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11th August 2007, 16:01 #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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11th August 2007, 16:24 #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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11th August 2007, 18:05 #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 xaxis).
It's obvious, that any substitution of e doesn't change the sense.
With respect,
Dmitrij

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15th August 2007, 13:29 #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?

15th August 2007, 17:50 #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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