# Re{ } component of a phase defined function

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 $\sin(\omega t)$?

The definition is:

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

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.

• Oh! Okay. Got it. Another question... does each cosine have a different period then? And is the fundamental frequency w = 1? Commented Mar 12, 2015 at 4:16
• 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) Commented Mar 12, 2015 at 4:27
• Got it. Thank you. Commented Mar 12, 2015 at 4:35
• Quick nitpick, but when you take the imaginary part, all the $j$s should disappear. For instance, $Im\{2cos(\omega t) + j 3 sin(\omega t)\} = 3 sin (\omega t) \neq j 3 sin(\omega t)$ Commented Mar 12, 2015 at 4:35
• You're right, I corrected it. Commented Mar 12, 2015 at 19:34