# Decomposition of $a\sin(\varphi t)+b\sin(\vartheta t)$ into AM and carrier

I feel like this should not be so hard, but I am somehow stuck.

I would like to decompose the signal $$a\sin(\varphi t)+b\sin(\vartheta t)$$ into an amplitude modulation and a periodic carrier signal. For $a=b$, the solution is $$\underbrace{2a\cos\left(\frac{\varphi - \vartheta}{2}t\right)}_{AM}\underbrace{\sin\left(\frac{\varphi +\vartheta}{2}t\right)}_{carrier}.$$

However, for $a\not=b$ I have problems deriving the closed form solution for the carrier. The AM in that case can be seen to be $$A(t)=\sqrt{a^{2}+b^{2}+2ab\cos\left(\left(\varphi-\vartheta\right)t\right)}.$$ This is the same as $$A(t)=\sqrt{\left(a-b\right)^{2}+4ab\cos^{2}\left(\frac{\left(\varphi-\vartheta\right)}{2}t\right)},$$ which reduces to the above case when $a=b$.

The carrier is obviously periodic, but after I did a few numeric simulations, I am not so sure anymore whether it is a single sinusoid. What I did is to look at the numeric Fourier spectrum of $$A(t)^{-1}\left(a\sin(\varphi t)+b\sin(\vartheta t)\right)$$ which clearly showed no single peak, but several peaks around the (sub)harmonics of $\varphi$ and $\vartheta$.

My questions are

1. Is there a close form solution of the carrier? Ideally I would like to have a function of a single carrier frequency: e.g. something like $f(\sin(\xi t+\eta))$. If would be great, of course, if $f$ turned out to be a polynomial.
2. If there is no close form solution, is there at least a close form term for the frequency of the carrier?
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There is no real modulation here (when $a=b$), because the modulation should vary slowly over a carrier period, but here they have the same frequency. – enzotib Jul 30 '14 at 17:51
Oh, I see, you have the same sign there, but it should be $\sin(x)+\sin(y)=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$. – enzotib Jul 30 '14 at 18:20
Thanks for spotting the typo. I corrected it. – fabee Jul 31 '14 at 6:10

If you write $a$ and $b$ as

$$a = {a+b \over 2} + {a-b \over 2}$$

and

$$b = {a+b \over 2} - {a-b \over 2},$$

then the signal have the form of the sum of two signals which you know how to analyze:

$${a+b \over 2} \left( \sin \varphi t + \sin \vartheta t \right) + {a-b \over 2} \left( \sin \varphi t - \sin \vartheta t \right).$$

If you still have questions after approaching the problem this way, let me know.

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This is a very neat trick. However, I don't see how this helps me to decompose it into a single carrier and AM. With your trick, I get two of them. $\left(a+b\right)\cos\left(\frac{ \varphi-\vartheta}{2}t\right)\sin\left(\frac{ \varphi+\vartheta}{2}t\right)+\left(a-b\right)\sin\left(\frac{ \varphi-\vartheta}{2}t\right)\cos\left(\frac{ \varphi+\vartheta}{2}t\right)$ – fabee Aug 2 '14 at 9:22
@fabee How do you figure? The signals have the same frequency, they are just out of phase with one another and have different amplitudes. Perhaps writing the $\cos$ and $\sin$ terms in complex form will help. – AnonSubmitter85 Aug 4 '14 at 22:40