# Solving the RMS of waveform

I was asked to find the form factor of this voltage wave:

But before getting the form factor, I should know the RMS value and the average value of the waveform. Knowing that the

$$V_{\text{rms}}=\sqrt{\frac{1}{3}\left(\int_{0}^{1}(4t)^{2}\:\mathrm{d}t + \int_{1}^{2} 4^{2}\:\mathrm{d}t + \int_{2}^{3} v^{2}\:\mathrm{d}t\right)}\tag{1}$$

My only problem is I can't solve the RMS by its formula because I don't know what to substitute on integral of $v^2$. Please teach me how to find the equation for the parabola so that I can substitute it into $(1)$ I also can't understand the "piecewise" mentioned on some pdf's. I found the value $4t$ by using ratio and proportion..

Any help will be truly appreciated. Thanks

• What's the formula for the parabola dude? Commented May 6, 2014 at 15:05
• according to google, the formula for the parabola that fits on the problem is x^2 = 4*a*y and f = (0,a) Commented May 6, 2014 at 15:09
• Hint: The integral of a 2nd order polynomial is a 3rd order polynomial. Polynomials are easy to integrate, since you can integrate each term separately and sum the results. Commented May 6, 2014 at 15:18
• i can't integrate the parabola part since i can't find its equation. can you tell me how to find it? Commented May 6, 2014 at 16:19
• nicy12 - I'm sorry to tell you this, but you are out of your depth. If you don't know how to integrate a function, you almost certainly don't know how to differentiate one, either. Without being able to do this, you will find yourself increasingly baffled by what your courses are asking for. I suggest you drop the EE course, and take a calculus course instead. THEN take the EE course. Commented May 6, 2014 at 16:51

$$\int_{0}^{1}v(t-2)^{2}\:\mathrm{d}t=\int_{2}^{3}v(t)^{2}\:\mathrm{d}t$$

Using this, and the generic formula for a parabola with it's vertex at the origin, we have that: $v(t-2)=4t^{2}$, we therefore have that:

$$\int_{2}^{3}(v(t))^{2}\:\mathrm{d}t=\int_{0}^{1}(4t^{2})^{2}\:\mathrm{d}t=\int_{0}^{1}16t^{4}\:\mathrm{d}t=\left[\frac{16t^{5}}{5}\right]_{0}^{1}=\frac{16}{5}$$

I hope this helps you!

EDIT: In order to clarify how I arrived at the equation $v(t-2)=4t^{2}$, we can examine the general equation of a parabola, given by: $$y(x)=a(x-b)^{2}+c$$

Where $a$ is a parameter which modifies the gradient of the parabola and $b$ and $c$ determine the vertex (bottom-/top-most point) of the parabola (at $(x,y)=(b,-c)$), in this case we have that the vertex of $v(t-2)$ is at the point $(0,0)$ (imagine shifting the graph 2 units to the left) and therefore $b=-c=0$ and therefore we have the equation: $$v(t-2)=a(t-2)^{2}$$

Using the fact that when $t-2=1$, $v(t-2)=4$, we get $a=4$.

• How did you conclude the first line in your algebra? Is it something related to the maths behind a parabola? Commented May 6, 2014 at 16:45
• i think you're asking in v(t-2)=4t^2 ? v(t-2)=4t^2 bothers me now. Commented May 6, 2014 at 16:46
• @Andyaka, you can get it through simple intuition by viewing the Riemannian integral $\int_{a}^{b}f(x)\:\mathrm{d}x$ as the area under the curve $f(x)$ in the interval $[a,b]$, does that help? Commented May 6, 2014 at 16:48
• @nicy12 What do you mean, you don't know how I arrived at: $v(t-2)=4t^2$? Commented May 6, 2014 at 16:49
• sad to say but yes sir shaktal. Commented May 6, 2014 at 16:51