Could someone explain for the following attachment (waveform image) how were exactly the values of "RMS" and "average" determined this way (for that curve of sine)? Given the duty cycle δ and period "Ts" (or frequency fs).

PS: I'm not asking about the [Total] RMS and average values of the given waveform but only about of that sine function curve/component of it!

a complex waveform including a sine function

Rms and avg of sine function curve between two points

  • \$\begingroup\$ Spencer - Hi, I see the images here are similar to those in your later question and have been copied / adapted from elsewhere. Therefore the same site rule on referencing applies here too. Those images need to be correctly referenced, so my comment on that later question should be followed here too. Please make the necessary edit to add those required references. TY \$\endgroup\$
    – SamGibson
    Commented Jun 20 at 21:48

3 Answers 3


The RMS value of an entire cycle of a sine wave (in relation to the peak) is the same for a 1/2 cycle or even 1/4 cycle starting at 0. You should be able to see that from symmetry and the fact that the waveform is squared.

So you can just integrate the sine wave \$\frac {i(t)} {I_{PK}} = \sin^2(\omega t) dt\$ from 0 to \$\pi/2\$ and and divide by the time \$\pi/2\$ (mean) and take the square root.

\$\frac{I_{RMS}}{I_{peak}} = \sqrt{\frac{2}{\pi}\int_0^{\pi/2}\sin^2(x)dx}\$ where \$x = \omega t\$

From trig identities, we know \$\sin^2(x) = \frac{1}{2}(1 - \cos(2x))\$

Evaluate the definite integral, the trig function disappears because of the beginning and end points, and you'll get the well-known relation for a sine wave:

\$\frac{I_{RMS}}{I_{peak}} = \frac{1}{\sqrt{2}}\$

You could as easily integrate to \$\pi\$, and it might be more straightforward.

The \$\sqrt{\delta}\$ factor comes in when you consider the blank time between the half cycles. When \$\delta =1 \$ you effectively have a whole sine wave. If \$\delta = 0.5 \$ you have 1/2 of the power of a sine wave and therefore 1/sqrt(2) the RMS current.

Average is similar, but easier since you don't have to square the sin. The \$\pi\$ shows up in the denominator as a result of integrating it over 1/2 (or 1/4) cycle.

P.S. in the general case mentioned in your title "sine wave defined between two points" you would, in general, find the trig functions don't disappear if you don't hit the integral multiples of pi/2 for both beginning and end points, but if you've worked out the integrals you just substitute the numbers for the actual beginning and end radians and the answer pops out without additional work. This way you can easily calculate the RMS and average values for a sine wave that is clipped at some voltage, phase control or reverse phase control waveforms, etc.


Well, the general formula for the RMS-voltage is given by:


Where \$\text{T}\$ is the period of the function.

For your case we get:

\begin{equation} \begin{split} \text{I}_\text{RMS}&=\sqrt{\frac{1}{\text{T}_\text{s}}\cdot\left(\int\limits_0^{t_1}\left(\text{I}_\text{a}\sin\left(2\pi\cdot\frac{1}{t_1}\cdot t\right)\right)^2\space\text{d}t+\int\limits_{t_1}^{t_1+t_2}0^2\space\text{d}t+\int\limits_{t_1+t_2}^{t_1+t_2+t_3}\text{I}_\text{b}^2\space\text{d}t+\int\limits_{t_1+t_2+t_3}^{t_1+t_2+t_3+\text{T}_\text{s}}0^2\space\text{d}t\right)}\\ \\ &=\sqrt{\frac{1}{\text{T}_\text{s}}\cdot\left(\int\limits_0^{t_1}\left(\text{I}_\text{a}\sin\left(2\pi\cdot\frac{1}{t_1}\cdot t\right)\right)^2\space\text{d}t+\int\limits_{t_1+t_2}^{t_1+t_2+t_3}\text{I}_\text{b}^2\space\text{d}t\right)}\\ \\ &=\sqrt{\frac{1}{\text{T}_\text{s}}\cdot\left(\text{I}_\text{a}^2\int\limits_0^{t_1}\sin^2\left(\frac{2\pi t}{t_1}\right)\space\text{d}t+\text{I}_\text{b}^2\int\limits_{t_1+t_2}^{t_1+t_2+t_3}1\space\text{d}t\right)}\\ \\ &=\sqrt{\frac{1}{\text{T}_\text{s}}\cdot\left(\text{I}_\text{a}^2\int\limits_0^{t_1}\sin^2\left(\frac{2\pi t}{t_1}\right)\space\text{d}t+t_3\text{I}_\text{b}^2\right)}\\ \\ &=\sqrt{\frac{1}{\text{T}_\text{s}}\cdot\left(\text{I}_\text{a}^2\cdot\frac{t_1}{2}+t_3\text{I}_\text{b}^2\right)}\\ \\ &=\sqrt{\frac{1}{\text{T}_\text{s}}\cdot\left(\frac{t_1}{2}\cdot\text{I}_\text{a}^2+t_3\text{I}_\text{b}^2\right)} \end{split}\tag2 \end{equation}


Could someone explain for the following attachment (waveform image) how were exactly the values of "RMS" and "average" determined this way (for that curve of sine)?


Fact: a half cycle of sine has the same RMS value as the full sinewave: -

$$I_{RMS} = \dfrac{I_{MAX}}{\sqrt2}$$.

When we want to consider how duty cycle affects the RMS value we must first convert to an equivalent power. As a thought experiment we can then convert the RMS to a power (making the assumption that it is passing through a 1 Ω resistor): -

$$P = \dfrac{I_{MAX}^2}{2}$$

This is the power dissipating in a 1 Ω resistor.

Then we know that if we only apply that power intermittently, the average power reduces as per the duty cycle so, we can modify the power formula by incorporating \$\delta\$ (the duty cycle):

$$P = \dfrac{\delta\times I_{MAX}^2}{2}$$

And finally, we can convert back to an RMS current (thus negating the 1 Ω resistor) by taking the square root: -

$$I_{RMS} = \sqrt{\dfrac{\delta\times I_{MAX}^2}{2}}\hspace{1cm} = \hspace{1cm} I_{MAX}\times \sqrt{\dfrac{\delta}{2}}$$


It is well known that the average of one half cycle is equivalent to the average of a full wave rectified sinewave: -

enter image description here

The image from here shows a voltage waveform but, it's the same for a current waveform. Hence: -

$$I_{AVG} = \dfrac{2\times I_{MAX}}{\pi}$$

And, because we are considering how duty affects the average value we get: -

$$I_{AVG} = \dfrac{2\times\delta\times I_{MAX}}{\pi}$$

In the previous RMS situation we had to convert to a power because power has an average intensity multiplied by duty but, because we have a simple formula for average current we can simply multiply directly by duty.


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