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A peak amplitude phasor is the complex amplitude that reconstructs a real single-frequency sinusoid without an extra amplitude multiplier:Its modulus is the sinusoid's peak amplitude, and its complex argument is the wave phase relative to the chosen cosine reference. This is the default convention in phasor. MIT's sinusoidal notes explicitly describe the cosine reference and phase/amplitude parameters.^[ref]

For an original example, take $v(t)=3\cos \omega t-4\sin \omega t$ volts. Expanding the real part shows that $\hat{V}=3+4i$ volts, with peak amplitude $5$ volts and wave phase $\arctan (4/3)\approx 53.13^{\circ }$. The corresponding root mean square phasor is $(3+4i)/\sqrt{2}$ volts, whose magnitude is about $3.536$ volts. The physical signal is the same under both conventions.

The peak phasor versus positive-frequency Fourier coefficient distinction follows fromThus the positive-frequency coefficient in this two-exponential expansion is $\hat{V}/2$, not $\hat{V}$. A real sinusoid requires its conjugate negative-frequency term as well. This algebra does not assign ordinary Fourier-transform function values to an infinite-duration sinusoid; those require the corresponding distributional interpretation.

A common time-origin change multiplies both voltage and current peak phasors by the same unit-modulus factor, leaving their relative phase and impedance ratio unchanged. Peak-versus-RMS normalization in complex electrical power is still essential when an amplitude convention changes rather than just the time origin.

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 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.