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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.

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 phasor versus positive-frequency Fourier coefficient distinguishes a real sinusoid's reconstruction amplitude from one coefficient in its two-sided exponential expansion. For $v(t)=\mathrm{Re}(\hat{V}{e}^{i\omega t})$,Thus the positive-frequency Fourier coefficient is half the peak amplitude phasor. The negative-frequency coefficient is its conjugate. A real sinusoid needs both; a complex exponential input has a different one-sided reconstruction.

The distinction also controls a sampled record. For $N$ evenly spaced samples of a real sinusoid with exactly $k$ cycles in the record and $0<k<N/2$, use the unscaled forward discrete Fourier transformOrthogonality of the discrete exponentials gives $X_k=N\hat{V}/2$. ConsequentlyThe default SciPy forward FFT has exactly this unscaled convention; its forward and orthonormal modes have different factors.^[ref]

For an original example, sample $v_m=3\cos (2\pi 5m/64)-4\sin (2\pi 5m/64)$ volts. Then $\hat{V}=3+4i$ volts and $X_5=96+128i$ volts. Dividing by 64 gives the positive-frequency coefficient $1.5+2i$ volts; doubling it recovers the peak phasor. The RMS amplitude is $5/\sqrt{2}\approx 3.536$ volts.

These exact relations assume coherent sampling of the stated sinusoid without aliased extra components. DC and, for even $N$, the Nyquist bin do not have distinct conjugate partners and must not be doubled. With a noninteger number of cycles or a windowed record, finite-record phasor extraction with coherent sampling must be distinguished from a generic bin readout; spectral leakage and the window's amplitude normalization alter the estimate. The parent peak amplitude phasor specifies the continuous-time convention before any sampled transform is chosen.

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.