Alternating signals, impedance and frequency response

State amplitude, period and reference phase

Write v(t)=Vp sin(2πft+φ). Vp is peak amplitude, f is cycles per second in hertz, and φ is a phase offset relative to a stated time reference. Period T=1/f. Angular frequency ω=2πf is measured in radians per second. A 90° phase difference corresponds to one quarter of a cycle, not a fixed delay independent of frequency.

For a pure sinusoid, rms voltage is Vp/√2. Rms means the square root of the time average of the squared signal. It connects a sinusoidal voltage across a fixed resistor to the same average heating as a DC voltage of that value. A nonsinusoidal waveform needs its own rms calculation; dividing every waveform’s peak by √2 is incorrect.

Combine magnitude with timing

In sinusoidal steady state, impedance describes the terminal voltage/current amplitude ratio and phase relation. A resistor has impedance R. An ideal inductor’s reactance magnitude is ωL, increasing with frequency; a capacitor’s is 1/(ωC), decreasing with frequency. The capacitor current leads its voltage by a quarter cycle; the inductor current lags its voltage.

Complex notation can encode both facts: ZR=R, ZL=jωL, ZC=1/(jωC), where j²=−1 represents a quarter-turn in the phase plane. These are relationships between sinusoidal representations, not instantaneous divisions that erase timing. You can follow the magnitude and phase comparisons before learning complex-number algebra.

Ask what reaches the output at each frequency

For an unloaded ideal RC low-pass model, output is the capacitor voltage. Its gain magnitude is 1/√[1+(2πfRC)²] and its phase relative to the input is −arctan(2πfRC). This follows from the resistor/capacitor voltage division in the frequency model. “Low-pass” means lower frequencies are transmitted more strongly here, not that higher frequencies abruptly vanish.

The characteristic frequency fc=1/(2πRC) gives gain 1/√2 and phase−45°. Our plot uses frequency relative to fc on a logarithmic horizontal axis: equal spacings represent factors of ten. This is steady sinusoidal response with ideal components and no additional load, not a claim about every filter or every transient input.

Connect a sinusoid to power and a filter scale

An original 2 V-peak sinusoid across 100 Ω gives Vrms=2/√2≈1.414 V, Irms≈14.14 mA and average resistor power Vrms²/R=0.020 W. The mean voltage is zero, while the mean of v²/R is positive.

For a separate RC model with R=1000 Ω and C=1 μF, RC=0.001 s and fc≈159.15 Hz. At f=fc, output amplitude is about 0.707 of the input and lags by 45°. At 10 fc, gain is 1/√101≈0.0995.

For sinusoidal voltage and current of the same frequency, average power is Vrms Irms cosθ under the passive reference, where θ=φv−φi is the voltage–current phase difference. Multiplying their rms values alone gives apparent power; it equals real power for the pure resistor, but not for every load.

Try the reasoning

1. Why does zero mean sinusoidal voltage not imply zero resistor heating?

   Show the answer and reasoning

   Power depends on voltage squared divided by resistance, whose average is positive.

2. Does a capacitor’s reactance become larger as frequency rises?

   Show the answer and reasoning

   No. Its ideal magnitude 1/(2πfC) decreases; inductive reactance increases instead.

3. Does a gain of 0.707 at fc mean that output voltage is always 0.707 of the input at every instant?

   Show the answer and reasoning

   No. It is the sinusoidal amplitude ratio at that frequency, with a phase shift. Instantaneous values also depend on time.

What to carry forward

Describe the waveform and phase before calculating power. Frequency response gives both gain and timing under specified conditions.