2}{1+ \pi^2}e^{j\pi t} + \frac{2}{1+\pi^2}e^{-j\pi t} [/itex]

    for k=+/-2

    [itex]\frac{2}{1+4 \pi^2}e^{j4\pi t} + \frac{2}{1+4\pi^2}e^{-j4\pi t}[/itex]

    and for k = +/-3

    [itex] \frac{2}{1+9 \pi^2}e^{j12\pi t} + \frac{2}{1+9\pi^2}e^{-j12\pi t} [/itex]

    and yes, i did forget to mention k=0 if 0 with mag of 1.
  5. Oct 8, 2013 #4

    rude man

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    You will, many times I'm sure.

    Reason: a + jb = √(a2 + b2) exp[ tan-1(b/a)]
    in other words, this is how you change from cartesian to polar & back.
    The second line does not follow from the first. Following what I wrote above,

    1/(1 + jπk) = 1/√[1 + (πk)2] exp[j tan-1(-πk)]

    so 1/(1 + jπk) exp(j4πkt) = 1/√[1 + (πk)2] exp[j tan-1(-πk)] exp(j4πkt)

    = 1/√[1 + (πk)2] exp j[4πkt + tan-1(-πk)].

    When you do any arc tan function, remember arc tan(-b/a) is not the same angle as arc tan(b/-a)
    so don't leave the expression as exp[-j arc tan(πk)] as you did in your post #1. It should be exp[j arc tan(-πk)].

    As I said, arc tan(-πk) is the phase angle, and arc tan(-πk/1) and arc tan(πk/-1) are 180 degrees apart!
  6. Oct 9, 2013 #5

    rude man

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    See correction above.
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