Finding the amplitude of the sum vector

[chilge]

Homework Statement

A signal E(t) is made up of three terms, each having the same frequency but differing in phase:

E(t) = E₀cos(ωt) + E₀cos(ωt + δ) + E₀cos(ωt + 2δ)

It is possible to find the amplitude of the sum vector by summing each vector described as a magnitude multiplied by a phase. The sum will therefore contain three terms. You can simplify this to express the sum as a real number times a phase. The real number is the amplitude of the sum vector. Make a plot of the amplitude of the sum vector as a function of δ as δ varies from 0 to 2∏.

Homework Equations

A signal can be represented as the real part of a complex number z,
z = Ae^i(wt+∅) = A[cos(wt+∅) + jsin(wt+∅)]

The Attempt at a Solution

E(t) = E₀cos(ωt) + E₀cos(ωt + δ) + E₀cos(ωt + 2δ)
E(t) = E₀exp(iωt) + E₀exp(i(ωt + δ)) + E₀exp(i(ωt + 2δ))
E(t) = E₀exp(iωt) * (1 + exp(iδ) + exp(i2δ))
E(t) = E₀cos(ωt) * (1 + exp(iδ) + exp(i2δ))

I can't think of a way to simplify it any further to put it in the form real number * a phase term. Am I missing something?

 

[Jolb]

First of all, this equation really makes no sense:

Ae^i(wt+∅) = A[cos(wt+∅) + jsin(wt+∅)]

because you are using two different notations for the imaginary unit in the same expression. If you use i to denote the imaginary unit, you should have:
Ae^i(wt+∅) = A[cos(wt+∅) + i sin(wt+∅)]

Second, between the first and second steps in your attempt at a solution, you incorrectly applied that formula. You have changed a real expression into an expression with a real part and a non-zero imaginary part.

If you do some simple algebra you should see that you can represent cos(x) as
cos(x) = (e^ix+e^-ix)/2
Alternatively, you could write cos(x) as the real part of e^ix, e.g.
cos(x)+cos(y) = Re[e^ix + e^iy]

Try applying those formulae and see if you can make any more progress. And feel free to post again if you get stuck. But you should make a better attempt before I give you any more hints.

 

[ehild]

Homework Statement

A signal E(t) is made up of three terms, each having the same frequency but differing in phase:

E(t) = E₀cos(ωt) + E₀cos(ωt + δ) + E₀cos(ωt + 2δ)

It is possible to find the amplitude of the sum vector by summing each vector described as a magnitude multiplied by a phase. The sum will therefore contain three terms. You can simplify this to express the sum as a real number times a phase. The real number is the amplitude of the sum vector. Make a plot of the amplitude of the sum vector as a function of δ as δ varies from 0 to 2∏.

Homework Equations

A signal can be represented as the real part of a complex number z,
z = Ae^i(wt+∅) = A[cos(wt+∅) + jsin(wt+∅)]

The Attempt at a Solution

E(t) = E₀cos(ωt) + E₀cos(ωt + δ) + E₀cos(ωt + 2δ)
E(t) = E₀exp(iωt) + E₀exp(i(ωt + δ)) + E₀exp(i(ωt + 2δ))
E(t) = E₀exp(iωt) * (1 + exp(iδ) + exp(i2δ))
[STRIKE]E(t) = E₀cos(ωt) * (1 + exp(iδ) + exp(i2δ))[/STRIKE]

I can't think of a way to simplify it any further to put it in the form real number * a phase term. Am I missing something?

You can not use the same notation for the real E(t) and the complex one. Write E(t)=Re(Z) where Z=E₀exp(iωt) + E₀exp(i(ωt + δ)) + E₀exp(i(ωt + 2δ)), and make it equal to Aexp(i(ωt+θ)), where A and θ ( both real) are to be determined.

You need to transform the factor 1 + exp(iδ) + exp(i2δ) into the exponential form, 1 + exp(iδ) + exp(i2δ)=Cexp(iψ). For that, expand the exponentials exp(iδ)=cosδ+isinδ. Note that exp(i2δ)=[exp(iδ)]². Collect the real and imaginary parts of the sum, and find the amplitude and phase.

ehild

 

[ehild]

the solution is much easier if you factor out E₀exp(iωt)exp(iδ):

\hat Z(t)=E_0 e^{i \omega t}+E_0 e^{i (\omega t+\delta)}+E_0 e^{i (\omega t+2\delta)}=E_0 e^{i (\omega t+\delta)} (e^{-i\delta}+1+e^{i\delta})

But e^{-i\delta}+1+e^{i\delta}=1+2\cos \delta, a real quantity. That makes

\hat Z(t)=E_0 (1+2\cos \delta)e^{i (\omega t+\delta)}

ehild