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A sawtooth waveform is fed to "Average reading (with full scale rectification)" ac electronic voltmeter. This voltmeter is calibrated for RMS value of pure sinusoidal input. What will be reading displayed on voltmeter? Also find percentage error in the reading. Take amplitude as 10 volts and time period of 1 second.

The given answers are:

  1. RMS indication = 6.5 V
  2. error = 12.7%

Here's what I have done so far

\$V_{RMS}\$ of sawtooth is 5.77 V and average is 5 V. \$V_{RMS}\$ of sinusoid is 7.07 V and average is 6.37 V. So the ratio of \$V_{RMS}\$ to \$V_{avg}\$ is 1.11 and 1.154 of sinusoid and sawtooth respectively. But the question here is that voltage of sawtooth waveform is being measured by sinusoidally calibrated voltmeter and I don't understand how to find the value that will be measured and what will be the error.

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    \$\begingroup\$ Show us what work you have so far \$\endgroup\$ – placeholder Oct 12 '14 at 12:59
  • \$\begingroup\$ I will find rms values and the resultant calculation. Just tell me what to do. Please just tell me how to start \$\endgroup\$ – shivam Oct 12 '14 at 13:22
  • \$\begingroup\$ I can find average and rms and peak reading of sinusoid and sawtooth waveform.That part I will do. Just tell me steps. I am not able to understand what I am supposed to do in this \$\endgroup\$ – shivam Oct 12 '14 at 13:24
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    \$\begingroup\$ Raw homework just dumped on us is off topic here. \$\endgroup\$ – Olin Lathrop Oct 12 '14 at 13:27
  • \$\begingroup\$ Really!? You have no idea whatsoever how to proceed? If so, then you shouldn't be in that class because surely the definition of RMS was discussed. What is the RMS of the sawtooth? What is the average after rectification? What factor does the meter apply to show RMS assuming the input is a sine? What will it therefore report as the RMS of the sawtooth? You really should have been able to come up with these questions on your own. \$\endgroup\$ – Olin Lathrop Oct 12 '14 at 13:31
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The key part in the question is that it's an "average reading (with full scale rectification)" rather than a true RMS calculation. What this means is that the input signal is rectified and the average value is then effectively multiplied by a constant with the assumption the input waveform was a pure sinusoid.

The definition of RMS for a periodic waveform is: $$ V_{RMS} = \sqrt{\frac{1}{T}\int_0^T v^2(t) dt}$$

The average value with full scale rectification would be: $$ V_{avg} = \frac{1}{T}\int_0^T |v(t)| dt$$

So the ratio of these for a pure sine wave would be the multiplication factor \$k\$ built into the meter. That is:

$$\begin{eqnarray} k &=& \frac{V_{RMS}}{V_{avg}}\\ k &=& \frac{\sqrt{\frac{1}{T}\int_0^T v^2(t) dt}}{\frac{1}{T}\int_0^T v(t) dt}\\ k &=& \frac{\sqrt{\frac{1}{2\pi}\int_0^{2\pi} \sin^2(t) dt}}{\frac{1}{2\pi}\int_0^{2\pi} |\sin(t)| dt}\\ k &=& \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\pi}}\\ k &=& \frac{\pi}{\sqrt{8}}\\ k &\approx& 1.11 \end{eqnarray}$$

In other words, the measured rectified average voltage will be multiplied by about 1.11 to get the indicated value.

To answer the first part of the question, you need to find \$V_{avg}\$ for the sawtooth waveform, which you have already correctly done and found it to be 5V.

What the meter will then do is multiply that by 1.11 and it will indicate 5.55V (not 6.5V -- your given answer is not correct, which was probably the source of your problem).

To determine the percentage of error is simply to compare that value with the true RMS value, which again, you have correctly calculated as 5.77V. $$err = \frac{5.55 - 5.77}{5.77} = -0.0381 = -3.81\%$$

So the percentage of error is -3.81%, with the negative sign meaning that the indicated value is lower than the actual. Here again, the given answer was simply not correct.

If this is from a textbook, you might want to see if there are published errata that corrects that. If not, you might send the author a note -- although it's too late for you, it could save countless hours of frustration for future students.

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  • \$\begingroup\$ I know that part. rms of sawtooth is 5.77 V. avg is 5 V. rms of sinusoid is 7.07 V and average is 6.37 V.Form factor is 1.11 and 1.154 of sinusoid and sawtooth respectively But the question here is that voltage of sawtooth waveform is being measured by sinusoidally calibrated voltmeter .So what will be the value that will be measured and waht will be the error? THatis what I want to ask. \$\endgroup\$ – shivam Oct 12 '14 at 13:36
  • \$\begingroup\$ You should add that to your question to clarify what you already know and what part is really the crux of your question. \$\endgroup\$ – Edward Oct 12 '14 at 13:45
  • \$\begingroup\$ I've updated my answer based on the revision of your question. \$\endgroup\$ – Edward Oct 12 '14 at 16:36
  • \$\begingroup\$ +1 The precise error will be 100 * (1-(pi/4)*(sqrt(3/2))) % \$\endgroup\$ – Spehro Pefhany Oct 12 '14 at 16:55
  • \$\begingroup\$ Edward Thanks very much, You are right, it may be an error in book \$\endgroup\$ – shivam Oct 13 '14 at 3:39
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It seems you know the AC voltmeter will remove the DC component, full wave rectify (find absolute value), average that, then multiply by 1.11 (your number, I haven't checked) to report RMS. This works fine when the voltage to be measured is a sine.

The meter will do the same thing with any signal. It will remove the DC component, absolute value, average, then multiply by 1.11. It seems you have already done the hard part. Follow this process for the sawtooth. That tells you what the meter will read. Since you already know the right answer for the sawtooth, you compare the meter reading to that to determine how much off it will be.

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