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The group delay introduced by a capacitive amplifier load is the negative angular frequency derivative of the extra wave phase produced by that load. For the resistive output model in reactive-load phase shift in a finite-output-impedance amplifier, divide the loaded transfer function by the unloaded one:The added group delay is thereforeThis subtraction of unloaded wave phase prevents the amplifier's intrinsic group delay from being attributed to the capacitor. The terminal model uses a Thévenin equivalent; the group delay definition differentiates an unwrapped wave phase, as in the SciPy signal-processing documentation. Its digital-filter function reports samples, whereas the analog derivative above uses radians per second and gives seconds.^[ref][ref]

For an original example, choose $R=200\Omega$ and $C=10\mathrm{nF}$, hence $\tau =2\mu \mathrm{s}$. At $\omega \tau =1$, corresponding to $f=1/(2\pi \tau )\approx 79.6\mathrm{kHz}$, the added group delay is $1\mu \mathrm{s}$. The same operating point has $-{\varphi }_{\mathrm{rel}}/\omega =\pi \tau /4\approx 1.571\mu \mathrm{s}$: a single sinusoid's phase displacement and the delay of a narrow frequency packet need not agree.

A practical test applies a weakly modulated carrier, varies its central frequency, and measures the output modulation displacement while comparing loaded and unloaded operation. The group delay describes that narrow-band envelope only when the frequency response varies slowly across its bandwidth. The corresponding causal impulse response begins at zero time, $h(t)=e^{-t/\tau }/\tau$ for $t\ge 0$, and vanishes for $t<0$; it is not a rigidly translated impulse. Thus group delay versus pulse-front arrival for an RC load matters when a measured pulse is broad enough to change shape.

Group delay versus pulse-front arrival for an RC load distinguishes a narrow-band modulation displacement from the earliest response to a newly applied input. For the relative amplifier-load model in group delay introduced by a capacitive amplifier load, the causal impulse response is $h(t)=e^{-t/\tau }/\tau$ for $t\ge 0$ and zero for $t<0$. Its convolution with an input that vanishes before zero also vanishes before zero. Nevertheless its group delay is $\tau /[1+(\omega \tau {)}^2]$, which is positive and frequency dependent. The phase-derivative definition concerns spectral components and their envelopes.^[ref]

An original finite-pulse calculation shows why a threshold measurement gives another time. For an initially relaxed load, apply a rectangular pulse of height $V_0$ and duration $T>0$. Convolution givesThe first mathematical response occurs immediately after zero. A rising threshold equal to fraction $a$ of the input pulse height is reached at $t_a=-\tau \log (1-a)$ only if the pulse lasts at least that long. This is the threshold-crossing time of a finite rectangular pulse through an RC load.

With $\tau =1\mu \mathrm{s}$ and $T=0.1\mu \mathrm{s}$, the largest output is only $(1-e^{-0.1})V_0\approx 0.0952V_0$: a $0.5V_0$ threshold is never reached. If $T=3\mu \mathrm{s}$ instead, that threshold is crossed at $0.693\mu \mathrm{s}$. Neither value is the $0.5\mu \mathrm{s}$ group delay evaluated at $\omega \tau =1$.

A pulse generator and oscilloscope can compare the threshold observation with a slow modulation on a carrier; the chosen threshold and modulation bandwidth must be recorded. MIT's laboratory likewise separates square-wave decay measurements from sinusoidal amplitude/phase measurements.^[ref] If wiring adds a transport delay $d$, shift the impulse response to $h(t-d)$: the front moves to $d$, while the modulation delay becomes $d+{\tau }_g$. The immediate onset of the unshifted lumped model is not a model of propagation along a finite-length cable.

A reactive-load phase shift in a finite-output-impedance amplifier occurs because the load and output impedance form a frequency-dependent divider. In the model of loaded voltage gain, assume a positive real output resistance $R_o$, an ideal capacitive load $C_L$ and open-circuit gain $A_0$ approximately constant over the band of interest. With $\omega =2\pi f$,The divider follows from the terminal equivalent described by the university circuit notes.^[ref]

Putting $\tau =R_o{C}_L$ givesThus even an ideal capacitor can change both amplitude and phase when the source driving it has finite resistance. Its reactive current and energy exchange are described by complex electrical power; zero average dissipation in the ideal capacitor does not mean zero output-current demand.

For an original example, choose $R_o=100\Omega$, $C_L=10\mathrm{nF}$ and $A_0=10$. Then $\tau =1\mu \mathrm{s}$ and the load pole is $f_p=1/(2\pi \tau )\approx 159.2\mathrm{kHz}$. At that frequency the gain magnitude is $10/\sqrt{2}\approx 7.071$ and the extra phase lag is $45^{\circ }$. A measured input of $10\mathrm{mV}$ root mean square gives about $70.7\mathrm{mV}$ root mean square at the capacitor, requiring approximately $0.707\mathrm{mA}$ root-mean-square capacitive current.

A bench test holds the actual amplifier input fixed while sweeping frequency and compares loaded and unloaded output amplitudes and phases. Include probe capacitance in $C_L$. The group delay introduced by a capacitive amplifier load also matters when comparing pulse waveforms.

This single-pole calculation assumes the stated constant open-circuit gain and resistive output impedance. The complete amplifier may have additional frequency dependence and feedback effects. The manufacturer's ideal feedback-gain examples are defined within their operating assumptions.^[ref] Check the actual output-current and linear operating range before interpreting a departure from this model as a capacitor-only effect.