Question

The maximum signal to quantization noise ratio (SQNR) for a \(16\)-bit analog-to-digital converter is approximately _________ dB.

• A. \(48\)
• B. \(68\)
• C. \(88\)
• D. \(98\)

Hint:

Use \(\text{SQNR}_{dB} = 6.02N+1.76\) with \(N=16\).

Answer 1

The Correct Option is D

Step 1: Recall the SQNR formula for an ideal ADC.
For an ideal \(N\)-bit analog-to-digital converter with a full-scale sinusoidal input, the ratio of the maximum signal power to the quantization noise power, expressed in decibels, is given by the standard result
\[ \text{SQNR}_{\text{dB}} = 6.02N + 1.76 \] This formula comes from treating the quantization error as a uniformly distributed random variable spread over one quantization step \(q\), which gives a noise power of \(q^2/12\), and comparing that against the power of a full-scale sine wave that just spans the converter's input range.

Step 2: Substitute the number of bits.
Here \(N = 16\), so
\[ \text{SQNR}_{\text{dB}} = 6.02(16) + 1.76 \] \[ 6.02 \times 16 = 96.32 \] \[ \text{SQNR}_{\text{dB}} = 96.32 + 1.76 = 98.08 \text{ dB} \]
Step 3: Round to the nearest option.
\(98.08\) dB rounds to approximately \(98\) dB, so the correct choice is option (D). A quick sanity check with the rougher rule of thumb of about \(6\) dB per bit plus \(2\) dB gives \(6(16)+2=98\) dB as well, confirming the result.

Step 4: Why the other options are too low.
Options (A), (B), and (C) would only be correct for converters with far fewer effective bits, or if the \(1.76\) dB offset that comes from the sine-wave crest factor were left out of the calculation by mistake. Since the question explicitly gives a full \(16\)-bit converter, none of these lower figures apply.

Final Answer:
The maximum SQNR of a 16-bit ADC is about 98 dB. \[ \boxed{\text{SQNR} \approx 98 \text{ dB}} \]