Wave Interference in complex notaion

[HeisenbergW]

Homework Statement

Four waves of equal frequency ω are made to interfere. The amplitudes of the first two waves are equal (A), but those of the other two are equal to 2A. Their phase angles relative to that of the first wave are 0, ∏/3, ∏/4, and 2∏/3. Find the amplitude, frequency and phase angle of the resulting wave.
Hint: Use complex notation

Homework Equations

A wave can be written as Acos(ωt+[itex]\varphi[/itex]) where [itex]\varphi[/itex] is the phase angle, A is the amplitude, and ω is the frequency

The Attempt at a Solution

Adding all the waves together, I get

A[cos(ωt)+cos(ωt+∏/3)+cos(ωt+∏/4)cos(ωt+2∏/3)]

for easy addition, and following the suggested hint, I turn this into complex notation, keeping in mind that only the real values apply to the wave.

A[e^(iωt)+e^(iωt+∏/3)+e^(iωt+∏/4)+e^{(iωt+2∏/3)}]

which turns to

Ae^(iωt)[1+e^(i∏/3)+e^(i∏/4)+e^(i2∏/3)]

Now here is where i reach a problem. I am unaware how to add these together to get back to a format where I can identify things like the phase angle of the resulting wave.
If i convert the e's inside the brackets back to their cos and sin forms, and then add up the values i get

Ae^(iωt)[(1+2√2)/2+i(3√3+2√2)/2]

which leaves me stumped

Any help is appreciated
Thank you in advance

 

[rude man]

Homework Statement

Four waves of equal frequency ω are made to interfere. The amplitudes of the first two waves are equal (A), but those of the other two are equal to 2A. Their phase angles relative to that of the first wave are 0, ∏/3, ∏/4, and 2∏/3. Find the amplitude, frequency and phase angle of the resulting wave.
Hint: Use complex notation

Homework Equations

A wave can be written as Acos(ωt+[itex]\varphi[/itex]) where [itex]\varphi[/itex] is the phase angle, A is the amplitude, and ω is the frequency

The Attempt at a Solution

Adding all the waves together, I get

A[cos(ωt)+cos(ωt+∏/3)+cos(ωt+∏/4)cos(ωt+2∏/3)]

Check your amplitudes!

EDIT:

OK, Plan B:
Plan B hint: acos(wt) +bcos(wt+φ) transforms to a + bexp(jφ)

= [a + bcosφ] + j[bsinφ] = [real part] + j[imaginary part]
= magnitude plus phase angle:

magnitude M = sqrt{(real part)^2 + imaginary part^2)}
= sqrt{(a + bcosφ)^2 + (bsinφ)^2}
= sqrt(a2 + 2abcosφ + b2)

and φ = arc tan{bsinφ/(a+bcosφ)}

and the final voltage is Mcos(wt + φ).

Extrapolate this method to all four terms.

PS - where does the great Werner Heisenberg come in?

 

[Simon Bridge]

I shall respectfully disagree with the rude man - the exponentials are usually easier to add than trig functions and often more usefully too.

In this specific case where the angular frequencies are the same for all the waves, it is trivial!

The angular frequency does not change - the phasors are all lock-step.
All you do is add the t=0 phasors like vectors: head-to-tail.
Tha amplitude of the wave is the length of the resultant phasor.
The phase of the resulting wave is just the angle of the resultant.

eg.
add two waves, same amplitude A, same frequency ω, relative phase is π/2.

Add head-to tail - forms a right-angled triangle.
The resulting amplitude is √2.A and the phase is π/4

[Note: in complex notation, the wave is [itex]Ae^{i(\omega t +\phi)}[/itex] - this is an arrow length A rotating anti-clockwise in the complex plane. The sine and cosines you are used to are projections of this onto the imaginary and real axis. The phase angle is measured anticlockwise from the positive-real axis.]

 

[rude man]

I wrote an edited version which essentially parallels your vector diagram method, except it proceeds exclusively algebraically. What I said about exponentials holds for real exponents but of course you're right, in the world of the complex plane Euler comes to the rescue and splits a complex exponential into real and imaginary parts on the plane.

Diagrammed vectors are an excellent heuristic tool but are soon forgotten in the real world.
I personally survived 45 years in the electrical engineering business without ever using them.
(Could never draw either!)

 

[Simon Bridge]

Diagrammed vectors are an excellent heuristic tool but are soon forgotten in the real world.
I personally survived 45 years in the electrical engineering business without ever using them.
(Could never draw either!)

We each have strengths - by comparison, I could draw well from an early age, and do everything visually.
If you do something a lot, you get used to just remembering the tricks associated with it. Though the vector/phasor method is exactly what you did - keeping the phasor representation turns the problem into basic-trig and the theorum of pythagoras.

The phasor method is all anyone does in electronic-physics. We'd never get anywhere without our complex impedences. It's also general to anything which rotates in some sense - such as quantum wave-mechanics and Foucault's pendulum. That's why it's taught - not to understand AC circuits, or whatever, exactly but because the method is generally applicable.

 

[HeisenbergW]

Just wanted to express my thanks to both of you for taking time to help me understand this problem.

 

[Simon Bridge]

No worries - did it help?

 

[HeisenbergW]

Yes, I would have never thought of using them as vectors when their frequencies are equal. Though I do wonder how I was meant to do the problem using the complex notation with the e's.

 

[rude man]

Sorry, I thought my post did that. I guess my weak point was talking about "transforms". It's quite straightforward but I didn't have the patience to derive the method. Any elementary network text has it. In case you missed my post the first time:

A[cos(ωt)+cos(ωt+∏/3)+cos(ωt+∏/4)cos(ωt+2∏/3)]

Check your amplitudes!
(I assume you caught that).

Then:
acos(wt) +bcos(wt+φ) transforms to a + bexp(jφ)

= [a + bcosφ] + j[bsinφ] = [real part] + j[imaginary part]
= magnitude plus phase angle:

magnitude M = sqrt{(real part)^2 + imaginary part^2)}
= sqrt{(a + bcosφ)^2 + (bsinφ)^2}
= sqrt(a2 + 2abcosφ + b2)

and φ = arc tan{bsinφ/(a+bcosφ)}

and the final voltage is Mcos(wt + φ).

Extrapolate this method to all four terms.

 

[HeisenbergW]

Rude Man
No need to be sorry (which isn't rude behavior, your name seems to be a lie), I was actually using both of your methods simultaneously, and they came up with the same answers, so I was pleased. The only question I have concerning your method is how it changes when the acos(wt) also has a φ so that it is acos(wt+φ)+bcos(wt+φ'). I was planning on using your method to add the first two, then last two together, then add those two remaining to obtain the final value, but did not know what to do once there were two separate angles.

And yes I did catch the amplitude blunder, thank you for the heads up on that silly mistake.

 

[rude man]

Rude Man
No need to be sorry (which isn't rude behavior, your name seems to be a lie), I was actually using both of your methods simultaneously, and they came up with the same answers, so I was pleased. The only question I have concerning your method is how it changes when the acos(wt) also has a φ so that it is acos(wt+φ)+bcos(wt+φ'). I was planning on using your method to add the first two, then last two together, then add those two remaining to obtain the final value, but did not know what to do once there were two separate angles.

And yes I did catch the amplitude blunder, thank you for the heads up on that silly mistake.

OK, let's take the following two, & I'm even throwing in different amplitudes a and b for you to be as general as possible:

v(t) = acos(wt+φ) + bcos(wt+φ').
Transformed: V = a*exp(jφ) + b*exp(jφ')
V = [acosφ + bcosφ'] + j[asinφ + bsinφ'] = (re) + j(im) ... (1)
= Mexp(jφ'')

magnitude M = sqrt{(real part)^2 + imaginary part^2)}
= sqrt{(acosφ + bcosφ')^2 + (asinφ +bsinφ')^2}
= sqrt[a2 + 2ab(cosφcosφ' + sinφsinφ') + b2)

and phase angle φ'' = arc tan(im/re) per the above equation (1).
So V = Mcos(φ'')
and the final voltage is v(t) = Mcos(wt + φ'').

You can check to see if I did it right by setting φ = 0 and φ' = φ from my earlier derivation. They should give you the same M and final phase angle.

PS - my name is Rudy and someone in my dept decided to call me rude man - I hesitate to speculate as to why!

 

[HeisenbergW]

Both of you have been extremely helpful. One FINAL question on a problem that seems to come up. When adding the first two waves, the ones with amplitude A, both of your methods seem to match up perfectly. But when adding the third and fourth waves together, there unfortunately still seems to be a difference. Amazingly the amplitude using both methods seems to be the same, but it is at the phase angle where a difference arises.

The two waves being added are 2Acos(ωt+∏/4)+2Acos(ωt+2∏/3)

Using Rude Man's method the phase angle should be
arctan((2sin(pi/4)+2sin(2pi/3))/(2cos(pi/4)+cos(2pi/3))) which gives me
1.288012

Using Mr. Bridge's method
I add the two final vectors tip to tail.
where the first vector is elevated ∏/4 counterclockwise, and the second starting at the tail is 2∏/3 both with a magnitude of 2A. this leaves the angles between the two vectors as 7∏/12
since ∏-2∏/3+∏/4=7∏/12
then I use the formula
2A/sina=2A/sinb=x/sin(7∏/12)
where a and b are the other two angles in the created triangle, and x is the resultant vector
I then get sina=sinb, so a=b
and 7∏/12-2a=∏ since this is a triangle
so a=5∏/24
which I assume I would add with ∏/4 since the angle is relative to the first vector, and I must find the phase angle relative to θ=0

Once again thank you
You both have been incredibly helpful

It is worth noting that BOTH methods got me the same answer when adding the first two components for both magnitude and angle

 

[rude man]

Both of you have been extremely helpful. One FINAL question on a problem that seems to come up. When adding the first two waves, the ones with amplitude A, both of your methods seem to match up perfectly. But when adding the third and fourth waves together, there unfortunately still seems to be a difference. Amazingly the amplitude using both methods seems to be the same, but it is at the phase angle where a difference arises.

The two waves being added are 2Acos(ωt+∏/4)+2Acos(ωt+2∏/3)

Using Rude Man's method the phase angle should be
arctan((2sin(pi/4)+2sin(2pi/3))/(2cos(pi/4)+cos(2pi/3))) which gives me
1.288012

Let me first do the entire problem so everyone can check if they like.
For the final phase (all four terms included) I got 1.126 rad., as follows:

V = A + Aexp(jπ/3) + 2Aexp(jπ/4) + 2Aexp(2π/3)
= A + A(.5 + j.866) + 2A(.707 + j.707) + 2A(-.5 + j.866)
Re = 1.914
I am = 4.012

Total phase = arc tan(Im/Re) = arc tan(4.012/1.914) = +1.126 radians.

OK, so for the last two terms only,
Im = (1.414 + 1.732)A = 3.146A
Re = (1.414 - 1.000)A = 0.414A
and phase for these two terms is arc tan(3.146/0.414) = 1.440 radians
which exactly agrees with Simon's method. So either I screwed up on the formulaic general derivation, or the formulas were misinterpreted (like where is the parenthesis etc.). In any case I think the procedure is obvious by now.

Good work all around, I'd say!

 

[Simon Bridge]

Without taking anything away from Rude Man - as you've seen his approach is actually the same thing... Formally, the phasor method goes like this:

For the sum of n arbitrary waves, all the same frequency:

[tex]\psi = \sum_n A_n e^{i(\omega t + \phi_n)} = e^{i\omega t}\sum_n A_n e^{i\phi_n}= Ae^{i\omega t}e^{i\phi}[/tex]

so:

[tex]Ae^{i\phi}=\sum_n A_n e^{i\phi_n}[/tex]

which is where you observe that the effect of the exponential is to rotate the amplitude ... leads to the vector treatment.

The first wave in your example has phase zero - this means that all the other phases have been measured with respect to this wave. In practice you always do this: there is no absolute zero for phase so we always measure it relative to something.

Have a look at this lab [pdf], I think it may be what your teacher had in mind and it should give you a deeper understanding.

 

[rude man]

Amen! Now if I could only convince Simon that addition is not multiplication ...

 

[Simon Bridge]

Hmmm ... I should have represented the RF, LO and IF signals (in the other thread) as complex phasors just to stir.

Aside: all those greek letters in the side-pane: you keep using the capitol pi, "∏", when you want the lower case "π". Not your fault: the lower case one looks like an n. The cap usually stands for a product ... nothing I can't handle.

You could also use the tex tags to write [itex]\pi[/itex] and [itex]\Pi[/itex].
In order that the letters come out, the latex parser uses a slightly bigger font.
Though now the cap pi looks like a roman numeral II. You can't win!

Just sayin :)