Sum of Sinusoids with Same Frequency = Sinusoid (proof)

To avoid any confusion, let us state that

• a (pure) sine has the form Asin(ωt),
• a (pure) cosine has the form Acos(ωt),
• a sinusoid has an arbitrary phase and one of the equivalent forms Asin(ωt+ϕ) or Acos(ωt+ψ) - where ϕ and ψ differ by a quarter turn.

So the sine and cosine are special cases of the sinusoid.

By the well-known addition formula, Asin(ωt+ϕ)=Asin(ωt)cos(ϕ)+Acos(ωt)sin(ϕ)=A′sin(ωt)+A″cos(ωt).

This means that

1. a sinusoid can be expressed as a linear combination of a sine and a cosine,
2. conversely, a linear combination of sine and cosine can be represented as single sinusoid∗,
3. a linear combination of two or more sinusoids can be expressed as a linear combination of a sine and a cosine, hence can be expressed as a single sinusoid.

These properties no more hold if you mix sinusoids of different periods.

∗This is done by solving the system Acos(ϕ)=A′Asin(ϕ)=A″, i.e. A=√A′2+A″2tan(ϕ)=A″A′.

You will soon discover that complex numbers are intensively used in harmonic analysis, based on Euler's formula eix=cos(x)+isin(x).

Sinusoids can be represented as the imaginary part of Aei(ωt+ϕ)=Aeiϕeiωt=Zeiωt, where Z is a complex number, that carries both the amplitude and the phase (Z=A is real for a sine, Z=iA is imaginary for a cosine).

Using this notation, adding sinusoids becomes a trivial matter:

Z0eiωt+Z1eiωt+Z2eiωt=(Z0+Z1+Z2)eiωt.

(I will use the notation M ∠ θ to represent a vector in polar form, with M the magnitude and θ the angle)

The general result (that I believe to be very useful) is:

If f(x)=A1 cos(ω x+ϕ1)+A2 cos(ω x+ϕ2) then f(x)=A3 cos(ω x+ϕ3) and A3 ∠ ϕ3=A2 ∠ ϕ2+A1 ∠ ϕ1

This also holds if cos is replaced by sin.

A proof of this can be done using Euler's representation of sinusoids. Given:

A3cos(ϕ3)=A2cos(ϕ2)+A1cos(ϕ1)

f(x)=A112(ei(ωx+ϕ1)+e−i(ωx+ϕ1))+A212(ei(ωx+ϕ2)+e−i(ωx+ϕ2))

Rearranging terms:

f(x)=12(A1eiϕ1+A2eiϕ2)eiωx+12(A1e−iϕ1+A2e−iϕ2)e−iωx

Here we use the fact that since A3 ∠ ϕ3=A2 ∠ ϕ2+A1 ∠ ϕ1, then A3eiϕ3=A2eiϕ2+A1eiϕ1 and A3e−iϕ3=A2e−iϕ2+A1e−iϕ1:

f(x)=12(A3eiϕ3)eiωx+12(A3e−iϕ3)e−iωx

f(x)=A312(ei(ωx+ϕ3)+e−i(ωx+ϕ3))

and back from Euler's representation:

f(x)=A3 cos(ω x+ϕ3)

Here is a less rigorous but (I consider) more intuitive proof:

Consider

• Point Y is rotating around point X at a frequency of ω2π
• Point Z is rotating around point Y at a frequency of ω2π
• The initial angle and magnitude of Y relative to X is φ1 and A1
• The initial angle and magnitude of Z relative to Y is φ2 and A2

As you can see, the inital vector Z relative to the origin is A1 ∠ φ1+A2 ∠ φ2. So if you can intuitively visualize that the motion Z makes is a circle around X, then you can see that the sum of the sinusoids (either horizontal or vertical value) is itself a sinusoid.

This is a small addition to the Answer 1.

The original question is about calculating A3 and ϕ3.

This can be done easily from the drawing in Answer 1. If Zx, Zy, Yx, and Yy are coordinates of points Z and Y than A3 and tan(ϕ3) can be easily expressed as a functions of Zx, Zy, Yx, and Yy. Than Zx, Zy, Yx, and Yy could be expressed as a functions of A1, ϕ1, A2, and ϕ2.

Another approach: To find A3 start from multiplying formulas for A3eiϕ3 and A3e−iϕ3 (see the Answer 1). (Euler's representation of expression [cos(ϕ1)cos(ϕ2)+sin(ϕ1)sin(ϕ2)] is required for conversions.) To find tan(ϕ3) start from dividing the same formulas.

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