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I'm attempting to evaluate $$f(t) = 2Re\left[(e^{j\omega t}) + \left(2e^{-j2\omega t}\right)\right] - 2$$

Would $$f(t) = 2\cos(\omega t) + 4\cos(2\omega t) - 2$$?

What would the Im[] of the same function be? When will I resolve to using \$\sin(\omega t)\$ instead of cosine? Is the \$e^{-j\omega t}\$ not \$\sin(\omega t)\$?

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The definition is:

$$ e^{j\phi}=\cos(\phi)+j\sin(\phi)$$

and your solution is correct.

What exactly do you mean by

What would the Im{} of the same function be?

Changing Re() to Im() inside the function gives $$2Im(e^{j\omega t}+2e^{-2j\omega t})-2=\{2\sin(\omega t) + 4\sin(-2\omega t)\}\rlap{\backslash}{j}-2$$

On the other side, the original f(t) is real, so Im(f(t))=0


EDIT: Had an extra j in the last equation. The result is real and does not contain any j.

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  • \$\begingroup\$ Oh! Okay. Got it. Another question... does each cosine have a different period then? And is the fundamental frequency w = 1? \$\endgroup\$
    – Omar Ayala
    Commented Mar 12, 2015 at 4:16
  • \$\begingroup\$ First of all, w is the angular frequency, not just the frequency: w=2*pi*f. The "fundamental" of your function is w, as it is the lowest "frequency", but it is not w=1. It is, what ever value is assigned to w.The right term has a "frequency" of 2w. (The quotes mean that there's that 2*pi missing) \$\endgroup\$
    – sweber
    Commented Mar 12, 2015 at 4:27
  • \$\begingroup\$ Got it. Thank you. \$\endgroup\$
    – Omar Ayala
    Commented Mar 12, 2015 at 4:35
  • \$\begingroup\$ Quick nitpick, but when you take the imaginary part, all the \$j\$s should disappear. For instance, \$Im\{2cos(\omega t) + j 3 sin(\omega t)\} = 3 sin (\omega t) \neq j 3 sin(\omega t)\$ \$\endgroup\$
    – Shamtam
    Commented Mar 12, 2015 at 4:35
  • \$\begingroup\$ You're right, I corrected it. \$\endgroup\$
    – sweber
    Commented Mar 12, 2015 at 19:34

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