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

[PenDraconis]

Homework Statement

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

(a.) 4e^jt+4e^-jt

Homework Equations

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

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^jθ not an e^jt.

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

 

[Simon Bridge]

Homework Statement

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

(a.) 4e^jt+4e^-jt

Homework Equations

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

The Attempt at a Solution

I attempted to use phasor analysis to simply have it be 4cos(ωt)+4cos(ωt),

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.

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_jθ not an e_jt.

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

 

[PenDraconis]

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

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!)?

 

[Simon Bridge]

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?

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

... and $Mcos(ωt+\theta)$ is interchangeable with $Me^{j\theta}$

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

 

[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?

 

[rude man]

If we use that knowledge we end up with:

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

Is that the correct thought process?

We have a winner!

You could also have gone

cosθ = [e^jθ + e^-jθ]/2

so e^jθ + e^-jθ = 2 cosθ

so 4[e^jθ + e^-jθ] = 8 cosθ

 

[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.

 

[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.