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Homework Help: [Complex exponential] Solutions must be wrong

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data

    I would like to get feedback if my z-plot is accurate for the following complex exponentiala:
    Code (Text):

    Further analysis:
    a= -2 because cos(∏) = -1 and sin(∏)=0
    b= actual complex number A*cos(∅)+j*sin(∅)
    c= -1 because cos(∏) = -1 and sin(∏)=0
    d=+j because -j*cos(∏)= +j and -j*j*sin(∏)=0

    However according to solutions from other source my analysis is wrong.

    I've decided to plot both graphs in MatLab so you can see the difference:
    (my solution on the left)/(solution from other source on the right)


    As stated in one, theory doesn't agree with solution found on the internet.
    Also, in whatever form I put the exponential in the function I keep getting totally different results.
    [2*exp(j*pi), 2*exp(j*pi*-1.25), 1*exp(j*pi), -j*exp(j*pi)] would not give me similar amplitude as above for some reason.

    P.S. Ignore the R{1} on the left graph.
  2. jcsd
  3. Nov 1, 2012 #2
  4. Nov 1, 2012 #3

    Ray Vickson

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  5. Nov 1, 2012 #4
    That make sense, however, so if e.g. 2 is multiplied by ejωt where ω is pi rad/sec then the result of this exponential is -1 (considering only real part of this complex exponential) thus the resulting value of this exponential is -2 but solution says 2.

    The first graphs (subplot 1 and 2) shows a synthesis of those 4 exponential using MatLab function which is based on this formula (sinusoidal synthesis):

    The question or the form of this exponentials is represented as follows in textbook:
  6. Nov 1, 2012 #5
    At what time are you supposing ej∏t is equal to -1? I'd agree with you it equals -1 at time t=1, eg, but at time t=0 it equals +1. By factoring out the time component we can plot the constant part of the phasor in the Z plane. If A = 2 ejwt and we agree to draw a diagram for t=0, then A will be plotted at (2,0) as in the solution. To find A at other times you have to multiply this position by ejwt, the part we factored out. This has the effect of rotating the point A counter clockwise by wt radians. For time t=1 and w=∏, eg, we would multiply 2 by ej∏ to find at time t=1, A=2*(-1) = -2 as you have been saying.

    On the phasor diagram, you can add those plotted points as vectors to find out what overall sum of the signal will be, then multiply by ej∏t and take the real part for the sinusoid.

    But I think your discrepancy is coming from evaluating ej∏t at time t=1 instead of t=0 if I've understood correctly.
  7. Nov 1, 2012 #6
    Just realized right now that I was over-thinking about that. Very basic, thanks for answer.
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