Root mean square (RMS) value
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Root mean square
Root mean square of a set of discrete numbers, which is commonly called by it’s abbreviation RMS is defined as:
\[X_{RMS} = \sqrt{\frac{1}{n}(x_1^2 + x_2^2 + x_3^2 + ... + x_n^2 )}\]Ex. Calculate the RMS of numbers \(4\), \(5\), \(2\) and \(7\).
Ans. \(X_{RMS} = \sqrt{\frac{1}{4}(4^2 + 5^2 + 2^2 + 7^2)} \approx 4.847\)
We’ve seen the definition of RMS for the cases we have discrete numbers. If we had a continuous function instead, we’d use the following definition:
\[f_{RMS} = \sqrt{\frac{1}{T_2 - T_1} \int_{T_1}^{T_2} [f(dx)]^2 dx }\]Ex. Calculate the RMS of function \(f(x) = x^3\) from \(x = 1\) to \(x = 4\).
Ans. Plugging it into formula:
\[f_{RMS} = \sqrt{\frac{1}{1 - 4} \int_{1}^{4} [x^3]^2 dx } \approx 27.9\]When do we use RMS?
When the source of electrical power in not a fixed DC value and it’s alternating over time, how would we calculate the average power consumed by a resistive element? From the formula of power we have:
\[P(t) = I(t).V(t)\]We also know from Ohm’s law that \(I(t) = \frac{V(t)}{R}\), therefore substituting it into the formula we get:
\[P(t) = \frac{V(t)^2}{R}\]This formula gives the power at each point in time. Since \(V(t)\) is alternating, the power varies over time as well and doesn’t have a fixed value. So to calculate the average power we should use integral for this.
\[P_{avg} = \frac{1}{R}.\frac{\int_{T_0}^{T_1} V(t)^2 dt }{T_1 - T_0}\]This is all we need for calculating the average power. However if we take the integral term (with it’s denominator \(T_1 - T_0\)) and take it’s square root, it turns into the RMS value we defined earlier.
\[V_{RMS} = \sqrt{\frac{\int_{T_0}^{T_1} V(t)^2 dt }{T_1 - T_0}}\]Now by having this value, to calculate the average power we should reverse the effect of taking the square root by raise it to power of \(2\) (and division by resistivity \(R\)):
\[P_{avg} = \frac{V^2_{RMS}}{R}\]Why do we take the square root of an expression just to raise it to the power of \(2\) later? Because if we somehow know the \(V_{RMS}\) of an AC signal, we would use the same formula for calculating average power as if we had a fixed DC voltage over time (\(P_{avg} = \frac{V^2_{dc}}{R}\)); Namely raising it to power of \(2\) then dividing it by \(R\). So RMS value of a voltage or current is a convenient way to describe an AC signal besides its peak value and frequency that we would normally do.
When we’re speaking about AC signals, we are most likely dealing with Sinusoidal AC signals. So let’s investigate and see if the RMS of a general sinusoidal AC signal yields anything constant. If so, then we’d omit the whole integration process every time we want to calculate the RMS of a sinusoidal signal and simply multiply that constant by some term.
General sinusoidal function has the following form in one cycle:
\[V(t) = V_{Max} \sin(\omega t)\]Plugging it into the definition of RMS,
\[V_{RMS} = \sqrt{\frac{\int_{T_0}^{T_1} (V_{Max} \sin \omega t) ^ 2 dt} {T_1 - T_0}}\]\(V_{Max}\) is just a constant and goes out of the square root. So we get:
\[V_{RMS} = V_{Max} \sqrt{\frac{\int_{T_0}^{T_1} (\sin \omega t) ^ 2 dt }{T_1 - T_0}}\]We also know from basic properties of trigonometric functions:
\[\sin^2 x = \frac{1 + \cos 2x}{2}\]Substituting the sine term with this property:
\[\begin{aligned} V_{RMS} = V_{Max} \sqrt{\frac{1}{\int_{T_0}^{T_1} \frac{1 + \sin \omega t dt}{2}}} = \\ V_{Max} \sqrt{\frac{1}{T_1 - T_0} \left[ \frac{t}{2} - \frac{\sin 2 \omega t}{4 \omega}\right]_{T_0}^{T_1} } \end{aligned}\]Now since the range \(T_0\) to \(T_1\) is one period of the sinusoidal function, the sine integral is just \(0\). Therefore:
\[\begin{gathered} V_{RMS} = V_{Max} \sqrt{\frac{1}{T_1 - T_0} \left[ \frac{t}{2} \right]_{T_0}^{T_1}} = \\ V_{Max}\sqrt{\frac{1}{T_1 - T_0} . \frac{T_1 - T2}{2}} = \\ V_{Max} \frac{1}{\sqrt{2}} \end{gathered}\]Which proves that for calculating the RMS of a sinusoidal signal, we can skip the integration process and simply divide its peak value by \(\sqrt{2}\).