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I've seen it done a few times, it's always roughly apparent that "oh yeah, okay, they've dropped this, pulled this there, yeah, okay, kind of convincing" - but I've never why or how to do it for myself.

I probably don't understand it enough to be searching in the right way, but I'm unable to find anything useful by Googling.

There's a really nice mathmo derivation that I barely understand a word of here.

And a thoroughly unhelpful Physicist "thus it is this" non-explanation here.

But how about a nice not-mathematically-proper-but-it-works method for Engineers?

To be clear, I want to know how to go from:

$$ \lim_{\Delta\omega\rightarrow0} \sum_{n=-\infty}^{\infty} \dfrac{G(n\cdot\Delta\omega)\cdot\Delta\omega}{2\pi} \cdot e^{jn\cdot\Delta\omega\cdot t} $$

to:

$$ \dfrac{1}{2\pi} \int_{-\infty}^{\infty}G(\omega) e^{j\omega t} \cdot d\omega $$

and vice versa, and know how to do this for other problems, since as I said, I can sort of see that we've kept the limits, integrated wrt the limit, and removed the summation variable \$n\$. But if I did that on another problem, I wouldn't know I was 'doing it right' - and going in the other direction, how do we know where to put the \$n\$ back in?

Community
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OJFord
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  • Typically we would go from integral to summation to implement the algorithm in firmware. I can't think of any reason to go the other way outside of academia, maybe reverse engineering. Maybe this question should be migrated to math.SE as is, and see if they can come up with a more intuitive explanation than what was given in your reference? – Matt Young Apr 05 '14 at 17:48
  • The example above for summation to integration I took from a derivation of the inverse Fourier transform for a signal $g(t)$ given the frequency domain $G(\omega)$. – OJFord Apr 05 '14 at 17:59
  • I meant the math.SE question. – Matt Young Apr 05 '14 at 18:01
  • You said you couldn't think of a practical reason to go from summation to integration, I was explaining where I got this example. – OJFord Apr 05 '14 at 18:03

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You just have to roll through the definitions. There are some small pitfalls but as long as all limits are defined, which they will be with appropriate adjectives on \$G\$ then we are ok. There are several definitions of the symbol \$\int\$. The one which you want to use is the Riemann integral. To simplify the notation I will set \$F(\omega) = \frac{1}{2\pi}G(\omega)e^{j \omega t}\$.

Now the Riemann integral \$\int_a^b F(\omega) d\omega\$ still has a pretty complicated definition involving sups and infs but when it does exist and \$a\$ and \$b\$ are integers then it is equal to $$\int_a^b F(\omega) d\omega =\lim_{j \rightarrow \infty} \sum_{n = aj}^{bj-1} F(n \frac{1}{j})\frac{1}{j}.$$

That is instead of taking \$\Delta \omega \rightarrow 0\$ it is good enough to take \$\Delta \omega = \frac{1}{j} \$ with \$j \rightarrow \infty \$. This is the definition you learned in Calculus class.

The improper integral is defined as $$\int_{-\infty}^{\infty}F(\omega) d\omega = \lim_{a \rightarrow -\infty , b \rightarrow \infty} \int_a^b F(\omega)d\omega.$$

Again, if this limit exists then this is just the limit (for \$k\$ an integer) $$\int_{-\infty}^{\infty}F(\omega)d\omega = \lim_{k \rightarrow \infty}\int_{-k}^k F(\omega)d\omega.$$

Combining all of this and not discussing when/how to changing the order of taking limits is valid we obtain $$\begin{eqnarray*} \int_{-\infty}^{\infty} F(\omega) d\omega &=& \lim_{k \rightarrow \infty} \int_{-k}^k F(\omega)d\omega\\ &=& \lim_{k \rightarrow \infty}\lim_{j \rightarrow \infty} \sum_{n = -kj}^{kj-1} F(n \frac{1}{j})\frac{1}{j}\\ &=& \lim_{j \rightarrow \infty}\lim_{k \rightarrow \infty} \sum_{n = -kj}^{kj-1} F(n \frac{1}{j})\frac{1}{j}\\ &=& \lim_{j \rightarrow \infty} \sum_{n = -\infty}^{\infty} F(n \frac{1}{j})\frac{1}{j}\\ &=& \lim_{\Delta \omega \rightarrow 0} \sum_{n = -\infty}^{\infty} F(n \Delta \omega)\Delta \omega \end{eqnarray*}$$

SomeEE
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  • I think I understand now (or at least, I'm a lot closer). Sorry to not accept earlier, your answer deserved more attention than I could give it at the time. Just to clarify - is there a distinction between $$lim_{a\rightarrow\infty, b\rightarrow\infty}$$ and $$lim_{a\rightarrow\infty}lim_{b\rightarrow\infty}$$ or did you just use the latter (with $k, j$) to make the following step(s) clearer? – OJFord Apr 05 '14 at 19:31
  • Yes, one is a two dimensional limit and the other is a sequential evaluation of two one dimensional limits. For example try the function $a/b$, your second limit will evaluate to 0 while the first does not exist. – SomeEE Apr 05 '14 at 21:12
  • Ah, got it. So the order matters too, since $\lim_{b\rightarrow\infty}\lim_{a\rightarrow\infty} a/b$ would not be zero, right? – OJFord Apr 06 '14 at 14:10