1
$\begingroup$

I'm reading the text book "Time-Frequency Analysis" by Leon Cohen and I've made my way through a decent portion. There is however a conceptual issue I keep coming back to. The book states:

If we consider $\lvert s(t)\rvert^2$ as a density in time, the average time can be defined in the usual way any average is defined:

$$ \langle t \rangle = \int t \lvert s(t)\rvert^2dt$$

I don't understand conceptually why multiplying by $t$ in the integration will provide us with an average. I didn't know that was the typical way to define an average either despite the books insistence that this is common knowledge (which it very may well be I'm just not interpreting it correctly). I've always thought about an average as dividing by some base, say by the total period of time $T$ NOT multiplying by $t$ inside the integral. It states earlier that:

$\lvert s(t)\rvert^2$ is the energy per unit time.

So I was thinking that perhaps mathematically multiplying by $t$ which has units of $s$ would get rid of the per unit time portion, but the units of $\lvert s(t)\rvert^2$ aren't actually $v^2/s$ as that would be the derivative, it's just $v^2$ so that doesn't make much sense (assuming the signal is an electrical signal and the unit time is seconds). Can someone point me in the right direction on how to think about this? I know it's a simple concept but I'm just having trouble making it over this hurdle. Thanks!

$\endgroup$
0
$\begingroup$

Take the real line (defined by the time location $t$). Weight it locally at time $t$ by the energy of the signal $s|(t)|^2$. The average time can then be seen as its center of mass. You can either look at it as a balance point (the signal's energy would be balanced putting the time line at the tip of your finger), or at a summary: it would be somehow equivalent to concentrate all the signal's energy at this point.

Remember the discrete version for the center of mass $M$for points $T_i$ with weights $w_i$, $\sum_i w_i \neq 0$: $$ \sum_i w_i\vec{T_i M} = \vec{0}. $$ Now look at a continuous formula like: $$ \int \lvert s(t)\rvert^2 \left( m - t \right) dt = 0$$ then the analogy is clear: $m$ would be the center of mass of "times" taken with an infinite summation. Solving for $m$ gives you: $$ m = \langle t \rangle = \frac{\int t\lvert s(t)\rvert^2 dt}{\int \lvert s(t)\rvert^2 dt}= \int t \left(\frac{\lvert s(t)\rvert^2}{\int \lvert s(t)\rvert^2 dt}\right) dt.$$

$\endgroup$
1
$\begingroup$

I think it is implied that $\int_{-\infty}^{\infty}|s(t)|^2 dt = 1$ (which is why the author refers to $|s(t)|^2$ as a density).

I agree with your sentiment "I've always thought about an average as dividing by some base", but in this case, the base would be 1, so no division is needed. If you were to plot $|s(t)|^2$ as a function of $t$, and think of this plot as a histogram, the the integral:

$$\int_{-\infty}^{\infty}t|s(t)|^2 dt $$ will tell you the average value, since each value of $t$ is weighted by the density function.

$\endgroup$
  • $\begingroup$ Thanks for your reply! I think I'm starting to grasp it but there are still a couple of things I want to reconcile. Just so that I think I understand, integrating $|s(t)|^2$ actually describes the energy of a signal (per wikipedia, though according to physics the units don't actually work out b/c we aren't dividing by an impedance. Consequently, we just take the density to essentially be a unit-less dimension whose maximum is always 1. Plotting this dimension though will show us how the relative "energy" at various times for $t$. $\endgroup$ – a_soy_milkshake Jul 27 '15 at 12:35
  • $\begingroup$ Integrating over the interval then (the duration of the signal) we can weight each time $t$ by the density function at that time $t$ which will give us the average, i.e., a characterization of where the density is concentrated in time. Is this a correct way to think about it? Sorry I had to type it in two comments, I reached the character limit. $\endgroup$ – a_soy_milkshake Jul 27 '15 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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