ADC - DAC — resolution, sampling rate, Nyquist: Worked examples

This page is the drill-ground for the parent topic. We take every kind of situation the ideas of resolution, sampling rate, and Nyquist can produce — and we work each one out loud, from zero. Nothing here uses a symbol the parent note did not build, and where a new one appears we re-earn it.

The scenario matrix

Every problem in this topic is really one of a small number of shapes. Here they are.

#	Cell (the kind of case)	What makes it tricky	Covered by
A	Resolution — normal	plain LSB / code arithmetic	Ex 1
B	Resolution — boundary / full-scale	the " $2^{N}$ vs $2^{N} - 1$ " trap at the top code	Ex 2
C	Resolution — degenerate (1 bit, tiny signal)	signal smaller than 1 LSB → dead code	Ex 3
D	Sampling — below Nyquist (aliasing)	frequency folds back	Ex 4
E	Sampling — above Nyquist (safe)	verify no folding, find margin	Ex 5
F	Sampling — exactly at $2 f$ (the limiting value)	boundary is NOT safe	Ex 6
G	Multiple tones / choosing $f_{s}$	must clear the highest frequency	Ex 7
H	Real-world word problem	translate words → numbers → filter choice	Ex 8
I	Exam twist (SNR ↔ bits, backwards)	solve the 6.02N formula in reverse	Ex 9
J	DAC reconstruction — staircase	zero-order hold, images above $f_{s} /2$	Ex 10

We will hit A → J in order. First, a shared picture of the two axes every problem lives on.

Figure — ADC - DAC — resolution, sampling rate, Nyquist

Recall The three formulas we will reuse (all from the parent)

Full-scale range $V_{r e f}$ split into $2^{N}$ codes with $2^{N} - 1$ gaps ::: $LSB = \frac{V _{r e f}}{2 ^{N} - 1} \approx \frac{V _{r e f}}{2 ^{N}}$ Code from a voltage ::: $code = round (\frac{V _{in}}{V _{r e f}} \times (2^{N} - 1))$ Nyquist rule ::: $f_{s} > 2 f_{ma x}$ ; Nyquist frequency $f_{N} = f_{s} /2$ Alias frequency ::: $f_{a l ia s} = f - f_{s} \cdot round (\frac{f}{f _{s}})$ SNR ceiling ::: $SNR_{m a x} = 6.02 N + 1.76 dB$

Here $N$ is the number of bits (the count of yes/no switches), $V_{r e f}$ is the reference voltage (the top of the input range), $f$ is a signal frequency in hertz, and $f_{s}$ is the sampling rate in samples per second. Nothing else is assumed.

Cell A — Resolution, the normal case

Worked example Ex 1 — Step size and code of a 10-bit ADC

A 10-bit ADC has $V_{r e f} = 5 V$ . (a) What voltage does one code step represent? (b) What code comes out for an input of $3 V$ ?

Forecast: guess — is one step closer to $5 mV$ or $5 μ V$ ? Will the code for $3 V$ be nearer $600$ or nearer $6000$ ?

Step 1. Count the codes: $2^{10} = 1024$ . Why this step? $N = 10$ independent binary switches count from $0$ to $2^{10} - 1$ , so there are $1024$ distinct levels.

Step 2. Find the LSB (the step between neighbouring levels): $LSB = \frac{V _{r e f}}{2 ^{N} - 1} = \frac{5}{1023} \approx 4.888 mV .$ Why $2^{N} - 1$ and not $2^{N}$ ? There are $1024$ fence-posts but only $1023$ gaps between them — the step size is a gap.

Step 3. Convert $3 V$ to a code: $code = round (\frac{3}{5} \times 1023) = round (613.8) = 614.$ Why round? The ADC is a mid-tread quantizer — it snaps the input to the nearest level, not the floor.

Verify: code $614$ represents $614 \times 4.888 mV \approx 3.001 V$ — back to $3 V$ within half an LSB ( $\approx 2.4 mV$ ). ✓ Units: volts in, volts out. ✓

Cell B — Resolution at the boundary (the $2^{N} - 1$ trap)

Worked example Ex 2 — What is the highest code, and what voltage is it?

Same 10-bit, $5 V$ ADC. What is the largest code, and what input voltage produces it? Show why using $2^{N}$ instead of $2^{N} - 1$ would misplace full scale.

Forecast: is the top code $1024$ or $1023$ ? Does it sit at $5 V$ or just below?

Step 1. Highest code $= 2^{N} - 1 = 1023$ . Why this step? Codes run $0, 1, \dots, 1023$ . Zero is a valid code, so the count $1024$ includes it; the maximum index is one less.

Step 2. Voltage of code $1023$ with the exact LSB: $1023 \times \frac{5}{1023} = 5.000 V .$ Why this matters: with the exact $LSB = V_{r e f} / (2^{N} - 1)$ , the top code lands exactly on full scale — clean.

Step 3. Now the trap — using the approximation $LSB \approx V_{r e f} / 2^{N} = 5/1024 \approx 4.883 mV$ : $1023 \times 4.883 mV \approx 4.995 V .$ Why show this? The approximate LSB places the top code $\approx 5 mV$ below $5 V$ — a full step short. For large $N$ that error is invisible; near full scale it is exactly one LSB of bias.

Verify: difference $5.000 - 4.995 = 0.005 V = 1 LSB$ . The gap between the two conventions is exactly one step, as claimed. ✓

Cell C — Degenerate: signal smaller than one LSB

Worked example Ex 3 — A 1-bit converter and a tiny signal

Extreme edge. $V_{r e f} = 3.3 V$ . (a) What is the LSB of a 1-bit ADC? (b) A sensor produces a swing of only $0.5 V$ around $0$ . What codes appear?

Forecast: how many levels does 1 bit give? Can a $0.5 V$ wiggle even be seen?

Step 1. Codes for $N = 1$ : $2^{1} = 2$ (just $0$ and $1$ ). Gaps $= 2^{1} - 1 = 1$ . $LSB = \frac{3.3}{2 ^{1} - 1} = 3.3 V .$ Why this step? One switch = two states = one giant step spanning the whole range. That is the coarsest possible converter.

Step 2. Feed in $0.5 V$ : $code = round (\frac{0.5}{3.3} \times 1) = round (0.1515) = 0.$ Why this result? The nearest level to $0.5 V$ is code $0$ (which represents $0 V$ ). The $0.5 V$ signal is smaller than the $\pm \frac{1}{2} LSB = \pm 1.65 V$ decision band, so it never flips the bit.

Step 3. The whole $0.5 V$ waveform stays at code $0$ — it is invisible. Why this is the degenerate case: any input change smaller than $\frac{1}{2}$ LSB lands in the same code. This is the "dead code" limit the parent warned about ("any change smaller than ~0.8 mV cannot be seen"), pushed to an extreme.

Verify: $\frac{1}{2} LSB = 1.65 V > 0.5 V$ , so no threshold is crossed → code stays $0$ . ✓ (Fix in real life: amplify the sensor first, or use more bits — see Sensor Interfacing on Microcontrollers.)

Cell D — Sampling below Nyquist → aliasing

Worked example Ex 4 — A tone that lies about its frequency

A $7 kHz$ pure tone is sampled at $f_{s} = 10 kHz$ . What frequency will you hear on playback?

Forecast: $7 kHz$ is above Nyquist — guess the fake frequency before computing. Higher or lower than $5 kHz$ ?

Step 1. Nyquist frequency $f_{N} = f_{s} /2 = 5 kHz$ . Since $7 > 5$ , the rule $f_{s} > 2 f_{ma x}$ is broken → aliasing happens. Why this step? Only content strictly below $f_{N}$ is safe. Anything above folds.

Step 2. Apply the alias formula: $f_{a l ia s} = 7 - 10 \cdot round (\frac{7}{10}) = ∣ 7 - 10 \cdot 1 ∣ = 3 kHz .$ Why round(7/10)=1? $0.7$ rounds to the nearest whole number of full sampling-rate hops, which is $1$ . The tone is $3 kHz$ on the other side of $f_{s}$ .

Step 3. Interpretation: a $7 kHz$ input is recorded as a false $3 kHz$ tone. Why it disguises: $cos (2 π \cdot 7000 t)$ and $cos (2 π \cdot 3000 t)$ hit identical values at every sample instant $t = n /10000$ .

Verify: $3 kHz < f_{N} = 5 kHz$ ✓ (a valid, representable frequency). And $7000$ and $3000$ differ by exactly $10000 = f_{s}$ — a folding partner. ✓ See the folding picture below.

Cell E — Sampling above Nyquist (safe) + margin

Worked example Ex 5 — Is this design safe, and by how much?

A signal contains nothing above $4 kHz$ . It is sampled at $f_{s} = 44.1 kHz$ . Is it aliased? What is the safety margin ratio $f_{s} / (2 f_{ma x})$ ?

Forecast: obviously safe — but guess: is the margin close to $2 \times$ , $5 \times$ , or huge?

Step 1. Nyquist frequency $f_{N} = 44100/2 = 22050 Hz$ . Highest content $= 4000 Hz$ . Why this step? We compare the highest real frequency to $f_{N}$ .

Step 2. $4000 < 22050$ , so $f_{s} > 2 f_{ma x}$ holds → no aliasing. Why: nothing pokes above the fold line, so nothing folds back.

Step 3. Margin ratio: $\frac{f _{s}}{2 f _{ma x}} = \frac{44100}{2 \times 4000} = \frac{44100}{8000} \approx 5.51.$ Why compute this? The parent recommends $\geq 2.5 \times$ for filter roll-off room; here we are oversampling $\sim 5.5 \times$ , very comfortable.

Verify: apply the alias formula as a sanity check — $round (4000/44100) = round (0.0907) = 0$ , so $f_{a l ia s} = ∣4000 - 0∣ = 4000 Hz$ : unchanged, no fold. ✓

Cell F — The limiting value: sampling at exactly $2 f$

Worked example Ex 6 — Why "

$f_{s} = 2 f$ " is a trap A $500 Hz$ sine is sampled at exactly $f_{s} = 1000 Hz = 2 f$ . Show that reconstruction can fail even though the arithmetic "just satisfies" $2 f$ .

Forecast: the parent says the Nyquist condition is strict ( $>$ , not $\geq$ ). Guess what goes wrong at the exact boundary.

Step 1. Sample instants: $t = n /1000$ seconds. The sine is $sin (2 π \cdot 500 t)$ . Why sine here? We deliberately pick a phase where the danger shows: choosing sine means the samples can hit the zero crossings.

Step 2. Evaluate at each sample: $sin (2 π \cdot 500 \cdot \frac{n}{1000}) = sin (π n) = 0 for every integer n .$ Why this step? $500/1000 = \frac{1}{2}$ , so each step advances the angle by exactly $π$ — always landing on a zero crossing.

Step 3. Every sample is $0$ . The recorded signal is silence — the amplitude is lost. Why this is the boundary lesson: at $f_{s} = 2 f$ the two samples per period can both sit on zeros, so amplitude (and hence the tone) is not guaranteed to survive. That is exactly why the condition is $f_{s} > 2 f_{ma x}$ , strict.

Verify: $sin (π n) = 0$ for $n = 0, 1, 2, 3$ → the sampled sequence is all zeros, confirming the failure. ✓

Cell G — Multiple tones: choose the sampling rate

Worked example Ex 7 — Three tones, one sampling rate

A signal is the sum of $300 Hz$ , $2.5 kHz$ , and $9 kHz$ tones. What is the minimum sampling rate to capture all three without aliasing?

Forecast: which tone sets the rule? Guess the minimum $f_{s}$ .

Step 1. Find $f_{ma x}$ = the highest frequency present $= 9 kHz$ . Why this step? Nyquist cares only about the top frequency — one tone above the fold ruins everything.

Step 2. Apply the strict rule: $f_{s} > 2 f_{ma x} = 2 \times 9000 = 18000 Hz .$ Why strict? From Ex 6, exactly $18 kHz$ risks the zero-crossing failure; pick anything above it.

Step 3. Add the practical margin the parent recommends ( $\geq 2.5 \times$ for filter roll-off): $f_{s} \approx 2.5 \times 9000 = 22500 Hz \Rightarrow pick a standard f_{s} \geq 22.5 kHz (e.g. 44.1 kHz).$ Why round up to a standard rate? Real hardware clocks come in standard values; overshoot is free safety.

Verify: at $f_{s} = 18000$ , the lower tones alias-check: $round (300/18000) = 0$ and $round (2500/18000) = 0$ , so both pass through unchanged ( $f_{a l ia s} = 300$ and $2500$ ). Only the $9 kHz$ tone sits at the boundary, confirming it is the binding constraint. ✓

Cell H — Real-world word problem

Worked example Ex 8 — Temperature logger design

You log a room temperature that ranges $0$ – $3.3 V$ from a sensor. You need to resolve steps of $1 mV$ or finer. (a) Minimum number of bits? (b) The signal's fastest meaningful variation is $2 Hz$ — what minimum $f_{s}$ , and do you need an anti-alias filter?

Forecast: guess the bit count — 8, 10, or 12? And is $2 Hz$ a demanding sampling problem?

Step 1. Bits from the resolution requirement. We need $LSB \leq 1 mV$ : $\frac{3.3}{2 ^{N} - 1} \leq 0.001 \Rightarrow 2^{N} - 1 \geq 3300 \Rightarrow 2^{N} \geq 3301.$ Why this inequality? Smaller LSB needs more levels; more levels need more bits.

Step 2. Solve: $2^{11} = 2048$ (too few), $2^{12} = 4096 \geq 3301$ (enough). So $N = 12$ bits. Why round up? You cannot buy a fractional bit; you take the next whole bit that clears the requirement.

Step 3. Check the achieved LSB: $LSB = 3.3/4095 \approx 0.806 mV \leq 1 mV$ ✓. Why re-check? Confirms the chosen bit count actually meets the $1 mV$ spec.

Step 4. Sampling: $f_{ma x} = 2 Hz$ , so $f_{s} > 4 Hz$ . Even $10 Hz$ is generous. Anti-alias filter? Temperature physically cannot change at kilohertz rates, so there is little high-frequency energy to fold — a simple RC low-pass (cutoff a few Hz) is enough. See Anti-Aliasing Filters.

Verify: $2^{12} = 4096 > 3301$ ✓ and $LSB = 3.3/4095 = 0.806 mV < 1 mV$ ✓; $f_{s} = 10 > 4$ ✓.

Cell I — Exam twist: run the SNR formula backwards

Worked example Ex 9 — How many bits for 80 dB of SNR?

An audio ADC must reach at least $80 dB$ of signal-to-noise ratio. Using the ideal ceiling $SNR_{m a x} = 6.02 N + 1.76 dB$ , find the minimum $N$ .

Forecast: roughly $6 dB$ per bit, so guess — 12 bits? 14? 16?

Step 1. Set the ceiling to meet the spec and solve for $N$ : $6.02 N + 1.76 \geq 80 \Rightarrow 6.02 N \geq 78.24 \Rightarrow N \geq 12.997.$ Why solve this way? SNR grows linearly in $N$ ; we invert the line to get the bit count.

Step 2. Round up to the next whole bit: $N = 13$ . Why up, not nearest? $N = 12.997$ is just short of $80 dB$ ; $N = 13$ is the smallest integer that clears the spec.

Step 3. Confirm what $13$ bits delivers: $SNR_{m a x} = 6.02 \times 13 + 1.76 = 80.02 dB \geq 80 dB .$ Why check? Ensures we did not under-round.

Verify: $N = 12$ gives $6.02 \times 12 + 1.76 = 74.0 dB < 80$ (fails), $N = 13$ gives $80.02 dB \geq 80$ (passes) — so $13$ is genuinely the minimum. ✓ (More on this in Quantization Noise & SNR.)

Cell J — DAC reconstruction & the staircase

Worked example Ex 10 — Zero-order hold staircase and its images

A DAC outputs codes at $f_{s} = 8 kHz$ , holding each sample constant for one period (zero-order hold). The wanted signal is a $1 kHz$ tone. (a) How long is each held step? (b) Where do the first spectral "image" copies land, and what cutoff should the reconstruction filter use?

Forecast: guess the step duration, and guess whether the junk images sit near $1 kHz$ or up near $8 kHz$ .

Step 1. Step (hold) duration: $T_{s} = \frac{1}{f _{s}} = \frac{1}{8000} = 125 μ s .$ Why this step? Each sample is held for exactly one sampling period, producing the flat treads of the staircase.

Step 2. Zero-order hold creates copies (images) of the signal spectrum around every multiple of $f_{s}$ . The nearest images of the $1 kHz$ tone appear at: $f_{s} - 1 kHz = 7 kHz and f_{s} + 1 kHz = 9 kHz .$ Why here? The staircase's sharp steps are built from high-frequency content — mathematically, mirror copies of the baseband tone sit at $∣ k f_{s} \pm f ∣$ .

Step 3. Reconstruction filter cutoff at the Nyquist frequency: $f_{c} \approx f_{N} = \frac{f _{s}}{2} = 4 kHz .$ Why $f_{s} /2$ ? It passes the wanted $1 kHz$ tone (below $4 kHz$ ) while blocking the $7 kHz$ and $9 kHz$ images — smoothing the staircase back to a clean sine. See Zero-Order Hold & Reconstruction.

Verify: wanted $1 kHz < 4 kHz$ (passes) ✓; first image $7 kHz > 4 kHz$ (blocked) ✓; step time $125 μ s \times 8000 = 1$ full second of samples per 8000 steps — consistent with $f_{s} = 8 kHz$ . ✓

Connections

ADC - DAC — resolution, sampling rate, Nyquist (index 5.5.4) — the parent topic
Quantization Noise & SNR — the $6.02 N + 1.76$ ceiling used in Ex 9
Anti-Aliasing Filters — the fix for cells D and H
Zero-Order Hold & Reconstruction — the staircase of Ex 10
Fourier Transform & Frequency Domain — why images and folding exist
Successive Approximation vs Sigma-Delta ADC — how these numbers become hardware
Sensor Interfacing on Microcontrollers — the dead-code fix from Ex 3