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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / Electrical foundations
LNGR · Learning workshop

AC voltage, frequency & phase

Voltage can change polarity. Frequency, amplitude and phase describe three different aspects of that change.

Try the experiment ↓
u(t)A · sin(ωt + φ)

Functional illustration · equations and topology in the text

Start without prior knowledgeUnderstand the model limitsDE / EN
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Circuit 09 · 5 small steps

RC low-pass: measure across C

RC low-pass: measure across CAn ideal sine source drives R1 and C1. Input A–B, output M–B. The output is unloaded. We compare peak amplitudes in steady state, not switch-on behaviour.~1 V peakAB · 0 VR1 = 1 kΩC1 = 100 nFM
An ideal sine source drives R1 and C1. Input A–B, output M–B. The output is unloaded. We compare peak amplitudes in steady state, not switch-on behaviour.

  1. Across which component is output measured?

    Hint

    Follow M and B.

    Answer and working

    Across C1. Choosing this output makes the arrangement a low-pass filter.

  2. What do you expect at very low frequency?

    Hint

    Capacitive reactance rises as frequency falls.

    Answer and working

    Almost the full input appears across C1: Uout approaches 1 V peak.

  3. Calculate cutoff frequency.

    Hint

    100 nF = 0.0000001 F; fc = 1/(2πRC).

    Answer and working
    Calculation notes
    C=100\,\mathrm{nF}=10^{-7}\,\mathrm{F}
    f_c=\frac{1}{2\pi RC}
    f_c=\frac{1}{2\pi\cdot1000\,\Omega\cdot10^{-7}\,\mathrm{F}}
    {f_c\approx1592\,\mathrm{Hz}}

  4. What are output amplitude and phase at fc?

    Hint

    Magnitude is 1/√2; a low-pass has negative phase.

    Answer and working
    Calculation notes
    |H(f_c)|=\frac{1}{\sqrt{2}}
    \hat U_{\mathrm{out}}=\frac{1\,\mathrm{V}}{\sqrt{2}}\approx0.707\,\mathrm{V}
    G=20\log_{10}\left(\frac{1}{\sqrt{2}}\right)\approx-3.01\,\mathrm{dB}
    {\varphi=-45^{\circ}}

    The voltage is not half of the input here.

  5. Double C1. Where does fc move? Check the filter model.

    Hint

    fc is inversely proportional to C.

    Answer and working
    Calculation notes
    C_{\mathrm{new}}=2C
    f_{c,\mathrm{new}}=\frac{1}{2\pi R\cdot2C}=\frac{f_c}{2}
    {f_{c,\mathrm{new}}\approx796\,\mathrm{Hz}}

    This calculation assumes no additional output load.

Think it through before opening the answer. Always write the formula, substituted values and units.

01

The mental picture

Two people can swing equally fast and equally high without reaching the top at the same instant. Repetition rate corresponds to frequency, height to amplitude and timing offset to phase. Audio is usually more complex than a single regular swing.

02

The principle

At 100 Hz a signal repeats one hundred times per second; one period takes 10 ms. Peak voltage is the maximum excursion from zero. RMS voltage gives the same average heating in a resistor as the corresponding DC voltage.

03

Try it yourself

Interactive experiment

AC voltage, frequency & phase

Set a phase shift of 180°. The two curves are inverted relative to each other. Then increase only frequency: the angle stays the same, but the corresponding time offset becomes smaller.

Model limits: Two pure sines of equal amplitude and frequency. The time axis automatically shows two periods. No harmonics or noise.

04

A worked example

A sine with a 5 V peak has about 3.54 V RMS. At 100 Hz, 90° is a quarter-period, or 2.5 ms. At 1 kHz it is only 0.25 ms. A fixed delay is therefore not a fixed phase angle across all frequencies.

A closer look · equations and assumptions
Formula notes
T=\frac{1}{f}
u(t)=\hat U\sin(2\pi ft+\varphi)
U_{\mathrm{RMS}}=\frac{\hat U}{\sqrt{2}}\quad(\mathrm{sine})
\varphi=360^{\circ}\cdot f\cdot\Delta t

The general RMS value is the square root of the mean of u²(t). The factor 1/√2 applies only to a sine. A symmetrical square wave has a higher RMS value at the same peak. Signal delay corresponds to negative phase; positive φ here makes curve B lead.

A common pitfall. A 180° shift at one frequency does not automatically mean reversed wiring. Reversing polarity changes the sign at every frequency; delay produces a frequency-dependent angle.

05

Workshop takeaway

Label every voltage as peak or RMS. Whenever you discuss phase, include the frequency.

Sources & transparency

Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

05 · Oscilloscope: 2 V per division; sine spans 3 divisions peak-to-peak. Find peak-to-peak and peak voltage.

Show worked answer
  1. Peak-to-peak = 2 · 3 = 6 V.
  2. For a symmetrical sine, peak is half that.
  3. Peak voltage: 3 V.
Calculation
U_{pp}=6\,V;\quad\hat U=3\,V

06 · Find RMS voltage of the sine in task 5, with no DC offset.

Show worked answer
  1. Divide peak by √2.
  2. 3/√2.
  3. RMS = 2.121 V.
Calculation
U_{rms}=\frac{\hat U}{\sqrt{2}}=2.121\,V

07 · Timebase 0.2 ms per division; one cycle spans 5 divisions. Find frequency.

Show worked answer
  1. Period = 5 · 0.2 ms = 1 ms.
  2. Frequency = 1/period.
  3. Frequency = 1 kHz.
Calculation
f=\frac{1}{0.001\,s}=1000\,Hz

08 · Two 1 kHz sine waves are offset by 0.25 ms. Find phase-difference magnitude.

Show worked answer
  1. Period = 1 ms.
  2. Offset is one quarter cycle.
  3. Phase magnitude = 90°.
Calculation
|\varphi|=360^\circ\frac{0.25}{1}=90^\circ

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