How to Find the RMS and THD of a +1, -1 volt square wave of fundamental frequency 60 Hz. The wave is symmetric in the sense +1 for 1/120 s, and -1 for 1/120 s
How to Find the RMS and THD of a +1, -1 volt square wave of fundamental frequency 60 Hz. The wave is symmetric in the sense +1 for 1/120 s, and -1 for 1/120 s
I think the RMS would be ((1^2)+(-1^2))^.5 or sq-rt of 2=1.4142.
I am not sure of the other term you are asking about. What is it not abreviated? Is it Total Harmonic Distortion?
[ February 03, 2006, 11:38 AM: Message edited by: spsnyder ]
I agree with 1.414 as being the RMS, and with the meaning of THD. But I do not know how to calculate THD.
Charles E. Beck, P.E., Seattle
Comments based on 2008 NEC unless otherwise noted.
THD = total harmonic distortion
Total harmonic distortion equals voltage (rms) distorted / voltage (rms) fundamental x 100
Ron
In this example, the RMS of the square wave is 1.414. The RMS of a pure sine wave is also 1.414. Are you saying that the THD of this signal is 100%?
Charles E. Beck, P.E., Seattle
Comments based on 2008 NEC unless otherwise noted.
Usually, THD is the ratio of the noise to the signal. For example, this is somewhat like the THD used for specifying the performance of audio equipment.
So if you want your signal to be a square wave, the THD is zero, because you have a square wave.
But if you are looking for a pure 60Hz sine wave, then THD would be defined as the ratio of the amplitude of all the noise to the amplitude of the 60 Hz sine wave. This is probably what you want. (I will assume we are talking about a voltage signal, and not power).
A perfect square wave is made up of an infinite number of sine waves, all odd multiplies of the fundamental. In your case, the fundamental sine wave is 60Hz, and it has a peak amplitude of 1 volt. The other frequencies are;
180 HZ at 1/3 volt
300 Hz at 1/5 volt
420 Hz at 1/7 volt
540 Hz at 1/9 volt
660 Hz at 1/11 volt
and on and on forever, with each one getting a smaller amplitude. (The nth term is n*60Hz in freq, and 1/n in amplitude.)
The RMS value of the square wave is actually 1. I'll leave the proof to Rattus, but basically you should have divided by 2 before taking the square root. Notice that the RMS value can't be larger than the peak value of the square wave.
So to get THD, we use the RMS value of our 60Hz wave, (which is a sine wave with a peak of 1v) to get .707. Subtract this from the total RMS of 1 to get a noise RMS value of 0.293. So I think the total THD is 0.293/.707 = 0.414 or 41.4% THD.
I'm not 100% sure of that answer. Maybe THD is sometimes defined as the noise over the total signal, which would be 29.3%.
Steve
To calculate THD, you have to decide what is the voltage rms fundamental
Ron
You are right. I missed that. The "M" word in "RMS" is "mean." You take 1 squared plus 1 squared, then divide by 2 to get the average before extracting the square root. One attaboy for Steve, one oops for me.Originally posted by steve66:The RMS value of the square wave is actually 1. I'll leave the proof to Rattus, but basically you should have divided by 2 before taking the square root.
Charles E. Beck, P.E., Seattle
Comments based on 2008 NEC unless otherwise noted.
The way I see it:
The peak value of the fundamental in this case is 4/pi = 1.273V, then the RMS value of the fundamental is 0.9V.
Total power is 1/R. Fundamental power is 0.81/R. Noise power is 0.19/R. Then,
THD = 0.19/0.81 = 0.235 or 23.5%.
An easy way to compute the RMS value of a square wave is to rectify it in your mind with an ideal full-wave rectifer which yields in this case 1Vdc which has by definition an RMS value of 1V.
Don't mess with B+!
(Signal Corps. Motto)
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