# Finding the average and RMS value of this signal

How can I find the RMS and the average voltage of the following diagram?:

My work:

• The RMS: $$\sqrt{\left(\sqrt{\frac{1}{3\times10^{-3}}\int_0^{3\times10^{-3}}\left(\frac{5000}{3}\cdot t\right)^2\space\text{d}t}\right)^2+\left(\sqrt{\frac{1}{10^{-3}}\int_0^{10^{-3}}5^2\space\text{d}t}\right)^2+\dots}\tag1$$
• The average: $$\frac{1}{10\times10^{-3}-0}\cdot\left\{\int_0^{3\times10^{-3}}\frac{5000}{3}\cdot t\space\text{d}t+\int_0^{10^{-3}}5\space\text{d}t+\dots\right\}\tag2$$

Am I right about my work?

• For the RMS values: (1) You are squaring each of the square-roots which cancels them out. If this was to remove any negatives then that was covered inside the integral term already. You've created an RSRMS equation! (2) The $\frac {5000}{3}t$ term is strange. Should that be $\frac {5}{3}t$? (I've only had a quick look.) – Transistor Mar 20 '18 at 15:31

I too would be inclined to sort this out graphically initially but the maths is required to back it up.

Remember that RMS is the root of the mean of the squares.

• The RMS: $$\sqrt{\left(\sqrt{\frac{1}{3\times10^{-3}}\int_0^{3\times10^{-3}}\left(\frac{5000}{3}\cdot t\right)^2\space\text{d}t}\right)^2+\left(\sqrt{\frac{1}{10^{-3}}\int_0^{10^{-3}}5^2\space\text{d}t}\right)^2+\dots}\tag {1a}$$

We don't need to square and root each portion of the curve. It does nothing to the result as it is already positive due to the squaring inside each integral term.

$$\sqrt{ \frac{1}{3 \times 10^{-3}} \int_0^{3\times10^{-3}} \left (\frac{5}{3 \times 10^{-3}}\cdot t\right)^2\space\text{d}t + \frac{1}{10^{-3}}\int_0^{10^{-3}} 5^2\ \text{d}t +\dots } \tag{1b}$$

We can make this much more readable by ignoring the fact that it's milliseconds. (Another way of looking at this is that the RMS value would be the same if the time scale was in seconds or minutes.)

$$\sqrt{ \frac{1}{3} \int_0^{3} \left (\frac{5}{3}\cdot t\right)^2\space\text{d}t + \int_0^{1} 5^2\ \text{d}t +\dots } \tag{1c}$$

• The average: $$\frac{1}{10\times10^{-3}-0}\cdot\left\{\int_0^{3\times10^{-3}}\frac{5000}{3}\cdot t\space\text{d}t+\int_0^{10^{-3}}5\space\text{d}t+\dots\right\}\tag{2a}$$

Again, ignoring the milliseconds we can simplify this to

$$\frac{1}{10}\cdot\left\{\int_0^3 \frac{5}{3}\cdot t \ \text{d}t \ + \int_0^1 5\space\text{d}t+\dots\right\}\tag {2b}$$

You are making this way too complicated. Maybe those equations are right, but I didn't bother looking at them.

Stop and actually think about the situation you have.

Start with the average. That's easy to do from inspection. You can immediately see that from t = 0 to 3 is a ramp, so the average is the average of its ends. The average of the first 3 ms is therefore 2.5. The next two ms cancel to average to 0. The next two is one full sine cycle, so the average is again 0. The remainder of this signal to 10 ms has two parts at 5 and 1 part at 10. So to recap, you have 3/10 of 2.5, 4/10 of 0, 2/10 of 5, and 1/10 of 10. You don't even need a calculator to average that.

Attack the RMS similarly. You can average the RMS of each 1 ms time segment. The only moderately tricky part is finding the RMS of the first 3 ms. After that, it's either provided outright, or comes from something you should already know without having to look it up.

Just do it.

I believe, during the linear periods (t 0-3 and 7-10), RMS and average are equivalent. During t = 3 to 7 voltage cancels to 0V. Therefore, RMS & average = (2.5 x 3) + 5 + 10 + 5 = 27.5; 27.5/10 = 2.75 volts. ...I think...

• This is wrong. No, the average and RMS are not the same from t = 0 to 3. – Olin Lathrop Mar 20 '18 at 16:32
• @ddurgin: Work out the RMS for 3 to 4: $V_{RMS} = \frac { \sqrt { {5^2} } }{1} = 5$. The RMS for 4 to 5: $V_{RMS} = \frac { \sqrt { {-5^2} } }{1} = 5$. While the average for those two periods is zero the RMS is definitely not. – Transistor Mar 20 '18 at 18:48