How to Derive the RMS Value of Pulse and Square Waveforms

The RMS value of a pulse waveform can be easily calculated starting with the RMS definition. The pulse waveform is shown in Figure 1. The ratio t1/T is the pulse signal duty-cycle. As shown in other articles in this website (MasteringElectronicsDesign.com:How to Derive the RMS Value of a Trapezoidal Waveform and MasteringElectronicsDesign.com:How to Derive the RMS Value of a Triangle Waveform), the RMS definition is an integral over the signal period as in equation (1).

pulse signalFigure 1

(1)

The pulse function, with the variable “time”, is a constant, which is the signal amplitude, between 0 and t1 and zero from t1 to T as in (2).

pulse waveform function (2)

where with u1(t) I noted the function of the waveform in Figure 1. After replacing u1(t) in equation (1) we can find the RMS value squared as in the following expression.

rms value calculation of a pulse signal (3)

Therefore, the RMS value of a pulse signal is

rms value of a pulse signal (4)

This expression can also be found as in (5)

rms value of a pulse signal as a function of duty-cycle (5)

where with D I noted the pulse signal duty cycle, D = t1/T.

What if the pulse signal is bipolar, as in Figure 2?

bipolar pulse waveform

Figure 2

In this case we should expect that the negative section of the signal to also contribute to the energy delivered to the load. To calculate its RMS value, let’s split the signal in two: from 0 to t1 and from t1 to T as in (6).

bipolar pulse waveform function (6)

where with u11(t) and u12(t) I noted the two sections of the waveform in Figure 2.

The RMS value of u11(t) is identical with the one shown in equation (3).

rms value of a bipolar pulse, the first section (7)

In a similar way, we can calculate the RMS value of u12(t):

rms value calculation of a bipolar pulse, second section (8)

The total RMS value of the bipolar pulse waveform is then calculated by applying the square root of the sum of squares of u11RMS and u12RMS.

rms value addition (9)

After calculations, the RMS value of a bipolar pulse waveform is

rms value of a bipolar pulse (10)

As you can see, the bipolar pulse RMS value does not depend on its duty-cycle, and it is equal with its amplitude.

Knowing the RMS value of a pulse waveform we can easily calculate the RMS value of a periodic square signal. The square wave in Figure 3 is a pulse signal with 50% duty-cycle. Its RMS value can be calculated from equation (5), where D = 1/2. Its RMS value is given in (11).

square wave

Figure 3

rms value of a square wave (11)

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42 thoughts on “How to Derive the RMS Value of Pulse and Square Waveforms”

  1. Hi,

    I found this page very usefull for a recap but i belive there is a mistake in equation 7, which refers to equation 4, it should refer to equation 3.

    At the moment it seems that the square of the Vp has vainish, to me it should be:

    U11^2rms = Vp^2 * t1/T — or as per equation 3, U11^2= Vp^2/T * t1

    Reply
  2. Salut. Stii cumva cum as putea afla valoare medie a unui semnal dreptunghiular nesimetric dandu-se doar factorul de umplere si frecventa?

    English translation: Hello. Do you know how can I find the average value of an asymmetric pulse signal, knowing just the duty-cycle and frequency?

    Reply
    • You can think of it like this: A pulse signal with its amplitude between 0 and Vp (its peak value) and a duty-cycle d, has the average value:

      VpulseAverage = d*Vp

      If the signal is asymmetric, it has a negative value Vneg and a positive value Vpos. So, it’s amplitude is Vpos – Vneg and it is shifted down with Vneg. Therefore, the average value is:

      VpulseAsymAverage = d * (Vpos-Vneg) + Vneg

      where Vneg is taken with its negative sign.

      Reply
  3. Hi Adrian.

    I have a question in relation to your post “Design a Bipolar to Unipolar Converter to Drive an ADC”.
    I noticed that the feedback is configured to attenuate, and was wondering how does this affect stability, and phase margin. If I configure an inverting amp that attenuates in a similar manner will it be stable?

    Also I can’t seem to launch an email from your web site. The verify for spam seems to be broken. Do you have an address I can send you a message?
    Regards

    Reply
  4. Hey Adrian,

    For the pulse in Figure 1, how do we find the AC rms value? I think I saw somewhere that the RMS^2=DC^2+ACrms^2 so ACrms^2=RMS^2-DC^2
    What would be the DC component in figure 1? Is it Vp/2? If so, then ACrms^2=Vp^2*D-Vp^2/4
    Let’s say D is 0.25, then ACrms=0 ??? Is this right?

    How do you find ACrms is the above method is not correct?

    Reply
    • For the pulse in Figure 1, how do we find the AC rms value? I think I saw somewhere that the RMS^2=DC^2+ACrms^2 so ACrms^2=RMS^2-DC^2
      Yes, correct.

      What would be the DC component in figure 1? Is it Vp/2?
      No, it is the average value, which is Vp * D, or Vp * (t1/T).

      Reply
  5. If I have a bipolar dc pulse that produces +6 volts and +6 amps and -6 volts and -6 amps would peak power be 141 watts?

    Reply
  6. Hi, Having trouble wrapping my brain around how a bipolar pulse rms can be independent of duty cycle. How can a signal with lower duty cycle(less on time), deliver the same power as a signal with higher duty cycle(more on time)?

    Reply
    • Actually, if it is a bipolar signal, there are no on/off states. The circuit that outputs the bipolar signal delivers power both during the time when the signal is positive as well as when it is negative. And RMS is a measure of power.

      Mathematically, raising the signal to the power of two eliminates the negative sign. Indeed, the RMS value depends only on the signal absolute value. It is like bringing the signal negative side to the positive side.

      If the signal is unipolar, then yes, you have an off time, when the signal caries no energy because it is zero, and so the RMS value depends on the duty-cycle as equation (5) shows.

      Reply
      • I’m sorry, you’re right. Instead of fig2, I was thinking of a pulse, with the same peak and width, both on pos and neg pulses, but with an off time involved. Just to be clear, if a signal has an off time, regardless of shape, then the duty cycle would have to be involved in the rms value, correct?

        Reply
  7. Hi,
    I have a sequence of 8 pulses with 0.5 duty- cycle. A T is 25 mks . Vp=120 V Then 20 ms follow 0VDC . And again 8 pulses like previously. RMS calculation throughout 20.2 ms observation window is close to 0.85 V . Is it right ?

    Reply
    • Your question is unclear. The 8 pulses repeat with a period T = 25 ms? Inside this 25 ms there is a 0 V amplitude with the duration of 20 ms? Or the 8 pulses have a duration of 25 ms, followed by 0 V with the duration of 20 ms?

      Better send me a link that points to a figure.

      Reply
  8. Hello, I have a bipolar rectangular wave with 20% Power Cycle. 10Vp (+-10V) 2KHz of frecuency and the multimeter indicates 8V, and with the same signal but 80% Power Cycle I got the same value of 8v, with 50% Power Cycle I got a value of 10V in the multimeter. Do you know why happens this?

    Reply
    • I am going to assume that you set your multimeter on RMS measurement. It measures correctly when the duty-cycle is 50% and then errors out as the duty-cycle goes to extremes. Most likely this is due to multimeter’s insufficient bandwidth. Comments invited.

      Reply
  9. In my smps design do I have to convert my squarewave pulses to rms before substituting it to the Bmax formula? Since Bmax formula is in rms value of sinewave.

    Reply
  10. Hi Adrian, I’m trying to determine the DC equivalent of a current pulse. Based on some articles I read, I need to use the RMS value, which brought me to this post. Is the RMS appropriate for this situation since the value will be the same regardless of period for a given Ipk – if I have a 50A pulse at 50% duty cycle, the RMS will be the same whether the period is 1 second or 1us? Do you think for this situation, the average value is more appropriate? Thanks.

    Reply
  11. Hi Adrian,

    I need to find the DC equivalent of a current pulse and based on some articles I read, it would be the RMS value. However, looking at the equation of RMS for a square wave, for a particular duty cylce say 50%, no matter what my period would be, the RMS will always be the same. For example, if I have a pulse of 100A, the RMS would be 70.7A whether the period is 1 sec. or 1 us. This doesn’t feel right because I would expect the DC equivalent for a 1sec period to be greater than a 1us period. What do you think? Is the average of the current pulse more appropriate? Thanks in advance for any input you can provide.

    Reply
    • The period of 1 sec has an on time that’s much longer than a period of 1usec, but then it also has a longer off time, so it cancels out.

      Reply
      • I am not sure what you are trying to say. The RMS value doe snot depend on time values, within reason. If you have a signal which is on for 3 days and then off for 3 days, then you can consider it as a DC signal depending on the corresponding processes in your system.

        For example: You try to determine what connector you can choose, based on the RMS value of the current that has to go through one pin. If you have a PWM signal of say, 100 kHz, then yes, you can use the RMS value. But if you have a signal with a period of 10 seconds, then you have to look at the maximum value of the current over 10 seconds. That spec becomes now DC, not AC.

        Reply

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