Sum of sin waves with same frequency and different amplitudes and phase

Question

For the equation:

$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$

I've been able to show that the amplitude of the sum is (I believe this is a standard problem):

$$ A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta_1 - \theta_2)} $$

and the new phase is:

$$ \theta_3 = \arctan \left(\frac{A_1 \sin \theta_1 + A_2 \sin \theta_2}{A_1 \cos \theta_1 + A_2 \cos \theta_2}\right) $$

My question is what happens when the phases $\theta_1$ and $\theta_2$ are zero (or just equal to each other). Most books and internet sites (eg http://mathworld.wolfram.com/HarmonicAdditionTheorem.html) state that under these conditions:

$$ A_3 = \sqrt{A_1^2 + A_2^2} $$

However, if I set $\theta_1$ and $\theta_2$ to zero one gets instead (Since $cos(0) = 1$):

$$ A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2} $$

I've been staring at this for a while and I don't understand why the published answer is different from mine. I am sure it is something simple but I've not been able to spot it (Addendum: Wikipedia site: https://en.wikipedia.org/wiki/List_of_trigonometric_identities has the same result under section Linear Combinations).

Answer 1

Be sure to correctly read your references.

The formula $\sqrt{A_1^2+A_2^2}$ applies to sinusoids in quadrature ($\frac\pi2$ phase difference), such as when summing a sine and a cosine.

Signals in phase just add up their amplitudes, $A_1+A_2$, while signals in opposition subtract, $|A_1-A_2|$.

Answer 2

The result given in Equation (21) of Wolfram site

for adding in-phase waves of same frequency with zero relative phase angle gives

$$A_3= \sqrt{A_1^2+A_2^2+ 2 A_1 A _2}= \pm (A_1+A_2) $$

The other result came about when the phase difference is $\pi/2$ obtained directly by Pythagoras thm as diagonal length of a rectangle of vectors or phasors.

The parallelogram of amplitudes can be drawn and the situation becomes clear. There are three situations. In-phase,add; Out of phase, subtract, or take the difference; when relative phase difference is $\pi/2$ take root square sum.

Answer 3

You wrote:

$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$

The Wolfram article you cite says:

It is always possible to write a sum of sinusoidal functions $$f(\theta)=acos(\theta)+bsin(\theta )$$
as a single sinusoid the form $$f(\theta)=ccos(\theta+\delta)$$ $$ A_3 = \sqrt{A_1^2 + A_2^2} $$

So you have sine plus sine, while Wolfram has sine plus cosine. To convert Wolfram's cosine to a sine, you need to shift the phase by $\frac{\pi}2$, which then makes the cosine of the phase difference equal to zero.

Answer 4

Waves with no phase difference (or even pi's) directly add up their amplitudes to form a new wave.

$$A_1 \sin (\omega t) + A_2 \sin (\omega t) = (A_1+A_2) \sin (\omega t )$$

The $A_3$ you prescribed is for waves with phase difference $(\theta_1 - \theta_2)=\frac{\pi}{2}$.

The equation you got putting $\theta_1=\theta_2=0$ is correct and simplifies to $A_3=(A_1+A_2)$.

Answer 5

My question is what happens when the phases θ1 and θ2 are zero (or just equal to each other)?

Short answer

If we use phasors (see explanation below), our starting point will be:

$\large A_3 e^{i \phi_3} = A_1 e^{i \phi_1} + A_2 e^{i \phi_2}$

We can easily test any hypothesis:

1. $\phi_1 = \phi_2$
   $\implies A_3 e^{i \phi_3} = (A_1 + A_2) e^{i \phi_1}$
   $\implies A_3 = A_1 + A_2$
   $\implies \phi_3 = \phi_1$

2. $\phi_1 = \phi_2 = 0$
   $\implies A_3 e^{i \phi_3} = (A_1 + A_2) e^{i.0}$
   $\implies A_3 = A_1 + A_2$
   $\implies \phi_3 = 0$

We see when phases are equal, amplitudes add. More generally phasors are complex numbers, which can be added by vectorial sum.

^{Phasors addition, source: Wikipedia}

Indeed in vectorial sum, the resulting amplitude is

$A_3 = \sqrt {A_1^2 + A_2^2 + 2A_1 A_2 \cos (\phi_1 - \phi_2)}$

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Additional explanation if you're not familiar with phasors

When combining sinusoids with the same frequency, the way to get the result is by transforming the trigonometric sum into a sum of complex constants known as phasors. The conversion simply takes advantage of Euler's formula:

$\large cos \, \phi + i \sin \, \phi = e^{i \phi}$

where $i$ is the imaginary unit (usually noted $j$ in sciences to avoid confusion with variable $i$). $sin \, \phi$ is the imaginary part ($\Im$) of $e^{i \phi}$, so the problem:

$\large A_3 \sin (\omega t + \phi_3) = A_1 \sin (\omega t + \phi_1) + A_2 \sin (\omega t + \phi_2)$

becomes:

$\large \Im \{ A_3 e^{i (\omega t + \phi_3)} \} = \Im \{ A_1 e^{i (\omega t + \phi_1)} + A_2 e^{i (\omega t + \phi_2)} \}$

or in expanded form:

$\large \Im \{ A_3 e^{i \omega t} e^{i \phi_3} \} = \Im \{ A_1 e^{i \omega t} e^{i \phi_1} + A_2 e^{i \omega t} e^{i \phi_2} \}$

$Ae^{i \phi}$ quantities above are constant phasors which contain both a maximum amplitude and an initial phase. We see the variable quantity $e^{i \omega t}$ can be removed from the equation, and we are left with constants only. We can also leave aside $\Im$ as it has no influence here, so the problem can be stated this way:

$\large A_3 e^{i \phi_3} = A_1 e^{i \phi_1} + A_2 e^{i \phi_2}$