Comments on: How to Derive the RMS Value of a Triangle Waveform https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/ Electronics Design and Modeling with Emphasis on Analog Design Fri, 19 Jan 2024 04:54:18 +0000 hourly 1 By: Adrian S. Nastase https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-52340 Fri, 19 Jan 2024 04:54:18 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-52340 In reply to simo.

It is the mathematical expression of a linear function that goes through 2 points:
– first point is the intersection of the function with the time axis at (t2, 0).
– second point is a random point (u52(t),t) between the apex and the intersection with the time axis, where with u52(t) I noted the value of the function at time t.

]]> By: simo https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-51195 Tue, 01 Aug 2023 16:30:39 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-51195 hi
please i’m seeking an answer to this one
in figure 5 how did we get that expression of vp.(t2-t)/(t2-t1)
and how we’re able to get one without an y intercept

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-47095 Sun, 03 Jan 2021 21:41:08 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-47095 In reply to nadir khan.

As I said in my previous posts, you use the function of a straight line that connects two points, as we learned in the basic algebra classes:
f(x) = ((y1-y0)/(x1-x0)) * (x-x0) + y0

In our case, u3(t) is the function and the variable is time. Replace y0 with Vp, x0 with 0, y1 with 0 and x1 with t1, as shown in the graph. The result is:

u3(t) = ((0-Vp)/(t1-0)) * (t-0) +Vp = -Vp * t/t1 + Vp = Vp * (t1-t)/t1

]]> By: nadir khan https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-46996 Tue, 15 Dec 2020 12:00:15 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-46996 In reply to Adrian S. Nastase.

hy adrian verry very explained article but …i tried 4,5 times for fig 3,U3 in eq (6) ..how u bring “t” ..and how to solve this ..plzzz..
ppllzz ..i will wait for ur reply

nadir.hpr@gmail.com

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-43537 Fri, 07 Sep 2018 23:23:14 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-43537 In reply to Aditya Sharma.

1. It is the function of a straight line connecting to points on the graph: (0,0) and (t1,Vp). A quick look in a math textbook will show that the function is defined as f(x) = (y1-y0)/(x1-x0) * (x-x0) + y0, where (x0,y0) and (x1,y1) are the two points connected by the straight line and (x,y) is an arbitrary point on the line.

2. Just replace t1 and t2 with your values in equation 13.

]]> By: Aditya Sharma https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-43168 Fri, 24 Aug 2018 02:35:30 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-43168 Hello Sir
Very nice post but please let me know how did you get equation (2). Also explain how to solve equation (5 2) for a graph with t1 and t2 to be equal to T/2 and T respectively.

Once again thank you for such a nice post
Good bye

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-42944 Tue, 17 Jul 2018 14:28:17 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-42944 In reply to Steve.

Good post, Steve. Most likely, not confusing but a different way of looking at the graphs.

]]> By: Steve https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-42906 Tue, 10 Jul 2018 13:24:37 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-42906 The last result is correct but confusing to the reader. A better comparison would be using waveforms of equal amplitude. Figure 6 has a peak-to-peak amplitude of Vp. Figure 7 has a peak-to-peak amplitude of 2*Vp. If the same Vp-p voltages are used for both, the equations are:

Figure 6: Vp/ROOT(3)

Figure 7: Vp/(2*ROOT(3))

Why this is important is that it illustrates the impact on RMS of adding a DC component to the waveform. There are many real-world systems that have sawtooth waveforms that maintain a constant peak-to-peak voltage/current but have varying DC components. The RMS value changes as a result.

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-42248 Sun, 18 Feb 2018 01:30:50 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-42248 In reply to Conan.

You need to be specific. Which linear functions?

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-triangle-waveform/#comment-42247 Sun, 18 Feb 2018 01:29:50 +0000 http://masteringelectronicsdesign.com/?p=1455#comment-42247 In reply to prash.

it is the expression of a liner function from basic algebra. Given two points in the plane x1,y1 and x2,y2, the linear function defined by these two points is
y(x) = (y2-y1)/(x2-x1)x + (x2y1-x1y2)/(x2-x1)

in our case x1 = 0, y1 = 0 because the function goes through zero. x2 = t1 and y2 = Vp. Replace in the above formula and you will find eq 2.

t is the variable time.

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