Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
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

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


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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.