So, I work at an ISP that uses wireless and DSL.

Something that keeps coming up is something called OFDM. It's used by some of our wireless solutions, and I understand (or at least think I understand) that DSL's G.dmt uses OFDM.

I'm trying to get a high level idea of what OFDM is.

From reading Wikipedia and the like, it sounds like you pick many different frequencies and transmit on them simultaneously. (I'm guessing you just switch the individual frequencies on and off, but I could be wrong on that, too).

I gather that it is not too hard to isolate one frequency from a signal mathematically.

From what I understand, the frequencies are picked so that they will cause the least interference with each other. How do they determine what frequencies are orthogonal?

I have tried to figure this out, but the problem is that most places seem like they are talking college textbook level with not enough context for me to catch on.


1 Answer 1


Caveat: I don't know this off the top of my head. I'm just trying to rephrase what's in Wikipedia to answer the question. I tried to answer this as simply as I can, but I realize I don't even know if you have learned calculus, so bear with me if and at the end I'll go back and just re-state the conclusions without the calculus.

First, for other readers, OFDM is basically breaking up into several parallel data streams and transmitting each one on a different frequency carrier. This means that the modulation on each carrier is much slower than would be needed to transmit all the data in "serial" on a single carrier.

Now, two signals \$g_1(t)\$ and \$g_2(t)\$ are orthogonal when measured over a certain period of time if

\$\int_{t_1}^{t_2} g_1(t) g_2(t)\ \mathrm{d}t = 0\$

In principle, any two non-equal frequencies are orthogonal if measured over a long enough time. That is,

\$\int_{-\infty}^\infty{\cos(2\pi{}f_1t + \phi_1) \cos(2\pi{}f_2t + \phi_2) \mathrm{d}t} = 0\$

so long as \$f_1 \neq f_2\$.

But if the limits on the integral are not +/- infinity, then the result will not always be zero. For example, if you take the integral over one bit period of your signal, it won't always be zero, and that's what's important for decoding the parallel data streams of OFDM.

So, what they choose is to space the frequencies used by \$\Delta{}f = \frac{k}{T_U}\$, where \$T_U\$ is the bit period (the period of time used to transmit a single bit on one of the subcarriers) and k is any positive integer and is usually 1.

Then the relevant integral becomes

\$\int_{t_0}^{t_0+T_U}{\cos(2\pi{}(f_0+m/T_U)t + \phi_1) \cos(2\pi{}(f_0+n/T_U)t + \phi_2) \mathrm{d}t}\$,

and this integral you can prove is always 0.

Now to get back to your specific questions,

How do they determine what frequencies are orthogonal?

As long as the spacing between the frequencies is \$\frac{k}{T_U}\$, then the subcarrier signals will be orthogonal when measured over any given bit interval. This means your frequencies are give by, for example, \$f_0\$, \$f_0+1/T_U\$, \$f_0+2/T_U\$, ...

But of course, you don't have to use all of these frequencies. You can simply not use some of them. This is what Wikipedia is talking about when they talk about "coping with severe channel conditions without complex equalization filters." It just means, if a certain one of your carrier frequencies is blocked for some reason, you simply don't use it and continue to send data on all the other carriers.

  • \$\begingroup\$ It's been a while since I did some calculus. (I might have to get out my book...), but your question answered it for me, especially the 'In principle, any two non-equal frequencies are orthogonal if measured over a long enough time.' and 'This means your frequencies are give by, for example, f0, f0+1/TU, f0+2/TU, ...' I may revisit it later and try to work the math for myself. Thanks for the very detailed answer! \$\endgroup\$
    – Azendale
    Mar 1, 2013 at 6:32
  • \$\begingroup\$ I was lost looking at the equation you gave. Wouldn't it be : ∫t0+TUt0cos(2π(f0+m/TU)t+ϕ1)cos(2π(f0+n/TU)t+ϕ2)dt \$\endgroup\$
    – user98362
    Jan 26, 2016 at 20:34

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