What is the RMS value of a periodic signal? When a periodic signal is generated by a source connected to a load, a resistor for example, the RMS value is the continuous signal, the DC value which would deliver the same power to the load as the periodic signal.

This article shows how to derive the RMS value of triangle waveforms with different shapes and duty cycles.

The triangle waveform in Figure 1 has a slower rise time than the fall time. In this case, the fall time is small so that it can be considered zero. If it is not zero, read further on deriving the RMS value of a triangle with comparable rise and fall times.

The RMS of this signal is calculated starting with the RMS definition, given in (1).

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(1) |

The linear function from 0 to t1 has the following expression,

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(2) |

where with u1(t) I noted the linear function of the signal in Figure 1. This event is repeated every T period. After replacing (2) in equation (1), the RMS is

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(3) |

or

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(4) |

In equation (4) t1/T is the signal duty-cycle. If the triangle signal duty-cycle is 100%, as in Figure 2

then t1 = T and the RMS value becomes

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(5) |

If we need to find the RMS value of a triangle with a fast rise time and slow fall time as in Figure 3

we can start from the same definition (1), taking into account that the linear function from 0 to t1 is as in equation (6).

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(6) |

After replacing (6) in (1), the RMS value is

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(7) |

Let’s change the variable to x = t1 – t. The integral becomes

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(8) |

Therefore, the triangle signal shown in Figure 3 has the same RMS value as the signal in Figure 1:

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(9) |

If the duty-cycle is 100%, as in Figure 4, then t1 = T and the RMS value is

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(10) |

What if the triangle signal has the rise time and fall time comparable, as in Figure 5?

In this case we can calculate the RMS value by splitting the waveform in two: from 0 to t1 and from t1 to t2. The waveform expression in the time domain is given in (11),

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(11) |

where with u51(t) I noted the waveform section from 0 to t1, and with u52(t) I noted the section from t1 to t2.

We will use the square-root of sum of squares to calculate the RMS value of the waveform in Figure 5. The RMS value squared of u51(t) is already calculated in (3), and the result is

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(12) |

Also, the RMS value squared of u52(t) is calculated in (7) and (8) with the difference that (t1 – t) / t1 is replaced by (t2 – t) / (t2 – t1). The calculations are the same. The variable change is x = t2 – t. Therefore, the RMS value of u52 squared is

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(13) |

We can calculate now the RMS value of the triangle waveform in Figure 5, by applying the square root of the sum of squares.

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(14) |

The result is

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(15) |

If the duty-cycle is 100%, then t2 = T and the RMS value of the waveform in Figure 6 is

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(16) |

For a bipolar triangle, the waveform looks like the one in Figure 7.

We could calculate the RMS value by splitting the signal in 3, from 0 to t1, then from t1 to t3, and then from t3 to T. However, we already know the RMS value of the waveform from 0 to t2. It is given in equation (15). The RMS value of the waveform from t2 to T, is the same as the one from 0 to t2, with the difference that we need to replace t2 with T-t2, as in (17).

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(17) |

The reason is because it does not matter whether the signal is positive or negative, the power delivered to the load is the same. You can verify this statement by applying the integral as I did for the other waveforms and calculate the RMS value starting with its definition. Therefore, the RMS value of the bipolar triangle waveform is

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(18) |

and the result

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(19) |

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