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?

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    $\begingroup$ Before getting everyone too deeply involved in answering your question, could you check the integral you have written above? On one side, $u$ is a function of two variables, on the other of one variable. Also, the integral just evaluates to some $x(t)$ making the right side just $h(t)x(t)$. Are you absolutely sure that the right side is not something like $$\int_{-\infty}^\infty h(\tau)u(t, \tau) \mathrm d\tau$$ which is a form of "convolution" integral that shows up in the analysis of time-varying linear systems? $\endgroup$ Commented Jul 3, 2012 at 1:28
  • $\begingroup$ @DilipSarwate: Thank you for your response. That's a really interesting observation. I've now added more information to my original question above. Is the convolution integral that you cite written as a summation as well? $\endgroup$ Commented Jul 3, 2012 at 2:12
  • $\begingroup$ @DilipSarwate: Yes, I've also checked the integral in the book, and it is correct. I assume this is what the author intended. $\endgroup$ Commented Jul 3, 2012 at 2:18
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    $\begingroup$ I share Dilip's confusion. The notation is really questionable. In the first two equations you have the function u() both on the left and the right side of the equation (which would indicate a non-homogeneous differential equation) . On the left side it's a function of one variable, on the right side it's a function of two. That doesn't make a lot of sense to me. $\endgroup$
    – Hilmar
    Commented Jul 3, 2012 at 12:32
  • $\begingroup$ @Hilmar: I agree, the book is challenging to read. I've tried to update my question to get a better understanding of what is really going on here. I think that in general, I am mostly interested in the following question: Given an equation for a continuous function integration, how does this relate to integration in the context of a discrete function? $\endgroup$ Commented Jul 3, 2012 at 14:34

2 Answers 2


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.

  • $\begingroup$ That's interesting. So for $\tilde U(\tau ,\omega ) = U(\tau ,\omega )\frac{1}{{\Lambda (\tau ,\omega )}}\exp \left[ {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\gamma (\tau ')}} - 1} \right)} \omega d\tau '} \right]$, the integration is always treated as a summation, if we are dealing with discrete sequences? $\endgroup$ Commented Jul 6, 2012 at 17:18
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    $\begingroup$ What is the integrand here? It does not appear to be a discrete signal... $\endgroup$ Commented Jul 6, 2012 at 17:21
  • $\begingroup$ The integrand is $\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\gamma (\tau ')}} - 1} \right)\omega$. You are right, I think this is a continuous function. But $U(\tau, \omega)$ is the Gabor transform (Short time frequency transform with Gaussian time-domain window) of a signal, and the multiplication implies a convolution (point-by-point multiplication between 2D matrices). I've found that in this case, numerical integration works well, since I think that what I want to find is a discrete approximation to a continuous function. $\endgroup$ Commented Jul 6, 2012 at 17:39
  • $\begingroup$ Could you update your answer to comment on when the discrete approximation of a continuous function is acceptable? It seems that the author of the monograph works primarily with continuous functions. Your answer provides great insight on discrete signals. $\endgroup$ Commented Jul 6, 2012 at 17:49
  • $\begingroup$ "When dealing with continuous signals (indexed by $\mathbb{R}$), use integration. When dealing with discrete signals (indexed by $\mathbb{N}$), use summation." To me, this is the crux of the answer. $\endgroup$ Commented Jul 7, 2012 at 1:27

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.


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