# Beating with signals of different amplitude

1. Nov 30, 2007

### odiakkoh

Okay: here's the question:

It's a well-known thing that when you add two sinusoids of equal amplitude and different frequency then you end up with a signal with the mean of the frequencies and a beat envelope. This just comes from the trig identity:

$$\cos(a) + \cos(b) = 2\cos(\frac{a+b}{2}) \cos(\frac{a-b}{2})$$

Now... here's the tricky bit: what happens if the two signals have different amplitudes?

I can get some kind of intuitive feel for what would happen by considering the limit of one signal being much stronger than the other, but I'd really like to be able to find some formula, similar to the above, perhaps with a form similar to this:

$$A\cos(a) + B\cos(b) = (A+B)\cos(x)(1-z\cos(y))$$

where x is the signal frequency, y is the beat frequency, and z is some number determining by how much the signal amplitude varies due to the beats.

Or perhaps this isn't possible? Anyhow: does anyone have any ideas?

2. Nov 30, 2007

### Ben Niehoff

If you write it in terms of complex exponentials,

$$Ae^{i\omega_1t}+Be^{i\omega_2t} = Ae^{i\omega_1t}\left(1+\frac{B}{A}e^{i(\omega_2-\omega_1)t}\right)$$

which might make it a little clearer.

3. Nov 30, 2007

### odiakkoh

Thanks... So now we're left with the problem of taking the real part of the RHS, i.e. evaluating this:

$$\Re\left( Ae^{iat}\left(1+\frac{A}{B}e^{i(b-a)t}\right)\right)$$

I've had a go, but I just can't manage to manipulate the result into the form I want.

4. Nov 30, 2007

### Ben Niehoff

You don't need to take the real part, actually. You can draw all the necessary conclusions from the formula as it is. In particular, remember

$$\left|e^{i\theta}\right|=1$$

So from that information, then, what conclusions can you draw by looking at the envelope term?

5. Dec 4, 2007

### coomast

There is a trick in rewriting the expression:
$$f=Acos(a)+Bcos(b)$$
as a product of the sum and difference of the two terms given.
First write the given expression as:
$$f=Acos\left(\frac{a+b+a-b}{2}\right)+Bcos\left(\frac{b+a+b-a}{2}\right)$$
Then you need to expand it, and recollect the terms, giving:
$$f=\left(A+B\right)cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)-\left(A-B\right)sin\left(\frac{a+b}{2}\right)sin\left(\frac{a-b}{2}\right)$$

In case A and B are equal, you get the expression you've allready mentioned. In case these are not equal, the second term comes in the picture.

6. Dec 4, 2007

### odiakkoh

Thanks, coomast, that's exactly the kind of expression I was looking for. Neat trick, too!

7. Dec 4, 2007

### odiakkoh

After further algebra

Taking this a little further, I've managed to rearrange this into an even more revealing form. The formula looks a little bulky to start, but please bear with me...

$$\left[A^2+B^2 + 2AB\cos\left(a-b\right)\right]^{1/2} \cos\left[\frac{a+b}{2} + \arctan\left(\frac{A-B}{A+B}\tan\left(\frac{a-b}{2}\right)\right)\right]$$

The term in the left square brackets is the beat envelope, oscillating slowly with frequency (a-b). I'm happy with that bit. The cosine term to the right of it is the signal, oscillating close to the (a+b)/2 frequency... but the tan and arctan parts on the right kind of ruin it as its not obvious at all what effect they have.

Can anyone think of a clever way of rewriting the arctangent term? It may be neccesary to make some approximation here. Perhaps A>>B or B<<A could be useful.

8. Dec 4, 2007

### coomast

Nice way of taking it a bit further :-) I forgot about this method. Anyway, why would you rearrange the arctan? It is (in case of slightly different frequencies) a small term. Approximating this could be done by considering for small x:

$$tan(x)\approx x$$

and

$$arctan(x)\approx x-\frac{x^3}{3}$$

This is however something you need to investigate. Are all terms small or not? In my opinion I would leave the formula as it is. It is in fact the solution to the original question. Besides, it's far too nice to be approximated :-)

9. Jan 13, 2011

### msas

Re: After further algebra

Absolutely brilliant derivation!
I'd really be interested how did you get all this done. A pointer to a book would suffice, no need to type the whole thing. The reason I'm asking is: I'm dealing with same functions, with only difference being that $$a,b,A,B$$ are all functions (we could say polynomials) of x. I'm more comfortable talking in terms of time, cause I'm into digital signal processing, but just as well. Could I just switch constants A,B with A(t),B(t)?

For the previous eq:
$$A\cos(a) + B\cos(b) = (A+B)\cos(x)(1-z\cos(y))$$
it is trivial to see that A,B can be swapped with functions A(t),B(t), so it's looking good so far. However I'm not sure if all the magic you did, that results in that aractan etc is also "immune" to such a swap.
Please let me know if you can. I'm googleing like crazy with no luck so far.

Thank you in any case!
Sash

10. Jan 13, 2011

### msas

Re: After further algebra

Absolutely brilliant derivation!
I'd really be interested how did you get all this done. A pointer to a book would suffice, no need to type the whole thing. The reason I'm asking is: I'm dealing with same functions, with only difference being that $$a,b,A,B$$ are all functions (we could say polynomials) of x. I'm more comfortable talking in terms of time, cause I'm into digital signal processing, but just as well. Could I just switch constants A,B with A(t),B(t)?

For the previous eq:
$$A\cos(a) + B\cos(b) = (A+B)\cos(x)(1-z\cos(y))$$
it is trivial to see that A,B can be swapped with functions A(t),B(t), so it's looking good so far. However I'm not sure if all the magic you did, that results in that arctan etc is also "immune" to such a swap.
Please let me know if you can. I'm googleing like crazy with no luck so far.

Thank you in any case!
Sash