I am learning control theory for school and I came across the basic transfer elements such as the integrator. We learned that the integrator has the transfer function F(s) = 1/s or if you use only the frequency F(ω)= 1/ω, so if the frequency doubles, the transfer function drops to a half and so on, as in this example:

Example of the transfor function of an integrator: Inductor

I have also researched and people wrote, that the amplitude decreases with a higher frequency: https://electronics.stackexchange.com/a/270517/311486

How can I adjust the size of an integrated signal?

Also, I plotted the functions, where it seems like for every period, the function with the higher freqency (green) fits two times into the function with the lower frequency (red):

Visualisation of sine of a frequency and double the frequency

So, I also tried to replicate these observations by calculating by myself, but I get the result that for double the frequency, the integral does not change. I would have expected the integral to become half of the integral of lower frequency, because I am still integrating over the same amount of time.

But when I calculate for $$2\pi f= 1$$ I get:

$$\int_{0}^{2\pi} \lvert sin(t) \rvert dt$$ $$= 2 \int_{0}^{\pi} \lvert sin(t) \rvert dt$$ $$= 2*(-cos(\pi)+cos(0))$$ $$= 4$$

and for $$2\pi f= 2$$ the same:

$$\int_{0}^{2\pi} \lvert sin(2t) \rvert dt$$ $$= 4 \int_{0}^{\pi/2} \lvert sin(t) \rvert dt$$ $$= 4*1/2*(-cos(\pi/2*2)+cos(0)$$ $$= 2*(-cos(\pi)+cos(0))$$ $$= 4$$

Does somebody maybe know, whether I made a mistake in my calculation, or if the integral actually behaves different from what I described? Thanks a lot for any ideas and help, please be welcome to leave a comment so that I can provide additional information!

  • 1
    \$\begingroup\$ Isn't integral of full sine wave periods always 0, because whatever the area of positive cycle adds up to, the negative area of negative cycle would cancel it back to 0? And the faster the sine wave, the less area the integrator accumulates on positive cycle compared to same integrator with slower sine wave. Makes sense even without math? \$\endgroup\$
    – Justme
    Jun 11 at 8:01
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    \$\begingroup\$ When dealing with transfer functions, use angular frequency, \$\small\omega\$ rad/sec. If necessary, convert to Hz at the end of the analysis. Having \$\small 2\pi f\$ floating around is really confusing. \$\endgroup\$
    – Chu
    Jun 11 at 9:41

2 Answers 2


You don't need equations to work this out. You can just think about it.

You already know that the area under the sine, from \$0\le\theta\le\pi\$ is 2. This is the sum of all the little tiny rectangular areas defined by, on their vertical side by a finite result of the sine function itself (the shape of which is invariant over a half-cycle), times the infinitesimal on the horizontal that is simply \$\text{d}\theta\$.

The integral is just the sum of an infinite number of infinitely thin rectangles stacked side to side.

Now, you can rescale the infinitesimal itself in any way you like. So, instead of an infinite number of \$\text{d}\theta\$ wide rectangles stacked side by side to span between \$0\$ and \$\pi\$, you could just as well instead use an infinite number of \$\text{d} t\$ wide rectangles stacked side by side to span between \$0\$ and \$\frac1{f}\$. The height of each is the same (defined by the sine.) It's just the width of each slice that varies. But since the relative width between them is some constant you create, \$\frac{\text{d}\theta}{\text{d}t}\$ (or \$\frac{\text{d}t}{\text{d}\theta}\$ if you prefer), then the area obviously scales, accordingly.

So it is no surprise to anyone that the sum across two half-cycles of a "squeezed" sine wave integral would be equal to the sum across one half-cycle of an "unsqueezed" sine wave integral. You could just as well take the two "squeezed" half-cycles, unpacking each tiny rectangle that are \$\text{d}t^{\,'}\$ wide, and then restack them so that two \$\text{d}t^{\,'}\$ wide rectangles (one from each of the two half-cycles) are combined to create a single rectangle that is \$\text{d}t\$ wide (where \$\text{d}t=2\:\text{d}t^{\,'}\$ or \$\frac{\text{d}\,t}{\text{d}t^{\,'}}=2\$) and use that new rectangle to add them back up in order to calculate the area for the resulting "unsqueezed" half-cycle.

The result isn't even hard to visualize.

  • \$\begingroup\$ OK, thanks a lot for that answer, but then its a bit tricky to understand, why the amplitude is supposed to decrease with higher frequency in the integrator: if I, for example, increase the frequency by one, then the area I lose, I gain back because I can put one more wave into the original interval (if I understood it right).. \$\endgroup\$
    – n328
    Jun 11 at 13:59
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    \$\begingroup\$ @n328 I'll probably need to see your definition of "the integrator" and where you find that "amplitude" is "supposed to decrease with higher frequency." Context is everything. It's possible that your "integrator" is only accepting a fixed number of half-cycles, for all I know. \$\endgroup\$
    – jonk
    Jun 11 at 14:06
  • \$\begingroup\$ okay right, we have this definition of the integrator from our script: $$x_a=\int_0^tx_edt$$.. ->Laplace-transform-> $$X_a=Xe/s \rightarrow F(\omega) = \frac{1}{\omega} $$ it says that $$F(\omega)$$ decreases every decade to 1/10 of its original amplitude, which I am trying to replicate, but a bit struggling, because it should in my view be in some kind related to the operation of integrating \$\endgroup\$
    – n328
    Jun 11 at 14:16
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    \$\begingroup\$ @n328 This is a very different question and I'm not interested in trying to answer it here. You actually already "kind of" have an answer from me on this in the earlier comment. But to explain it means an entirely different discussion. You need to shake off something in your perception. But the way to do that is just completely different than what I already wrote. I'd be mixing apples and oranges. \$\endgroup\$
    – jonk
    Jun 11 at 14:35
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    \$\begingroup\$ @n328 I'll toss a clue, though. Think in terms of dimensional analysis and consider what taking an integral does from that perspective. (Which also applies to Laplace, as well, which itself is an integral operator.) \$\endgroup\$
    – jonk
    Jun 11 at 15:09

Does somebody maybe know, whether I made a mistake in my calculation

Yes, I think so. Using \$|\sin ()|\$ is wrong here : -

$$\int_{0}^{2\pi} \lvert sin(t) \rvert dt$$

And, although the above does equal this: -

$$2 \int_{0}^{\pi} \lvert sin(t) \rvert dt$$

It doesn't if you remove the magnitude braces.

The integral of a full sinewave is zero.


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