Is the sum of two sine waves at the same frequency always a sine wave?

[drkirkby]

If you have two sine waves with peak amplitdes V1 and V2, at the same frequency f, with an arbitrary phase difference between them, is the sum always a sine wave? Can anyone provide a proof or reference to this.

Dave

 

[iukhan]

If you have two sine waves with peak amplitdes V1 and V2, at the same frequency f, with an arbitrary phase difference between them, is the sum always a sine wave?

Yes, the sum of two sine wave having different amplitudes and phase is always sinewave.

Can anyone provide a proof or reference to this.

For mathimatical proof, see **broken link removed**

 

[drkirkby]

Yes, the sum of two sine wave having different amplitudes and phase is always sinewave.

For mathimatical proof, see **broken link removed**

Thank you. I see a derivation of something in a book, and I could see the proof relied on the fact that the sum of two sine waves would be a sine wave, but it was not stated. So I got a bit concerned perhaps the proof was invalid, but it seems not the case.

That was most helpful.

Dave

 

[albbg]

If you have two generic sinusoid with amplitude "A" and "B" and same argument "x" and arbitrary phase "θ" between them, the sum will be a sinusoid having amplitude let say "M" and phase "φ" so that:

Asin(x)+Bsin(x+θ)=Msin(x+φ)

using trigonometric rules:

Asin(x)+Bsin(x)cos(θ)+Bcos(x)sin(θ)=Msin(x)cos(φ)+Mcos(x)sin(φ)

that is:

[A+Bcos(θ]sin(x)+Bcos(x)sin(θ)=Msin(x)cos(φ)+Mcos(x)sin(φ)

equating now the terms sin(x) and cos(x):

[A+Bcos(θ]=Mcos(φ)
Bsin(θ)=Msin(φ)

the ration between them is

Bsin(θ)/[A+Bcos(θ]=tan(φ)

from wich:

φ=arctan{Bsin(θ)/[A+Bcos(θ]}

of course now you can find M as:

M=Bsin(θ)/sin(φ)

This means that a generic sum of two sinusoid can be written as a single sinusoid having amplitude "M" and phase "φ".

 

[opopson]

Note that the sum of two sine wave that are out of phase by 180 degrees is zero ( more like a destructive interference)

 

[drkirkby]

Note that the sum of two sine wave that are out of phase by 180 degrees is zero ( more like a destructive interference)

Well, in general it can still be considered a sine way - it just happens to have an amplitude of zero.

I must admit, intuately I though that if there was some phase difference between them, then the sum would no longer be a pure sine wave, but it seems my intuition was wrong.

 

[keith1200rs]

Note that the sum of two sine wave that are out of phase by 180 degrees is zero ( more like a destructive interference)

Only if the amplitudes are equal.

Keith

 

[W_Heisenberg]

In signal processing field, usually we decompose signals into sine wave form. This in other view to see, or actually more professionally, is in the form of exponential function.

Thus, here I will use the method of exp function. This simplifies the calculation since it avoids two pure sine form wave summation. Also this is why Euler solves a very important question with his Euler equation.





Hopefully, this helps a bit.

 

[albbg]

In effect exp(jwt) doesn't represent exactly sin(wt) nor cos(wt). It is instead a cisoidoidal signal:

exp(jwt) = cos(wt)+j*sin(wt)

 

[W_Heisenberg]

if u want to write sin(t) = (exp(jwt)-exp(-jwt))/(2j), it will be more accurate.

I agree.