Finding Vrms over time

Question

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\$

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?

Answer 1

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

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: -

Answer 2

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.

Answer 3

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:

$$\text{V}_\text{RMS}=\sqrt{\frac{1}{15}\cdot\left\{\frac{8}{3}+32+\frac{20}{3}\right\}}=\frac{2}{3}\sqrt{\frac{31}{5}}\approx1.65999\space\text{V}\tag4$$