Distinguish DC vs AC signals: Worked examples — Hardware deep dive

Before we start, one reminder of the symbols we lean on the most. If a symbol here is unfamiliar, it was defined in the parent — but we re-anchor the key ones so you can follow from line one:

Definition The quantities and symbols we keep reusing

$v (t)$ — the voltage at time $t$ , measured in volts (V). Think of it as "how hard, and in which direction, charge is being pushed right now." Positive = pushing one way, negative = pushing the other way.
$V_{p e ak}$ — the biggest value $v (t)$ ever reaches (the top of the wave).
Amplitude — the size of the swing away from the centre of a wave, i.e. how far up (or down) from its middle line the wave reaches. In $v (t) = V_{d c} + A sin (\dots)$ the amplitude is $A$ . For a wave centred on zero, amplitude $= V_{p e ak}$ .
Offset (DC offset), symbol $V_{d c}$ — the constant flat level the wiggle rides on. It is the value the wave would have if you removed the wiggle: the centre line. In $v (t) = V_{d c} + A sin (\dots)$ it is $V_{d c}$ .
Ripple — a small unwanted AC wiggle sitting on top of a DC level (common after imperfect smoothing in a power supply). Its ripple frequency is simply the frequency $f$ of that wiggle — the number that multiplies $2 π t$ inside the sine.
$T$ — the period: how long one full up-and-down cycle takes, in seconds.
$f = 1/ T$ — the frequency: how many full cycles happen in one second, in hertz (Hz).
$ω = 2 π f$ — the angular frequency, measured in radians per second (rad/s). It is how fast the generator's angle sweeps round; one full turn is $2 π$ radians, and there are $f$ turns each second, so $ω = 2 π f$ .
$ϕ$ (phi) — the phase offset (or phase shift), measured in radians. It is the starting angle of the wave at $t = 0$ : it says "where along the cycle the wave already was" before the clock started.
$⟨ \cdot ⟩$ — the time-average over one full period. So $⟨ v^{2} ⟩$ means "take $v^{2}$ at every instant across one cycle and average all those values."

The scenario matrix

Every DC/AC question is really one of these cells. The last column names the worked example that covers it.

#	Case class	What makes it tricky	Covered by
A	Pure DC (flat line)	no reversal, peak = rms = average	Example 1
B	Pure AC sinusoid	peak↔rms↔period conversions	Example 2
C	DC offset + AC ripple, stays positive	never crosses zero → still DC	Example 3
D	DC offset + AC ripple, crosses zero	offset too small → becomes AC	Example 4
E	Read period/freq off a scope grid	translate divisions → seconds	Example 5
F	Phase offset $ϕ$ / value at a given time	plug a time into the sine	Example 6
G	Degenerate: $f = 0$ or $V_{p e ak} = 0$	limiting cases → collapse to DC / to nothing	Example 7
H	Real-world word problem (mains + insulation)	rms vs peak safety margin	Example 8
I	Exam twist: non-sinusoid (square and triangle)	the $2$ rule fails	Examples 9 & 10

Rows A, C, D, G test the DC/AC boundary (does it reverse?). Rows B, E, F, H, I test the numbers (peak, rms, period, frequency, phase). Together they fill the table.

Worked Examples

Worked example 1 — Pure DC (cell A)

A fresh AA battery reads a steady $v (t) = 1.5$ V. Give its peak, average, rms, and frequency.

Forecast: guess before reading — for a flat line, are peak, average and rms different or all equal?

Peak $= 1.5$ V. Why this step? Peak means the largest value; a flat line's largest value is itself.
Average $= 1.5$ V. Why this step? Averaging a constant over any time just gives that constant.
RMS $= ⟨ v^{2} ⟩ = 1. 5^{2} = 1.5$ V. Why this step? Here we use the master rms formula $⟨ v^{2} ⟩$ ; for a constant, $⟨ v^{2} ⟩ = 1. 5^{2}$ . Square, average, square-root → back to the constant. For DC, rms = the value itself.
Frequency $= 0$ Hz. Why this step? Frequency counts reversals per second; DC never reverses, so zero cycles.

Verify: all four numbers equal $1.5$ V (except $f = 0$ ). That is the signature of pure DC — the $2$ factor does not appear because there is no wave.

Worked example 2 — Pure AC round-trip (cell B)

A pure sinusoid has $V_{r m s} = 12$ V and frequency $f = 400$ Hz. Find $V_{p e ak}$ , the period $T$ , and the angular frequency $ω$ .

Forecast: will $V_{p e ak}$ be bigger or smaller than $12$ V?

$V_{p e ak} = V_{r m s} 2 = 12 \times 1.41421 \dots = 16.97$ V. Why this step? Because this is a pure sine, we may use the $2$ shortcut. rms is always below the peak by exactly $2$ for a sine, so to climb up to peak we multiply.
$T = 1/ f = 1/400 = 0.0025$ s $= 2.5$ ms. Why this step? Period and frequency are reciprocals — see Frequency and Period.
$ω = 2 π f = 2 π \times 400 = 2513.3$ rad/s. Why this step? $ω$ (angular frequency, defined above) is how fast the generator angle sweeps; one turn is $2 π$ radians, and there are $f$ turns per second.

Verify: invert step 1 — $V_{p e ak} / 2 = 16.97/1.414 = 12.0$ V ✓. Units: rad/s for $ω$ , s for $T$ , both consistent with $ω = 2 π / T$ .

Worked example 3 — Offset that stays positive: DC with ripple (cell C)

A poorly smoothed power supply outputs $v (t) = 5 + 1.5 sin (2 π \cdot 120 t)$ V. Here the offset is $V_{d c} = 5$ V and the amplitude of the ripple is $A = 1.5$ V. Is it AC or DC? What is its ripple frequency?

Forecast: it has a wiggle — does "wiggle" automatically mean AC?

Find the minimum: $v_{m i n} = V_{d c} - A = 5 - 1.5 = 3.5$ V. Why this step? The sine dips lowest at $- 1$ , so the smallest value is (offset $-$ amplitude). This is where zero-crossing would happen if it ever did.
Check the sign: $3.5$ V $> 0$ , so $v (t)$ never reaches zero. Why this step? The DC/AC test is polarity reversal. No zero-cross → current never flows backward → it is DC (with ripple), not AC.
Ripple frequency: from $2 π \cdot 120 t$ we read $f = 120$ Hz. Why this step? Whatever multiplies $2 π t$ inside the sine is the frequency by definition — that is the ripple frequency.

The figure shows: the green trace is the full signal $5 + 1.5 sin (\dots)$ ; the two dashed grey lines mark its ceiling $6.5$ V and its floor $3.5$ V; the thick red line is the zero level. Notice the green curve stays entirely above red — that gap between the floor and zero is the visual proof it never reverses.

Verify: the whole trace sits between $3.5$ V and $6.5$ V — entirely above the red zero line. No reversal → DC. ✓

Symmetric mirror case: a purely negative offset like $v (t) = - 5 + 1.5 sin (\dots)$ has maximum $- 5 + 1.5 = - 3.5$ V, which is below zero. It never rises to zero, so it also never reverses — it is DC too (just flowing the other way). The zero-crossing test is symmetric: stay wholly above or wholly below zero → DC.

Worked example 4 — Same shape, smaller offset: now it's AC (cell D)

Same wiggle amplitude but a weaker offset: $v (t) = 1 + 1.5 sin (2 π \cdot 120 t)$ V. Now the offset is $V_{d c} = 1$ V and the amplitude is still $A = 1.5$ V. AC or DC now?

Forecast: we only lowered the flat part from 5 to 1 — can that flip the whole classification?

Minimum: $v_{m i n} = V_{d c} - A = 1 - 1.5 = - 0.5$ V. Why this step? Same recipe (offset $-$ amplitude). Now the amplitude beats the offset.
Since $v_{m i n} = - 0.5$ V is below zero, the signal does cross zero and go negative. Why this step? Negative $v$ means charge is being pushed the other way → genuine polarity reversal → AC (with a DC component of $+ 1$ V).
Threshold insight: the boundary is when |offset| $=$ amplitude. Here $∣1∣ <$ amplitude $1.5$ , so it must dip below zero. Why this step? This gives you a one-line test: reverses ⇔ amplitude > |offset|.

The figure shows: the blue trace now swings from $- 0.5$ V up to $+ 2.5$ V; the shaded orange patch is the part of the wave that has dropped below the red zero line. That orange region — absent in Example 3 — is the polarity reversal that makes this AC.

Verify: trace ranges $- 0.5$ V to $+ 2.5$ V, straddling the red zero line ✓. Compare with Example 3: identical amplitude, only the offset changed, yet the classification flipped — that is the whole point.

Worked example 5 — Read frequency off an oscilloscope (cell E)

An oscilloscope shows one full cycle spanning 2.5 horizontal divisions, and the timebase is set to 2 ms per division. Find $T$ and $f$ .

Forecast: more divisions per cycle → higher or lower frequency?

$T = 2.5 div \times 2 ms/div = 5$ ms $= 0.005$ s. Why this step? Divisions are just a ruler; multiply how many divisions the cycle takes by the time each division represents.
$f = 1/ T = 1/0.005 = 200$ Hz. Why this step? Frequency is periods-per-second, the reciprocal of the period.

The figure shows: the blue sine, with vertical dotted grey lines marking each 2 ms division of the screen. The orange double-headed arrow spans exactly one cycle and lands on 2.5 divisions — that is the measurement you read off the glass, giving $T = 5$ ms.

Verify: $f \cdot T = 200 \times 0.005 = 1.0$ ✓ (one cycle per period, always). Units: ms/div × div = ms → seconds after conversion.

Worked example 6 — Phase and value at a chosen instant (cell F)

A signal is $v (t) = 10 sin (2 π \cdot 50 t + \frac{π}{6})$ V. Find (a) the voltage at $t = 0$ , (b) the first time after $t = 0$ that $v$ reaches its peak.

Forecast: because of the $+ π /6$ head-start, will $v (0)$ be zero or already climbing?

At $t = 0$ : $v (0) = 10 sin (π /6) = 10 \times 0.5 = 5$ V. Why this step? $ϕ = π /6$ is the phase offset (starting angle, defined above); the wave did not start at zero, it started partway up. $sin (π /6) = 0.5$ .
Peak condition: $sin$ peaks when its angle $= π /2$ . Set $2 π \cdot 50 t + π /6 = π /2$ . Why this step? We ask "which angle gives the maximum?" — that angle is $π /2$ , so we solve for the time that reaches it.
Solve: $2 π \cdot 50 t = π /2 - π /6 = π /3$ , so $t = \frac{π /3}{100 π} = \frac{1}{300}$ s $\approx 3.33$ ms. Why this step? Isolate $t$ ; $2 π \cdot 50 = 100 π$ cancels the $π$ neatly.

Verify: plug back: $100 π \times \frac{1}{300} + π /6 = π /3 + π /6 = π /2$ ✓, and $10 sin (π /2) = 10$ V = peak ✓.

Worked example 7 — Degenerate limits (cell G)

Two edge cases. (a) $v (t) = 8 sin (2 π \cdot 0 \cdot t)$ (frequency $f = 0$ ). (b) $v (t) = 0 \cdot sin (2 π \cdot 50 t)$ (peak $= 0$ ). Classify each.

Forecast: with $f = 0$ , does the "AC" wave still alternate?

(a) $f = 0$ : the angle $2 π f t = 0$ for all $t$ , so $v (t) = 8 sin (0) = 0$ V — a flat line at zero. Why this step? Zero frequency means the generator never turns; no rotation → no reversal → this collapses to DC (a trivial $0$ V DC).
(b) $V_{p e ak} = 0$ : $v (t) = 0$ for all $t$ regardless of $f$ . Why this step? An amplitude of zero means no swing at all; the "frequency" is meaningless because nothing moves. Also DC ( $0$ V).
Lesson: AC requires both $f > 0$ and $V_{p e ak} > 0$ . Kill either and the wave degenerates to DC. Why this step? This closes the boundary — you now know the exact conditions for a signal to be AC.

Verify: rms of case (a) $= ⟨ 0^{2} ⟩ = 0$ V; rms of case (b) $= 0$ V. Both give zero power ✓, consistent with "flat line, no reversal."

Worked example 8 — Real-world: mains insulation (cell H)

A device is rated for a $230$ V rms, $50$ Hz mains supply. An engineer must pick a capacitor. (a) What peak voltage must it survive? (b) What is the period? (c) If capacitors come in $250$ V and $400$ V ratings, which is safe?

Forecast: is the "230 V" on the label the biggest voltage the capacitor ever feels?

Peak: $V_{p e ak} = 2302 = 325.3$ V. Why this step? The rms label hides a higher peak; hardware fails at the peak, so we must convert up by $2$ (valid because mains is a pure sine).
Period: $T = 1/ f = 1/50 = 0.02$ s $= 20$ ms. Why this step? Reciprocal of frequency — see Frequency and Period.
Choice: $325.3$ V is above $250$ V but below $400$ V, so the 400 V capacitor is required. Why this step? A $250$ V part would be exceeded on every cycle and eventually break down. Always rate above the peak, not the rms.

Verify: $325.3/ 2 = 230.0$ V ✓ recovers the label. Safety margin of the 400 V part: $400 - 325.3 = 74.7$ V of headroom ✓.

Worked example 9 — Exam twist: a square wave breaks the

$2$ rule (cell I) A symmetric square wave swings between $+ V_{p e ak}$ and $- V_{p e ak}$ , spending equal time at each. It has $V_{p e ak} = 6$ V. Find its rms. Does $V_{r m s} = V_{p e ak} / 2$ apply?

Forecast: will the answer be $6/ 2 = 4.24$ V like a sine, or something else?

Square the signal: at every instant $v^{2} = (\pm 6)^{2} = 36$ V², a constant. Why this step? We go back to the master formula $⟨ v^{2} ⟩$ . Squaring removes the sign, and since the magnitude is always $6$ , the square never varies.
Average the square: $⟨ v^{2} ⟩ = 36$ V² (average of a constant is itself). Why this step? Unlike a sine where $⟨ sin^{2} ⟩ = \frac{1}{2}$ , here there is no "in-between" — the value is always at full magnitude.
Root: $V_{r m s} = 36 = 6$ V. Why this step? For a square wave rms equals the peak — the $2$ factor is a sine-only result.

The figure shows: the orange square wave sits flat at $+ 6$ and $- 6$ V; its rms line (dotted) rides right at $6$ V. The blue dashed sine of the same $6$ V peak has its rms much lower, at $4.24$ V. The gap between the two dotted rms lines is exactly the point of this example.

Verify: compare to the sine value $6/ 2 = 4.24$ V. The square's $6$ V $\neq = 4.24$ V ✓ — proving the shape matters.

Worked example 10 — Exam twist continued: a triangle wave (cell I)

A symmetric triangle wave ramps straight up to $+ V_{p e ak}$ and straight down to $- V_{p e ak}$ , with $V_{p e ak} = 6$ V. Find its rms.

Forecast: will it be closer to the square's answer ( $6$ V) or the sine's ( $4.24$ V), or different again?

Over one rising ramp, $v$ goes linearly from $- 6$ to $+ 6$ ; by symmetry $⟨ v^{2} ⟩$ over the ramp equals $⟨ v^{2} ⟩$ over the whole cycle. Why this step? Again the master formula $⟨ v^{2} ⟩$ . Squaring makes every quarter-cycle identical, so we can average over just one straight ramp.
For a straight line rising from $0$ to $V_{p e ak}$ , the mean of the square is $\frac{1}{3} V_{p e ak}^{2}$ (standard result: $\int_{0}^{1} x^{2} d x = \frac{1}{3}$ ). Why this step? The square of a rising line spends most of its time small, so its average is only a third of the peak-square — smaller than the square wave's full $V_{p e ak}^{2}$ .
Root: $V_{r m s} = \frac{1}{3} V_{p e ak} = \frac{V _{p e ak}}{3} = \frac{6}{3} = 3.46$ V. Why this step? The triangle's factor is $1/ 3 \approx 0.577$ , different from both the sine ( $0.707$ ) and the square ( $1.0$ ) — so the answer is $3.46$ V, below both other shapes.

The figure shows: the green triangle wave ramping between $+ 6$ and $- 6$ V; the two dotted green lines at $\pm 3.46$ V mark its rms band. Notice the rms sits well inside the peak — even lower than a sine would — because a straight ramp spends so much time near zero.

Verify: three shapes, three factors — square $1.0$ , sine $0.707$ , triangle $0.577$ — all with the same $6$ V peak give rms $6$ , $4.24$ , $3.46$ V respectively ✓. The shape decides the factor.

Recall Quick self-test across the matrix

A signal never goes negative — AC or DC? ::: DC (no polarity reversal), possibly with ripple. A signal stays wholly below zero — AC or DC? ::: Still DC — it never reverses; the zero-crossing test is symmetric. $v (t) = 1 + 1.5 sin (ω t)$ — AC or DC and why? ::: AC, because amplitude 1.5 > |offset| 1, so minimum = −0.5 V crosses zero. A cycle spans 2.5 divisions at 2 ms/div — what is $f$ ? ::: $T = 5$ ms, so $f = 200$ Hz. Peak voltage a 230 V rms circuit reaches? ::: $2302 \approx 325$ V. RMS of a 6 V-peak square wave? ::: $6$ V (the $2$ rule is sine-only). RMS of a 6 V-peak triangle wave? ::: $6/ 3 \approx 3.46$ V. When does an AC formula collapse to DC? ::: When $f = 0$ or $V_{p e ak} = 0$ — no reversal, flat line.

Connections

Distinguish DC vs AC signals — the parent topic these examples drill.
Frequency and Period — the $f = 1/ T$ conversions in Examples 2, 5, 8.
RMS and Power in Resistors — the squaring/rooting logic behind Examples 1, 9, 10.
Rectifiers and Power Supplies — where the ripple of Examples 3–4 comes from.
Oscilloscopes — reading period off divisions, Example 5.
Electromagnetic Induction — why the sine shape (and its $2$ ) exists at all.
Current, Voltage and Charge — polarity = which way charge moves.
🇮🇳 Hinglish version

Worked examples — Distinguish DC vs AC signals

The scenario matrix

Worked Examples

Connections

Concept Map