mysterious equation e^jwt

1. 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. •

2. 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.... •

3. 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 •

4. e(jwt)

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

But, how is it souce code? 5. 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..... --[[ ]]--