Conceptually you could do the convolution thing, but you can implement that by computing the FFT of both signals and taking the product point by point. I would have to think a little bit about the details, but I am pretty sure this is workable.

If you have two signals X(f) and Y(f) which are the Fourier transforms of the signals, then the quotient X(f)/Y(f) will have a constant magnitude and a linear phase if the two signals are equal. The phase shift corresponds to the delay between the two signals, and the constant amplitude is the scale difference between them. Noise will of course ensure that the magnitude won't be exactly constant, nor exactly linear phase, so you have to decide when the two signals are considered to be equivalent although they don't compare as exactly equal.

edit: I messed up some info in this post and it has been edited a few times. I think it should be correct now, in case someone read it between my edits.

Prefect and Brother Bob are pointing out essentially the same idea, which is simpler and more direct than what you or I were thinking.

Instead of dividing point by point (because I don't like potentially dividing by zero), you can try to compute what number you would have to multiply one of them to get the other, using the canonical Hermitian inner product, as follows. Let X and Y be the complex vectors that describe the FFT of your two signals. If Y = k * X, then <Y,X> = k * <X,X>. So compute the complex number k = <Y,X>/<X,X> which is trivial to compute and will tell you how much you have to magnify and shift one signal to match the other. Let's define Z:=Y-k*X. If the match is good, <Z,Z> will be small.

I think the procedure above should work well.

[quote name='Brother Bob' timestamp='1324499874' post='4896296']
If you have two signals X(f) and Y(f) which are the Fourier transforms of the signals, then the quotient X(f)/Y(f) will have a constant magnitude and a linear phase if the two signals are equal. The phase shift corresponds to the delay between the two signals, and the constant amplitude is the scale difference between them. Noise will of course ensure that the magnitude won't be exactly constant, nor exactly linear phase, so you have to decide when the two signals are considered to be equivalent although they don't compare as exactly equal.

That sounds like what I'm looking for. Would I just do a point-wise X(f)/Y(f) for each element in X and Y (i.e. x[sub]i[/sub](f)/y[sub]i[/sub](f) for all i), and just compare the results to see if they're (mostly) constant?
[/quote]
I don't know what your i is supposed to represent. When you have a signal x(n) in time n, you get a signal X(f) in frequency f when you take the Fourier transform. There are no other subscripts or indices, unless there is some information about the signal you're leaving out in your description. But what I mean with division is the element-wise division of each frequency f.

I'm not sure how to check if the phase is linear once I do X(f)/Y(f)...

I did leave out some implicit information that is kind of given for signal processing guys like myself, sorry. Linear phase simply means that the phase is a linear function of the frequency; that is, phase=constant*frequency. Thus, if you take the phase of the quotient X(f)/Y(f) and divide by the frequency f, you get a constant value if the phase is linear.

Also something you have to be aware of with the phase is that phase wraps around the unit circle, and so multiples of 2*pi appear as the same instantaneous phase. You need to do something called phase unwrapping, which unwraps the phase and removes this 2*pi-ambiguity. You can search for phase unwrapping if you want to know more about it. It effectively means that, starting from f=0 and walking up in frequency, you add 2*pi once the phase has made one turn around the unit circle, and then 4*pi after the second turn, and so on.

Prefect and Brother Bob are pointing out essentially the same idea, which is simpler and more direct than what you or I were thinking.

Instead of dividing point by point (because I don't like potentially dividing by zero), you can try to compute what number you would have to multiply one of them to get the other, using the canonical Hermitian inner product, as follows. Let X and Y be the complex vectors that describe the FFT of your two signals. If Y = k * X, then <Y,X> = k * <X,X>. So compute the complex number k = <Y,X>/<X,X> which is trivial to compute and will tell you how much you have to magnify and shift one signal to match the other. Let's define Z:=Y-k*X. If the match is good, <Z,Z> will be small.

I think the procedure above should work well.

Your solution may be easier than mine. You just have to complex conjugate the second operand of the inner product though.

Your solution may be easier than mine. You just have to complex conjugate the second operand of the inner product though.

True. However, I am following the notation in the link I posted (which is also what I would normally use), which defines <a,b> as sum(a_i * conj(b_i)).

[quote name='Brother Bob' timestamp='1324504577' post='4896321']
Your solution may be easier than mine. You just have to complex conjugate the second operand of the inner product though.

True. However, I am following the notation in the link I posted (which is also what I would normally use), which defines <a,b> as sum(a_i * conj(b_i)).
[/quote]
Yeah, sorry, I missed that detail. In hindsight, it is sort of obvious to me from the name of the link alone that you had a complex conjugate there already.