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  1. #1
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    e^jwt

    As you know, e^jwt=cos(wt)+j*sin(wt)
    I can plot its diagram by mathlab.
    And get the result of a sinusoidal.
    But, when I try another numers, 2 or 3 instead of e in above equation.
    The result is still like sinusoidal.

    How to explain it?

    j is key point, and j=sqrt(-1).
    It is really beyond my imagination.

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  2. #2
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    e jwt

    if you visualise the numbers u substitute say 2 or 3 as e^k you will realise that you are just multiplying e^jwt by a constant and hence you are just multiplying a sine wave with a constant and hence will obtain a sinusoid only....



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  3. #3
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    mysterious equation

    Well, this question, you are anxious about, is not so strange as it probably seems to you. The explanation of this circumstance is the followin:

    1) Remember the main logarithmic identity, studied in the school course of algebra:

    e^(ln(a)) == a, if a>0. Now let's apply this expression to your example:

    2^(jwt) = e^(ln(2^(jwt))) = e^(iwt*ln(2)) = exp(jwt*ln(2)).

    2) Now we simply adjust the Euler formula, which you mentioned, to exp(jwt*ln(2)) and get the following:

    exp(jwt*ln(2)) = cos (wt*ln(2)) + j * sin(jwt*ln(2)).

    So as you see again we obtained harmonic functions, which represent real and imaginary parts correspondingly. The 90 degrees phase shift preserves, the only difference - is that these functions are scaled (compressed, respectively the x-axis).

    It's obvious, that any substitution of e doesn't change the sense.

    With respect,

    Dmitrij



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  4. #4
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    e(jwt)

    When I type 2^(sqrt(-1)) in mathlab, I can get sinusoid figure.

    But, how is it souce code?



  5. #5
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    e^-jwt

    well it is just based on the individual function i.e.sqrt and the exponentiation(^)

    the code of can be seen by using help.....



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