Root mean square phasor ($V_{\mathrm{rms}}$)

A root mean square phasor represents a single-frequency sinusoid by a complex amplitude whose magnitude is its root mean square value. With the convention $v(t)=V_{\mathrm{pk}}\cos (\omega t+\varphi )$ and $\omega >0$, defineIndeed, the period average of $v^2$ is $V_{\mathrm{pk}}^2/2$, so $|\underline{V}|=V_{\mathrm{rms}}$. The factor $\sqrt{2}$ in reconstruction distinguishes this convention from the peak-amplitude phasor convention in the parent article. The DOE handbook derives the same effective value by comparing the heating from alternating current and direct current.^[ref]

Using RMS phasors for both electric potential difference and electric current, the passive-sign convention gives complex electrical powerTo see the normalization directly, multiplying the two reconstructed sinusoids and averaging over a period leaves $P=\mathrm{Re}(\underline{V}{\underline{I}}^{\ast })$. Peak-amplitude phasors would instead require the prefactor $1/2$. This peak-versus-RMS normalization in complex electrical power must be stated before combining a source model with measured amplitudes.

For an original example, let $\underline{V}=10e^{i0}\mathrm{V}$ and $\underline{I}=2e^{-i\pi /3}\mathrm{A}$. The electric current lags the electric potential difference by $60^{\circ }$, givingThe corresponding time-domain peaks are $10\sqrt{2}\mathrm{V}$ and $2\sqrt{2}\mathrm{A}$. Treating those peaks as RMS phasor magnitudes would falsely double the computed complex electrical power. This representation assumes a sinusoid at one common frequency; a nonsinusoidal signal can have a root mean square value without being represented by one RMS phasor.

Peak-versus-RMS normalization in complex electrical power determines the coefficient multiplying the product of voltage and conjugate current phasors. A peak amplitude phasor reconstructs a sinusoid directly, while a root mean square phasor requires a factor of $\sqrt{2}$. The DOE effective-value derivation relates these amplitudes by comparing heating from AC and DC currents.^[ref]

To track any combination, writewhere each reconstruction scale is $1$ for a peak phasor or $\sqrt{2}$ for an RMS phasor. A period average of the product leaves the real powerUsing the same positive-frequency and passive-sign convention, the corresponding complex electrical power isThus the coefficient is $1/2$ for two peak-normalized amplitudes, $1$ for two RMS-normalized amplitudes, and $1/\sqrt{2}$ for a mixed pair. The last case is why merely recognizing a product as a power calculation is not enough to choose its normalization.

For an original example, let the electric potential difference be represented by peak amplitude $V=100\mathrm{V}$ and the electric current by RMS amplitude $I=2e^{-i\pi /6}\mathrm{A}$, at the same frequency. ThenMultiplying the two numbers as though both were RMS phasors overestimates the complex electrical power by $\sqrt{2}$. Inserting a peak-pair factor $1/2$ instead underestimates it by $1/\sqrt{2}$. The time-domain reconstruction, $v(t)=100\cos \omega t$ and $i(t)=2\sqrt{2}\cos (\omega t-\pi /6)$, independently verifies the period average.

The mixed peak and RMS gain-calibration error also matters when source and receiver amplitude reports are combined: matching units does not establish matching amplitude conventions. This calculation concerns one sinusoidal component. A nonsinusoidal waveform requires its actual period average or an appropriately normalized harmonic decomposition, not one unspecified phasor product.

A mixed peak and RMS gain-calibration error occurs when input and output amplitude reports use different reconstruction scales but their numerical ratio is treated as the physical gain. Let the true peak amplitude phasors obey ${\hat{V}}_{\mathrm{out}}=G{\hat{V}}_{\mathrm{in}}$. If a reported amplitude $A$ reconstructs the signal with scale $a$, so that $\hat{V}=aA$, thenThe scale is $1$ for a peak phasor and $\sqrt{2}$ for an RMS phasor. Equal physical units do not establish equal amplitude conventions. The sinusoidal RMS-to-peak conversion follows from the period average of the squared signal.^[ref]

For a peak-normalized input and RMS-normalized output, the raw ratio is $G/\sqrt{2}$; for the reverse pairing it is $\sqrt{2}G$. The amplitude-gain error in decibels is therefore minus or plusThese positive real normalization factors do not themselves change the relative phase. They change the measured gain magnitude.

For an original calibration example, let a sinusoidal input be reported as $0.2\mathrm{V}$ peak and its output as $1.0\mathrm{V}$ RMS. The raw amplitude ratio is $5$, or about $13.9794\mathrm{dB}$. Converting the output to peak amplitude gives actual gain $5\sqrt{2}\approx 7.0711$, or about $16.9897\mathrm{dB}$. A source-to-receiver comparison that ignores the conventions would falsely report a $3.0103\mathrm{dB}$ gain deficit.

A further factor-of-two issue occurs when one report is a positive-frequency Fourier coefficient rather than a peak phasor; peak phasor versus positive-frequency Fourier coefficient derives that distinction. It produces a different $6.0206\mathrm{dB}$ amplitude factor and must not be treated as an RMS conversion. Likewise, peak-versus-RMS normalization in complex electrical power uses a product of reconstruction scales rather than the ratio used here. A calibration record should therefore state the sinusoidal amplitude convention and, for sampled estimates, the FFT normalization before numerical gains are compared.