Distinguish DC vs AC signals: Exercises — Hardware deep dive

Before we start, one shared picture. A signal is just a graph of voltage $v$ (up-down) against time $t$ (left-right). The horizontal line $v = 0$ is the "no-push" line. If the wiggle stays entirely on one side of that line → DC. If it crosses to the other side → AC.

Level 1 — Recognition

Exercise 1.1

Classify each as DC or AC, and say why using the zero-crossing test:

(a) A flat line at $v = 12$ V.
(b) $v (t) = 8 sin (2 π \cdot 50 t)$ .
(c) A square wave that jumps between $0$ V and $+ 5$ V.
(d) A square wave that jumps between $- 5$ V and $+ 5$ V.

Recall Solution 1.1

The single test is: does the voltage ever change sign (cross below zero)?

(a) Flat at $+ 12$ V. Never negative → DC. (Constant is the simplest DC.)
(b) A pure sine centred on zero: it spends half its time positive, half negative → crosses zero → AC.
(c) Bounces between $0$ and $+ 5$ : the lowest value is $0$ , it never goes negative → DC (a pulsing DC logic signal).
(d) Bounces between $- 5$ and $+ 5$ : it genuinely reverses polarity → AC.

What it looks like: in the figure above, (a) is the flat yellow line, (b)/(d) cross the pink zero-line, (c) hugs the top and only touches zero.

Exercise 1.2

A 9 V battery and a 230 V rms wall socket. For each, is the peak value equal to the labelled number? Answer yes/no with one sentence of reasoning.

Recall Solution 1.2

Battery, 9 V: DC and constant, so peak $=$ average $=$ rms $=$ 9 V. Yes — the label is the peak.
Wall socket, 230 V rms: the "230 V" is the rms value, not the peak. Peak $= 2302 \approx 325$ V. No — the peak is higher than the label.

Level 2 — Application

Exercise 2.1

An oscilloscope shows one full cycle spanning 5 ms. Find the period $T$ and frequency $f$ .

Recall Solution 2.1

What: read one period off the screen, then convert to frequency. Why $f = 1/ T$ : frequency means "how many whole cycles fit into one second." If one cycle takes $T$ seconds, then $1/ T$ of them fit in a second. That is the definition. $T = 5 ms = 0.005 s, f = \frac{1}{T} = \frac{1}{0.005} = 200 Hz .$

Exercise 2.2

A sinusoid has peak voltage $V_{p e ak} = 170$ V. Find its rms value.

Recall Solution 2.2

What: apply the peak→rms factor for a pure sine. Why divide by $2$ : power depends on $v^{2}$ ; averaging $sin^{2}$ over a cycle gives exactly $\frac{1}{2}$ , and the square root of $\frac{1}{2}$ is $1/ 2$ . So the "equivalent steady voltage" is smaller than the peak by that factor. $V_{r m s} = \frac{V _{p e ak}}{2} = \frac{170}{1.4142} \approx 120.2 V .$ (This is why US mains, ~120 V rms, peaks near 170 V.)

Exercise 2.3

Mains at $f = 50$ Hz. How long is one cycle in milliseconds, and how many times per second does the current reverse direction?

Recall Solution 2.3

$T = \frac{1}{f} = \frac{1}{50} = 0.02 s = 20 ms .$ Reversals: one full cycle goes positive→negative→positive, so there are two direction reversals per cycle. With 50 cycles per second: $50 \times 2 = 100 reversals per second .$

Level 3 — Analysis

Exercise 3.1

Classify $v (t) = 3 + 4 sin (2 π \cdot 60 t)$ as pure AC, pure DC, or DC-with-ripple. Give the maximum and minimum voltage.

Recall Solution 3.1

What: find the extremes of the signal to see if it ever crosses zero. Why the extremes: the sine term $sin (\dots)$ swings between $- 1$ and $+ 1$ , so $4 sin (\dots)$ swings between $- 4$ and $+ 4$ . Add the constant $3$ : $v_{ma x} = 3 + 4 = 7 V, v_{min} = 3 - 4 = - 1 V .$ Because $v_{min} = - 1$ V is negative, the signal does cross zero and reverse. So this is genuinely AC riding on a DC offset — but since it reverses, it is not pure DC. What it looks like: a sine centred at $+ 3$ that dips just under the zero-line.

Exercise 3.2

Contrast with $v (t) = 5 + 2 sin (2 π \cdot 60 t)$ . Is this one DC-with-ripple (never reverses) or AC?

Recall Solution 3.2

$v_{ma x} = 5 + 2 = 7 V, v_{min} = 5 - 2 = 3 V > 0.$ The minimum is $+ 3$ V, still positive → it never crosses zero → current never reverses. This is DC with ripple (a small 60 Hz wiggle on a steady 5 V), which behaves as DC for direction purposes.

The key comparison: same shape, same frequency as 3.1 — the only difference is whether the DC offset is big enough to keep the whole wave above zero. Compare offset vs amplitude: if $∣ offset ∣ \geq$ amplitude → no reversal (DC-with-ripple); if $∣ offset ∣ <$ amplitude → it dips negative (AC).

Exercise 3.3

A resistor $R = 100 Ω$ carries an AC voltage of peak $V_{p e ak} = 20$ V. Find the average power dissipated.

Recall Solution 3.3

Why we cannot use peak directly: power is $P = v^{2} / R$ , and $v$ swings, so the instantaneous power swings too. The meaningful number is the average power, and average power uses the rms voltage — that is exactly what rms is built to give. $V_{r m s} = \frac{V _{p e ak}}{2} = \frac{20}{2} \approx 14.14 V .$ $P = \frac{V _{r m s}^{2}}{R} = \frac{( 20/ 2 ) ^{2}}{100} = \frac{200}{100} = 2 W .$ (Neat: $(V_{p e ak} / 2)^{2} = V_{p e ak}^{2} /2 = 400/2 = 200$ , then $/100 = 2$ W.)

Level 4 — Synthesis

Exercise 4.1

A generator's output is $v (t) = 325 sin (2 π \cdot 50 t)$ volts. From this single expression, extract: (i) peak voltage, (ii) rms voltage, (iii) frequency, (iv) period, (v) angular frequency $ω$ , and state whether it is AC or DC.

Recall Solution 4.1

Compare term-by-term with the standard form $v (t) = V_{p e ak} sin (2 π f t)$ .

(i) $V_{p e ak} = 325$ V — the number multiplying the sine.
(ii) $V_{r m s} = 325/ 2 \approx 229.8$ V (this is "230 V mains").
(iii) The factor multiplying $t$ inside the sine is $2 π f = 2 π \cdot 50$ , so $f = 50$ Hz.
(iv) $T = 1/ f = 1/50 = 0.02$ s $= 20$ ms.
(v) $ω = 2 π f = 2 π \cdot 50 \approx 314.2$ rad/s.
It is a symmetric sine centred on zero → crosses zero → AC.

Exercise 4.2

Two heaters give the same heat. Heater A runs on 12 V DC. Heater B runs on AC. What peak AC voltage must B have to match A (same resistor)?

Recall Solution 4.2

What "same heat" means: rms is defined as the DC voltage delivering equal heating power. So heater B must have $V_{r m s} = 12$ V to match a 12 V DC heater. Why then convert to peak: the question asks for the peak, and peak $= V_{r m s} \times 2$ for a sine: $V_{p e ak} = V_{r m s} 2 = 12 \times 1.4142 \approx 16.97 V .$ So an AC source peaking at ~17 V heats the resistor exactly like steady 12 V DC.

Exercise 4.3

A signal is $v (t) = 6 sin (2 π \cdot 100 t)$ . Sketch-reason: at what times in the first cycle does it (a) hit its peak $+ 6$ V, (b) cross zero going downward, (c) hit its minimum $- 6$ V? Give times in ms.

Recall Solution 4.3

What: the sine's shape over one period tells us where each feature sits. One period is $T = 1/100 = 0.01$ s $= 10$ ms. A sine goes: $0 \to$ up to peak at a quarter turn $\to$ back to zero at a half turn $\to$ down to minimum at three-quarters $\to$ back to zero at a full turn.

(a) Peak at a quarter cycle: $t = T /4 = 2.5$ ms.
(b) Zero going down at a half cycle: $t = T /2 = 5$ ms.
(c) Minimum at three-quarter cycle: $t = 3 T /4 = 7.5$ ms.

What it looks like: see the figure below — the quarter-cycle markers land exactly on peak, zero, trough.

Level 5 — Mastery

Exercise 5.1

Prove from scratch that for a pure sinusoid $v (t) = V_{p e ak} sin (ω t)$ , the rms value is $V_{p e ak} / 2$ . Show every step and say why each is allowed.

Recall Solution 5.1

Step 1 — square it (why): power is $P = v^{2} / R$ , always positive; the plain average of $v$ is zero and useless, so we work with $v^{2}$ . $v (t)^{2} = V_{p e ak}^{2} sin^{2} (ω t) .$ Step 2 — flatten $sin^{2}$ (why): $sin^{2}$ is hard to average by eye, but the identity $sin^{2} θ = \frac{1 - cos ( 2 θ )}{2}$ turns it into a constant plus a cosine we can average. $v (t)^{2} = V_{p e ak}^{2} \cdot \frac{1 - c o s ( 2 ω t )}{2} .$ Step 3 — average over a whole cycle (why): over any whole number of cycles, $cos (2 ω t)$ spends equal time positive and negative → averages to $0$ . Only the constant $\frac{1}{2}$ survives. $⟨ v^{2} ⟩ = V_{p e ak}^{2} \cdot \frac{1}{2} .$ Step 4 — take the root (why): "rms" = root of the mean square; undo the squaring from Step 1. $V_{r m s} = ⟨ v^{2} ⟩ = \frac{V _{p e ak}^{2}}{2} = \frac{V _{p e ak}}{2} \approx 0.707 V_{p e ak} . ■$

Exercise 5.2

A device draws 2 A rms at 230 V rms mains. (a) What average power does it use? (b) What is the peak instantaneous voltage the insulation must survive?

Recall Solution 5.2

(a) For rms quantities, average power is simply $P = V_{r m s} \times I_{r m s}$ (this is why rms is defined — it makes the DC power formula work for AC): $P = 230 \times 2 = 460 W .$ (b) Insulation must survive the peak, not the rms: $V_{p e ak} = 2302 \approx 325.3 V .$ Design to 325 V, never to 230 V.

Exercise 5.3

Design task. You must build a signal that is technically DC (never reverses) but has the largest possible AC ripple at 60 Hz on top of a chosen DC offset $D$ , using a sine of amplitude $A = 3$ V. What is the smallest allowed offset $D$ , and what are the resulting max/min voltages? What happens exactly at $D = A$ ?

Recall Solution 5.3

What: we want $v (t) = D + 3 sin (2 π \cdot 60 t)$ to never go negative, i.e. $v_{min} \geq 0$ . Why this condition: "technically DC" means no polarity reversal, i.e. the wave never crosses below zero. The lowest point is $v_{min} = D - A = D - 3$ . $D - 3 \geq 0 \Rightarrow D \geq 3 V .$ Smallest allowed offset: $D = 3$ V. Then: $v_{ma x} = 3 + 3 = 6 V, v_{min} = 3 - 3 = 0 V .$ At $D = A = 3$ (the boundary): the wave just touches zero once per cycle but never goes below. This is the degenerate/limiting case — it is still DC (no reversal), but with zero margin. Any smaller $D$ (say 2.9 V) dips to $- 0.1$ V → it becomes AC. So $D = 3$ is the exact knife-edge between DC-with-ripple and AC.

Answer key (quick check)

#	Answer
1.1	DC, AC, DC, AC
1.2	9 V yes; 230 V no (peak 325 V)
2.1	$T = 0.005$ s, $f = 200$ Hz
2.2	$\approx 120.2$ V
2.3	$T = 20$ ms, 100 reversals/s
3.1	AC; max 7 V, min $- 1$ V
3.2	DC-with-ripple; max 7 V, min 3 V
3.3	$P = 2$ W
4.1	325 V, 229.8 V, 50 Hz, 20 ms, 314.2 rad/s, AC
4.2	$\approx 16.97$ V peak
4.3	2.5 ms, 5 ms, 7.5 ms
5.2	460 W; 325.3 V peak
5.3	$D \geq 3$ V; max 6 V, min 0 V

Connections

Frequency and Period — every $f = 1/ T$ step above.
RMS and Power in Resistors — L3/L5 power problems.
Rectifiers and Power Supplies — how DC-with-ripple (Ex 5.3) is actually produced.
Oscilloscopes — reading $T$ off the screen (Ex 2.1, 4.3).
Electromagnetic Induction — why Ex 4.1's generator gives a sine.

Recall Feynman check: one line each

DC vs AC test in one word ::: Reversal. Peak from rms ::: multiply by $2$ . Average AC power ::: $V_{r m s} I_{r m s}$ (or $V_{r m s}^{2} / R$ ).