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

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#### drkirkby

##### Full Member level 6
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

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

### drkirkby

Points: 2
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.

Dave

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".

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

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.

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

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.

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)

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

I agree. Status
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