Complex Numbers converting from Polar form to Acos(wt + x)

1. Sep 5, 2014

PenDraconis

1. The problem statement, all variables and given/known data

"Put each of the following into the form Acos(ωt+θ)..."

(a.) 4ejt+4e-jt

2. Relevant equations

Euler's Identity: e = cos(θ)+jsin(θ)
Phasor Analysis(?): Mcos(ωt+θ) ←→ Me
j = ej π/2
Trignometric Identities

3. The attempt at a solution

I attempted to use phasor analysis to simply have it be 4cos(ωt)+4cos(ωt), however, the given problem doesn't seem to follow the same form as the "examples" for phasor analysis.

I also attempt to use Euler's identity (thus changing it into the cos()+jsin() form) but that still confused me because the example given wth Euler's is for a e not an ejt.

I basically need a solid jump off point so I can correctly reason the answer out.

Last edited: Sep 5, 2014
2. Sep 5, 2014

Simon Bridge

No it doesn't. In the phasor form, the expression is the sum of two vectors - the resultant vector is what you want to express in cos form.
Careful - the minus sign is important.

... put $e^{j\theta}=e^{jt}$ and solve for $\theta$.

3. Sep 5, 2014

PenDraconis

So, basically $j\theta=jt$, thus $\theta=t$?

So then we'd have $4e^{j\theta}+4e^{-j\theta}$?

How do I account for the negative then?

If that's the case (and $Mcos(ωt+\theta)$ is interchangeable with $Me^{j\theta}$) wouldn't the end result be:

$4cos(ωt+\theta) + 4cos(ωt+(-\theta))$?​

I also know: $Acos(ωt + θ)=\frac{1}{2}(X+X^∗)$ where $X=Ae^{j(ωt+θ)}=Ae^{jθ}e^{jωt}$.

But I'm unsure how to use that info, is that even applicable in this situation (I feel a little ridiculous for not understanding this but I'm trying my best!)?

Last edited: Sep 5, 2014
4. Sep 5, 2014

Simon Bridge

Put $e^{-j\theta}=e^{j(-\theta)}$ ... what is special about negative angles?

... it isn't. $Me^{j\theta}=M(\cos\theta + j\sin\theta)$ ... you know this.

5. Sep 6, 2014

PenDraconis

Using that information we have:

$4e^{j\theta}+4e^{j(-\theta)}$​

Expanding that out we have:

$4(\cos(\theta) + j\sin(\theta)) + 4(\cos(-\theta) + j\sin(-\theta))$​

We know that $\cos(\theta)+\cos(-\theta)$ is $2\cos(\theta)$ due to trigonometric identities.

We also know that the opposite is true for sine, rather that $sin(-\theta) = -sin(\theta)$. So $sin(\theta)+sin(-\theta) = 0$.

If we use that knowledge we end up with:

$8cos(\theta)+0 → 8cos(\theta) → 8cos(t)$​

Is that the correct thought process?

6. Sep 6, 2014

rude man

We have a winner!

You could also have gone

cosθ = [e + e-jθ]/2

so e + e-jθ = 2 cosθ

so 4[e + e-jθ] = 8 cosθ

7. Sep 6, 2014

PenDraconis

Hah!

It was so easy but for some reason I was battling myself the whole time thinking it was much more complicated.
Thank you for the assistance Simon and Rude Man.

8. Sep 6, 2014

Simon Bridge

The other approach was to draw the phasors.
The "jt" version is an arrow, length 1, rotating anti-clockwise; while the "-jt" version is an arrow length 1 rotating clockwise. At t=0 they are both aligned on the real axis - so they add to 2.
As the rotate in opposite directions, their imaginary components (the y-components) will always cancel.
The real components (the x components) are the cosine of the angle.

Well done though.