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I wonder why under assumption that \$\omega \gg \frac{1}{T}\$ then \$\int_{0}^{T} \sin(\omega t)dt \approx 0\$?

Since the integral should be like \$\frac{\cos(\omega t)}{w}\$ from \$0\$ to \$T\$ and after plugging the valued we will end up with :

$$\frac{-\cos(\omega T)+1}{\omega}$$

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    \$\begingroup\$ I'm voting to close this question as off-topic because it does not relate to electronics and is a pure math based question, and so should belong on math.stackexchange.com \$\endgroup\$ – efox29 Dec 3 '15 at 8:22
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    \$\begingroup\$ Absolutely not. This estimation is used in all communication system and is not pure math question since in terms of mathematics only this integral is not always zero \$\endgroup\$ – user59419 Dec 3 '15 at 8:23
  • \$\begingroup\$ Do you mean \$ \frac{1}{T}\int ...\$? \$\endgroup\$ – Chu Dec 3 '15 at 8:34
  • \$\begingroup\$ No. there is no \$\frac{1}{T}\$. If \$\frac{1}{T}\$ is present it makes sense and I have seen it in various places. \$\endgroup\$ – user59419 Dec 3 '15 at 8:40
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If you are talking about telecommunications, I assume we are talking about high frequencies. If that is the case:

  • \$\frac{1}{T} = f\$
  • \$\omega \gg \frac{1}{T}\$

\$−\cos(\omega T)+1\$ ranges from \$0\$ to \$+2\$, if you divide this by a big number you get approximately zero.
To give you an idea: for a frequency around \$1\;\text{kHz}\$ (which is considered "ultra low"), the result will be AT MAXIMUM \$0.002\$.

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    \$\begingroup\$ Much better explanation than my brute force approach. \$\endgroup\$ – Arsenal Dec 3 '15 at 9:24
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    \$\begingroup\$ I don't think this is the full answer: it's possible for even small values of \$\omega\$ to satisfy \$\omega \gg \frac1T\$, if \$T\$ is large enough. \$\endgroup\$ – Ilmari Karonen Dec 3 '15 at 18:20
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    \$\begingroup\$ @IlmariKaronen T is never large enough in the telecommunications. \$\endgroup\$ – FMarazzi Dec 3 '15 at 18:48
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By increasing the frequency, we're putting more oscillation periods in the integration interval.

Since the integral of a sine over one period is zero, we should only consider the "incomplete" period at the end of the integration interval.

When we increase the frequency, the area of this incomplete period becomes thinner and thinner (explaining the \$\omega\$ in the determinator).

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If I plug in some values, I get the following:

\$T = 1\$

\$\omega \rightarrow\$ result

\$10^0 \rightarrow 0.460\$

\$10^1 \rightarrow 0.184\$

\$10^2 \rightarrow 0.001\$

\$10^3 \rightarrow 4.376E-04\$

\$10^4 \rightarrow 1.952E-04\$

\$10^5 \rightarrow 1.999E-05\$

\$10^6 \rightarrow 6.325E-08\$

Now I'm not sure which order of magnitude \$>>\$ signifies and how small the result must be to be considered \$\approx 0\$, but it tends to get zero if it is much larger.

What are the typical values for \$\omega\$ and T you are looking at?


Update (because of the comments):

As FMarazzi has explained quite well there is an upper boundary for the case that \$\cos(\omega T)\$ is -1, so you'll have \$\frac{2}{\omega}\$, which is the absolute maximum you will ever get for any T.

So if you choose the value for T, in a way you get the maximum for a given \$\omega\$ the table turns into:

\$\omega \rightarrow\$ maximum possible value

\$10^0 \rightarrow 2\$

\$10^1 \rightarrow 0.2\$

\$10^2 \rightarrow 0.02\$

\$10^3 \rightarrow 2E-03\$

\$10^4 \rightarrow 2E-04\$

\$10^5 \rightarrow 2E-05\$

\$10^6 \rightarrow 2E-06\$

And so on. I don't know in which context the approximation is used, but as pointed out by the comments it is for communication systems, and my guess would be that those are not about some UART at 9600 baud but something like ethernet or faster things, so \$\omega\$ is in the order of \$10^7\$ or higher, for which the result of the integral gets small and probably doesn't contribute to the other terms of interest.

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  • \$\begingroup\$ Thanks. Your question definitely makes sense and that's exactly my problem because range of T and w is not given and only condition that wT>>1 is mentioned. I was thinking what if T=1000 and w=1 then the integral is not zero. \$\endgroup\$ – user59419 Dec 3 '15 at 9:11
  • \$\begingroup\$ If T is arbitrary, the area under sin(wt) will, generally, be non-zero. There must be another constraint. \$\endgroup\$ – Chu Dec 3 '15 at 10:00
  • \$\begingroup\$ @Chu I'm not saying that it will be 0, it just tends to be very close to 0, so close that for practical purposes it can be neglected (this is a common simplification to make things solvable for humans). FMarazzi has actually given a better analysis of the upper bound of the result. \$\endgroup\$ – Arsenal Dec 3 '15 at 10:36
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    \$\begingroup\$ @Arsenal, but you've assumed a value for T. There is no such specification in the original question - both w and T are free to wander. So the integral could be a long way from zero \$\endgroup\$ – Chu Dec 3 '15 at 11:02
  • \$\begingroup\$ @Chu yeah that was a bit short-sighted in hindsight. I've updated my answer to make the point clear. It cannot be a long way from zero for higher omegas. \$\endgroup\$ – Arsenal Dec 3 '15 at 12:19
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In the equation as written a larger \$\omega\$ will result on average in a smaller value of the integral but a larger \$T\$ will not.

I suspect more context is needed to properly understand what is meant.

In particular we need to think about what exactly we mean by "\$\approx 0\$". "\$\approx 0\$" should probablly be intepreted as "negligable" but what "negligable" means is highly dependent on context. If there is some related value that increases with increasing values of \$T\$ then it may be that the result of the integral when large \$T\$ is large but \$\omega\$ is small can still be considered negligable.

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