
Determine the RMS amplitude of the squarewave signal.
Came across this question on a practice test. Here is the corresponding picture and full question.
Determine the RMS amplitude of the square wave signal, as displayed by an oscilloscope with a vertical sensitivity of 0.5 volts per division.
The answer is 0.5V apparently, but I thought that the answer would be 0.35V. My thought process:
The peak to peak voltage would be 1V, and the positive peak would be 0.5V. Thus if I was asked for the peak voltage, I would have stated the answer was 0.5V. Since I was asked for the RMS voltage, I instead took that 0.5V and divided it by the square root of 2, which is 0.35V.
Am I correct, or is the test answer correct? If I am wrong, could someone please explain where I went wrong in my thought process?

Essentially, the RMS value is the DC equivalent value which will produce the same heating effect. The RMS value is the square root of the mean (average) value of the squared function of the instantaneous values.
After some calculus, which you can look up if interested, a sine wave results in V_{RMS}=V_{PEAK}/√2
For a square wave this results in V_{RMS}=V_{PEAK}
Thinking about it in basic terms of a heating effect, you have a constant unchanging magnitude, with only polarity changing periodically, so it is a direct equivalent of a DC voltage.

Originally Posted by
Erased
Essentially, the RMS value is the DC equivalent value which will produce the same heating effect. The RMS value is the square root of the mean (average) value of the squared function of the instantaneous values.
After some calculus, which you can look up if interested, a sine wave results in V_{RMS}=V_{PEAK}/√2
For a square wave this results in V_{RMS}=V_{PEAK}
Thinking about it in basic terms of a heating effect, you have a constant unchanging magnitude, with only polarity changing periodically, so it is a direct equivalent of a DC voltage.
Ah, this makes a lot of sense. Thank you for the clarificafion!

Originally Posted by
Erased
Essentially, the RMS value is the DC equivalent value which will produce the same heating effect. The RMS value is the square root of the mean (average) value of the squared function of the instantaneous values.
After some calculus, which you can look up if interested, a sine wave results in V_{RMS}=V_{PEAK}/√2
For a square wave this results in V_{RMS}=V_{PEAK}
Thinking about it in basic terms of a heating effect, you have a constant unchanging magnitude, with only polarity changing periodically, so it is a direct equivalent of a DC voltage.
A square wave is the result of a bridge rectifier, which many technicians use to simulate DC, then smarter technicians apply a capacitor to get a smoother DC signal, even smarter techs use a regulated DC power supply.

Originally Posted by
Kalbi_Rob
A square wave is the result of a bridge rectifier, which many technicians use to simulate DC, then smarter technicians apply a capacitor to get a smoother DC signal, even smarter techs use a regulated DC power supply.
I agree with the rest of what you said, but a bridge rectifier doesn't produce a square wave. Bridge rectifiers produce a pulsed DC. https://en.wikipedia.org/wiki/Pulsed_DC
Subscribe
Related Topics

By Ronwilson1801 in forum NETA Level 2 Exam
Replies: 4
Last Post: July 26, 2020, 04:14 PM

By whiskers in forum NETA Level 3 Exam
Replies: 9
Last Post: August 17, 2019, 06:22 PM

By rasilva in forum NETA Level 3 Exam
Replies: 5
Last Post: December 21, 2018, 06:25 PM

By Machine Gun Kelly in forum Electrical Testing Talk
Replies: 3
Last Post: April 15, 2016, 12:09 PM
Tags for this Thread