Estimation of Instantaneous Amplitude

I'm reading a paper on EMG analysis. The formulas are all clear to me, but the paper refers to the signal amplitude as "instantaneous" amplitude.

I know what instantaneous mean, but what does it mean in the context of signal processing?

The instantaneous amplitude (or envelope) is usually defined as the magnitude of the (complex-valued) analytic signal $x_a(t)$ associated with the given signal $x(t)$:

$$x_a(t)=x(t)+j\mathcal{H}\{x(t)\}\tag{1}$$

where $\mathcal{H}$ denotes the Hilbert transform. So the instantaneous amplitude (envelope) of $x(t)$ is given by $|x_a(t)|$.

As a very simple example, take $x(t)=\cos(\omega_0t)$. Its Hilbert transform is $\sin(\omega_0t)$, and, consequently, its associated analytic signal is

$$x_a(t)=\cos(\omega_0t)+j\sin(\omega_0t)=e^{j\omega_0t}\tag{2}$$

The instantaneous amplitude (envelope) of $x(t)$ is $|x_a(t)|=1$.

Instantaneous amplitude is usually done in conjunction with analytic signals and the Hilbert transform. See this answer on this site.

For signal $s(t)$, given its Hilbert Transform $\hat{s}(t)$ it is defined as a composition:

$$s_A(t)=s(t)+j\hat{s}(t)$$

The Analytic Signal that we obtain is complex valued, therefore we can express it in exponential notation:

$$s_A(t)=A(t)e^{j\psi(t)}$$

where:

$A(t)$ is the instantaneous amplitude (envelope)

$\psi(t)$ is the instantaneous phase.

• You were a few secs faster, so we have a lot of redundancy now ... Good that you quoted that other answer, I didn't realize we already had such a good explanation here. +1 – Matt L. Jan 19 '16 at 13:41
• @MattL. : Yes, I was tempted to close this question as a duplicate of the other one, but it's not: the other one asks explicitly about the Hilbert Transform, and this one is more general. I felt just giving a link to the other answer was poor form, but I wanted to give credit --- hence the quotation. – Peter K. Jan 19 '16 at 14:11