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

• $N$ ::: number of bits — how many yes/no switches the converter has.
• $2^N$ ::: total number of distinct output codes (levels).
• $V_{ref}$ ::: full-scale voltage span, e.g. $0$ to $3.3$ V.
• $\text{LSB}=V_{ref}/(2^N-1)$ ::: the smallest voltage step (one "rung" of the amplitude ladder).
• $f_s$ ::: sampling rate in Hz; $f_N=f_s/2$ is the Nyquist frequency (the highest honest frequency).

WHAT: each bit is one on/off switch. WHY $2^N$: $N$ independent switches count $0$ up to $2^N-1$, which is $2^N$ different values. $2^{10}=1024\text{ codes}$

• Resolution ($N$ bits) → amplitude precision (how finely we slice voltage).
• Sampling rate ($f_s$) → time precision (how often we measure). They are independent knobs. One does not help the other.

$f_s>2f_{max}$ The inequality is strict (a bare $>$, not $\ge$). The boundary $f_s/2$ is the Nyquist frequency $f_N$.

WHAT: the smallest voltage the ADC can distinguish. WHY $2^N-1$: the span is divided by the number of gaps. $\text{LSB}=\frac{5}{2^{12}-1}=\frac{5}{4095}\approx 1.221\text{ mV}$ Any input change smaller than $\approx 1.22$ mV lands in the same code and is invisible.

WHY round: the ADC snaps to the nearest level (mid-tread quantizer). $\text{code}=\text{round}\left(\frac{2.0}{5}\times 4095\right)=\text{round}(1638.0)=1638$

Nyquist frequency $f_N=2500/2=1250$ Hz. Since $1800>1250$, the tone is above Nyquist, so it aliases. $f_{alias}=\left|1800-2500\cdot \text{round}\left(\frac{1800}{2500}\right)\right|=|1800-2500\cdot 1|=700\text{ Hz}$ The 1.8 kHz tone masquerades as a fake 700 Hz tone.

WHY this formula: each extra bit halves the quantization step, and halving noise amplitude adds $\approx 6.02$ dB. ${\text{SNR}}_{\max }=6.02\cdot 16+1.76=96.32+1.76=98.08\text{ dB}$

$f_N=1000$ Hz.

• $300$ Hz is below Nyquist → stays at $300$ Hz.
• $1700$ Hz is above Nyquist → $f_{alias}=|1700-2000\cdot \text{round}(0.85)|=|1700-2000|=300$ Hz. Both land on 300 Hz. They are now indistinguishable — a permanent, irreversible collision. Look at Figure s01: the fast wave folds down and sits exactly on the slow one.

No later processing can un-mix them. This is why the anti-alias filter must live before the ADC.

No. Bits control amplitude precision; aliasing is a time/frequency corruption. The 16-bit part will record the fake 300 Hz alias more precisely — a sharper picture of the wrong thing. The only fixes are: (a) raise $f_s$ above $2\times 1700=3400$ Hz, or (b) low-pass filter out the 1700 Hz tone before sampling. See Anti-Aliasing Filters.

WHAT we compute: the sample values at each instant $t=n/1000$. $\sin \left(2\pi \cdot 500\cdot \frac{n}{1000}\right)=\sin (\pi n)=0\text{for every integer }n$ Every sample is zero. Figure s02 shows the dots landing exactly on the zero-crossings.

A flat line of zeros is indistinguishable from a silent signal — the amplitude is lost. This is precisely why Nyquist is a strict inequality $f_s>2f_{max}$, not $\ge$.

Step 1 — sampling rate. Nyquist minimum is $2\times 8000=16000$ Hz. Add $25\%$: $f_s\ge 1.25\times 16000=20000\text{ Hz}$ Choose $f_s=20$ kHz (or a standard 22.05 kHz if a codec chip forces it).

Step 2 — bits from SNR. Need $6.02N+1.76\ge 90$: $N\ge \frac{90-1.76}{6.02}=14.66\Rightarrow N=15\text{ bits}$

Step 3 — bits from LSB. Need $V_{ref}/(2^N-1)\le 0.5$ mV, i.e. $2^N-1\ge 3.3/0.0005=6600$, so $2^N\ge 6601\Rightarrow N\ge 13$ (since $2^{13}=8192$).

Step 4 — combine. Take the larger bit requirement: $N=15$ bits (rounds up to a real 16-bit part). Check LSB at 16 bits: $3.3/65535\approx 0.0504$ mV — comfortably under $0.5$ mV. ✓

Step 5 — filter. Anti-alias low-pass cut-off just below Nyquist $f_s/2=10$ kHz (pass the 8 kHz signal, kill everything above 10 kHz). See Zero-Order Hold & Reconstruction for the matching DAC-side smoothing filter.

Final design: $f_s=20$ kHz, $N=16$ bits (16-bit part), anti-alias cut-off $\approx 9$–$10$ kHz.

The DAC holds each sample for $T_s=1/f_s=50\mu$s (zero-order hold), making a staircase. Those sharp steps inject spectral images — copies of the signal centred at multiples of $f_s$ (20 kHz, 40 kHz, ...). A reconstruction low-pass filter with cut-off $\approx f_s/2=10$ kHz erases the steps, keeping only the original band. See Fourier Transform & Frequency Domain for why hard edges = high-frequency content.

Part A — is the 9 kHz tone safe? Nyquist $=10$ kHz. Since $9<10$, the tone is below Nyquist → not aliased, it stays at 9 kHz. ✓

Part B — the noise fold. Anything above 10 kHz folds into 0–10 kHz. Noise at 50 kHz aliases to $|50-20\cdot \text{round}(50/20)|=|50-40|=10$ kHz worth of fold — it lands right at the band edge. So the filter must attenuate the noise between 10 kHz and 50 kHz before it folds.

Part C — filter roll-off. From the filter's corner at $10$ kHz to the worst noise at $50$ kHz is a frequency ratio of $50/10=5\times$, i.e. ${\log }_{10}(5)\approx 0.699$ decades. We need $60$ dB of attenuation across that span: $\text{slope}=\frac{60\text{ dB}}{0.699\text{ decades}}\approx 85.8\text{ dB/decade}$ A standard filter gives $20$ dB/decade per order ($n$), so: $n=\frac{85.8}{20}\approx 4.29\Rightarrow n=5\text{ (round up)}$ A 5th-order anti-alias filter. See Anti-Aliasing Filters.

WHAT: SNR scales with how much of full scale the signal uses. WHY: quantization noise is fixed by the LSB (it doesn't shrink when your signal shrinks), so a small signal has a worse ratio.

Full-scale SNR at 14 bits: $6.02\cdot 14+1.76=84.28+1.76=86.04$ dB.

The signal swing is $\pm 0.2$ V out of $\pm 1$ V (half of $V_{ref}$), i.e. fraction $0.2/1=0.2$ of full scale. Amplitude loss in dB: $20{\log }_{10}(0.2)=20\cdot (-0.699)=-13.98\text{ dB}$ Effective SNR: $86.04-13.98=72.06\text{ dB}$ Bits lost: each $6.02$ dB $=1$ bit, so $13.98/6.02\approx 2.32$ bits lost. You threw away roughly 2.3 effective bits by under-driving the ADC — a 14-bit part behaving like an $\approx 11.7$-bit one. Fix: add gain before the ADC to fill the range. See Sensor Interfacing on Microcontrollers.