The plot of voltage over time is given in below plot.

As you can see,

\$V_{max}=2v \text{ and } V_{min}=0v\$

\$t1=2ms, t2=8ms, t3=5ms, \text{ and period, } T = 15ms\$

Question is to find \$V_{rms}\$ over \$T\$

Voltage over time

We know that $$V_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}V(t)^2dt}$$

Here's my attempt at finding \$V_{rms}\$

$$\begin{alignedat}{0} \require{cancel} V_{rms}&=\sqrt{\frac{1}{T}\int_{0}^{T}V(t)^2dt}\\ &=\sqrt{ \frac{1}{T}\int_{0}^{\frac{2}{15}T}(\frac{V_{max}}{\frac{2T}{15}}t)^2dt + \frac{1}{T}\int_{\frac{2}{15}T}^{\frac{10}{15}T}(V_{max})^2dt + \frac{1}{T}\int_{\frac{10}{15}T}^{\frac{15}{15}T}(\frac{-V_{max}}{\frac{5T}{15}}t)^2dt } \\&=\sqrt{ \frac{1}{T}\frac{V_{max}^2\cdot 15^2}{4T^2}\frac{t^3}{3}\rvert_{0}^{\frac{2T}{15}} + \frac{1}{T}(V_{max})^2(\frac{10T}{15}-\frac{2T}{15}) + \frac{1}{T}\frac{V_{max}^2\cdot 15^2}{25T^2}\frac{t^3}{3}\rvert_{\frac{10T}{15}}^{\frac{15T}{15}} } \\&=\sqrt{ \frac{1}{T}\frac{V_{max}^2\cdot 15^2}{4T^2}\cdot\frac{1}{3}\cdot\frac{2^3T^3}{15^3} + \frac{1}{T}(V_{max})^2(\frac{8T}{15}) + \frac{1}{T}\frac{V_{max}^2\cdot 15^2}{25T^2}\cdot\frac{1}{3}\cdot\frac{15^3T^3-10^3T^3}{15^3} } \\&=\sqrt{ \frac{1}{ \cancel{T} }\frac{V_{max}^2\cdot 15^2}{4\cancel{T^2}}\cdot\frac{1}{3}\cdot\frac{2^3\cancel{T^3}}{15^3} + \frac{1}{\cancel{T}}(V_{max})^2(\frac{8\cancel{T}}{15}) + \frac{1}{\cancel{T}}\frac{V_{max}^2\cdot 15^2}{25\cancel{T^2}}\cdot\frac{1}{3}\cdot\frac{15^3\cancel{T^3}-10^3\cancel{T^3}}{15^3} } \\&=\sqrt{ \frac{V_{max}^2\cdot 8}{4\cdot3\cdot15}+\frac{V_{max}^2\cdot 8}{15}+\frac{V_{max}^2\cdot(15^3-10^3)}{25\cdot3\cdot15} } \\&=\sqrt{ \frac{V_{max}^2\cdot8}{180}+\frac{V_{max}^2\cdot8}{15}+\frac{V_{max}^2\cdot2375}{1125} } \\&=\sqrt{ \frac{V_{max}^2\cdot8}{180}+\frac{V_{max}^2\cdot8}{15}+\frac{V_{max}^2\cdot2375}{1125} } \\&=\sqrt{ V_{max}^2\cdot0.04444+V_{max}^2\cdot0.53333+V_{max}^2\cdot2.1111 } \\&= V_{max}\cdot\sqrt{2.68887}=2\cdot1.639=3.28 \end{alignedat} $$

The book answer, however, shows the following method: $$\begin{alignedat}{0} \require{cancel} V_{rms}&=\sqrt{\frac{1}{T}\int_{0}^{T}V(t)^2dt}\\ &=\sqrt{ \frac{1}{T}\int_{0}^{\frac{2}{15}T}(\frac{V_{max}}{\frac{2T}{15}}t)^2dt + \frac{1}{T}\int_{0}^{\frac{8}{15}T}(V_{max})^2dt + \frac{1}{T}\int_{0}^{\frac{5}{15}T}(\frac{-V_{max}}{\frac{5T}{15}}t)^2dt } \\&=\sqrt{ \frac{V_{max}^2\cdot8}{180}+\frac{V_{max}^2\cdot8}{15}+\frac{V_{max}^2\cdot125}{1125} } \\&=\sqrt{ V_{max}^2\cdot0.04444+V_{max}^2\cdot0.53333+V_{max}^2\cdot0.1111 } \\&=V_{max}\cdot\sqrt{0.68885111}\\ &=2\cdot0.829971=1.6599 \end{alignedat} $$

Upon closer inspection, it looks like the integral for \$t3\$ is incorrect in my method, but for the life of me I can't figure out why it won't work. Where exactly did I do wrong?

  • 2
    \$\begingroup\$ \$V_{RMS}^2 = \dfrac{(\frac{4}{3}\times 7) + (4\times 8)}{15} = 2.75555 \text{ thus, } V_{RMS} = 1.65998661307\$ <-- use your knowledge about triangle waves and DC to compute it without all the rigour of the math. \$\endgroup\$
    – Andy aka
    Commented Dec 8, 2022 at 10:36
  • \$\begingroup\$ "Ideally", integration should be done on the (relative) interval [0,x] ... (3 functions) \$\endgroup\$
    – Antonio51
    Commented Dec 8, 2022 at 12:15
  • \$\begingroup\$ I hope you are planning on using a faster method if you ever have to answer a time limited (eg. exam) question like this one. Triangle squared integrated forward or backward are the same, so either kind of sawtooth and a triangle wave (even if asymmetrical) are the same.. \$\endgroup\$ Commented Dec 8, 2022 at 12:29

3 Answers 3


This can be solved in a far simpler (and practical EE) way: -

enter image description here

To get the RMS value of the composite waveform, you: -

  • Square the individual parts (triangle and DC) to get the respective powers into a 1 Ω resistor
  • Weight them individually with their duty cycle
  • Add the two weighted powers together and finally,
  • Take the square root to get back to RMS voltage and lose the 1 Ω dependency.
  1. For the triangle section, it's weighted power is \$\frac{4}{3}\times 7\div 15\$
  2. For the DC part it's just \$4\times 8\div 15\$
  3. Add them to get 2.755555
  4. Take the square root to get 1.65998661307

Proof of triangle waveform RMS: -

enter image description here


Your function definition of the third integral is indeed incorrect. $$ \frac{1}{T}\int^{\frac{15T}{15}}_{\frac{10T}{15}}\left(\frac{-V_{max}}{\frac{5T}{15}}t\right)^2 $$ should be $$ \frac{1}{T}\int^{\frac{15T}{15}}_{\frac{10T}{15}}\left(\frac{-V_{max}}{\frac{5T}{15}}(t-T)\right)^2 $$ You can see why this is the case by plugging in \$t=T\$ into the equation. This does not result in zero for your definition.

  • \$\begingroup\$ I tried it, and it works, but where does the -T come from? Is it because it did not start at T=0? How come it is not -10, since it starts at t=10msec? \$\endgroup\$
    – user97662
    Commented Dec 8, 2022 at 10:50
  • 1
    \$\begingroup\$ There are two known points for the function: \$f(t=T)=0\$ and \$f(t=\frac{10T}{T})=V_{max}\$. Solve for the unknowns in the target function \$f(t) = a(t-b)\$ and you will find the answer. \$\endgroup\$
    – jdum
    Commented Dec 8, 2022 at 11:00
  • \$\begingroup\$ Looks like my 'book' answer for the 3rd term is also incorrect, it somehow got to the right answer due to the symmetric property of the waveform. The correct way to write the 3rd term if one were to evaluate from 0 to \$\frac{5T}{15}\$ should be: \$\frac{1}{T}\int_{0}^{5T/15}(\frac{-V_{max}\cdot15}{5T})^2\cdot(t-5T/15)^2\$ \$\endgroup\$
    – user97662
    Commented Dec 8, 2022 at 11:12

The answer given by @Andyaka is good in it's rightfull way. I'll show the mathematical way of solving this.

Well, your voltage is piecewise defined:

$$ \text{V}\left(t\right):=\begin{cases} t&\space\text{if}\space0\leq t<2\\ \\ 2&\space\text{if}\space2\leq t<10\\ \\ 6-\frac{2}{5}\cdot t&\space\text{if}\space10\leq t<15 \end{cases}\tag1 $$

So, using the standard RMS function we find:

\begin{equation} \begin{split} \text{V}_\text{RMS}&=\sqrt{\frac{1}{\text{T}}\int\limits_0^\text{T}\left(\text{V}\left(t\right)\right)^2\space\text{d}t}\\ \\ &=\sqrt{\frac{1}{15}\int\limits_0^{15}\left(\text{V}\left(t\right)\right)^2\space\text{d}t}\\ \\ &=\sqrt{\frac{1}{15}\cdot\left\{\int\limits_0^2t^2\space\text{d}t+\int\limits_2^{10}2^2\space\text{d}t+\int\limits_{10}^{15}\left(6-\frac{2}{5}\cdot t\right)^2\space\text{d}t\right\}}\\ \\ &=\sqrt{\frac{1}{15}\cdot\left\{\int\limits_0^2t^2\space\text{d}t+4\int\limits_2^{10}1\space\text{d}t+\int\limits_{10}^{15}\left(6-\frac{2}{5}\cdot t\right)^2\space\text{d}t\right\}}\\ \\ &=\sqrt{\frac{1}{15}\cdot\left\{\left[\frac{t^{2+1}}{2+1}\right]_0^2+4\cdot\left[t\right]_2^{10}+\int\limits_{10}^{15}\left(6-\frac{2}{5}\cdot t\right)^2\space\text{d}t\right\}}\\ \\ &=\sqrt{\frac{1}{15}\cdot\left\{\frac{1}{3}\cdot\left[t^3\right]_0^2+4\cdot\left[t\right]_2^{10}+\int\limits_{10}^{15}\left(6-\frac{2}{5}\cdot t\right)^2\space\text{d}t\right\}}\\ \\ &=\sqrt{\frac{1}{15}\cdot\left\{\frac{2^3-0^3}{3}+4\cdot\left(10-2\right)+\int\limits_{10}^{15}\left(6-\frac{2}{5}\cdot t\right)^2\space\text{d}t\right\}}\\ \\ &=\sqrt{\frac{1}{15}\cdot\left\{\frac{8}{3}+32+\int\limits_{10}^{15}\left(6-\frac{2}{5}\cdot t\right)^2\space\text{d}t\right\}} \end{split}\tag2 \end{equation}

Now, let \$x=6-\frac{2}{5}\cdot t\$ this gives:

$$\int\limits_{10}^{15}\left(6-\frac{2}{5}\cdot t\right)^2\space\text{d}t=-\frac{5}{2}\int\limits_2^0x^2\space\text{d}t=\frac{5}{2}\int\limits_0^2x^2\space\text{d}t=\frac{5}{2}\cdot\left[\frac{x^{2+1}}{2+1}\right]_0^2=\frac{20}{3}\tag3$$

So, we end up with:



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