Integration with infinite/finite limits as a form of summation in DSP notation for discrete signals

Question

While reading some DSP literature, I've noticed integrals of the following form. An example integral (Equation 7.17) is given on pg. 124 of a book that I've found very challenging to read (Seismic Inverse Q filtering):

$u(t) = h(t)\int\limits_{ - \infty }^\infty {u(\tau ,t)d} \tau$

This is exactly how the author has written the equation. Gauging the author's intent, I think that $u(\tau ,t)$ is really another function, so in the above, let's re-write the equation as the following:

$u(t) = h(t)\int\limits_{ - \infty }^\infty {p(\tau ,t)d} \tau$

In the above, $u(t) \neq p(\tau ,t)$, so I think that this clears up some of the notation.

This appears to be notation associated with a continuous signal $p(\tau, t)$ that is of infinite length. For a discrete signal $p[\tau, t]$ with a finite length $N$, how do I evaluate the integral?

Case #1

Is the integration treated as a summation?

$u[t] = h[t]\sum\limits_{i = 1}^N {p[i,t]}$

Case #2

Is the integration treated as a "classical" integration using the trapezoidal rule?

How do I know which of these cases apply for the discrete signal case? Is this really an issue of notation?

Another Example

Another example can be found on pg. 128 of the same book (Equation 7.21):

$\tilde U(\tau ,\omega ) = U(\tau ,\omega )\frac{1}{{\Lambda (\tau ,\omega )}}\exp \left[ {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\gamma (\tau ')}} - 1} \right)} \omega d\tau '} \right]$

When the integration is being done over $\tau'$, is this integration a summation or a "classical" integration?

Yet another example

On pg. 9 of "R. Wang, Introduction to Orthogonal Transforms: With Applications in Data Processing and Analysis. Cambridge University Press, 2012", the author states that the energy in a continuous signal $x(t)$ is given by:

$E = \int\limits_{ - \infty }^\infty {{{\left| {x(t)} \right|}^2}} dt$

The energy in a discrete signal $x[n]$ is:

$E = \sum\limits_{n = - \infty }^\infty {{{\left| {x[n]} \right|}^2}}$

But shouldn't a discrete signal have finite lower and upper bounds of summation? What is really meant here using infinite lower and upper bounds? Shouldn't the trapezoidal rule be used here?

So here is the gist of my question: Using this notation, is integration done using the trapezoidal rule for discrete signals, or is it simply a form of summation? What is the general use of this type of notation in DSP?

Given an equation for a continuous function integration, how does this relate to integration in the context of a discrete function?

Answer 1

First of all, I apologize for all the hand-waving, but it might be helpful in this case...

Discrete signals are sequences

I don't really understand why you are complicating things with the trapezoidal rule. A discrete signal is not a map from $\mathbb{R}$ to $\mathbb{R}$ (function), but a map form $\mathbb{N}$ to $\mathbb{R}$ (sequence) so there is nothing between samples, and it is not a good idea to try to making up stuff between samples when reasoning about them. Let's blame it on the many pop science articles in which discrete signals are shown with "steps" - but a correct graphical representation of a discrete signal uses "stems".

If you stumble at bridging the continuous and discrete world, it might help to consider a "continuous equivalent" of discrete signal $x(n)$ as a sum of diracs:

$x(t) = \sum_{n = -\infty}^{\infty} \delta(t - n) x(n)$

Note that in this case, $x(t)$ is no longer a function - it is a distribution and your usual insights about the integral of functions no longer apply.

You can use this to compute the energy:

$$ \begin{eqnarray}E &=& \int_{-\infty}^{\infty} |x(t)^2| dt \\ &=& \int_{-\infty}^{\infty} \left|\sum_{n = -\infty}^{\infty} \delta(t - n) x(n)\right|^2 dt \\ &=& \sum_{n = -\infty}^{\infty} \int_{-\infty}^{\infty} \delta(t - n) |x(n)|^2 dt \\ &=& \sum_{n = -\infty}^{\infty} |x(n)|^2 \\ \end{eqnarray}$$

When dealing with continuous signals (indexed by $\mathbb{R}$), use integration. When dealing with discrete signals (indexed by $\mathbb{N}$), use summation. If for some reason you need to work with both domains, it might help to represent the discrete signal as a sum of $\delta$ and use $\int$ notation, but then remember that you are integrating/derivating distributions instead of functions and that special rules might apply...

Discrete signals are mathematical objects and do not have to be physically realizable

Regarding the integration bounds: discrete signals don't need to have a finite support. There's nothing that states that discrete signals should be physically "realizable". Thus there is nothing wrong in defining, for example, the discrete signal $s(n) = \sin (2 \pi \frac{440}{48000} n)$. It has an infinite support and infinite energy. $s(n) = e^{-n^2}$ is another example of discrete signal with infinite support but this time finite energy.

Answer 2

In the absence of any further information, we must assume that tradition plays a role in the numerical implementation. The language of mathematics implies that an integration is an integration, so we can assume that the trapezoidal rule holds:

$\int_a^b f (x){\mkern 1mu} dx \approx \frac{1}{2}\sum\limits_{k = 1}^{N} {\left( {{x_{k + 1}} - {x_k}} \right)} \left( {f({x_{k + 1}}) + f({x_k})} \right)$

However, the intent of the author is another mitigating factor, so careful reading is required. There are often some pathological issues of notation, but these are strange.

After some numerical experimentation, I've found that in general, an integration is an integration, no matter which field of inquiry. The language of mathematics is universal.